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(C) g , 2 2 3 g . (D) 2g , 2 3g . Sol: amax = .2A or g = .2A or g = .2 2 2 1 2 A . A or g = .2 12 . 3 or g = 2.2 . g 2 . . and if occurs at extreme at upper extreme position. . A= sin .t + 3 cos .t or 2 = sin .t0 + 3 cos .t0 or 2 = 2sin(.t0 + .) But tan 3 3 1 . . . . 60º 3 . . . . or 0 2 2sin t 3 . .. . .. . . . . or 0 t 3 2 . . . . . or 0 t 2 3 6 . . . . . . . or 0 t 2 6 6 g . . . . . Hence option (A) is correct. SIMPLE HARMONIC MOTION www.physicsashok.in 81 THINKING PROBLEMS 1. Which ofthe following functions are (a) aperiodic (b) periodic but not simple harmonic (c) simple harmonic : (i) sin 2.t (ii) 1 + cos 2.t (iii) a sin .t + b cos .t (iv) sin .t + sin 2.t + cos 2 .t (v) sin3 .t (v) log (1 + .t) (vii) exp ( .t) ? 2. Can everyoscillatorymotion be treated as simple harmonicmotionin the limit of small amplitude ? 3. What would happen to themotionof an oscillating systemif the sign of the forc e termin equation F = kx is changed ? 4. (a) Can amotion be periodic but not oscillatory ? (b) Can amotion be oscillatory but not simple harmonic ? If your answer is yes, give an example and if not, explainwhy ? 5. Can a bodyhave accelerationwithout having velocity ? 6. Determine whether or not the following quantitie scan be in the same directio nfor a simple harmonicmotion : (a) displacement and velocity, (b) velocityand acceleration, (c) displacement an d acceleration. 7. (a) Canwe ever construct a simple pendulumstrictlyaccording to its definition ? (b) Is themotion of a simple pendulumlinear SHMor angular SHM? 8. Agirl is swinging in a sitting position.Howwill the period of swing be affect ed if : (a)The girl stands upwhile swinging ? (b)Another girl of samemass comes and sits next to her ? 9. Ahollowmetal sphere is filledwithwater and a small hole ismade at its bottom. It is hanging bya long thread and ismade to oscillate.Howwill the period ofoscillation change ifwater is allow ed to flowthrough the hole till the sphere is empty ? 10. The resultant of two simple harmonicmotions at right angles and of the same frequency is always a circular motion.True or false?Explain. SOLUTION OF THINKING PROBLEMS 1. (a) Functions (vi) log (1 + .t) and (vii) exp ( .t) increase (or decrease) cont inuouslywith time and can never repeat themselfso are aperiodic. (b) Function (ii) (1+ cos 2.t), (iv) sin .t + sin 2.t +cos 2.t and (v) sin3 .t a re periodic [i.e., f(t + T) = f(t)] with periodicity (./.), (2./.) and (2./.) respectively but not simple harmonic as for these functions (d2y/dt2) is not . y. (c) Functions (i) sin 2.t and (iii) a sin .t + b cos .t, i.e., (a2 + b2)1/2 sin [.t + tan 1 (b/a)] are simple harmonic [with time period (./.) and (2./.) respectively] as for these (d2y/dt2) .. y. 2. No, not at all. In the limit of small amplitudes onlythose oscillatorymotions can be treated as simple harmonic for which the restoring force (or torque) becomes linear. For example, the oscil latorymotion of a simple or spring pendulumormotion of atoms inamolecule becomes simple harmonic in the limi t ofsmall amplitude as restoring force (or torque) becomes linear while in case of oscillatorymotion of a ballbetween two inclined planes or two perfectly elasticwalls, themotion does not become simple harmonic evenfor vanishingysmall amplitude. 3. If the sign is changed in the force equation, accelerationwill not be opposit e to displacement and hence the particlewillnot oscillate, butwillaccelerate inthedirectionofdisplaement. So the motionwillbecome accelerated translatory. However, equations ofmotion cannot be applied to analyse themotion as acceleration (= .2y) is not constant. Mathematical analysis shows that in this situation both velocity and displacemen t will increase exponentially withtime. SIMPLE HARMONIC MOTION www.physicsashok.in 82 4. Yes;Uniformcircularmotion, (b)Yes;When a ball is thrown froma height on a per fectlyelastic plane surface themotion is oscillatory but not simple harmonic as the restoring force F =mg = constt. and not F . x] 5. Yes; In SHMat extreme position velocityis zero but acceleration ismaximum= .2 A. 6. (a)Yes;whenthe particle ismoving fromequilibriumposition to extreme position, (b)Yes;whenthe particle is moving fromextreme to equilibriumposition, (c)No; as inSHMdisplacement is always opposite to acceleration.] 7. (a) No, (b)Angular SHM. 8. (a) decreases, (b) unchanged 9. Time periodwill first increase, reaches amaximumand thenwill decrease. 10. False. The generalmotion is elliptical.Acircle is a particular case of ellip sewhen itsmajor axis is equal to its minor axis. In general, themotion is described by 2 2 2 2 2 d y 2xy cos sin a b ab . . . . . When a = b and . = ./2, x2 + y2 = a2 ASSERTION-REASON TYPE Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements,mark the correct answer as (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct e xplanation ofStatement-1. (C) If Statement-1 is true but Statement-2 is false. (D) If both Statement-1 and Statement-2 are false. (E) IfStatement-1 is false but Statement-2 is true. 1. Statement-1 : The force acting on a particle moving along x axis is F = kx + v0t , where k is a constant. Statement-2 :To an observermoving along x-axiswith constant velocity v0, it repr esents SHM. 2. Statement-1 : Aparticle executing simple harmonicmotioncomes to rest at the e xtreme positions. Statement-2 :The resultant force on the particle is zero at these positions. 3. Statement-1 : Soldiers are asked to break stepswhile crossing the bridges. Statement-2 :The frequencyofmarchingmaybe equalto the naturalfrequencyof bridge andlead to resonance which can break the bridge. 4. Statement-1 : The functionY= cos2 .t + sin .t does not represent a simple har monicmotion. Statement-2 : Sumof two harmonic functionmaynot be harmonicmotion. 5. Statement-1 : Ablock ofmassmis attached to a spring inside a trolley of mass 2mas shown in figure, all surfaces are smooth. Spring is stretched and released. Both trolleyand block oscillate simple harmonicallywith same time period and same amplitude. m 2m Statement-2 : Inabsence ofexternalforce,centreofmass ofsystemofparticles does not accelerate. SIMPLE HARMONIC MOTION www.physicsashok.in 83 MATCH THE COLUMN 1. Asimple harmonic oscillator consists ofa block attached to a spring with k= 200N/m.Theblock slides ona frictionless horizontal surface, with equilibriumpoint x = 0.Agraph of the block s velocity v as a function of time t is shown. Correctlymatch the required information in the column Iwith the values given in the column II. (use .2= 10) 2 2 0 0.10 0.20 t(s) V(m/s) Column-I Column-II (A) The block smass in kg (P) 0.20 (B) The block s displacement at t = 0 inmetres (Q) 200 (C) The block s acceleration at t = 0.10 s inm/s2 (R) 0.20 (D) The block smaximumkinetic energyin Joule (S) 4.0 2. Matchthe following Column-I Column-II (A) Linear combination of two SHM s (P) T = 2. R g (R is radius of the earth) (B) y =Asin .1t +Asin(.2t + .) (Q) SHMfor equal frequencies and amplitude (C) Time periodof a pendulumofinfinite length. (R) Superpositionmay not be a SHM always. (D) Maximumvalue of time period ofan oscillating (S) amplitude will be 2 A for . 1 = .2 and pendulum. phase difference of ./2. 3. Matchthe following Column-I Column-II (A) Aconstant force acting along the line of (P) The time period SHMaffects (B) Aconstant torque acting along the arc of (Q) The frequency angular SHMaffects. (C) Aparticle falling on the block executing SHM (R) themeanposition when the later crosses themean position affects (D) Aparticle executing SHMwhen kept on a (S) The amplitude uniformly accelerated car affects. SIMPLE HARMONIC MOTION www.physicsashok.in 84 LEVEL 1 1. Two particlesP andQdescribe SHMof same amplitude a, same frequencyf along the same straight line.The maximumdistance between the two particles is a 2 . The phase difference between the particle is : (A) zero (B) ./2 (C) ./6 (D) ./3 2. Arod of length l is inmotion such that its endsA and B aremoving along x-axis and y-axis respectively. it is given that d 2 dt . . rad/s always. P is a fixed point on the rod. LetMbe the projection ofPon x-axis. For the time interval inwhich . changes from 0 to ./2, choose the correct statement, y x P M l A B (A) The acceleration ofMis always directed towards right (B)Mexecutes SHM (C)Mmoveswith constant speed (D)Mmoveswith constant acceleration 3. The coefficient of frictionbetween block ofmassmand 2mism= 2 tan .. There is no friction between blockofmass 2mand inclined plane. Themaximumamplitude of two block systemfor which there is no relativemotion between both the blocks. k m 2m (A) g sin . k m (B) mg sin k . (C) 3mg sin k . (D) None of these 4. Graph shows the x(t) curves for three experiments involving a particular spri ng-block systemoscillating in SHM. The kinetic energy of the systemismaximumat t = 4 sec. for the situation : 1 2 4sec. t(in sec) 3 0 x (A) 1 (B) 2 (C) 3 (D) Same in all 5. Aparticle ofmassm= 2 kg executes SHMin xy-plane between pointsAand B underaction of force x y F . F i . F j . .Minimumtime taken by particle to move fromAto B is 1 sec.At t = 0 the particle is at x =2 and y=2.Then Fx as function of time t is A(2, 2) y x B( 2, 2) (A) 4.2 sin .t (B) 4.2 cos .t (C) 4.2 cos .t (D) None of these 6. The speed v of a particlemoving along a straight line,when it is at a distanc e (x) froma fixed point of the line is given by v2 = 108 9x2 (allquantities are in cgs units) : (A) themotionis uniformlyaccelerated along the straight line (B) themagnitude of the acceleration at a distance 3cmfromthe point is 27 cm/sec 2 (C) themotionis simple harmonic about the given fixed point. (D) themaximumdisplacement fromthe fixed point is 4 cm. SIMPLE HARMONIC MOTION www.physicsashok.in 85 7. The time period of an ideal simple pendulumis given by : T = 2. l / g The time period of actual simple pendulum(T´) is slightlydifferent due to smalldam ping (friction). Then (A) T´ > T (B) T´ < T (C) T = T´ (D) None 8. Asmallbob of a simple pendulumcontainswater. Its periodic time isT.During osc illation, the temperature of surrounding is lowered so that water gets freezed. The newtime period is T´. Then : (A) T > T´ (B) T < T´ (C) T = T´ (D) None 9. Two simple pendulaAand B are shown in the figure. The phase difference betwee n A and B is : (A) . (B) ./2 l l (C) 0 (D) ./4 A B 10. Abodyexecutes simple harmonicmotionunder the action of a force F1witha time period (4/5) second. If the force is changed to F2 it executes SHM with a time period (3/5) second. If both the forces F1 and F2 act simultaneously in the same direction on the body, its time period in second is : (A) 12/25 (B) 24/25 (C) 35/24 (D) 25/12 11. What should be the displacement of a simple pendulumwhose amplitude is A, at which potential energy is 1/4th of the total energy ? (A) A/ 2 (B)A/2 (C)A/4 (D) A/ 2 2 4)2} J.Totalmechan 12. The potential energyUof a particle is given byU= {20 + (x ical energy of the particle is 36 J. Select the correct alternative(s) (A) The particle oscillates about point x= 4m (B) The amplitude of the particle is 4m (C) The kinetic energy of the particle at x = 2mis 12 J (D) Themotion of the particle is periodic but not simple harmonic 13. Arod ofmassMand lengthL is hinged at its centre ofmass so that it can rotate in a vertical plane. Two springs each of stiffness k are connected at its ends, as shown in the figure. The time period of SHMis L, M k Hinge k (A) 2 M 6k . (B) 2 M 3k . (C) 2 ML k . (D) M 6k . 14. Aparticlemoves along the x-axis according to the equation x =Asin2 .t (A) The particle oscillates about the origin (B) The particle oscillates about t he point x =A (C) The particle oscillateswith a period T = ./. (D) The particle oscillateswith amplitudeA/2 15. In the arrangement shown in figure the pulleys are smooth and massless. The spring k1 and k2 aremassless. The time period of oscillation of themassmis m k k1 (A) 2 1 2 2 m k k . . (B) 1 2 2 2m k k . . (C) 1 2 1 2 m(k k ) 2 2k k . . (D) 1 2 1 2 m(k k ) 2 k k . . SIMPLE HARMONIC MOTION www.physicsashok.in 86 16. The equationofSHMofa particle oscillating alongthe x-axis is givenbyx=3 sin t 6 . .. . . . . . . cm.The acceleration of the particle at t = 1 s is (A) 1.5 .2 cms 2 (B) 2.6 .2 cms 2 (C) 2.6 .2 cms 2 (D) 1.5 .2 cms 2 17. For the systemshown in the figure, initiallythe spring is compressed bya distance a fromits natural lengthandwhenreleased, itmoves to a distance b fromits equilibriumposition. The dicrease in amplitude for one half cycle ( a to +b) is k µ m a b (A) µmg k (B) 2µmg k (C) µmg 2k (D) none of these 18. Themotion of a particle is given x =Asin .t +B cos .t. Themotion of the part icle is (A) not simple harmonic (B) simple harmonicwith amplitudeA+B (C) simple harmonicwith amplitude (A+B)/2 (D) simple harmonicwith amplitude A2 . B2 19. Two massesm1 and m2 are suspended together by amassless spring of force cons tant K.When the masses are in equilibriumis removedwithout disturbing the system. The amplitude of osci llation is (A)m2g/K (B)m1g/K (C) (m1+m2)g/K (D) (m1 m2)g/K 20. Two bodiesMandNof equalmasses are suspended fromtwo seperatemassless springs of spring constants k1 and k2 respectively. If the two bodies oscillate vertically such that theirma ximumvelocities are equal, the ratio of the amplitude of vibration ofMto that onNis : (A) k1/k2 (B) 1 2 k / k (C) k2/k1 (D) 2 1 k / k 21. Aparticle executing SHMwhilemoving fromone extremityis found at distance x1, x2, x3 fromthe centre at the ends of three successive seconds, the period is (A) . (B) ./2. (C) 2./. (D) 1/. where ..= cos 1(x1 + x3)/2x2 22. One end of a spring of force constant k is fixed to a verticalwall and the o ther to a body ofmassmresting on a smooth horizontal surface. There is anotherwall at a distance x0 fromthe body. The spring is then compressed by 2x0 and released. The time taken m x0 to strike thewall is (A) 1 k 6 m . (B) m k (C) 2 m 3 k . (D) k 3 m . 23. In the above problemthe velocitywithwhich the body strikes the other wall is (A) 0 x m k (B) 0 3k x m (C) 0 k x m (D) 0 2x m k 24. Aparticlemoves intheX Yplane according to the equation: r . .2 i . 4 j. . sin .t. Themotion ofthe particle is: (A) parabolic (B) circular (C) straight line (D) None of these SIMPLE HARMONIC MOTION www.physicsashok.in 87 25. The speed v of a particlemoving along a straight line, when it is at a dista nce x froma fixed point on the line is given byV= (108 9x2)1/2.All quantities are in SI units. (A) Themotionis uniformlyaccelerated along the straight line. (B) Themagnitude of acceleration at a distance 3mfromthe fixed point is 27m/s2. (C) Themagnitude of acceleration at a distance 3mfromthe fixed point is 9m/s2. (D) Themaximumdisplacement of the particle fromthe fixed point is 4m. 26. The potential energyUof a particle is given byU= (2.5X2 + 100) Joule. If the mass of the particle is 0.2 kg, then: (A) themotion of the particle is SHM. (B) themean position isX= 0. (C) angular frequency of the oscillation is 5 rad/s (D) the time period of oscil lation is 1.26 sec. 27. In SHM: (A) displacement and velocitymaybe in the same direction. (B) displacement and velocity can never be in the same direction. (C) velocity and accelerationmay be in the same direction. (D) displacement and acceleration can never be in the same direction. 28. Aparticlemoves in theX Yplane according to the equation r . .3 i . j.cos5 t . . Themotionof theparticle is: (A)Along a straight line (B) along an ellipse (C) periodic (D) along a parabola 29. The potential energyof a particle ofmass 2 kg,moving alongthe x axis is given byU(x) = 16(x2 2x) J,where x is inmetres. Its speed at x = 1 mis 2 ms 1 : (A) Themotion of the particle is uniformlyaccelerated (B) Themotion of the particle is oscillatory fromx = 0.5mto x= 1.5m. (C) Themotion of the particle is simple harmonic (D) The period of oscillation of the particle is ./2 s. 30. If a SHMis given by y= (sin .t + cos .t)m,which of the following statements is/are true ? (A) The amplitude is 1m (B) The amplitude is 2 m. (C) Particle starts itsmotion fromy= 1m. (D) Particle starts itsmotion fromy= 0m . 31. Three simple harmonic motions in the same direction having the same amplitud e a and same period are superposed. If each differs in phase fromthe next by45º, then : (A) the resultant amplitude is (1 + 2 ) a. (B) the phase of the resultantmotion relative to the first is 90º. (C) the energy associatedwith the resultingmotion is (3 + 2 2 ) times the energy associatedwith any single motion. (D) the resultingmotion is not simple harmonic. 32. Alinear harmonic oscillator offorce constant 2 × 103N/mand amplitude 0.01mhas a totalmechanical energy of 160 J. Its : (A)maximumpotentialenergyis 100 J. (B)maximumkinetic energy is 100 J. (C)maximumpotential energy is 160 J. (D)minimumpotentialenergyis zero. SIMPLE HARMONIC MOTION www.physicsashok.in 88 33. Auniformcylinder of lengthLandmassMhaving cross-sectional areaAis suspended, with its lengthvertical, froma fixed point by amassless spring, such that it is half-submerged in a liqui d of density . at equilibrium position.When the cylinder is given a small downward push and released it starts oscillating verticallywith small amplitude. If the force constant of the spring is k, the frequencyof oscil lation of the cylinder is : (A) 1/ 2 1 k A g 2 M . . . . . .. .. (B) 1/ 2 1 k A g 2 M . . . . . .. .. (C) 1 k gL2 1/ 2 2 M . . . . . .. .. (D) 1/ 2 1 k A g 2 A g . . . . . .. . .. 34. A particle of mass m is executing oscillations about the origin on the x-axi s. Its potential energy is V(x) = k|x|3where k is a positive constant. If the amplitude of oscillations is a, then its time period Tis (A) proportional to 1/ a (B) independent of a [JEE, 98] (C) proportional to a (D) proportional to a3/2 35. Aparticle free to move along the x-axis has potential energy given by exp( x2)] for . < x < +., where k is a positive constant of appropriate U(x) = k[1 dimensions. The (A) at point awayfromthe origin, the particle is in unstable equilibrium. (B) for any finite nonzero value of x, there is a force directed awayfromthe ori gin. (C) ifits totalmechanical energyis k/2, it has itsminimumkinetic energyat the or igin. (D) for smalldisplacements fromx= 0, themotion is simple harmonic. [JEE, 99] 36. Three simple harmonic motions in the same direction having the same amplitud e a and same period are superposed. If each differs in phase fromthe next by45º, then [JEE, 99] (A) the resultant amplitude is (1. 2) a (B) the phase of the resultantmotion relative to the first is 90º (C) theenergyassociatedwiththeresultingmotionis (3. 2 2) timesthe energyassociat edwithanysinglemotion. (D) the resultingmotion is not simple harmonic. 37. The period of oscillation of simple pendulumof lengthL suspended fromthe roo f of a vehiclewhichmoves without friction down an inclined plane of inclination . is given by [JEE, 2000] (A) 2 L gcos . . (B) 2 L gsin . . (C) 2 L g . (D) 2 L g tan . . 38. Aparticle executes simple harmonicmotionbetween x= Aand x= +A.The time taken for it to go from0 to A/2 is T1 and to go fromA/2 toAisT2. Then [JEE(Scr), 01] (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2 39. Aparticle is executing SHMaccording to y = a cos .t. Thenwhich of the graphs represents variations of potential energy: [JEE(Scr), 03] I II i P.E. III IV P.E. x (A) (I)&(III) (B) (II)&(IV) (C) (I)&(IV) (D) (II)&(III) C = 2B. Ablock Pofmassmis placed on a frictionless horizontal surface. M .in 89 40.Another blockQ of same mass is kept on P and connected to the wallwith the help of a spring of spring constant k as shownin the figure. M (C) 1 6k 2.2 sec 10 5mm (B) 5mm 0. C = 2B.t + C sin . 06] (A) for any value ofA.T and he c ommits a humanerror of 0. Astudent performs an experiment for determination of g = (4.2/32 cm/s2 (C) . Its two ends are attached to two springs of equal spring constants k.2 sec 20 5mm (C) 5mm 0.t cos .1 sec. M (D) 1 24k 2. Asimple pendulumhas time period T1. Themaximumvalue of the friction force between P andQis [JEE. amplitude = | B | 43.2cm/s2 [JEE. 06] .T n Amplitude of oscillation (A) 5mm 0.t represents SHM [JEE. Thex tgraphofaparticleundergoing simpleharmonicmotionis shown below.l. The springs are fixed to ri gid supports as shown in the figure. µS is the coefficient of friction Q P k µS between P andQ. Auniformrod of lengthLandmassMis pivoted at the centre. 09] 45.1 sec 20 1mm (D) 1mm 0. 04] (A) kA (B) kA/2 (C) zero (D)µSmg 41. The blocksmove together performing SHMof amplitudeA. l .For the takes the time on n oscillationswiththe stopwatch of least count .t + B cos2.physicsashok.SIMPLE HARMONIC MOTION www. the frequencyof oscillation is [JEE. C = 0 (D) ifA= B. 05] (A) 5/6 (B) 11/10 (C) 6/5 (D) 5/4 42. and the rod is free to oscillate in the horizon tal plane.1 sec 50 1mm 44. M (B) 1 k 2. The rod is gently pushed through a small angle .2/32 cm/s2 (D) 3 / 32 . Function x =Asin2. For which of the following data.When the period of suspensionmoves vertic ally up according to the equation y= kt2where k = 1m/s2 and t is time then the time period of the pendulu mis T2 then (T1/T2)2 is [JEE(Scr). the acceleration of the particle at t = 4/3 s is 1 1 0 4 12 t(s) x(cm) 8 (A) 3 / 32 . amplitude = | B 2 | (C) ifA= B.2l/T2)l = 1mand he commits anerror of . themeasurement of gwillbemost accurate ? [ JEE. in one direction and released. B and C (expect C = 0) (B) ifA= B. 09] (A) 1 2k 2.2cm/s2 (B) . The amplitude of the point Pis [JEE.46. k (D) 2 1 2 k A k . ThemassMshownin the figure oscillates in simple harmonicmotionwith amplitude A. k . 09] P M (A) k1 k2 1 2 k A k (B) 2 1 k A k (C) 1 1 2 k A k . . . . . . . . .. .... . (B) The block-bullet systemperforms oscillatorymotionbut not SHMabout y=mg/k. 3. . . . . . The originO is considered at distance equal to natural length of spring fromceiling and vert ical downward direction as +veYaxis.. .. .SIMPLE HARMONIC MOTION www. vibrations/sec.When the sytemis in equilibriuma bullet ofmass m/3moving in verticalupward directionwith velocityv0 strikes the block and embed s into it. (C) The block-bullet systemperforms SHMabout y = 4mg/3k. .. (D) 2 2 mv0 4mg 6k 3k ... v0 m m/3 Y O k l Based onabove information. (C) 4m sin 1 4mg sin 1 mg 6k 3kA 3kA . (B) 3k cos 1 mg cos 1 4mg 4m 3kA 3kA . . . ... . .physicsashok. . . . . with an amplitude of 5 cm a nd a frequency of 10/.Ablock is placed on the platformat the lowest point of its path.. . . . answer the following equations : 1. .. . . ... (D) None of the above Passage 2 A platform is executing SHM in a vertical direction... . . (B) 2 20 mv mg 12k 3k . . 2. .. .. (C) 2 20 mv mg 6k k .. . Mark the correct statement(s).As a result theblock (withbullet embeddedinto it)moves upand starts osci llating. ... (D) The block-bullet systemperformoscillatorymotion but not SHMabout y= 4mg/3k. .. . The amplitude ofoscillationwould be (A) 2 20 4mg mv 3k 12k . . .. . .in 90 PASSAGE Passage 1 Ablock ofmassmis suspended fromone end of light spring as shown. . . The time taken by block-bullet systemto move fromy=mg/k (initial equilibriump osition) to y= 0 (natural length of spring) is [Arepresents the amplitude ofmotion] (A) 4m cos 1 mg cos 1 4mg 3k 3kA 3kA . . . . . . . . (A) The block-bullet systemperforms SHMabout y=mg/k. . ... . 3 cmabove equilibriumposition (B) 1. the normal contact force betwe en the block and platformincreases.[Take g = 10m/s2] Answer the following questions based on above information : 4. the block returns to the platform? (A) 1.5 cmfrommean positionwhen platformismoving up. (C) 2. Mark the correct statement(s). 6. At what point will the block leave the platform? (A) 2. (D) 2. (D) Both (B) and (C) are correct. (C)As platformmoves up awayfrommean position. At what point. the normalcontact force between th e block and platform decreases.5 cmabovemean positionwhen platformismoving down.5 cmfrommean positionwhen acceleration is acting down and velocity is in u pward direction.3 cmbelowequilibriumposition . 5.3 cmabove equilibriumposition (D) 4.5 cmbelowthemean position. (B)As platformapproachesmean position frombottom. (A) Normal contact force between the platformand block is constant. (B) 2.3 cmbelowequilibriumposition (C) 4. K1. Fig.Thering is symmetrically attached with two springs.Neglecting gravity. Equation of position of the block in co-ordinate systemshown in Figure is x = 10 + 3 sin(10t).. Two identical blocks Aand B ofmassm= 3 kg are attached with ends of an ideal spring of force constant K = 2000 Nm 1 and rest over a smooth horizontal floor. If the ring is slightly displaced vertical ly. p . each of stiffness k. where t is in second a nd x in cm.. Calculate 1 kg 3 kg x (i) newamplitude of oscillations. as shown in the fig.physicsashok.5 kg is attachedwith upper end of a vertical spring of fo rce constant K= 1000Nm 1 as shown in fig. and (iv)maximumcompressionof the spring.6ms 1 as shown in fig. (iii) oscillation energyof the system.moving towards the originwith velocity 30 cm/sec collides with the block performing SHM at t = 0 and gets stuck to it. calculateperiodofsmalloscillationsoftheparticle alonga line perpendicul ar to the plane of the figure. attachedwith four identical springs.Another identical block A falls froma height h = 49. shows a particle ofmassm= 100 gm. O (ii) newequation for position of the combined bodyand (iii) loss of energyduring collision. 5. In the arrangement shown in fig. One end of an ideal spring is fixed to awall at originOand axis of spring is parallel to x-axis. The combined body starts to per form verticaloscillations. 100 strikes the blockA and gets stuck to it.Another identical block C mo ving with velocity v0 = 0. Another block ofmassM=3 kg.SIMPLE HARMONIC MOTION www. pulleys are small and light and springs are ideal. determine its time period. eachoflengthl =10 cm. InitialtensionineachspringisF0 =25 newton. Calcula te for subsequent motion m V K 0 (i) velocityofcentre ofmass of the system.Ablock ofmass m= 1 kg is attached to free end of the spring and it is performing SHM. with the horizontal. K2. (g= 10ms 2) 3.in 91 LEVEL 2 1. Ablock B ofmassm= 0. m m (ii) frequencyof oscillations of the system. Calculate period ofsmall vertical oscillations of block ofmassm. Each sprin g makes an angle . K2 K4 K1 K3 m 6. h A B Calculate amplitude of these oscillations. Aringofmassmcanfreelyslide ona smoothverticalrod.Neglect friction. k k m rod 2.5 cmon the block B and gets stuckwith it. K3 and K4 are force constants of the springs. 4. In the arrangement shown in Fig.A m B C D 7.Ahorizontal light spring of force constant K= 100 Nm 1 fixed at one end keeps the systemin static equilibrium. bodyB is a solid cylinder of radius R = 10 cm withmassM= 4 kg.. (g = 10ms 2) . It can rotatewithout friction about a fixed horizontal axisO Ablock A ofmass m = 2 kg suspended by an inextensible thread is wrapped around the cylinder. Calculate A R O (i) initial elongationinthe spring and B (ii) period of small vertical oscillations of the block. Find the period of small oscillations in a vertical plane performed by a ball ofmassmfixed at themidle of a horizontally stretched string of length l. and the spring stiffness is k. The mass of the thread and the spring are negligible. Find the naturalfrequenc yof oscillation.physicsashok. (a)Ablock ofmassmis tied to one end of a stringwhich passes over a smoothfix ed pulleyAand under a light smoothmovable pulleyB. Find the amplitude. 13.t +4 cos . the thread does not slip over the pulleyand there is no f riction in the axis of the pulley. Aparticlemoves along the x-axis according to the lawx = a cos . Find also itsmaximumvelocitywhen the amplitude of vibration is 1 cm. m R 14. a m a O 2 (b) theminimumvalue of awhenthe bodystarts falling behind the plank. 11.SIMPLE HARMONIC MOTION www. itsmoment of inertia relative to the axis of rotation is I.= 11 rad s 1 between the levels 1 and 2 separated by a distance 2a as shown infig. 15. Theother end of the string is attached to the lower end o fa spring ofspring constant k2.t. Arigid rod ofmassmwith a ballofmassMattached to the free end is restrained to oscillate in a verticalplane as shown in the figure. m A B k2 . 1 (c) the amplitude of oscillation at whichthe body jumps up to a height h= 50 cmr elative to level 1.t.in 92 8. frequencyand epoch ofthe simple harmonicmotionrepresente d byx=3 sin . Ahelical spring elongates 10 cmwhen subjected to a tension of 5N. Find : (a) the force that the body exerts on the plank as it moves fromlevel 1 to 2 whe n a = 4 cm. 12. Find thema sswhich should be attached to the bottomof the spring so that when pulled down and released themasswillvibr ate twice per second. The radius of the pulley isR. Find the frequency of small oscillations of the arrangement illustrated in f igure. 10. Aplankwith a body ofmassmon it executes simple harmonicmotionof cyclic frequency . M 3/4L L/4 k m 9. the mass of the body ism. Find the d istance that the particle covers during the time interval t =0 to t. The tension of the string is assumed to be constant and equal to F >> mg. Find the period of smalloscillations ofmassmabout its equilibriumposition. The other end of a s tring is fixed to a ceiling. Find angular frequency of small oscillation of the system. .k1 Figure (a) m A B k2 k1 Figure (b) (b)Ablock ofmassmis attached to one end ofa light inextensible string passing ov er a smooth light pulleyB and under another smooth light pulleyAas shownin the figure.A and B are held by springs of spring constants k1 and k2. (i) if endAismoved down througha smalldistance d and released. Find the angular frequencyofvibrationfor smalloscillationof the system.t).physicsashok.SIMPLE HARMONIC MOTION www. B D C k b 1 k2 m 17. M l b m k1 k2 C 20. Find the angular frequencyofmotion of blockmfor smallmotion of rodBDwhenwe neglect the inertial effects of rod BD. Find the angular frequency for small oscillation of block ofmassmin the arra ngement shown in the figure. Aplankwith a body ofmassmplaced on it start moving straight up according to cos . ArodAB ofmassMis attached as shown belowto a spring of constant k. b m k B A l (ii) determine the largest allowable value of d if the blockmis to remain at all times in contactwith the rod. .= 11 rad/s. Auniformbroad of lengthL andweightWis balanced on a fixed semi-circular cylinder of radiusR as shown in the figure. the lawy = a(1 where y is the displacement fromthe initial position. l/2 l/2 k2 k1 m k3 18. Two springs having the spring constants k1 and k2 respectively are attached to the rod at the distances b and c fromthe hinge as shown in the figure. Aparticle ofmassmfree to move in the x y plane is subjected to a forcewhose co mponents are Fx = kx and . Find : (a) The time dependence of the force that the body exerts on the plank. A rod ofmass mand length l is hinged at its upper end and carries a block of mass M at its lower end. l r 21. Spring constant are k1 and k2. (b) Theminimumamplitude of oscillation ofthe plank at which the bodystarts falli ng behind the plank. Neglect themass of the rod. If the plank is tilted slightlyfromi ts equilibriumposition.in 93 16.Neglect fr iction forces. 19. determine the period of vibration. determine the period ofits oscillations. 22.A small block ofmassmis placed on the rod at its free endA. (c) If the condition obtained in (b) ismet. 23. m m 1 2 k1 k2 (a) Frequencyof such oscillations. In the shown arrangement. they together perform simple harmonicmotion.Obtaion. There is no friction betweenm1 and the surface. 3). The particle is released when t = 0 at the point (2. what canbemaximumamplitude of their oscillations ? . Prove that the subsequentmotion is simple harmonic along the straight line 2y 3x = 0. (b) The conditionif the frictional force on blockm2 is to act in the direction o f its displacement frommean position. If the blocks are displaced slightly. where k is a constant. The coefficient of friction betweenm2 andm1 isµ. both the springs are in their natural lengths.Fy = ky. What is the vibrational frequencyof themolecule ? [REE.96] 3.physicsashok. find the frequencyofthese oscillations. Position ofmassm1 at time t is given bythe equation. Asmallbodyattached to one end of a verticallyhanging spring is performing SHM about it s mean position with angular frequency . This systemwhilemoving with a velocity v0 along +ve x axis pass through th e origin at t = 0. [REE.SIMPLE HARMONIC MOTION www. (b) l0 in terms ofA. The atoms are compressed towards eachother fromtheir equilibriumposit ions and released. 2000] 4.t) Calculate : [JEE.t + cos2. [JEE. The potentialenergyof themolecu le for the interatomic separation r is given byV(r) = A+B(r r0)2. If the bob executes SHMin the ver ticaldirection. and amplitude a.The Young smodulus of thematerial of the string isY./4] and x2 = 5 2 (sin 2.in 94 LEVEL 3 1.At this position the string snaps. [REE.t) their amplitudes are in th e ratio 1 : 2 [REE. Ablock is kept on a horizontal table. Find themaximum amplitude of the table at which the block does not slip on the surface. andAandBare positive constants. 6. 03] (a) Position of the particlem2 as a function of time. 05] . 01] 5.96] 2. the table is undergoing simple harmonic motion of frequency 3Hz in a horizontalplane. Two massesm1 andm2 connected bya light spring ofnatural length l0 is compress ed completelyand tied bya string. Adiatomicmolecule has atoms ofmassesm1 andm2. Statewhether true or false Two simple harmonicmotions are represented by the equations x1 = 5sin [2.72.The coefficient of static frictionbetweenblock and the table sur face is 0. Abob ofmassMis attached to the lower end of a vertical string of lengthL and cross sectional areaA. If at a height y* fromth e mean position the bodygets detached fromthe spring. calculate the value of y* so that the height m y0 H attained by the mass ismaximum. x1(t) = v0t A(1 cos .2 > g). The body does not interact with the spring dur ing it s subsequent motion after detachment.t + . (a.where r0 is the equilibriumseparation. 2m/sec. . (i) 20 cm. C C B C B ABC ACD AC BCD B Q. 31 32 33 34 35 36 37 38 39 40 Ans. . (B QRS).physicsashok. (i) 0.135 Joule 4. . (C PQS). f 1 3k 2 27M 7m . (ii) 5 10 Hz . (iii) 0. 5 cm 3. [(A R). l 3sin(5t) cm. (C Q). . (ii) 0. (B RS). C ABD D D C D PASSAGE Q. T 2 m 2k . 1 2 3 4 5 6 7 8 9 10 Ans. (D P)] 3.)cmor x = 10 ) 0. . 8. 1 2 3 4 5 6 Ans. 7. (D RS)] LEVEL 1 Q. 41 42 43 44 45 46 Ans. . C ABC A CD D D B D B D Q. . [(A QR). (iv) 3 10 mm. (iii . 9. C B A A C D LEVEL 2 1. 0. [(A RS). . (B P). sec. . T m F . 11 12 13 14 15 16 17 18 19 20 Ans. 2. 6. B B C A B BC A B A A Q. sec.4 . .in 95 Answer Key ASSERTION-REASON TYPE Q. . . (i) 3 cm.SIMPLE HARMONIC MOTION www. 1 2 3 4 T 4 m 1 1 1 1 k k k k .. . AC BC B A D AC A A A B Q.09 Joule. (ii) x = 10 + 3sin(5t + .02 . 21 22 23 24 25 26 27 28 29 30 Ans. (C P). 5. . (D S)] 2. A C A B D MATCH THE COLUMN 1. . 1 2 3 4 5 Ans. S = a[n + 1 cos (. (a) a 2 N mg 1 cos t g . .t n 2 . . l . m= 310 g. g . . . 2 1 2 2 2 1 2 k k C m k C k (a b) . 14.5 cm/sec. (c) a = 0. . 2 3 1 2 3 1 4k k k m k 4k . (b) min 2 a . .physicsashok./2. . 13. . . 2 f 1 k 2 m 1/R . . . 17.2 m. . f = . 18. . . . . and epoch = 53º 12. . (b) 1 2 1 2 1 k k 2 m(k k ) . . vmax = 12. (a) 1 2 1 2 m(k k ) T 2 k k . 16.SIMPLE HARMONIC MOTION www. A= 5. . .t n 2 . 15. . )] when n is even. . . )] when n is odd. . . . 11. . . . (i) 2 2 T 2 (m M/ 3) kb . . . S = [n + cos (. . . (ii) 2 .in 96 10. . . T 2 L 3 gr 4 . .max 2 g (m M/ 3) d kb . . (b) 1 0 2 m 1 A m . . . . l 19. 3 LEVEL 1. . . l l 20. .2 cos. (a) 1 2 1 2 1 k k 2 m m . . .t). (c) 1 2 2 1 2 2 1 µ(m m )m g m k m k . . . . 1 2 1 2 1 m m 2 f 2B(m m ) . . 2 2 1 2 2 (k b k c ) (M/m/ 2)g (M m/ 3) . . 5. 2 y mg g a . (a) N = m(gta. . . 21. (b) 8 km. 23. ML 4. . . . . True 2. 2cm 3. . 1 YA 2. (b) 1 1 2 2 k m k m . (a) 1 2 0 2 m x v t A(1 cos t) m . . l 6. . .k . . . TIRUPATI. 9440025125 .OPTICS DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. PH NO. (xii) Visual region and intensity of image depend on size of mirror.in 1 Review of Concepts (a) Due to reflection. Real Image P Virtual Object P n n (ix) If reflected beam or refracted beam from an optical instrument is diverging in nature. image is real. wavelength and speed of light change. P P Real Object Virtual Object n n . . the reference frame is chosen in which optical inst rument (mirror. (vii) The converging point of reflected or refracted beam from an optical instru ment behaves a image. the obje ct is real. (iii) Image size = Object size. image is real. lens.OPTICS www. object is virtual. lens etc) is converging in n ature. reflected ray and normal on inident point are coplanar. (v) If incident beam on optical instrument (mirror. image is virtual. (ii) For virtual object.physicsashok. (x) For solving the problem. none of frequency. (B) Law of reflection : (i) Incident ray. . (iv) The converging point of incident beam behaves as object. (xi) The formation of image and size of image is independent of size of mirror. . . image is virtual. (vi) If incident beam on the optical instrument is diverging in nature. Some important points : In case of plane mirror : (i) For real object.) is in rest. (viii) If reflected beam or refracted beam from an optical instrument is converg ing in nature. (ii) The angle of incidence is equal to angle of reflection. etc. then the number of images is euqal to n. Here. v cos. object is lying on that circle. . is even number. . is odd and object is not situated on bisector of angle between mirrors. v cos. (iii) If . (ii) If .in 2 (xiii) If the plane mirror is rotated through an angle . Image Rest Object (xviii) v vm 2 vm . Also. 360. .physicsashok. . 360. the reflected ray and image is rotated through an angle 2 .OPTICS www. Image Rest Object (xvi) v Image Rest Object (xvii) v sin. v sin. then the focal length does not c hange. n . (D) Law of reflecteion in vector form : Let e 1 = unit vecotr along incident ray. 360 where . 360. the number of image is n 1. = angle between mirrors. (xiv) If mirror is cut into a number of pieces. . is odd number and object is placed on bisector of angle between mirrors. (xv) The minimum height of mirror required to see the full image of a man of hei ght h is h/2. in the same sense.v Object Image In rest (xix) vm 2 vm Image Object (xx) v vm 2 vm + v (C) Number of images formed by combination of two plane mirrors : The images for med by combination of two plane mirrors are lying on a circle whose centre is at the me eting points of mirrors. then number of images is n 1. (i) If . .Let e 2 = unit vector along reflected ray n n 1 . 2 . 2 . e 2 .e 1. n = unit vector along normal on point of incidence Then.n . e 1 .n (e) Spherical mirrors : (i) It is easy to solve the problems in geometrical optics by the help of co-ord inate sign convention. . If object and image both are real. 1 u 1 1 Also. 2. .1 . = . 4. = . moon or distant object is formed at focus of mirror. These formulae are only aplicable for paraxial rays. . is negative.OPTICS www. Laws of Refraction 1. . is positive.. (iii) All distances are measured from optical centre. . ..1 sin. 1. (B) . . (a)The incident ray. If object is real but image is virtual.1 . = the angle subtended by sun or moon s disc Then tan .2 sin.2 . 5. If object is virtual but image is real. . It means optical centre is taken as origin. . is in radian. is positive. the refracted ray and normal on incidence point are copl anar. . is negative 3. (iv) The sign convention are only applicable in given values. (v) The transverse magnification is object size u image size . . d y D Here. = focal length of the mirror d = diameter of the image . If y = the ddistance of sun or moon from earth. R = 2 . If object and image both are virtual. . Image of star. . D = diameter of moon or sun s disc.in 3 x x y y x x y y x x y y x x y y x x y y (ii) The mirror formula is . D F d Sun .physicsashok. . ..cons tant . . n Let.1 (C) Snell s law in vector form: .2 .2 . ..1 . e 1 = unit vector along incident ray e 2 = unit vecotr along refracted. 1 .2 2 . optical path remains constant. .2 x2 .. .e. Then .. 1 2 1 2 2 1 c c . .1 . (ii) The value of refractive index depends uponmaterial of medium.. 1 2. colour of lig ht and temperature of medium. real depth (B) When object is in air and observer is in Air Observer Denser medium .1 c1 . . Some important points : (i) The value of absolute retractive index . 1 (vi) The frequency of light does not depend upon medium.e 2 . c1 . refractive index decreases. . (iv) Optical path is defined as product of geometrical path and refractive index . c .2 .2 c2 . .x (v) For a given time.e. .e 1 .physicsashok. is always greater or equal to one. i. . . 2 1 1 2 c . n .1 .1 x1 . (iii) When temperature increases. . . i.OPTICS www. (a) When observer is in rarer medium and object is in denser medium: Then apparent depth . . . . . .in 4 n = unit vector along normal on incidence point. .1 . c . c2 . .. .e. . . dt dx dt dx 2 2 1 . .2 .. . optical path = . . n . cons tant . i. . ... . t1 . Here..ti = t1 + t2 + . .. . t 1. .. . . . x .. (D) The equivalent rerfractive index of a combination of a number of slabs for n ormal incidence is i i i t t . (ii) The object shiftness does not depend upon the P P Q Object shiftness = x t position of object.) P Object Apparent depth P Real depth denser medium: real position . .(. .1 . . .. 1 (i) This formula is ony applicable when observer is in rarer medium.. apparent position (C) The shift of object due to slab is .. . . 2 2 1 1 i ti t t .2 t2 . . . . (iii) Object shiftness takes place in the direction of incidence ray. . 2 ) to rarer medium ( .in 5 (e) The apparent depth due to a number of media is i ti . . . . i t r d . ) 1 i r c i i i < C i = C Denser medium (. total internal reflection takes place. . . . Mathematically. . 3.1). . . 2 sinC 1 . . . . the c Rarer . the corresponding deviation is .2 90° corresponding angle of incidence is critical angle.. i graph is (i) Critical angle depends upon colour of light. . then for 90° angle of refraction../2 and temperature of medium. . material of medium. . the corresponding deviation is sin sini i 1 1 2 . 2i when i > C 4. for i = C Rarer medium (.1 Denser . ) 2 (ii) When angle of incidence is greater than critical angle. (a) Cricital angle : When a ray passes from denser medium ( .physicsashok. . (f) The lateral shifting due to a slab is d = t sec r sin (i r). . . refraction takes place. . . The .OPTICS www. . (B) (i) When angle of incidence is lesser than critical angle. c i . . . (g) . ..i.A. 2 A A B C n n r r i i (f) For limiting angle of prism. . . i graph for prism: (h) For minimum deviation. > cosec the limiting If angle rays . i = i = 90°. of prism exceeds the limiting values. (i) i = i and r = r (ii) 2 sin A 2 A sin m . .m of incidenct.. i = 90° (e) For not transmitting the ray from prism. angle of prism = 2C where C is critical angle. . i = 90° (D) For grazing emergence. . i . then the are totally reflected.. .i.(ii) Critical angle does not depend upon angle PRISM (a) Deviation produced by prism is . (B) r + r = A (C) For grazing incidence. . . r = . y 1 r (n) .. 1 u 1 1 .D . .OPTICS www. .y . . . LENS 1. 0 x A x B O C . . . (B) Transverse magnification.1. ray is passing through prism symmetrically. . . i = 90° or i = 90° (j) For thin prism. Lens formula : . .1 . (i) For maximum deviation .. . A (l) Angular deviation.. . . . .. . 2 r y (o) For dispersion without deviation. . .r . . .. .y . . . . = image distance. .1.in 6 In the case of minimum deviation. ...physicsashok. . u r 2 1 . r = radius of curvature of spheri cal surface (a) For plane surface.A (m) dispersive power = . Here. . . . . . . . .1 .2 Refractive surface formula. . . 0 (p) For deviation without dispersion. . . . . .y . .. . . u = object distance. . object size u m Image size 2 1 . .2 . A (k) Angular dispersion. (C) Refractive surface formula is only applicable for paraxial ray.. . . D .max . . .. (f) f(+ve) (i) f(-ve) (ii) f(-ve) (iii) f(+ve) (iv) f(-ve) (v) f(+ve) (vi) (g) This lens formula is applicable for converging as well diverging lens. . . (D) Magnification formula is only applicable when object is perpendicualr to opt ical axis. . . .1 .1 . .. .. . . . (C) object size u m image size . . . 1 1 2 2 1 r 1 r 1 1 . . (e) Lens formula and the magnification formula is only applicable when medium on both sides of lenses are same. .(a) Lens formula is only applicable for thin lens. .Thin lens maker s formula : 2 . . . formula is not applicable for lens. (B) r = 2 . . . . .e. focal length does not change.5 . (g) If lens is cut by a vertical. (e) When medium on both sides of lens are not same.physicsashok. (C) Intensity is proportional to square of aperture. .OPTICS www.2 .e.. (k) If a number of lenses are in contact. . Then number of images of an object formed by the lens is + + + + + + . .. . . (a) Thin lens formula is only applicable for paraxial ray. Then both focal lengths are not same to each other. (f) If a lens is cut along the diameter. . . (j) The equivalent focal elngth of co-axial combination of two d<f1 d<f2 o1 o2 f1 f2 d lenses is given by 1 2 1 2 1 1 d F 1 .in 7 2. . Then converging lens behaves as diverging lens and vice versa. 1 2 1 1 1 .6 equal to number of different media.1 . .4 .3 . (D) When lens is placed in a medium whose refractive index is greater than that of lens. . i.. it converts into two lenses of different + f f1 f2 focal lengths. . (B) This formula is only applicable when medium on both sides of lens are same. . . . . . (i) The minimum distance between real object and image in is 4 . 2 . i.1. . then . (h) If a lens is made of a number of layers of different refractive index (shown in figure). . F = focal length of equivalent mirror. . . . .. 1 Here. .. . 1 (ii) Power of mirror is F P . 1 (m) If a lens is silvered at one surface. . 1 1 2 2 1 r 1 r 1 Pm = Power of silvered surface Fm . . . then the system behaves as an equivale nt mirror. . F P . . . . 2 F r2 m .. . where r2 = radius of silvered surface. PL = Power of lens = . . .. . F P . . . whose power P = 2PL + Pm Here.1 2 1 1 F 1 (l) (i) Power of thin lens. . 1 Here. OPTICS www.physicsashok.in 8 ASSERTION & REASION THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R ) is given just below it of the statements, mark the correct answer as (A) If both A and R are true and R is the correct explanation of A. (B) If both A and R are true but R is not the correct explanation of A. (C) If A is true but R is false. (D) If both A and R are false. (E) If A is false but R is true. 1. Assertion (A) : A single ray can t be isolated from a source however small it m ay be. Reason (R) : The concept of single ray is hypothetical. (A) (B) (C) (D) (E) 2. Assertion (A) : Virtual images can be photographed. Reason (R) : Rays from virtual images are diverging. (A) (B) (C) (D) (E) 3. Assertion (A) : Virtual object can t be seen by human eye. Reason (R) : Virtual object is formed by converging rays. (A) (B) (C) (D) (E) 4. Assertion (A) : A Convex mirror is used as rear view mirror. Reason (R) : The Convex mirror always forms virtual, erect and diminished image. (A) (B) (C) (D) (E) 5. Assertion (A) : The behavior of any lens depends on surrounding medium. Reason (R) : A lens can be looked upon as a collection of small prism with varyi ng prism angle. (A) (B) (C) (D) (E) 6. Assertion (A) : Human eye can see virtual object. Reason (R) : Virtual object is formed by apparent intersection of incident rays. (A) (B) (C) (D) (E) 7. Assertion (A) : Real image is formed by real intersection of reflected or ref racted rays. Reason (R) : Real image can t be obtained on screen. (A) (B) (C) (D) (E) 8. Assertion (A) : If a portion of lens or mirror is blocked or removed, then in tensity of image reduces. Reason (R) : As every portion of lens or mirror forms image, hence blocking or r emoving a portion will result in intensity reduction. (A) (B) (C) (D) (E) 9. Assertion (A) : A rectangular glass slab produces no deviation and no dispers ion. Reason (R) : Dispersive power of glass is zero. (A) (B) (C) (D) (E) 10. Assertion (A) : A double convex lens .. .1.5. has focal length 10 cm. When i mmersed in water 4 3 ... . .. . . , its focal length becomes 40 cm. Reason (R) : 1 2 1 1 1 l m m f R R . . . . . . . . . . . . (A) (B) (C) (D) (E) OPTICS www.physicsashok.in 9 11. Assertion (A) : A convex lens of glass .. .1.5. behave as diverging lens whe n immersed in carbon disulphide of higher refractive index .. .1.65.. Reason (R) : A diverging lens is thinner in the middle and thicker at the edges. (A) (B) (C) (D) (E) 12. Assertion (A) : A biconvex lens of focal length 10 cm is split into two equa l parts by a plane parallel to its principal axis. The focal length of each part will be 20 cm. Reason (R) : The focal length depends on how many parts the convex lens has been split. (A) (B) (C) (D) (E) 13. Assertion (A) : Radius of curvature of a convex mirror is 20 cm. If a real o bject is placed at 10 cm from pole of the mirror, image is formed at infinity. Reason (R) : When object is placed at focus, its image is formed at infinity. (A) (B) (C) (D) (E) 14. Assertion (A) : For a prism of refracting angle 60° and refractive index 2 , m inimum deviation is 30°. Reason (R) : At minimum deviation, 1 2 30 2 r . r . A . . (A) (B) (C) (D) (E) 15. Assertion (A) : Image formed by concave lens is not always virtual. Reason (R) : Image formed by a lens if the image is formed in the direction of r ay of light with (A) (B) (C) (D) (E) 16. Assertion (A) : Minimum deviation for a given prism does not depend on the r efractive index . of the prism. Reason (R) : Deviation by a prism is given by . . 1 2 . . i . i . A and does not have the term . . (A) (B) (C) (D) (E) Level # 1. Objective Type Question Multiple Choice Question with ONE correct answer : 1. Two plane mirrors M1 and M2 are inclined to each other at 70°. A ray incident o n the mirror M1 at an angle . falls on M2 and is then reflected parallel to M1 for (A) . = 45° (B) . = 50° (C) . = 55° (D) . = 60° 2. A light ray is incident on a horizontal plane mirror at an angle of 45°. At what angle should a second plane mirror be placed in order that the reflected ray finally be reflected horizontally from the second mirror, as shown in figure. (A) . = 30° (B) . = 24° (C) . = 22.5° (D) . = 67.5° 3. A plane mirror is placed in y-z plane facing towards negative x-axis. The mir ror is moving parallel to yaxis with a speed of 5 cm/s. A point object P is moving infront of the mirror with a velocity (3 cm/s)i + (4 cm/s) j + (5 cm/s)k . Find the velocity of image with respect to mirror (A) ( 3 cm/s)i + (4 cm/s) j + (5 cm/s)k (B) (3 cm/s)i + (4 cm/s) j + (5 cm/s)k (C) (3 cm/s)i (4 cm/s) j (5 cm/s)k (D) none of the above. OPTICS www.physicsashok.in 10 4. The size of the face of a dancer is 24 cm x 16 cm. Find the minimum size of a plane mirror required to see the face of dancer completely by (i) one eyed dancer. (ii) two eyed dancer. (Distance between the eyes is 4 cm.) (A) (i) 12 x 8 cm2 (ii) 12 x 6 cm2 (B) (i) 8 x 10 cm2 (ii) 12 x 2 cm2 (C) (i) 10 x 12 cm2 (ii) 9 x 8 cm2 (D) (i) 12 x 2 cm2 (ii) 6 x 13 cm2 5. A bullet of mass m2 is fired from a gun of mass m1 with horizontal velocity v . A plane mirror is fixed at gun facing towards bullet. The velocity of the image of bullet formed by the plane m irror with respect to bullet is (A) . .. . . .. . . 1 2 m 1 m (B) . . .. . . .. . . 1 1 2 m m m (C) . . . . 1 2 1m 2 m m (D) none of these 6. In the given figure, the angle of reflection is (A) 30° (B) 60° (C) 45° (D) none of these. 7. Two plane mirrors A and B are aligned parallel to each other, as shown in figure. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with the plane of figure. The 0.2 m A B 30° 2 3 m maximum number of times the ray undergoes reflections (excluding the first one) before it emerges out is (A) 28 (B) 30 (C) 32 (D) 34 8. A point source of light B is placed at a distance L in front of the centre of a mirror of width d hung vertically on a wall as shown. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown . d B L 2 L The greatest distance over which he can see the image of the light source in the mirror is (A) ½ d (B) d (C) 2d (D) 3d 9. A plane mirror having a mass m is tied to the free end of a massless spring of spring constant k. The other end of the spring is attached to a wall. The spring with the mirror held vertically to the floor can slide along it smoothly. When the spring is at its natural length, the mirror Wall is found to be moving at a speed of v cm/s. The separation between k the images of a man standing before the mirror, when the mirror is in its extreme positions (A) k v m (B) k m 2 v (C) k 2v m (D) k 4v m 10. Two spherical mirrors M1 and M2, one convex and other concave having same ra dius of curvature R are arranged coaxially at a distance 2R (consider their pole sepa ration to be 2R). A bead of radius a is placed at the pole of the convex mirror a shown . The M1 M2 ratio of the size of the first three images of the bead is (A) 1 : 2 : 3 (B) 1 : 3 : 1 2 1 (C) 41 : 1 11 : 1 3 1 (D) 3 : 11 : 41 11. An object is placed in front of a convex mirror at a distance of 50 cm. A pl ane mirror is introduced covering the lower half of the convex mirror. If the distance between the object and the plane mirror is 30 cm, there is no parallax between the images formed by the two mirrors. The radiu s of curvature of the convex mirror (in cm) is (A) 60 (B) 50 (C) 30 (D) 25 OPTICS www.physicsashok.in 11 12. A rectangular glass slab ABCD of refractive index n1, is immersed in water o f refractive index n2 (n1 > n2). A ray of light is incident at the surface AB of the slab as shown. The maximum v alue of the angle of incidence .max , such that the ray comes out only from the other surface CD is g iven by (A) .. .. . .. .. . . . . . . . . . . .. . . .. . . . 1 1 2 2 1 1 n cos sin n n sin n (B) .. .. . .. .. . . . . . . . . . . .. . . .. . . . 2 1 1 1 n sin n cos sin 1 (C) . .. . . .. . . 2 1 1 n sin n (D) . .. . . .. . . 1 1 2 n sin n 13. Two thin slabs of refractive indices . 1 and . 2 are placed parallel to each other in the x-z plane. If the direction of propagation of a ray in the two media are along the vectors j b i a r 1 . . and j d i r 2 . c . x y .2 .1 then we have (A) .1a . .2b (B) 2 2 2 2 2 1 c d c a b a . . . . . (C) . 1 (a2 + b2) = . 2 (c2 + d2) (D) none of these 14. A man stands on a glass slab of height . and inside an elevator accelerated upwards with a . The bottom of the slab appears to have shifted with respect to the man by a distance (if the R. I. of the glass is . g) (A) less then . .g (B) greater than . .g (C) equal to . .g (D) can t be said. 15. A ray of light travels from a medium of refractive index . into air. If the angle of incidence at the plane surface of separation is . and the corresponding angle of deviation is D, the va riation D with . is shown correctly by the figure. C C C D D D D D1 D1 D1 D2 D2 D2 (0, 0) . ./2 (0, 0) . (0, 0) . ./2 (0, 0) . ./2 (A) (B) (C) (D) 16. An observer can see through a pin hole at the top end of a thin rod of heigh t h placed as shown in the figure. Beaker height is 3h and its radius is h. When the beaker is filled with a liquid up to a height 2h, he can see the h 2 h 3 h Eye lower end of the rod. Then refractive index of the liquid is (A) 2 3 (B) 2 3 (C) 2 5 (D) 2 5 . (Assume that the distance between rod and the wall is negligible). 17. A glass sphere of radius 5 x 10 2 m has a small bubble 2 x 10 2 m from its centr e. Bubble is viewed along the diameter of the sphere, from the side on which it lies. If refractive index of glass is 1.5 then how far from the surface will the bubble appear? (A) 2.1 cm (B) 2.5 cm (C) 1.5 cm (D) 2.0 cm OPTICS www.physicsashok.in 12 18. A ray of light travelling in a transparent medium falls on a surface separat ing the medium from air at an angle of incident 45°. The ray undergoes total internal reflection. If . is the re fractive index of the medium with respect to air, select the possible value(s) of . from the following : (A) 1.3 (B) 1.4 (C) 1.5 (D) 1.7 19. A tank contains a transparent liquid of refractive index n the bottom of whi ch is made of a mirror as shown. An object O lies at a height d above the mirror. A person P vertically above the object sees O and its image in the mirror and d O P finds the apparent separation to be (A) 2nd (B) n 1 2d . (C) n 2d (D) .1 n. n d . 20. A fish looks up at the surface of a perfectly smooth lake. The surface appea rs dark except a circular area directly above it. The plane angle . that this illuminated region subtends is (A) 48.6° (B) 24.3° (C) 97.2° (D) 12.15° 21. A ray of light enters an anisotropic medium from vacuum at grazing incidence . If . is the angle made by the reflected ray inside the medium with the interface and n( . ) is the refract ive index of the medium then, (A) n( . ) sin . = 1 (B) n( . ) cos . = 1 (C) 1 sin n( ) . . . (D) 1 cos n( ) . . . 22. The slab of a material of refractive index 2 shown in figure has a curved surface. APB of radius of curvature 10 cm and a plane surface CD. On the left of APB is air and on the right of CD is water with refractive indices as given in figure. An object O is placed at a distance of 15 cm from pole P as shown. The distance of the final image of O from P, as viewed from the left is (A) 20 cm (B) 30 cm (C) 40 cm (D) 50 cm 23. An object is placed at a distance of 12 cm from a convex lens on its princip al axis and a virtual image of certain size is formed. On moving the object 8 cm away from the lens, a real ima ge of the same size as that of virtual image is formed. The focal length of the lens in cm is (A) 15 (B) 16 (C) 17 (D) 18 24. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at the point O and PO = OQ. T he distance PO is equal to (A) 5 R (B) 3 R (C) 2 R (D) 1.5 R 25. A lens of focal length . is placed in between an object and screen fixed at a distance D. The lens forms two real images of object on the screen for two of its different positions, a di stance x apart. The two real images have magnifications m1 and m2 respectively (m1 > m2). (A) m1 m2 x . . . (B) m1 m2 = 1 (C) 4D D2 . x2 . . (D) all the above OPTICS www.physicsashok.in 13 26. A liquid of refractive index 1.33 is placed between two identical plano-convex lenses, with refractive index 1.50. Two possible P Q arrangement P and Q are shown. The system is (A) divergent in P, convergent in Q. (B) convergent in P, divergent in Q. (C) convergent in both (D) divergent in both. 27. A lens of refractive index . is put in a liquid of refractive index .. . If the focal length of the lens in air is . , its focal length in liquid will be (A) . . .. . . . ... . .1 (B) . . .. .. .1. . .. . . (C) . . . ... . .. .. . .1 (D) .. . ... ... . 28. A convergent lens is placed inside a cell filled with a liquid. The lens has a focal length +20 cm when in air and its material has a refractive index 1.50. If the liquid has a refractive index 1.60, the focal length of the system 24 cm (D) 80 cm (D) + 80 cm (A) 160 cm (B) 29. A double convex lens, made of glass of refractive index 1.5, has focal lengt h 6 cm. The radius of curvature of one surface is double than that of other surface. The small radius of curvatu re has value (A) 4.5 cm (B) 6 cm (C) 4 cm (D) 9 cm 30. If the distance between a projector and screen is increased by 1%, then illu mination on the screen decreases by (A) 1 % (B) 2 % (C) 3 % (D) 4 % 31. A lens forms a sharp image of a real object on a screen. On inserting a para llel slide between the lens and the screen with its thickness along the principal axis of the lens it is fou nd necessary to shift the screen parallel to itself d away from the lens for getting image sharply focused o n it. If the refractive index of the glass relative to air is ., the thickness of slab is (A) . d (B) .d (C) 1 d . . . (D) . . . . .1 d 32. A thin convex lens in used to form a real image of a bright point object. Th e apeture of the lens is small. A graph, shown is obtained by plotting a suitable O X Y -1 . parameter Y against another suitable parameter x. If . = the focal length of the lens u = object distance v = image distance and Real Positive Convention is used then (A) (uV) . x; (u + V) . y (B) (u + V) . x; (uV) . y (C) u . x; v u . y (D) u 1 . x; v 1 . y OPTICS www.physicsashok.in 14 33. Which of the following best represents object distance u vs image distance v graph for a convex lens. (A) y . (B) y . (C) y . (D) y . 34. Three thin prisms are combined as shown in figure. The refractive indices of the crown glass for red, yellow and violet rays are . r, . y and . v respectively and those for the flint glass are . r, . y and . v respectively. The ratio A /A for which there is no net angular dispersion. (A) . . 1 2 1 y y . . . . (B) y y y y 2 . . . . . . . (C) . . . . y y y y 1 1 . . . . . . . . (D) y y 2 y . y . . . . . . . 35. A point object is placed at distance of 0.3 m from a convex lens of focal le ngth 0.2 m cut into two equal halves, each of which is displaced by 0.0005 m, as show n in figure. If C1 and C2 be their optical centres then, (A) an image is formed at a distance of 0.6 m from C1 or C2 along principal axis . (B) two images are formed, one at a distance of 0.6 m and other at a distance O C1 C2 of 1.2 m from C1 or C2 along principal axis. (C) an image is formed at a distance of 0.12 m from C1 or C2 along principal axi s. (D) two images are formed at a distance of 0.6 m from C1 or C2 along principal a xis at a separation of 0.003 m. 36. A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face AB is totally reflected to reach on the face BC if (1983) A B . (A) sin 8 9 . . (B) 2 sin 8 3 9 . . (C) sin 2 3 . . 37. A ray of light from a denser medium strike a rarer medium at an angle of inc idence i (see Figure). The reflected and refracted rays make an angle of 90° with each other. The angles of r eflection and refraction are r and r The critical angle is i r r' (A) sin.1 .tan r . (B) sin.1 .tani. (C) sin.1 .tan r.. (D) tan.1 .sin i. are superpos ed. the lens will diverge a parallel beam of light if it is filled with (A) air and placed in air (B) air and immersed in L1 (C) L1 and immersed in L2 (D) L2 and immersed in L1 43. . and 3 . . A hollow double concave lens is made of very thin transparent material. . (C) 9 . and . (D) 9 . (B) 5 .44. .5 has both surfaces of same radi us of curvature R. . refractive index 1. A ray of light passes through four transparent media with refractive indices 1 .in 15 38. . The angles of incidence of the two extreme rays are equal. . (C) 3 4 . 0 R (C) divergent lens of focal length 3. A diverging beam of light from a point source Is having divergence angle . . . falls symmetrically on a glass slab as shown. This can be achieved by appropriately placing (A) a concave mirror of suitable focal length (B) a convex mirror of suitable fo cal length (C) a convex lens of focal length less than 0.OPTICS www.. the surfaces of all media are parallel. . and .physicsashok. . . A diminished image of an object is to be obtained on a screen 1. and . we must have (A) 1 2 .5 R (D) divergent lens of focal length 3. 2 . . .5 R (B) convergent lens of focal length 3. (C) sin 1 1 n . . . . (B) 2 3 . . If the emergent ray CD is parallel to the incident ray AB. and 4 .1. . 39. it will behave as a (A) convergent lens of focal length 3. . . If the thickness of the glass slab is t and the refractive index n. . (D) 2sin 1 1 n .2 respectively . On immersion in a medium of refractive index 1.1 . Two coherent monochromatic light beams of intensities . as shown in the figure. . The maximum and minimum possible intensities in the resulting beam are (A) 5 .2 . The rays emerge from the oppos ite faces 120° (A) are parallel to each other (B) are diverging (C) make an angle 2 [sin 1 (0.72) with each other 40. The parallel mon ochromatic rays enter the prism parallel to each other in air as shown. and 3 .72) 30°] with each other (D) make an angle 2 sin 1 (0. . . It c an be filled with air or either of two liquids L1 and L2 having refractive indices 1 . (D) 4 1 . 44. and 4 . . An isosceles prism of angle 120° has a refractive index 1.25 m (D) a concave lens of suitabl e focal length 41.0 R 42. 3 . then the divergence angle of the eme rgent beam is (A) zero (B) . A concave lens of glass.75.0 m from it. A ray of light is incident at the glass-water interface at an angle i.5 cm (C) 1.25 cm (B) 2. Two plane mirrors A and B are aligned parallel to each other. Ad ditional prism Q and R of identical shape and of the same material as P are now added as shown in the figu re. The size of the image of an object. which is at infinity. would be (A) . it em erges finally parallel to the surface of water. as formed by a con vex lens of focal length 30 cm is 2 cm. (A) R1 R2 (B) R (C) R R (D) R 47.sini (B) 1 sin i (C) 4 3 (D) 1 50.in 16 45. then the value of g . The maximum number of times the ray undergoes reflections ( including the first one) before it emerges out is (A) 28 (B) 30 (C) 32 (D) 34 48.05 cm (D) 2 cm 49. Orange.4 3. If a concave lens of focal length 2 0 cm is placed between the convex l ens and the image at a distance of 26 cm from the convex lens. Indigo. The colors of the light which will come out to air ar e (A) Violet. Which one of the following spherical lenses does not exhibit dispersion? The radii of curvature of the surfaces of the lenses are as given in the diagrams.OPTICS www. as shown in th e figure A light ray is incident at an angle 30° at a point just inside one end of A. The plane of inciden ce coincides with the plane of the figure. The ray will now suffer P Q R (A) greater deviation (B) no deviation (C) same deviation as before (D) total internal reflection 46. A beam of white light is incident on glass air interface from glass to air s uch that green light just suffers total internal reflection. Blue (B) All colors except green (C) Yellow. (A) 1. A given ray of light suffers minimum deviation in an equilateral prism p. calculate the new size of the image. Red (D) White light .physicsashok. Their equivalent focal length is 30 cm. 10 (B) 10.67 dioptres 57.5. An equilateral prism is placed on a horizontal surface. A convex lens of focal length 40 cm is in contact with a concave lens of foc al length 25 cm. A converging lens is used to form an image on a screen. A source emits sound of frequency 600 Hz inside water.1..OPTICS www.physicsashok. . The frequency heard i n air will be equal to (velocity of sound in water = 1500 m/s. 58. A point object is placed at the centre of a glass sphere of radius 6 cm and refractive index 1. (B) 1 2 b f u f . The size of the image is approximately equal to (A) 1 2 b u f f . (C) . . .5 dioptres (B) 6. The distance of virtual image from the surface is (A) 6 cm (B) 4 cm (C) 12 cm (D) 9 cm 54.25 cm. For minimum deviation P Q R S (A) PQ is horizontal (B) QR is horizontal (C) RS is horizontal (D) Any one will be horizontal 52. A concave mirror is placed 15 cm above the water level and the image of an object placed at the bottom is formed 25 cm below the water level.33. What are their individual focal lengths? (A) 15. velocity of sound in air = 300 m/s) (A) 3000 Hz (B) 120 Hz (C) 600 Hz (D) 6000 Hz 53. The power of the combination is (A) 1. . . upto a height of 33. 50 (D) 75. . the magnitude of the ratio of their focal length is 2 3 . . A convex lens is in contact with concave lens. A ray pQ is incident onto it. 31 cm (D) 10 cm Multiple Choice Question with ONE or MORE THAN ONE correct answer: 56. . Focal length of the mirror is (A) 15 cm (B) 20 cm (C) 18. . . .5 diopres (D) +6. A short linear object of length b lies along the axis of a concave mirror of focal length f at a distance u from the pole of the mirror. 15 (C) 75.5 dioptres (C) +6. . When the upper half of the lens is covered by an opaque screen (A) half the image will disappear (B) complete image will be formed (C) intensity of the image will increase (D) intensity of the image will decrease. 50 55. A container is filled with water . .in 17 51. figure. A beam of light consisting of red. . . The refractive indices of the material of the prism for the above red. . The prism will . . . . . . . 59. . 1.44 and 1. .39.b u f f . .47 respectively. (D) 2 b f u f . green and blu e wavelengths are 1. green and blue colours is incident on a r ight angled prism. Which of the following form(s) a virtual and erect image for all positions o f the object? (A) Convex lens (B) Concave lens (C) Convex mirror (D) Concave mirror. . .72 to produce dispersion wit hout deviation. 60. . . .33° (B) 4° (C) 3° (D) 2. . for a parallel beam of rays coming from the left. . Taking the origin of coordinates O. (D) . f f d f d f d x y f f d f f d . . Two thin convex lenses of focal lengths f1 and f2 are separated by a horizon tal distance d (where d . y f f . are given by: (A) 1 2 1 2 x f f .OPTICS www. 1 2 1 2 1 2 . 0 f f d f d x y f f d .54 is combined with another thin prism P2 made from glass of refractive index 1.physicsashok. as shown in Figure. f1 . (B) . . (C) . .6° 61. . 1 2 1 1 1 2 1 2 .in 18 45° (A) separate part of the red colour from the green and blue colours (B) separate part of the blue colour from the red and green colours (C) separate all the three colours from one another. . A thin prism P1 with angle 4° and made from glass of refractive index 1. . 1 2 1 1 2 . (D) not separate even partially any colour from the other two colours. . d . f2 ) and their centres are displaced by a vertical separation . . . . The angle of the prism P2 is (A) 5. . . . f f d x y f f d f f . . at the center of the first lens the x and y coordinates of the focal point of this lens system. . . . . 62. . . . . It has a real image.63. also located at C. The ray just undergoes total internal reflection. (C) virtual. Let O be the pole of the mirror and C its centre of curvature. with its axis directed ver tically upwards.3 (B) 1.4 (C) 1. If n is the refractive index of the medium with respect to air. and located at a point between C and . the image will be. A point object is placed at C. and located at a point between C and O (D) real. A concave mirror is placed on a horizontal table. If the mirror is now filled with water. and will remain at C (B) real. and located at a point between C and O .5 (D) 1.6 64. (A) real. A ray of light travelling in a transparent medium falls on a surface separat ing the medium from air at an angle of incidence of 45°. select the possible value(s) of n from the follo wing: (A) 1. A ray of light undergoes deviation of 30° when incident on an equilateral prism .. The magnification is ... On the left of APB is air and on the rig ht of CD is water with refractive indices as given in the figure......5 has 7a focal length of 15 cm in air. just touches the rod.OPTICS www..... In the medium the velocity of the light wave is ........ 3. (1985) 4...... ....physicsashok. In young s double-slit experiment.. cm.. has a curved surf ace APB of radius of curvature 10 cm and a plane surface CD.. A light wave of frequency 5 x 1014 Hz enters a medium of refractive index 1...... A point source emits sound equally in all directions in a non-absorbing mediu m. (1987) 6.... . In the medium its wavelength is . The dista nce of the final image of O from P............ then d is equal to ..... In another experiment with the same set-up the two slits are source s of equal amplitude A and wavelength . the two slits act as coherent sources of equ al amplitude A and of wavelength ...5........ A thin lens of refractive index 1.. If a parallel beam of light falling on A leaves B as a parallel beam... when the lens is placed in a medium of refractive index 4 3 .......... A convex lens A of focal length 20 cm and a concave lens B of focal length 5 cm are kept along the same axis with a distance d between them......... its frequency is .. (2 Marks) 2.... The ratio of the intensity of light at the midpoint of the screen in the first case to that in the second case is .0 n2=2. as viewed from the left is ... and its wavelength is .. A B C D P C O n1=1... (1991) 9... A thin rod of length 3 f is placed along the optic axis of a concave mirror of focal length f such that i ts image which is real elongated.... (1986) 5.5 . .in 19 Fill in the blanks: 1... cm. but are incoherent.. The ra tio of amplitudes of the waves at P and Q is ... (1989) 7.....0 20 cm 15 cm 3 4 3 n = An object O is placed at a distance of 15 cm from the pole P as shown. Two points P and Q are at a distance of 9 meters and 25 meters respectively from the source...... A monochromatic beam of light of wavelength 6000Å in vacuum enters a medium of refractive index 1..... its focal length will become ... (1991) 8.... A slab of a material of refractive index 2 shown in Figure....... . ... The angle made by the ray inside the prism with the base of the prism is ...... (1992) 10.of refractive index 2 ... Å. A light of wavelength 6000Å in air.. Inside the medium its frequency is . enters a medium with refractive index 1...... ...... Hz and its wavelength is ......... (1997) .5.... II-C.in 20 .. R1 R2 A. R1 R2 B. R1 = 30 cm. Theintensityof light at adistance r from theaxisof alongcylindrical sourceisinve rselyproportional to r . IV-A www.. A convex lens of focal length 1 meter and a concave lens of focal length 0. II-A. Match List I and List II and select the correct answer using the codes given below the lists: The arrangement shows different lenses made of substance of refractive index 1. (1981) 14. A parallel beam of light first passes through the convex lens. A ray of light is incident normally on one of the faces of a prism of apex angle 30° and refractive index 2 .... IV-C (B) I-C. Two thin lenses.. then through the concave lens and moves to a focus 0. III-C. m and . . III-A... A beam of white light passing through a hollow prism give no spectrum. (1983 ) 16. when in contact.25 m apart. R2 = 60 cm. degrees.. Their focal lengths are 15 cm and 30 cm respectively for the mean wave length in white light.. Match the focal lengths Table I Table II I. produce a combination of power +10 diopters. R1 R2 C. On the other side of the lens system. W hen they are 0. (1983) 15. A parallel beam of white light fall on a combination of a concave and a conv ex lens. 40 cm R1 R2 IV. one sees coloured patterns with violet col our at the outer edge. II-B.75 meter apart. the power reduces to +6 diopters. +40 cm III.. IV-B (D) I-B. 120 cm II. III-D. m. The angle of deviation of the ray is .physicsashok.2 5 meter are kept 0. II-D. D. III-B... (1997) True / False : 13.. +120 cm (A) I-A.. (1997) 12. IV-D (C) I-D.OPTICS 11. The focal length of the lenses are .5 m away from the concave lens.. (1988) Table Match 17.5 and kept in air. both of the same material. in 21 18.5 cm for a grownup person. erect and virtual (A) A . (ii) Plane mirror C. B . For a concave mirror of focal length 20 cm. (iii) B . II. III-A. (iii) Concave mirror (A) A . III. The ima ge distance is therefore fixed for clear vision and it equals the distance of retina from eye-lens. I. I. PASSAGE TYPE QUESTIONS THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The ciliary muscles of eye control the curvature of the lens in the eye and henc e can alter the effective focal length of the system. IV-B (C) I-C. C . (i) B . (ii) B . C . IV. III-D. D . 40 cm (iii) Smaller. III-C. II-D. 10 cm (i) Magnified. Magnification is 1 before a convex mirror. IV. II. IV. III (C) A . II. 30 cm (ii) Equal size. . Magnification is +0. For a clear vision. C . 50 cm (iv) Magnified. IV (B) A . I. D . inverted and real D. It is about 2. B . (iii) C . Magnification is +1 a concave mirror. (i) C . D . (iii) (D) A . An object is placed at centre of curvature D. (ii) C . match the followings: Table I Table II Objective distance Nature of image A. IV-E (D) I-B. D . An object is placed at focus before C. inverted and real B. III. An object is placed at the centre of B. (A) I-B. (ii) B .physicsashok. (i) 20. An object is placed at focus before A. Table I Table II I.5 curvature before a concave mirror III. IV-C 19.OPTICS www. III-A. . I. B . Magnification 3 2 m . IV-E (B) I-A. (iii) (C) A . the image must be on retina. Magnification m = +1 (i) Convex mirror B. Match the followings: Table I Table II A. inverted and real C. When the muscles are fully relaxed. . Magnification is . Magnification 2 3 m . C . II-D. (ii) C . the focal length is maximum. When the muscles are strained the curvature of lens increases (that means radius of curva ture decreases) and focal length decreases. (i) (B) A . B . II-B. II (D) A . II-E. IV. III. a convex mirror II. then number of lens will be +3.in 22 A person can theoretically have clear vision of objects situated at any large di stance from the eye. The smallest distance at which a person can ciliary muscles are most strained in thi s position.OPTICS www. then we can s ay that the person is (A) normal sighted person (B) near-sighted person (C) far-sighted person (D) a person with exceptional eyes having no eye defect. For an average grown-up person. A person suffering for eye defects uses spectacles (eye glass). is given by sec . For example. The number of spectacle lens used for the remedy of eye defect is decided by the power of the lens required and the number of spectacle-lens equal to the numerical value of the power of lens with sign.5 cm (C) 25 9 cm (D) 25 11 cm 2. ' as shown in figure. The function of lens of spectacles is to form the image of the objects within the range in which person can see clearl y.5 cm (C) 25 9 cm (D) 25 11 cm 3. minimum distance of object should be around 25 cm. Assume that the eye lens is equiconvex lens. 1. The image of the spectacle lens becomes object for eye-lens and whose image is formed on retina. Neglect the distance between eye lens and the s pectacle lens. .physicsashok. For all the calculations required you can use the lens formula and lens maker s fo rmula. Minimum focal length of eye-lens of a normal person is (A) 25 cm (B) 2. power of lens required is +3 D (converging lens of focal length 100 3 cm). The number of the spectacles lens necessary for the remedy of this defect will be (A) + 1 (B) 1 (C) + 3 (D) 3 4. A near-sighted man can clearly see object only upto a distance of 100 cm and not beyond this. The number of the spectacles lens that will make his range of clear vision equal to an average gro wn up person is (A) + 1 (B) 1 (C) + 3 (D) 3 5. Maximum focal-length of eye lens of normal person is (A) 25 cm (B) 2. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Spherical aberration in spherical mirrors is a defect which is due to dependence of focal length ' f ' on angle of incidence '. A person who can see objects clearly from distance 10 cm to . A far-sighted man cannot see object only upto a distance of 100 cm from his e yes. is the angle of incidence. where R is radius of curvature of mirror and . As a result of above dependence different rays are brought to focus at dif ferent points and the image of a point object is not a point. . . R .2 f . C F Principal axis Pole (P) f . R . The ra ys which are closed to principal axis are called paraxial rays and the rays far away from principal axi s are called marginal rays. If angle of incidence is 60°. fm (B) p m f . If p f and m f represent the focal length of paraxial andmarginal rays respec tively. then focal length of this rays is: (A) R (B) 2 R (C) 2R (D) 0 8. A person observing the drops will see different colours of the spectrum at differe nt angles.physicsashok. The light sur ffers only one total linternal reflection in drops C and D forming primary rainbow.OPTICS www. Secondary rainbow . The rainbow which results from single total internal reflection is called primary rainbow an d secondary rainbow is formed due to two total internal reflections suffered by rays falling on water d rops. then correct relationship is: (A) f p . When rays of the sun fall on rain drops. A B C D V R R V Red Red violet violet Rays from sun Secondary rainbow Primary rainbow Figure shows formation of rainbow due to four drops A. f (C) p m f . The total deviation suffered by the ray falling on mirror at an angle of inci dence equal to 60° is: (A) 180° (B) 90° (C) Can t be determined (D) None 9. For paraxial rays.in 23 6. C and D. B. Which of the following statements are correct regarding spherical aberration : (A) It can be completely eliminated (B) it can t be completely eliminated but is can t be minimised by allowing either p araxial or marginal rays to hit the mirror (C) It is reduced by taking large aperture mirrors (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Rainbow is formed during rainy season due to refraction and total internal refle ction of rays falling on suspended water droplets. the rain drop s disperse the light and deviate the different colours by refraction and total internal reflection to the eye of the observer. focal length approximately is: (A) R (B) 2 R (C) 2R (D) None 10. f (D) None 7. is and 11. Rainbow is an arc of: Circle (B) Ellipse Parabola (D) Can t be determined The visibility of the rainbow is due to: All rays (B) Rays undergoing maximum deviation Rays undergoing minimum deviation (D) None . (A) (C) 12. (A) (C) formed by drops A B where light suffers two total linternal reflections. x2 where x is measured from on e end.in 24 13. If the light passes through a number of media. The necessary condition for the observer to see rainbow is: (A) Sun. According to this principle a ray of light travels from one point to another such that the time ta ken is at a stationary value (maximum or minimum). .. . The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is: A B P (A) Maximum (B) Minimum (C) Constant (D) None 18. If c is the velocity of light in a vacuum. so that Fermat s principle states that the path of a ra y is such that the optical path in at a stationary value. . Now. The optical path length followed by ray from point A to B given that laws of reflection are obeyed as shown in figure is A B . If refractive index of a slab varies as . and 1 n dl c . the colour of inner edge is: (A) Red (B) Violet (C) Indigo (D) None 15. laws of reflection and refraction can be summarised in one fundamental also known as Fermat s principle. then optical path length of a slab of thickness 1 m is: (A) 4 3 m (B) 3 4 m (C) 1 m (D) None 17. the total time taken is 1 nl c . observer s eye and the centre of the rainbow arc lie on the same line (B) Sun. the velocity in a medium of refractive index n is c n . observer s eye and the centre of the rainbow arc lie on the different lin e (C) From any position provided sun is at the back of the observer (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The laws governing the behavior of the rays namely rectilinear propagation.nl is the total optical path. If refractive index varies continuously.1. This principle is obviously in agreement with the fact that the ray are straight lines in a homogenous isotropic medium. In secondary rainbow. . . . It is found that it also agrees with the classical laws of reflection and refraction.OPTICS www.physicsashok. hence time taken to travel a distance l is nl c . . 16. . In primary rainbow. the colour of outer edge is: (A) Blue (B) Violet (C) Red (D) None 14. (A) Maximum (B) Minimum (C) Constant (D) None . inverted and diminished 23. virtual and small (C) erect. Taking another look at the gypsy you decide to take your chances with the wolves. 21. where is the image of the gypsy in focus as you walk towards her? (A) 6.physicsashok. the gypsy lady is 1. virtual and magnified (D) real. You are about to make a hasty exit when you hear the howl of wolves outside.I. The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is A B (A) Maximum (B) Minimum (C) Constant (D) None 20.9 cm from the crystal ball (C) 8. At some point you can no longer get an image of her.2 m from the central of ball. Resign ed to bad situation your approach her slowly. www. = 1.OPTICS OPTICS 19. If the crystal ball is 20 cm in diameter with R.in 25 . inverted an enlarged (B) erect. wondering just what is the focal length of that nifty crystal ball. but the door is jammed shut.5. The image of old lady is: (A) real. T he refractive index of a material is a measure of the amount by which light is bent upon entering th e material. The optical path length followed by ray from point A to B given that laws of refraction are obeyed as shown in figure is A and B are focii of ellipse A B (A) Maximum (B) Minimum (C) Constant (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE One hard and stormy night you find yourself lost in the forest when you come upo n a small hut.9 cm from the crystal ball (B) 7. The old lady moves the crystal ball closer to her wrinkled old face. The transmittance range is the range of wavelengths over which the material is trans parent. At what object distance will there be no change of the gypsy fo rmed? (A) 10 cm (B) 5 cm (C) 15 cm (D) None THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The table below contains some physical properties of common optical materials. Entering it you see a crooked old woman in the corner hunched over a crystal bal l.9 cm from the crystal ball (D) None 22. which material(s) will transmit light at 25 . 24.35-2 Excellent Cesium iodide 1.12-12 5-9.12-6 2. Total internal reflection could o ccur if light were travelling from (A) Lithium fluoride of flint glass (B) potassium bromide to cesium iodide (C) quartz to potassium bromide (D) flint glass to calcium fluoride 27.4 Good Sodium chloride 1. how is the transmittance range relate d to the useful prism range (A) The transmittance range is always narrower than the useful prism range (B) The transmittance range is narrower than or equal tot he useful prism range (C) The tranmittance range increases as the useful prism range decreases (D) The tranmittance range is wider than and includes within it the useful prism range 28.m Transmittance range (.3 7.35 2.79 0.66 0.43 0.56 0.m) Chemical resistance Lithium fluoride 1. According to the table.in 26 Physical Properties of Optical Materials Material Refractive index for light of 0.2 0.3-17 8 16 Poor Quartz 1.20-2.589 . When light travels from one medium to another. total internal reflection can occur if the first medium has a higher refractive index than the second.5 Poor Calcium fluoride 1.OPTICS www.m.0 15-55 Poor *Flint glass is lead oxide doped quartz.20-3.54 0.3-29 15 28 Poor Flint glass* 1.7 Excellent Potassium bromide 1. The properties of which of the following materials contra dicts this hypothesis (A) Lithium fluoride (B) Flint glass (C) Cesium iodide (D) Quartz 26.54 0.7-5.m) Useful range for prisms (. The addition of lead oxide to pure quartz has the effect of (A) decreasing the transmittance range and the refractive index (B) decreasing the transmittance range and increasing the refractive index (C) increasing the transmittance range and the useful prism range . Based on the information in the table.m (A) Potassium bromide only (B) Potassium bromide and cesium iodide (C) Lithium fluoride and cesium iodide (D) Lithium fluoride and flint glass 25. A scientist hypothesizes that any material with poor chemical resistance wou ld have a transmittance range wider than 10.physicsashok.39 0.3 0. like those used on .(D) increasing the transmittance range and decreasing the useful prism range. The periscope. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE A periscope viewing system is to be used to observe the behavior of primates in a large environmentally controlled room on the upper floor of a large research facility. (A) The focal length of the eyepiece lens is too short. magnified (C) virtual. inverted. (D) virtual.33 m (D) 7 m 30. because the critical angle for crown glass is 47° (B) yes. What is the flaw? (A) The focal length of the eyepiece lens is too short. 31. upright.67x (B) 1x (C) 3x (D) 300x 33. can be moved forward or back in order to focus on the primates a s they move closer to or further away from the objective lens. As the primate moves rather closer to the telescope. is essentially a large. The telescope is focused on a primate rather far away on the farside of the larg e habitat. 32. Describe the properties of the image that one sees with this preliminary design (A) real. because the critical angle for crown glass is 41°. inverted. The eyepiece. 34. (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. Like all Newtonian telescopes. A visitor seeing the sketch points out an important flaw that will require a design change. The distance between the lenses is approximately equal to the sum of their focal lengths.52 for the prism? (A) yes. The prisms (45 45 90° prisms) turn the light path through 90° by total internal refle ction from the inside hypotenuse faces of the prisms when the incident angle is 45° as in the ske tch. (D) The prisms cannot be used in this way. What will be the approximate magnification of this periscope? (A) 0. 29. folded-path. (C) Can one use crown glass with an index of refraction of 1. upright. The total tube length of the three sections is to be 4 m. because the critical angle for crown glass is exactly 41°.OPTICS OPTICS submarines. (D) The prisms cannot be used in this way. The objective lens available has a focal length of 3 m. in this design. because the critical angle for crown glass is exactly 47° (D) No. A sketch of the preliminary design appears below. what is the flaw? (B) The images of the primates will be inverted (C) The objective lens should be a diverging lens. magnified 35. same size as object. in the same man ner that one would use a magnifying glass. it uses a relatively long focal length objective lens to form a real image in front of the eyepiece lens ( of shorter focal length). A visitor seeing the sketch points out an important flaw that will require a design change.75 m (B) 1 m (C) 1. What should the focal length of the eyepiece lens be? (A) 0. magnified (B) real. what must the observer do to see t he primate clearly? . No. low power telescope (using pris ms to fold the light path). The observer looks through the eyepiece lens to see the final image. www. (D) Use an inverting eyepiece because the image flips. (B) Move the eyepiece away from the objective.in 27 . (C) Move the eyepiece closer to the objective.(A) No change.physicsashok. the image remains clear. When using this equation. in diopters. The range over which clear vision is possible is bounded by the far point and th e near point. light from an object is refracted by the cornea-lens sy stem at the front of the eye and produces a real image on the retina at the rear of the eye. 36. cm 10 7 20 10 30 14 40 22 50 40 60 200 In the myopic (nearsighted) eye. The near point is farther away than normal. In normal vision the far point is infinity and the near point depends on the radius of cur vature of the lens. f .A corrective lens will put a virtua l image of a distant object at the position of the actual far point of the eye. the lens flattens out. AGE. all distances are given in centimeters. The lens system of the myopic eye is best described as . II. The power of corrective lenses is usually given in units called diopters. This causes rays from an object at infinity to focus at a point in front of the retina. and virtual images have n egative distances from the lens. which has a fixed curvature that is convex with respect to incoming light. Unfortunately.OPTICS www. For a given eye.5 cm. the lens losses elasticity with age and the ability to alter curvature decreases. The far point is closer than normal . years NEAR POINT. The lens is surrounded by the ciliary muscle.1 i .Converging lenses have positive focal lengths. The importance of the lens is that its radius of curvature ccan be changed. allowing the lens to fine-tune the focus. the lens-to-retina length is too short and/or the radius of curvature of the cornea is not great enough. its lens-to-retina distance is fixed at about 2. A corrective lens will p ut a virtual image of the close object at the position of the actual near point. is the reciprocal of the focal length in meters : 1 diopter meter P . By convection I. The relation among the object (o) and image (i) distances from the eye and the f ocal length (f) of the lens is given by the lens-distance rule : 1 o .1 f . Power. the lens-to-retina length is too long and/or th e radius of curvature of the cornea is too great. This causes rays from an object at infinity t o focus at a point behind the retina. decreasing tis radius of curvatu re. For normal eyes the average near point for reading is 25 cm. Most of the focusing of an image is done by t he cornea. Real images have positive distances from the lens. In the hyperopic (farsighted) eye. and diverging lenses have negat ive focal lengths. when the muscle relaxes. Contraction of the muscle decrease s tension on the lens. This allows the natural elasticity of the lens to produce an increase in t he radius of curvature.physicsashok.in 28 THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the normal human eye. (B) producing too little convergence.50 diopters (B) 2.0 diopters (C) +2. The farthest object he can clearly focus on with his right eye is 50 cm away.0 diopters (D) +5. What is the power of the contact lens required to correct the vision in his right eye (A) 0.0 diopters .(A) producing too much convergence. (D) producing too little divergence. 37. An optometrist examined John s eyes. (C) producing too much divergence. and several exa mples of commercial optical equipment. (D) relaxing the ciliary muscles and decreasing the radius of curvature. lighted object sources. In a mildly hyperopic eye. 40. such as lens holders . such as microscopes and telescopes. a laser is used to f latten the cornea by placing as series of hairline cuts around the perimeter of the cornea.00 diopters in one eye to be ab le to clearly focus on an object 2. A lighted object is placed 18 cm in front of the lens and it is found that a clear image can be focused on a screen placed 36 cm behind the lens. meter sticks and tapes. What is the focal lengths of his eyeglasses (A) 50 cm (B) +50 cm (C) 48 cm (D) +48 cm 41. Based on the vision in this eye. the focal length of the eye s natural lens can be corre cted by (A) contracting the muscle and increasing the radius of curvature. (D) RK corrects hyperopia by increasing the focal length of the eye. (B) contracting the ciliary muscle and decreasing the radius of curvature (C) relaxing the ciliary muscle and increasing the radius of curvature. which of the following is the most likely age range for Jane (A) Less than 40 years old (B) From 40 to 49 years old (C) From 50 to 59 years old (D) 60 years or older (B) RK corrects myopia by increasing the focal length of the eye. (B) 8 cm behind the lens: a real image (C) 16 cm in front of the lens: a real image (D) 16 cm behind the lens. One lens is placed in a holder. what is the focal length of this lens? (A) 8 cm (B) 12 cm (C) 27 cm (D) 46 cm 43. 39. Optical benches. In a surgical procedure called radial keratotomy. a virtua . image screens.0 cm in front of his eyes. 42. (A) 2x is the image and which kind of image is it? A lighted object is placed 6 cm in front of the second lens. (C) RK corrects hyperopia by decreasing the focal length of the eye. (RK).5 cm in front of the eye. Which statement i s most accurate (A) RK corrects myopia by decreasing the focal length of the eye. THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Student are given a variety of lenses and optics equipment. His incorrect far point is 50 cm. George wears eyeglasses that sit 2. What magnification is produced by the above lens when the object is 18 cm in front of the lens and the image is 36 cm behind the lens? (B) 3x (C) 4x (D) 6x 44. which has a focal l ength of +24 cm.OPTICS OPTICS 38. A student is given a short focal length converging lens and long focal lengt h converging lens. They are to work in an op en-ended optics lab in order to learn the general principles of lenses and the optical devices that can be constructed using lenses. Where (A) 8 cm in front of the lens: a virtual image. Jane must wear a contact lens with a power of +3. l image. 45.in 29 .physicsashok. What is the magni fication of this simple refractor? (A) 0.6x (B) 3x (C) 4x (D) 6x 46. A commercial microscope is examined by the student. what power objective should replace the above objective so that t he microscope s magnification will be 400x (A) 5x (B) 10x (C) 40x (D) 100x www. The 24 cm focal length lens is used as the objective of a simple refracting tele scope and a third converging lens of focal length +8 cm is used as the eyepiece. The objective is marked 20x and the eyepiece is marked 10x. or different refractive index.OPTICS www.5 cm (B) 3. The image that he sees through his new contact lens is a virtual image because h e looks through the lens to see the image. A lens of focal length +24 cm is used to view an object placed 12 cm in fron t of the lens. Th e image formed by this lens is the object for a second lens of +24 cm focal length. and the angles refer to the angle of incidence in the first mediu m . and the angle of refraction in the second . where is the final image with respect to the second lens? (A) 24 cm in front of # 2 (B) 24 cm behind # 2 (C) 36 cm in front of # 2 (D) 48 cm behind # 2 48. . 1 . It is worth noting that prisms break up white light into t he seven colors of the rainbow because each color has a slightly different velocity in the medium. 2 refers to the second medium. where I refers to the first medium through which the ray pas ses. such as the starts? (A) 20 cm (B) 30 cm (C) 4 cm (D) 25 cm THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The phenomenon of refraction has long intrigued scientists and was actually used to corroborate one of the major mysteries of early science: the determination of the speed of l ight. From this ratio. From the information in the passage. (D) This cannot be determined from the information . A near sighted student cannot see objects clearly unless they are as close a s 80 cm (his far-point ).3 cm (C) 7.physicsashok. The refractive index for sea water is 1. they shone a beam of light down into the water using a high intensity white light source as s hown in Figure. How tall is the image? (A) 2. It is given by: n1 sin.5 cm (B) 3. . The second lens is place d 72 cm behind the first lens. The refractive index of a transparent material irrelated to a number of the phys ical properties of light.in 30 47. what focal length lenses does he need in order to see ver y distance objects. The object is 5 cm tall.2 .5 cm (D) 10 cm 49. How tall is the image if the object is 5 cm tall? (A) 2. In terms of velocity. how would you expect the speed of light in air to compare with the speed of light in a vacuum (which is given by c )? (A) It would be the same (=c) (B) It would be greater than c.33 while that for air is 1.00. A ship went out on a search for a sunken treasure chest. A diverging lens of focal length 24 cm is now used with the object 12 cm in f ront of the lens. it can be seen that light is retarded when it passes through most types of matter. 2 . A lighted object is placed 24 cm in front of a +12 cm focal lengths lens. the refractive index represents the ratio of the velocity of light in a vacuum to its velocity in the material. Snell s law allows one to follow the behavior of light in terms of its path when m oving from a material of one refractive index to another with the same. . (C) It would be less than c. 51.3 cm (C) 8 cm (D) 10 cm 50.1 . n2 sin. In order to locate the chest. given. . 4° (C) 45. . It is unseemly that such a relatively simplistic apparantus took generations to be developed. particularly the field of cellular and molecular biology.75 (C) 0. nred (B) violet red n . n (D) This depends on the relative speeds of the different co lors in a vacuum. in the previous question.in 31 52.4° 53.physicsashok. What is the value of the s ine of this angle when the ray moves from water towards air? (A) 2 (B) 0. How does the refractive index in water light compare with that of red light given that violet light travels more slowly in water than red light? (A) nviolet . THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE The invention of the compound microscope by Jansen in the late 1500 s truly revolu tionized the world of science.50 (D) 0 55. n (C) violet red n . The discov ery of the cell as the fundamental unit of living organisms and the insight into the bacterial world ar e two of the contributions of this instrument to science. This microscope has a resolution about a hundred times that of the light microsc ope. x 40 and x 60. The two lenses act independentl y of each other when bending light rays. Photographic film must be used otherwise no image would be for med on the retina.2° (B) 30. 54. Light from the object (O) first passes thought he objective and an enlarged. The actual lens set-up depicted in Figure. The eyepiece then magnifies this image.6° (D) 63. if the be am of light was travelling from water to a substance with a greater refractive index than air. Its main component are two convex lenses: one acts as the main magnifying lens and is ref erred to as the objective. inv erted first image is formed.OPTICS www. Usually the magnification of the eyepiece is fixed (either x 10 or x 10) and three rotating objective lenses are used : x 10. what must the approximate value of 2 . The most recent development in microscope technology is the electron microscope which uses a bea m of electrons instead of light. Using the information in the passage. Which of the following would you expect to remain constant when light travel s from one medium to another and the media differ in their refractive indices? (A) Velocity (B) Frequency (C) Wavelength (D) Intensity. Total internal reflection first occurs when a beam of light travels from one medium to another medium which has a smaller refractive index at such an angle of incidence that the angl e of refraction is 90°. This angle of incidence is called the critical angle. and another lens called the eyepiece. but a lower refractive index than water? (A) It would increase (B) It would decrease (C) It would remain the same (D) Total internal reflection would not be possible . 56. be such that it hit the chest as shown in Figure? (A) 15. What would happen to the critical angle. 57. Based on the passage. what type of image would have to be produced by the ob jective magnification? (A) Either virtual or real (B) Virtual (C) Real (D) It depends on the focal length of the lens. . (C) The eyepiece and objective positions were reversed in A. A student attempted to make a compound microscope. (D) 5 7 .04 mm 62. 60. The student used a converging lens as her objective (A) I. III and IV (B) I. (D) The eyepiece and objective positions were reversed in B. d . III. IV 61. Now a second lens is placed exactly at the same position where first was kept. Both had objectives and eyep ieces with the same magnification but A gave an overall magnification that was greater than tha t of B. II. Where would the first image have to be produced by the objective relative to the eyepiece such that a second. The magnification by this second lens is 3. The object distance for eyepiece lens as her eyepiece.60 mm (C) 0. II. Which of the following is a plausible explanation? (A) The distance between objective and eyepiece in A is greater than the corresp onding distance in B. F (D) 2 i e d . (B) The distance between objective and eyepiece in A is less than the correspond ing distance is B. What will be the new magnification.38 mm (D) 0. F . What is the focal length of the combination when both lenses are in contact.physicsashok. However. The student used a diverging lens as her eyepiece. II. IV.67 mm (B) 0. no image was seen. Fe (B) i e d . Now both the lenses are kept in contact at the same place. enlarged image would be generated on the same side of the eyepiece as th e first image (first image distance = d1)? (A) di . The object distance = focal length of objective. 64. IV (D) II.0 diopters (C) 20 diopters (D) 200 THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Magnification by a lens of an object at distance 10 cm from it is 2. Which of the following could explain t he mishap? I. III. II. (B) 12 7 . What is the object height? (A) 1.OPTICS www. III (C) I. The magnification of the eyepiece of a compound microscope is x15. without changing the distance object and lens. What is the refractive power of an objective lens with a focal length of 0. Two compound microscopes A and B were compared. F 59. F (C) 2 e i e .2 diopters (B) 2. 63. (C) 6 11 . . . when she tried t o view an object through the apparatus. The image height is 25 mm and the magnification of the objective is x40. (A) 13 5 .5 0 cm? (A) 0.in 32 58. when object is moved from f to 2f. its image is real . It moves from f to infinity on other side. (A) will increase (B) will decrease (C) will first increase then decrease (D) will first decrease than increase. the magnitude of linear magnifications. inverted and magnified.(A) 60 17 cm (B) 5 17 cm (C) 12 7 cm (D) 13 9 cm THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE In the case of convex lens. Focal length of a convex lens is 10 cm. 65. When the object is moved from 15 cm to 25 cm. . 5 for glas s). Both of these problems can be corrected by intr oducing another lens in front of the eye so that the two lens system produces a focused image on the retina. 69. . sometimes referre d to as nearsightedness. is the focal length of th e correcting lens. What kind of lens would be suitable to correct myopia and hyperopia respecti vely? (Note : Assume that the correcting lens is at the focal point of the cornea so that c x . If an object is so far away from the lens system that its distance may be taken as infinite. occurs when the cornea focuses the image of a distance object i n front of the retina. where c f is the focal length of the cornea. For a distant object. x is the distance from the correcting lens to the cornea. l f .) . and v is the ima ge distance measured from the cornea. Hyperopia.5 cm (B) 1. f . The retina then transmits electrical impulse along the optic nerve to the brain. 67.OPTICS www. (Note : The index of refraction is 1.in 33 66. the n the following relationship holds: 1 1 1 c l f f x v . .physicsashok.0 cm 68. The cornea is a converging lens located at the outer surface of the eye with fixed focal length approximately equal to 2 cm. cornea retina Two common defects of vision are myopia and hyperopia.0 cm (D) 4. . occurs when the cornea focus es the image of a nearby object behind the retina. Parallel light rays coming from a very distant object are refracted by the cornea to produce a focused image on the ret ina. How far away should the retina be from the cornea for normal vision? (A) 0.0 for air and 1. Image of object AB shown in figure will be like: 2F A F B (A) F 2F A' B' (B) F 2F A' B' (C) F 2F A' B' (D) F 2F A' B' THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE Figure shows a simplified model of the eye that is based on the assumption that all of the refraction of entering light occurs at the cornea.0 cm (C) 2. Myopia. sometimes referred to as farsightedness. the image produced by the cornea is: (A) real and inverted (B) real and upright (C) virtual and inverted (D) virtual and upright. 5 cm (C) 2. diverging (C) Diverging. What is the image distance v measured from the cornea for a distant object? (A) 1.5 cm .5 cm from her cornea.8 cm. converging 70.(A) Converging. converging (B) Converging.0 cm (B) 1.5 cm at a distance x = 1. The focal length of a woman s cornea is 1. diverging (D) Diverging. and she wears a correcting len s with a focal length of .0 cm (D) 2.16. A point source of light S is placed on the major optical axis of concave mirr or at a distance of 60 cm.. A pile 4 m high driven into the bottom of a lake is 1 m above the water.physicsashok. Prove that the mark will appear nearer than it really is. = 30 cm from it. (C) be the same as the old value (D) be twice the old value. a fish water watches a fish through a 3. 4. 2.e.OPTICS www. has a small mark on its interi or surface which is observed from a point outside the sphere on the side opposite the center. If the focal length of both the cornea and the contact lens ar e doubled.0 cm thick glass wall of a fish tank.0 cm Wall Water 3.8 cm (a) To the fish.in 34 71. In the case of contact lens. the index of refraction of the glass is 8/5 and that of the water is 4/3. . In figure. In what direction should a beam of light be sent from point A (Figure) contai ned in a mirror box for it to fall onto point B after being reflected once from all fou r walls? A B Point A and B are in one plane perpendicular to the walls of the box (i. with the source in a certain position two real image. Determine the length of the shadow of the pile on the bottom of the lake if the sun rays make an angle of 45° with the water surface. A concave mirror has the form of a hemisphere with a radius R = 55 cm. The refractive index of water is 4/3. how far away does the fish appear to be? 6. . 5. A hollow sphere of glass of refractive index . 3.0 cm 6. one of which (formed by direct reflection) coincide with source and the other is at a distance of . then the image distance v for a distant object would: (A) be 1/4 the old value (B) be 1/2 the old value. of the liquid. The watcher is in level with the fish. Level # 2 1. the cornea and the correcting lens are actually touching and act together as a single lens. Find the refractive index . by a di stance . how far away does the watcher appear to be? (B) To the watcher. A thin layer of an unknown transparent liquid is poured into this mirror. and it was found that the given o ptical system produces. The inner cavit y is concentric with external surface and the thickness of the glass is every where equal where equal to the radius of the inner surface. in t he plane of the drawing). Observer 8. At what distance from the concave mirror should a flat mirror be placed for the ray s to converge again at the point S having been reflected from the concave mirror and then from the flat one?Will the position of the point where the rays meet change if they are first reflected from the flat m irror? The radius of the concave mirror is 80 cm. . where R is the radius of the inner surface. .3 1. . 1 R .. . . Find the . Then find distant point that he can see on the road. . A man of height 2. 1. 2 respectively. when viewed through the paperweight. ay .0 cm from the centre of the sphere and perpendicular to a radius of the sphere that passes through the 8.0 m is standing on level road where because of temperature variation the refractive index of air is varying as .OPTICS www. where y is height from road. A ray of light is incident on a composite slab at a angle of incidence i as shown in the figure. A paperweight is constructed by slicing through the sphere on a plane that is 2. The refractive indices of the mediums on the left and right sides are .0 cm 3. A glass sphere has a radius of 5.in 35 7. of the medium varies as 1 . A portion of straight glass rod of diameter 4 cm and refractive index 1. .5 is bent into an arc of circle of radius R.0 cm from the table top as shown in figure.1 . The paperweight is placed on a table and viewed directly above by an observer who is 8. Find the smallest value of R which permits all the light to pass around the arc.2 . (a) Determine the x-coordinate of the point A.physicsashok. 9. Find the lateral shift x of the ray when it comes out from the otherside. (B) Write down the refractive index of the medium at A. 1 and . how far away does the tabletop appear to the observer? 11. . 10. 8. where . i from the side of medium of refractive index . . 12. A ray of light is incident A . (C) Indicate the subsequent path of the ray in air. The refractive index of air is 1. the refractive index .0 cm Observer center of the circle formed by the intersection of the plane and the sphere. 1 at an angle i and comes out from the other side as shown in the figure.. where the ray intersects the upper surface of the slab-air boundary.0 cm 5.0 x 10 6 m 1. A prism of apex angle A is made up of a material of refractive index .6. A parallel beam of light is incident on R it as shown in the figure. 0 . A long rectangular slab of transparent medium of thickness d is placed on a table with length parallel to the x-axis and width parallel to the y-axis.0 and A Y O X medium d r (> 1) are constants. If a = 2.0 cm and a refractive index of 1. . A ray of light is travelling along y-axis at origin.x r. angle of deviation. . From the details given in the figure. . arranged with their principal axes inclined at an angle . are formed due to feeble internal reflections. The incident ray is parallel to a horizontal diameter and b R u the distance between the incident ray and the horizontal diameter is b. suffered by the ray. A point object lies on the principal axis of the O X Y . .OPTICS www. (a) Find the coordinates of the final image formed by the system of lenses taking O as the origin of coordinate axes. with the normal in the xy plane and comes out through a point P on its curved surface. A ray of light is incident at the point O at an angle .physicsashok. . Multiple extra images F1. . Find the coordinate of the point P if . Find the position of the first flare spot. calculate the value of h. 14. 0 . F2. 16. A hemisphere of radius a/2 and made up of a material of variable refractive index is placed with its base centre O at the origin as shown in the figure... The separation between the optical centers of the lenses is 2 . The image of the object shown in the figure is formed at the bottom of the t ray filled with water. An intense beam parallel to the principal axis is incident on a convex lens.5. as shown in the figure. A ray of light is incident on the sphere of radius R and refractive index . convex lens at a large distance to the left of convex lens./4 . 15. = 30 cm 17.. .in 36 13.. called flare spots as shown in the figure. . The radius of curvature of the lens is 30 cm and Principal axis F1 F2 F0 60 cm and the refractive index is 1. Find the angle of deviation ... and (B) Draw the ray diagram. The refractive index of the material of the hemisphere varies as a x a . 2. O 36 cm 1 m h 85 cm . . In the given figure there are two thin lenses of same focal length . How can the coincidence of the images be established by direct observation? 20. The height of a candle flame is 5 cm. and cos .5 cm awa y from the lens. A lens with a focal power of 5 diopters . ). A thin converging lens of focal length .OPTICS www. (b) A ray of light passes through a prism in a principal plane the deviation bei ng equal to angle of incidence which is equal to 2. . 1. is the an gle of minimum deviation and . Show t hat 8 . A glass hemisphere of radius 10 cm and . (a) A prism has refracting angle equal to . The distance between the candle and the screen is d (> 4 . . + 1/ .physicsashok. A concave mirror forms the real image of a point source lying on the optical axis at a distance of 50 cm from the mirror. two different images can be obtained on the screen. Prove that sin . is moved between a candle and a scr een. How will the images formed by the halves of the mirror be arranged? 21. The mirror is cut in two and its halves are drawn 1 cm a distance of 1 cm apart in a direction perpendicular to the optical axis.5 is silvered over tis curved s urface.cos . 22. Show that for two different positions o f the lens. . It is given that . /2. 25. If the ratio of dimensions of the image is . = 1. 19. ). cos 12 2 . Find the positi on of the images of this bubble seen by observer looking along the axis into the flat surface of the hemi sphere. There is an air bubble in the glass 5 cms from the plane surface along the axis. A thin flat glass plate is placed in front of a convex mirror. a and 4a respectively are placed in order along the axis so that the distance between consecutive lenses is 4a. where . the candle is moved over a distance of . = 1. At what distance b from the plate should a point source of light S be placed so that its image produced by the rays reflected from the front surface of the plate coincides with the image formed by the rays reflected from the mirror? a b S The focal length of the mirror is . . It is given that . Determine the main focal length of the lens. 24. Three convergent thin lenses of focal lengths 4a. A converging bundle of light rays in the shape in the shape of a cone with t he vertex angle of 40° falls on a circular diaphragm of 20 cm diameter. is the refractive index of the material of prism. = 20 cm and the distance from the plate to the mirror a = 5 cm. 23. The focal length of the mirror is 25 cm. = . Without touching the lens.. and a sharp image of the flame 10 cm high is obtained again after shifting the screen. find the value of ( . is the angle of prism.in 37 18. is the deviation of the ray entering at grazing incidence. = sin2 . A lens produces an image of this flame 15 cm high on a screen. Prove that this comb ination simply inverts every small object on the axis without change of magnitude or position. . After refraction the ray intersects the axis at a point. The medium ont he left side of the spherical surface is air. The distance of the incident ray from the axis of the spherical surface is b.. of the point F from the pole O. A ray of light is incident on the spherical surface of radius of curvature R as shown in the figure.is fixed in the diaphragm. F. What will the new cone angle be? 26. Therefractive index on the right side of spherical surface is . Find the distance . (You may take the distance of the near surface of the slab from the mirror to be 1 cm).5 is then placed close to the mirror in th e space between the object and the mirror. An equi biconvex lens of focal length 10 cm in AIR and made up of material of refractive index 3/2 is polished on one side.in 38 27. B and C along the diameter. The thickness of the middle layer C is 1 cm. Level # 3 1. Consider an arrangement of two equibi convex lenses of focal length in air 10 cm. Consider an equilateral prism ABC as shown in the figure. 28. The space between 10 cm 10 cm the two lenses is filled with a liquid of refractive index 4/3. 30. Find the distance between the two images formed. A paraxial beam of light is incident along tyhe axis of the part C. A small object O is placed on the axis at a distance of 10 cm from the first lens in air as shown in the figure. The distance of separation between the two lenses is 10 cm. [IIT 1980] 2. Find the refractive index . 29. A ray of light in . The x-y plane is the boundary between two transparent media. The total A B C 60° 60° 60° angle of deviation is 120°.g . w = 4/3. A ray of light is incident on the face AB and gets transmitted into the prism. has a refractive index 3 . 0.physicsashok. The refractive index of the glass of which the lenses are between the made is . A glass slab of thickness 3 cm and refractive index 1.g the space two lenses is . g = 3/2 and the refractive index of water filling 10 cm 10 cm O AIR AIR . Medium 1 with z . The middle layer is now removed and the two parts A nad B are put together to form a composite lens. A thin convex lens of focal length 1m is cut into three parts A. Find the position of the final image formed. Find the position and magnification of the final image. 0. An object is placed 21 cm in front of a concave mirror of radius of curvature 10 cm. 0 has refractive index 2 and medium 2 with z . Then total internal reflection takes place at the face BC and the ray comes out of prism through the face AC. Then the part C is also placed infront of this A A B B C C 1 cm composite lens symmetrically as shown in the figure. Another identical lens (not polished) is placed infront of the polished lens at a distance of 10 cm as shown in the figure. An object O is placed infront of the unpolished lens at a distance of 10 cm. Find the final position of the image. of the material of the prism.OPTICS www. . What is the radius of curvature of the convex mirror? [IIT 1973] .medium 1 given by the vector k 10 j 3 8 i A . Find the unit vector in the direction of the refracted ray in medium 2. A pla ne mirror is introduced covering lower half of the convex mirror. 6 3 . . it is found that there is no parallux between the images formed by two mirro rs. is incident on the plane of separation. An object is placed in front of a convex mirror at a distance of 50 cm. If the distance between the object and the plane mirror is 30 cm. [IIT 1999] 3. What will be its focal length in w ater. What is the velocity of light in glass of refractive index 1. A glass lens has focal length 5 cm in air. [IIT 1995] 11. What wil l happen if the angle of incidence exceeds the angle? (refractive index of glass is 1.0 m (see figure). (b) Draw a neat ray diagram. where the ray intersects the upper surface of the slab-air boundary. Find the value of the refractive index of glass for which the letters on the page are not visible from any of the vertical faces of the block. A light ray is incident on an irregular shaped slab of refractive index . A point object P is kept at a distance of mR from it.5? (Velocity of light in air = 3 x 1010 cm/sec.51 and that of water is 1. A ray of light is travelling form diamond to glass.5 is place d over a small coin.47)[IIT 1977] 7.0 (meter) 3/2 The refractive index of air is 1. The point of incidence is the origin A (0. (c) If the eye is slowly moved away from the normal at a certain position the ob ject is found to disappear due to total internal reflection. the size of the film in the camera is 18 cm x 18 cm.physicsashok. (d) Indicate the path of the ray subsequently. A quarter cylinder of radius R and refractive index 1. A rectangular glass block of thickens 10 cm and refractive index 1. [IIT 1977] 6. (Refractive index of glass is 1. Find the value of m for which a ray from P will emerge parallel to the table as shown in the figure. (b) Obtain an equation for the trajectory y (x) of the ray in the medium. y1) of the point P.51 and that of diam ond is 2.5 is placed on a table.) [IIT 1976] 8. [IIT 1979] 5. A ray of light travelling in air is incident at grazing angle (incident angle = 90°) on a long rectangular slab of a transparent medium of thickness t = 1.33). A beaker filled with water of refractive index 3 4 to a height of 10 cm and is placed over the glass block.in 39 4. At which surface does this happen and why? [II T 1975] 10. 0).OPTICS www. (c) Determine the coordinates (x1. Calculate the minimum ang le of incidence of the ray as the diamond glass interface such that no light is refracted into glass. y) in the medium and the incident angle at that point. [IIT 1999] 12. (a) Obtain a relation between the slope of the trajectory of the ray at a point B (x. The medium has a variable index of refraction n (y) given by n(y) = [Ky3/2 + 1]½ where K = 1. Photographs of the ground are taken from an aircraft flying at an altitude of 2000 meters by a camera with a lens of focal length 50 cm. What area of the ground can be photographed by this camera at any one time? [IIT 1976] 9. (a) Find the apparent position of the object when it is viewed at near normal in cidence. A rectangular block of glass is placed on a printed page lying on a horizonta l surface.0. 4 m. = 1. [IIT 2004] .514 and passes through point E. If the radius of curved surface is equal to 0.2 at an angle of 45° with the normal on the incline face as shown in the figure. find the distance OE correct up to two decimal places. the ray finally emerges from the curved surface in the medium of the refractive index . An object is approaching at thin convex lens of focal length 0. [IIT 2001] 18. On repeating with another liquid. [IIT 2004] 17.2 cm is placed on the optic axis PQ of the lens at a distance of 20 cm from the lens. Also locate position of A and B with respect to the optic axis RS. the object coincides with its own image. [IIT 1974] 16. find the distance of A B from the pole of the mirror and obtain its magnification. A thin biconvex lens of refractive index 3 is placed on a horizontal 2 plane mirror as shown in the figure. If A B is the image after refraction from the lens and reflection from the mirror. Find the magnitudes of the rates of change of position and lateral magnification of image when the object is at a distance of 0. An upright object AB of height 1. [IIT 2000] www. The distance between the lens and mirror is 30 cm. 3 It is found that when a point object is placed 15 cm above the lens on its principle axis.01 m/s. the object and the image again coincide at a d istance 25 cm from the lens. An object of height 4 cm is kept to the left of and on the axis of a converging lens of focal length 10 cm as shown in figure. A convex lens of focal length 15 cm and a concave mirror of focal length 30 cm are kept with their optic axes PQ and RS parallel but separated in vertical direction by 0.3 m with a speed of 0. On the other side of the lens. [IIT 1972] 15. a convex mirror is placed at a distance of 10 cm from th e lens such that the image formed by the combination coincides with the object itself.4 m from the lens. The space between the lens 4 andthemirroristhenfilledwithwaterofrefractiveindex . An object is placed at 20 cm left of the convex lens of focal length 10 cm. Calculate the refractive index of the liquid.in 40 . A point object O is placed at a distance of 12 cm on the axis of a convex lens o f focal length 10 cm. Find the position and size of the image formed by the lens and mirror combination. If a concave mirror of focal length 5 cm is placed at 30 cm to the right of the lens find the magnification and the nature of the final image. trace the rays forming the image. What is the focal le ngth of the convex mirror? [IIT 1976] 14.OPTICS OPTICS 13.physicsashok. A plane mirror is placed inclined at 45° to the lens axis 10 cm to the right of the lens (see figure). Draw the ray diagram and locate the position of the final image.6 cm as shown. The radii of curvature of both the surfaces are equal to R. A thin equiconvex lens of glass of refractive index 2 . is split in to two halves: one of the halves is shifted along the optical axis (see figure).physicsashok.in 41 19.OPTICS www. [IIT 2001] . as shown in figure.5 has a radius of curvature 20 cm. Find the focal length of the lens shown in the figure. The convex side is silvered and placed on a horizontal surface. . . A beam of white light is incident at a small angle of this prism. A small object is placed outside the t ank in front of the lens at a distance of 0.8 m. 3 and of focal length 0. The refractive indices of the crown glass for blue and red lights are 1. 3 4 . [IIT 1997. . Find the focal-length of the lens and separation between the two halves. The other flin t glass isosceles prism is combined with the crown glass prism such that there is no deviation of the incid ent light. [IIT 1984] 22. [IIT 1996] 21.73 respectively. A thin plano-convex lens of focal length .. . a mirror is placed inside the tank on the tank wall perpendicular to the lens axis. Calculate the net dispersion of the combined sys tem. . the apparent depth of the centre of the plane face is found to be 25/8 cm. Draw the ray diagram for image formation. The magnification of the image formed by one of the half-lenses is 2. If the lens is inverted such that the plane face is in contact with the table. The convex surface of a thin concavo-convex lens of glass of refractive index 1. find the distance through which the pin should be moved so that the image of the pin again coincide with the pin. [IIT 1981] 23.9 m from the lens along its axis. A plano convex lens has a thickness of 4 cm. Determine the angle of the flint glass prism. The concave surface has a radius of curvature 60 cm.. [IIT 2003] 24. (a) Where should a pin be placed on the optic axis such that its image is formed at the same place? (b) If the concave part is filled with water of refractive index 3 4 . May] 20.51 a nd 1. The separation between object and image planes is 1.77 and 1. When placed on a horizontal tab le with the curved surface in contact with it.. Find the position (relative to the lens) of the image of the object formed by the system. An isosceles prism of angle 6° is made of crown glass. The separation between the lens and the mirror is 0.8 m. Find the focal length of the lens. the apparent depth of the bottom most point of the lens is f ound to be 3 cm.49 respectively and those of the flint glass are 1. On the opposite side of the lens.3 m in air is sealed into an opening at one end of a tank filled with water . find (a) the angle of emergence. A light of wavelength 5500A. A prism of refracting angle 30° is coated with a thin film of transparent mate rial of refractive index 2. [Given refractive index of the material of the prism is 3 ].25. .2 on face AC of the prism. is incident on face AB such that angle of incidence is 60°. Light is incident at an angle . It is desired that a ray of light incident on the face AB emerges parallel to the incident direction after two internal reflections. calculate the angle of incidence at AB for which the refracted ray passes through the diagonal face undeviated. [IIT 1986] 29. A ray i s incident on AB (see figure).352. (a) Calculate the angle of incidence at AB for which the ray strikes the diagonal face at the critical angle. (b) Assuming n = 1. The assembly is in air. (a) find the position of the image due to refraction at the first surface and th e position of the final image.physicsashok. show that for any value of n3 all light will ultimately be reflected back again into medium II. Assuming the light rays to be paraxial. as shown in figure. (iv) Calculate the image distance when the object is placed as in (iiii).. A ray of light is incident at an angle of 60° on one face of prism which has a . [IIT 1988] 30. A). The radius of curvature of the convex face of a plano convex lens is 12 cm a nd its refractive index is 1. Determine the least value of n so that the light entering the rod does not emerg e from the curved surface of the rod irrespective of the value of . (a) What should be the minimum refractive index n for this to be possible? (b) For n = 3 5 is it possible to achieve this with the angle B equal to 30 degrees ? [IIT 1987] 28. Consider separately the cases. (b) draw a ray diagram showing the positions of both the images. (ii) The plane surface of the lens is now silvered. Monochromatic light is incident on a plane interface AB between two media of refractive indices n1 and n2 (n2>n1) at an angle of incidence . (i) Find the focal length of this lens. A parallel beam of light travelling in water (refractive index = 3 4 ) is refracted by a spherical air bubble of radius 2 mm situated in water.OPTICS www. A right angle prism (45° 90° 45°) of refractive index n has a plate of refractive index n1 (n1 < n) cemented to its diagonal face. A right angled prism is to be made by selecting a proper material and the angles A and B (B . on one planar end of a transparent cylindric al rod of refractive index n. (i) n3 < n1 and (ii) n3 > n1. (iii) Sketch the ray diagram to locate the image. [IIT 1992] 31. as shown in the figure. [IIT 1 979] 32.5. [IIT 2003] 26. 20 cm from the lens (polished). Now if a transparent slab DEFG of uniform thickness and of refractive index n3 is introduced on the interface (as shown in the figure). [IIT 1996] 27.in 42 (b) the minimum value of thickness of the coated film on the face AC for which t he light emerging from the face has maximum intensity. At what distance from the le ns will parallel rays incident on the convex face converge. when a point object is placed on the axis. The angle . is infinitesimally greater than the critical angle for the two media so that total internal reflection takes place. [IIT 1978] .n angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through wh ich it emerges and calculate the refractive index of the material of the prism. [IIT 1973] . The refractive index of the material of a prism of refracting angle 45° is 1. Determine the position of the final image.6 for a certain monochromatic ray. The surface of the lens farther away from the pin is silvered and has a rad ius of curvature are 22 cm. 0 for which rays incident at any angle on the int erface BC pass through without bending at that interface. (b) For light of wavelength . [ I I T 1976] 35.5.. trace the path of the emergent rays for the same incident ray.. . . What is the minimum value of the refractive index of the material of the prism? When the prism is immersed in water. The angles of the prisms are as shown n1 and n2 depend on . . A ray of light incident normally on one of the faces of a right angled isosceles prism is found to be totally reflected as shown in the figure. and 2 4 2 n 1 45 1 80 10 .in 43 33. made of a material of refractive index 1. the wavelength of light. . . . . indicating the values of all the angles. . A prism of refractive index n1 and another prism of refractive index n2 are stuck together without a gap as shown in the figure. . It throws an image of a 2 cm x 1 cm slide on a screen 5 metre from the lens. What should be minimum angle of incidence of this ray on the prism so that no total internal reflection takes place as the ray comes out of the prism. [IIT 1998] 36. according to 2 4 1 n 1. A projector lens has a focal length 10 cm. .OPTICS www.. 0.20 10. find the angle of incidence i on the face AC su ch that the deviation produced by the combination of prisms is minimum. Is the image real as virtual? [IIT 19 78] 34. Find : (a) the size of the picture on the screen and (b) ratio of illuminations of the slide and of the picture on the screen. . is in nm. . . .. (a) Calculate the wavelength . [IIT 1 975] 37. .physicsashok. where. . 3 4 .8 10 . A pin is placed 10 cm in front of a convex lens of focal length 20 cm. in 44 . C BC CD D 7 8 9 10 B D D C 15 16 17 B CD C C 25 26 27 A A A B 35 36 37 A C D C 45 46 47 B B B C 55 56 57 BD D A C 18 19 20 28 29 30 38 39 40 48 49 50 58 59 60 Fill in the Blanks / True False / Match Table 1. 15° Que.4 x 10 6 m 2. A A A A B E C A C A Que. 11 12 13 14 Ans. 21 22 23 24 Ans. B B B A D C Que. 60 cm 6. C C C A D A Que. 2 x 108 m/s. 13 14 15 16 17 18 19 20 Ans. T F T T D A C B www.5 m 12 . 0. 4000Å 11. 1 2 3 4 5 6 7 8 9 10 Ans. 5 x 1014 Hz 4. B C A A C C Que. A D B D C C Que. 11 12 13 14 15 16 Ans. d = +15 cm 3. 7. 31 32 33 34 Ans. B C A A B A Que. 1 2 3 4 5 6 Ans. 51 52 53 54 Ans. B D D B B D Assertion & Reasion Level # 1 Objective Type Que.125 m. D A B C A D Que. 4000Å. 41 42 43 44 Ans. 1. 61 62 63 64 Ans.physicsashok. 0. Zero 10. 2 25 5.OPTICS OPTICS Answer Key Assertion & Reasion Que.3 9 9. 35 cm 8. 0. 5 x 1014 Hz. C D D B C A B C B A B C A A B Que. (a) 13. 12 cm . 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans. . .88 m 5. 4. .. . .in 45 Que. . . . . . . . 61 62 63 64 65 66 67 68 69 70 71 Ans.3 cm (b) 14. 2 Km. . . 10.. 2. . 1. 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 Ans. ... . 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 Ans. A B A B C A D B B A D D B B B Que.6 3. . .42 cm 11.OPTICS www.physicsashok. (c) Ray will become parallel to y-axis. . 9. . . 8. . . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans. Yes.. . . 2. 7. D B B C D B D D B A A C C A A Que. ..975 cm 7. . 2 2 0 20 A 1 r x r 1 d (b) 2 1 2 20 20 0 A 1 r d . . . . 90 cm. sin i sin i t 1 cosi sin i x t 1 cos i 2 2 2 2 2 1 1 . . . (a) . D D C A B C C A D C D Passage Type Level # 2.... R . .. C C B D B A B C C C B B A A B Que. . . . . . . . . . .. . . 20 cm 17. . . . . . . . . . i . . . . . . sin A sin2 i cosAsin i 2 1 1 1 13. . . . The first spot is at 12 cm on left side from the optical centre. . . . . . . . . . . . . . . sin. . . . a 8 a 14. . . 2 2 2 1 2 R 1 b R b R 2sin b 15. . .. . . . 12. . . 16. . . . . . . . . . . .. . . . A . .. 15 8 .. . . ... . . 2 2 d 2 . .. . . 21. from each other. At a distance of 15 cm. . . . . . . . 28. 9 cm 23. . 20. . . . . 7 .. . . 0. . . 2 2 2 2 2 R 1 b R b R 1 1 27. . . . . 3 . . .33 cm from flat surface and the second at inf inity. . . .. b = 15 cm. 22. from the mirror 2 cm. 25. First Image at a distance of 3. .. . 81°40 26. . .5 cm 29. . 0 cos 1 2cos 1 19. . . .. . . 30.. 25 cm from the second lens on the right side magnification m = 2. . 6 cm back side of unpolished lens. . . . 32.physicsashok.84 cm 6.. 19. 2. (a) cot i dx Slop dy . (b) No glass water interface. 29. . At the position of the object magnification = 1. . . (b) 4 2 4 x K y . 7. Distance of A above RS is 0.6 18.9 m from the lens (0..6 m. 0. . 5 2 k 5 j 4 i 3 . 24. = sin 1 0. 16. .3 per second . OE = 6. 2 sin 1 1.16 cm below the water surface. . Distance of B below RS is 1. .OPTICS www.in 46 Level # 3 1. At a perpendicular distance of 20 cms from the lens axis 8 cm is size orient ed parallel to lens axis. (a) . 4 12. (a) sin B 1 (b) No. 0. . 0.4 m. . 8. .67 cm from the mirror. . Distance of A B from pole of mirror a15 cm. magnification = 1.6115 = 38° No ray is refracted into glass. 18.04° 25.. 7. 30. . . .5. 25 cm. = 0. 720 m x 720 m 9.. (a) 15 cm (b) 1.06 m 13. 1m) (d) Ray emerges parallel to the positive x-axis. (a) Image due to first surface at a distance of 6 mm before the first surfac e final image at a distance of 1mm before the first surface. . (a) 14. 73. . . 21. 75 cm 22. 15. . 17.352 27. . 3. 0.3 cm. 2 5. 3 m .. 2 x 1010 cm per second. 1 2 1 1 n2 n n 2 sin 1 (b) .15 cm towards the lens 23. d = 0. 1. 11. 2 31.09 m/s.1 m behind the mirror) 20. (c) (4m. 10.5 cm. . 3 1 3 R . (a) 24 cm (b) 12 cm (d) 80 cm. (a) . . 4°. (a) 0 (b) 1250Å 26. . 14. 25 cm 4.. 3 . . . 2 .1.1. . Angle of refraction in water r = sin. 35.33. . 37. X X X X . .0 . 17 cm infront of lens. 600 nm (b ) i . 34. 10. sin. . (a) 100 cm x 50 cm (b) 2401 picture slide .0.1 ¾ 36. Real.176. (a) . sin.1 ¾ . TIRUPATI. 9440025125 . PH NO.ELECTROSTATICS DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. . ... 1 coulomb 1 ab coulomb 3 109 stat coulom b. if a body in the first column is rubbed against a body in the second column. . .e. . .. ± 1 e 3 and . i. . it has been discovered that elementary particles such as proton or neutron are composed of quarks having charge .. But in SI systemthe unit of charge is coulomb. the charge on anybodywill be some integralmultipl e of e. Woollen cloth Plastic objects Electric Charge : Electric charge canbewritten as newhere n is a positive or negative integer and e is a constant ofnature called the elementarycharge (approximately1.e.5e etc. Woollen cloth Rubber shoes 5.physicsashok. . FirstColumn SecondColumn 1. the (algebraic) net charge of any isolated systemcannot be changed. 2. where n = 1.. . .while that in the second columnwilla cquire negative charge. The study of electrical effects of charge at rest is called electrostat ics. q =± ne. In the table given below.No..whereas onewith an imbalance is electricallycharged.. Charge on any body can never be 1 e 3 . ± 2 e 3 .. Regarding charge following points areworthnothing : (a) Like charges repel each other and unlike charges attract each other.. Glass rod Silk rod 2. the quantum of charge is still e.. .. Electric charge is conserved. 1. 10 . The strength ofparticle s electric interactionwithobjects around it depends on its electric charge.. .. NOTE : Recently.whichcan be either positive or negative.. Woollen cloth Amber 4. Flannes or cat skin Ebonite rod 3. (c) Charge is quantized. i. as quarks do not exist in free state. . .. (d) The electrostatic unit of charge is stat-coulomb and electromagnetic unit is ab-coulomb inCGS system.. . TABLE S.in 1 ELECTROSTATICS ELECTRIC CHARGE Charge is the propertyassociatedwithmatter due towhichis produces and experience s electrical andmagnetic effects.. 3. the body in first columnwill acquire positive charge.. positive or negative.An object withequal amounts of two kinds of charge i s electricallyneutral.ELECTROSTATICS www. However. (b) Charge is a scalar and can be of two types.60 × 10 19C). . Example-1 : Howmanyelectrons are there in one coulomb of negative charge ? Sol. n = 6.6 × 10 19 C. the number of electrons 19 n q 1. . . : The negative charge is due to presence of excess electrons. since they ca rry negative charge. Because an electron has a charge whosemagnitude is e = 1. .6 10.0 e 1.25 × 1018 . .Mathematically. (invacuum) 1 2 e 2 0 F' 1 .. F3 . Thus. .physicsashok. . Fn . holds for anynumber of charges. called the principle of superposition of forces. q1 q2 Fe Fe r In vacuum Thus. .ELECTROSTATICS www.0 = permittivityof free space .0 10 Nm /C 4 .854 × 10 12 C2/N m2 RegardingCoulomb s lawfollowing points areworthnoting : (a) When two charges exert forces simultaneously on a third charge.. F2 . where .in 2 COULOMB S LAW The force acting between two point charges is directly proportional to the produ ct of the charges and it is inverselyproportional to the square of distance between them. .... 1 2 e 2 0 F 1 .. the net f orce acting on eachcharge is altered because charges are induced in themolecules of intervenningmedium. .. K is called permittivityof themedium (c) The coulomb s lawexpresses the force betweentwo point charges at rest. . Inapply . Here 0.. 1 2 2 F kq q r . (Inmedium) or 1 2 e 2 F' 1 . . . (b) If some dielectric (K) is placed in the space between the charges. Fnet . . q q 4 k r . 1 2 2 F q q r . q q 4 r . . the total fo rce acting on that charge is the vector sumof forces that the two chargeswould exert individually. 9 2 2 0 k 1 9. . .0 = 8. F1 . q q 4 r .This important property.. . . This statement is called Earnshow s theorem.. . . then electric force on charges q1 due to charge q2 is 1 2 . .. Suppose the position vectors of two charges q1 and q2 are 1 r . IMPORTANT FEATURE 1. 12 3 1 2 0 1 2 F 1 . . . electric force on q2 due to charges q1 is 1 2 21 3 2 1 0 2 1 F 1 . . ... . q q (r r ) 4 |r r | .. and 2 r . q q r r 4 r r . . (d) Acharged paritcle under the action of coulombian forces onlycan never be stab le. .. . . . that is too for external p oint. . ..ingit to the case of extended bodies of finite size care should be taken in assuming the whole charge of a bod y to be concentrated at its centre as that is true onlyfor sphercal charged body. Similarly. . where 1 1 1 (x . . 19 1 2 q .ELECTROSTATICS www.. y1 j. . . . k r Thus.in 3 Here q1 and q2 are to be subsitiuted with sign. y . z1k . 2 2 2 2 r . we see that for the electric force between two charges at distance r in a dielectricmediumis equivalent to a distance 0 r . . . q q 4 k r . . 1 2 1 2 2 2 0 0 0 1 .. .0 10 )(1. 1 2 e 2 0 F 1 . e .6 10 )(1. . . 2.physicsashok. . r1. or 0 r .6 . x i . remains the same or F = F0 Then. Now. q . . For Fe to bemimimumq1 q2 should bemimimum. . .0 m? Sol. Example-2 :What is the smallest electric force between two charges placed at a d istance of 1. q q 4 r . y j. z k .10. q q 4 r .. z ) are the co-ordinates of charges q1 and q2.0) . q q 1 q q 4 k r 4 r . 1.. the same charges are placed in a dielecricmediumof dielectric constant K at distance r (<r0) such that the electric force between them 1 2 2 0 F 1 . k r in vacuum. and . C Substituting in above relationwe have 9 19 19 e 2 (F ) (9.. x1i . Suppose two charges q1 and q2 are placed in vacuumat a distance r0 and the el ectric force between themis 1 2 0 2 0 0 F 1 . . .z ) and 2 2 2 (x .6 10 ) (1. y . the direction of the resultant force is along the negative x-direction. . Hence the correct choice is (b)..0m.28N Example-3 :Three point charges +q.304 . The angle between these two forces is 120º. The charge at R exerts a repulsive force on charge at P along PS ofmagnitude F.As shown in the figure. If 2 2 0 1 q F . 4 r . . Refer to Fig. Sol. 2. From parallelogramlaw. q and +q are placed at the vertices P. q2 = 2µc and q3 = 3µc are placed on the vertices of an equilateral triangle of side1. The charge at Q exerts an attractive force F on charge at P along PQ. Q and R of an equilateral triangle as shown in fig. the magnitude of the resultant force is 2 2 2 2 r F = F + F + 2F cos 120º +q y x r q r +q Q R 120º P Fr S F F = 2F2 F2 = F2 or Fr = F. +q y P x r r q r +q the force on charge at P due to charges at Q and R is : Q R (A) F along positive x-direction (B) F along negative x-direction (C) 2 F along positive x-direction (D) 2 F along negative x-direction..Find the net electric force acting on charge q1.10. Example-4 : Three charges q1 = 1µc. where r is the side of the triangle. .[(0 .0.in 4 Sol. q2 and q3 in this co-ordinate systemare (0. . .. .0). . 0.. q q (r r ) 4 |r r | . .87).. .5m.87). (0.10. force on q1 due to charge q3 1 3 3 1 3 0 1 3 1 .. called electric field strength or a . ..j.349. . (0. .. 0. 0)k..2N ELECTRIC FIELD Electirc field is the region around an electric charge (or a group of electric c harges) inwhich the electric force can be experienced. Electric field at a point can be defined in terms of either a vector functionE.0 10 ) (1. q q (r r ) 4 |r r | . 0. 0)k. . . 1 F .10. (0 .349.87m. . .physicsashok. 0) respectively.0 10 )(3. . . 2. q3 q1 q2 x y Now.i . 0. F1 .0) .] . . .0) and (0.0) . . Let us assume a co-ordinate axeswith q1 at origan as shown infigure. 9 6 6 3 (9. (1m.[(0 .. 1 2 3 1 2 0 1 2 1 .5)..5).0 10 )(1. 0..ELECTROSTATICS www. (0 .j) . .0 10 )(3. . The co-ordinates of q1.8 .0 10 ) (1.j.2N Therefore net force on q1 is F . 9 6 6 3 (9.1.j) .i .i . . (1. . . 2. ?(0 . .10. (.35.0 10 )(1.] . .45. = force on q1 due to charge q2. ..2 i)N and F2 . If an electric charge is placed in such region.i . . F2 . it experienc es either an attractive or a repulsive force. 0. . The electric field can also be visual ised graphicallyinterms of lines of force.What ar e themagnitude and direction of the force that acts on a charge of + 2µC and 5µC at this spot ? . Electric Field Intensity : The intensity of electric field at anypoint is defined as the force acting on a unit positive charge placed at that point. If the electrostatic force experienced by a small test charge q0 is e F . then intensity of electric field e 0 E F q . on a positive test charge. The SI unit of electric field isN/C. . Example-5 : An electric field of 105N/C points duewest at a certain spot. . The electric field is a vector quantityand its directionis the same as the direc tion of the force e F ..scalar functionVcalled electric potential. . cos 60º q. . .physicsashok. .866 = 2. . or = 4.2.8 . Ifmassmis released fromthe positionshown infigure find the angular velocityof the rodwhenit passes through .732 0. .. . Example-7 : In space horizontalElectric field (E= (mg)/q) exist as shown in figu re and a massm attached at the end of a light rod.cos 30º Thus acceleration is or a F g sin 30º µg cos30º µq cos 60º q cos30º m m m . The different forces on the particle are shown in figure. . . of 100 V/m(see figure).345 sec. . Sol.9 1. find the time it will take the particle to reach the bottom.cos 30º q 30º mg sin 30º q cos 30º e qe µN q sin 30º e mg cos 30º N 30 º 30º or F = mg sin 30º mg cos 30º µq.98 × 1.10 0. . . N = mg cos 30º + q. . .551.9 0. . .01 100 0. .in 5 Sol. . .8 0. .2 9. Force on + 2µC = qE = (2 × 10 6) (105) = 0. distance travelled in time t is s 0 1 at2 2 = + or t 2 2 a . 0.5 N (due east) Example-6 : An inclined planemaking an angle 30º with the horizontal is placed in a uniformhorizontal electric field .5 0.732 or = 4. Particle of mass 1 kg and charge 0.. .237 Now. cos 60º Friction f = µN = µ mg cos 30º + µ q. . . Fromfigure.. .2 N (due west) Force on 5µC = (5 × 10 6) (105) = 0. .01 C is allowed to slide down 1m e= 100 V/m q 30º fromrest froma height of 1 m. If the coefficient of friction is 0. .01 100 3 1 1 2 .10 0. cos 60º Now the total force F acting along the inclined plane is F = mg sin 30º µ N q. [As s = 1/sin30 = 2] or 4 1.ELECTROSTATICS www. 3 / 2.5 0. .697 0. . . . 2. or a 9.2 0.237 . = 1 2 mv2 { E mg q .) = 1 2 mv2 fromeqn (1) mg l sin.. According towork energytheorm: w = .) m m +q q q=45º l mg E q = +q l sinq qE l cosq l l cosq qE l sin .. +mg l mg l cos. Wg = mg (l l cos .m m +q q q=45º mg E q = the bottommost position (A) g l (B) 2g l (C) 3g l (D) 5g l Sol.T WE + Wg = 1 2 mv2 0 .(1) WE = qE l sin .. + mg (l l cos. = } . .2 2 3/ 2 0 E 1 . . q 4 r . . v = 2gl v 2g 2g w = = = l l l l . . . 4. Adisc of charge 2 2 0 E 1 x 2 x R . ..ELECTROSTATICS www. Aring of charge . . . q r p Isolated charge E 2. r . 3. .l = . . . 5.No. Isolated charge 2 0 E 1 . . Infinite sheet of charge 0 E 2 . . .. .in 6 g g g 1 mv2 2 2 2 l + l . w = 2g l Ans. TABLE : Electric field Intensity of Various System S. System Electric Field Intensity 1.. .physicsashok. Infinitelylongline ofcharge 0 E 2 r . . . qx 4 R x . in 7 6. | | 0 E cos cos 4 x . R. Electric lines offorce are graphical representationof electric field. .. .. . R 2 0 E q 4 r . (b) Outside r . . . . Finite line of charge . 8. . Charged spherical shell (a) Inside 0 . 0 E sin sin 4 x . 7. . .. r . .0. E = 0 (b) Outside r . . . where q = charge on sphere Electric Lines of Force : Faraday gave a newapproach for representationof electric field in the formof ele ctric lines of force. Solid sphere of charge (a) Inside 0 .physicsashok. R 2 0 E q 4 r . . Electric lines of forcemay be fraction. EP EQ Q P Thismodelofelectric field has the following characteristics : (i) Electric lines of force are originated frompositive charge and terminated in to negative charge. (iii) The number oflines per unit area that pass througha surface perpendicular to the electric field lines isproportional .. Anelectric line of f orce is animaginarylineor curve drawn through a region of space so that its tangent at any point is in the direc tion of the electric field vector at that point.ELECTROSTATICS www. r . . R 0 E r 3 . (ii) The number of electric lines of force originates froma point charge q is q/ . . . . . to the strength of field inthat region. . If the size of plates are infinitelylarge.thencharged particle move along an electric line offorce. (xi) If a metallic plate is introduced between plates of a charged capacitor.physicsashok.Itshows that field inside the parallel plate capacitor is uniform. If two electric lines of force cross each other. (x) Electriclinesofforceinsidetheparallelplatecapacitorisuniform.ELECTROSTATICS ELECTROSTATICS (iv) No electric lines of force cross each other. (viii) No electrostatic lines of force are present inside a conductor. thenthe charged particle mayormaynor followtheelectriclinesofforce. (vi) Electric lines offorce for point positive charge and a nearby negative poin t charge that are equal in magnitude are said to have rotational symmetry about an axis passing through both charges in the plane of the page.Also elect ric lines of force are perpendicular to the surface of conductor. (ix) Ifachargedparticleisreleasedfromrestinregionwhereonlyuniformelectricfieldis present. This is knownas fringing effect. But if charged particle has initia lvelocity. th en electric lines of force can be discontinuous.in 8 . E1 E2 Lineofforce Lineofforce (v) Electric lines of force for two equal positive point charges are said to hav e rotational symmetryabout the axis joining the charges. then fringing effect can be neglected . It shows electric lines of force at the edge of plates is non-uniform. This is not physically possible. then induced charges are developed on the sphere. (vii) Electric lines of force due to infinitely large sheet ofpositive charge is normalto the sheet. For example if a conducting sphere is placed in a regi on where uniform electric field is present. it means electric field has two directions at the point of cross. electric lines of force are curved. But at the edge of plates. www. qq . then potential at the point b is b b 0 W V q . Since. (xiv) Lines of force have tendency to contract longitudinally like a stretched e lastic string producing attraction between opposite charges and repel each other laterally resulting in. .. the potential is also a scalar quantit y. . then number of lines of forces in dielectric is lesser than that in case of vacuumspace. SupposeWwork be done inbringing a small test charge q0 fromthe point a to a point b against the repulsive force acting o n it.Also if a line offorce is a closed curve.work done round a closed pathwil lnot be zero and electric field willnot remain conservative. Following three formulae are veryusefulin the problems related to work done i nelectric field. to the value of test charge. Obviously. IMPORTANT FEATURES 1. then potential difference between the points is a b b a 0 V V W q . repulsion between similar charges and edge-effect (curving of lines of force near the edges ofa charged conductor).ELECTROSTATICS www. Electric Potential Difference : The potentialdifference between two points in an electric field is equal to the ratio ofwork done inmoving a test charge fromone point to the other. potentialdifference is also a scalar quantity.Wbe thework required in bringing a test charge q0 frominfinity to a poin t b against the repulsive force F acting on it.in 9 (xii) If a dielectric plate is introduced between plates of a charged capacitor. to the value oftest charge. (xiii) Electrostatics electric lines of force can never be closed loops. Suppose. as a li ne can never start and end on the same charge. (Wa b)electric force = q0 (Va Vb) (Wa b)external force = q0 (Vb Va) = (Wa b)electric force (W.Wand q0 both are scalar quantities. a)external force = q0Va 2. . ELECTRIC POTENTIAL AND ELECTRIC POTENTIAL DIFFERENCE : Electric Potential : Electric potential at anypoint in a electric field is equalto the ratio of thewor k done in bringing a test charge frominfinityto that point.physicsashok. Electric potential due to a point charge q : Fromthe definition of potential. 0 0 0 0 1 . . . .U 4 r V q q .. Electric potential due to a systemof charges : Just as the electric field due to a collection of point charges is the vector sumof the fields produced by each charge.. a)external force = (2 × 10 6) (104) = 2 × 10 2 J . if q is negative...physicsashok. using the relation.6 ×10 19)[(20) ( 40)] = 9. Here. In either case. Sol. it produces a potential that is negative everywhere.. Here. if the whole charge is at equaldistance r0 fromthe point where Vis to be evaluated.. . i. the electric potential d ue to a collection of point charges is the scalar sumof the potentials due to each charge. the test charge is an electron. .. Sol. In the equation i 0 i i V 1 q 4 r .6 × 10 18 J) = 9. q 4 r . r is the distance fromthe point charge q to be point at which the potentia l is evaluated. . If q is positive. where qnet is the algebraic sumof all the charges ofwhich the systemismade. (W. a)external force = q Va We have (W.6 × 10 19 C VA = 20V and VB = 40 V Work done by external force (WB A)external force = q0(VA VB) = ( 1. . net 0 0 V 1 . the potential that it produces is positive at allpoints.e. . i 0 i i V 1 q 4 r .6 × 10 18 J Work done by electric force (WB A)electric force = (WB A)external force = ( 9. 3. Example-8 : The electric potential at pointAis 20Vand at Bis 40V.ELECTROSTATICS www. 4. q0 = 1. Find thework do ne by a external force and electrostatic force inmoving an electron slowly fromBtoA. thenwe canwrite. q 4 r . . Vis equal to zero at r = .in 10 or 0 V 1 . .6 × 10 18 J Example-9 : Find thework done bysome external force inmoving a charge q = 2µCfromi nfinityto a point where electric potential is 104V. . . (0. Sol. 0. 2m. . . 0) and (0. 3m) respectively. . 1 2 3 0 1 2 3 1 q q q V 4 r r r . 0. The net electric potential at origin is. Find the electric potential at origin. . 0). .we have .. . q2 = 2µC and q3 = 3µC are placed at (1m.Example-10 : Three point charges q1 = 1µC. . Substituting thevalues. in 11 V .0 3. . Isolated charge 0 V q 4 r . .. . p x ++++++++ . . .. . . Adisc of charge 2 2 0 V R x x 2 . . . V = 9.physicsashok.ELECTROSTATICS www. Infinite sheet of charge Not defined 5. 1 2 3 10 6 1. Finite line of charge 0 V ln sec tan 4 sec tan . . Aring of charge 2 2 0 V q q 4 R x . . .. . . Infinitelylongline ofcharge Not defined r 6. .0 . q p r 2. 4. . . .0 109 . . 3. ... . .0 × 103 volt TABLE : Electric Potential of Various Systems S. . . . .0 2. .9. FirstColumn SecondColumn 1. .No. r .. R 0 V q 4 r . (b) Outside r . . . . . 8. R 2 2 2 0 V R 3 r 6 R .in 12 7. (b) Outside r > R 0 V q 4 r . . . . . . R 0 V q 4 R . Example-11 : In a regular polygon of n sides. Solid sphere of charge (a) Inside 0 .ELECTROSTATICS www. Identicalcharges are 1) corners. r . . . Charged spherical shell (a) Inside 0 . . . (A) r n (B) r(n 1) (C)(n 1)/r (D) r(n 1)/n Sol.. 2 0 E q 4. r . placed at (n The ratioV/E hasmagnitude.. . . . . and 0 v (n 1)q 4. . . .At the centre. . . each corner is at a distance r fro mthe centre. .physicsashok. 0 2 0 (n 1)q v 4 r q (n 1)r E 4 r . r . the intensityis E and the potential isV. . 2 2 0 V q 0 4 R x . one of radius a and the other of radius b have the charges +q and (2/5) 3/2 q respectively as shown in the figure..Example-12 : The figure shows a nonconducting ring which has positive and negati ve charge non uniformly distributed on it such that the total charge is zero. Sol. . (D) If the ring is placed inside a uniformexternal electric field then net torqu e and force acting on the ring would be zero. Find the ratio b a q =+q A q = (2/5) q B 3/2 z=a b/a if a charge particle placed on the axis at z = a is in equilibrium. .Which of the following statements is true ? (A) The potential at all the points on the axis will be zero. Example-13 : Two concentric rings. . . axis O + + + + + + + + + + (B) The electric field at all the points on the axiswill be zero. ( . (C) The direction of electric field at all points on the axis will be along the axis. .q = 0) There for the potential at all the points on the axis will be zero. O ++ + + + + + + + + x R2 + x2 Hence (A) is correct. . . . . .. . . . 3 3 2 2 2 2 3 2 2 a b 5 2a 2 . . . . .in 13 Sol. . 3 3 2 2 2 2 3 2 2 2 a b 2 a a 5 . . . . .ELECTROSTATICS www. . . . 2 2 2 a b 5 2a 2 . .physicsashok. . . . . .. . . . . . E = 0 A B 3 3 2 2 2 2 2 2 0 0 q z q z 0 4 (z a ) 4 (z b ) . . . . . . . . . . . . . . QE = 0 . . . . . . At equilibrium F = 0 . . . . . . . . A B 3 3 2 2 2 2 2 2 0 0 q z q z 4 (z a ) 4 (z b ) . 3 2 3 3 2 2 2 2 2 2 0 0 2 qa qa 5 4 (a a ) 4 (a b ) . Example-14 : Acircular ring of radiusRwith uniformpositive charge density.with an initial kinetic en ergy 0 q 4 .b a q =+q A q = (2/5) q B 3/2 z=a 2a2 + 2b2 = 10a2 . b 2 a .Aparticle ofmassmand positive charge q is pr ojected fromthe point P(R 3 . . per u nit lengthis located in the yz planewith its centre at the originO. Where q1 is the charge on ringh q + + + + + ++ + + + + + O R l R 3 R2 + 3R2 = x ..(i) 1 i 0 U q q 4..R). . q1 = (2. . . According tomachenical energyconser Ui + Ti = Uf + Tf . Sol. Ans. i 0 U q(2 R ) 4 (2R) . (B) The particle returns to P (C) The particlewill just reachO. . 2b2 = 8a2 2 2 b 4 a . (x) .O) on the positive x-axis directly towardsO. . O. (A) The particle crossesOand goes to infinity. (D) The particle crosses Oand goes to R 3 . . . i 0 T q 4 . . . . 2 f T 1 mv 2 = where v is a velocityofcharge putting the volues in eqn (1) . 1 f 0 U q q 4. . R . f 0 0 U q(2 R ) q 4 R 2 . . . . . . . . i 0 U q 4 . . . . .physicsashok.in 14 2 0 0 0 q q q 1 mv 4 4 2 2 . . 1 mv2 0 2 . . . . . Example-15 : Find the electric field at centre of semicircular ring shown in fig ure. . . . But 0 K 4 l . 2 2q R .ELECTROSTATICS www. . . . . . + ++ ++ + 45º 45º 45º q +q E0 E0 O X E Y 45º . . . . . E . . 2 2 0 0 0 E . . v . E . . 2 E But 0 0 E sin 2 R 4 . 0 The particlewill just reach centre of ring. + ++++ + Y R X q q Sol. . . . 2 0 0 q q 1 mv 2 2 2 . cos. The electric field intensitydue to nonconducting sheet is 0 E 2 . 2 sin2 qE 2 . sin. directed negative x-axis.) = 0 qE .R . . . . . Ifsuddenlya charge +q is giv en to the bob&it is released fromthe position shown in figure. 2 E 4Kq . .) qE 1 cos mg sin .. . =mg. Example-16 : Asimple pendulumof length l and bobmassmis hanging in front ofa lar ge nonconducting sheet having surface charge density. mg... sin . Sol.. .) Wg = mg ( ..(2) WE = qE ..k = 0 Since initial and finalvelocity ofbob is zero Therefore.(1) According towork energy theorem Wnet = . . Find the maximum + + + + + + + + + l s angle throughwhich the string is deflected fromvertical. . . Putting the values in eqn (2) + + + + + + + + + cosq s mg v=0 qE q sinq qE . sin. (1 cos. WE + Wg = 0 . (1 cos. . . . . . . . . . the value ofE fromeqn (1) . . . tan 2 mg . . . Ans. . qE . . . . cos 1 qE . . . . . . . 2 tan 1 mg . . . .mg 2sin 2 2 . . 1 . . . . 0 2 tan q 2 mg . . . . . qE tan mg 2 . . . . . Putting 1 0 2 tan q 2 mg . . . Find themaximumdistance fromAon sheet where the particle can strike. ..u sin ..(1) where ux = u sin. Electric field intensitydue to nonconducting sheet 0 E 2 .ELECTROSTATICS www. T 2umsin qE . . as shown in figure. x a qE m = + + + + + + + + u A ucos usin X Y 2 x x x u T 1 a T 2 = + where x = 0 2 x x 0 u T 1 a T 2 = + . 0 . . . . .T 1 qE T2 2 m . uy = u cos..T 1 qE T2 2 m .u sin . . + + + + + + + + u A Sol.physicsashok.in 15 Example-17 : A particle ofmassmand negative charge q is thrown in a gravityfree spacewith speed u fromthe pointAon the large non conducting charged sheet with surface cha rge density . . .5m 5 m 5 m +s +2s s Sol. 2 0 y 2u msin 2 q . = 1 Example-18 : The figure shows three infinite non-conducting plates of charge per pendicular to the plane of the paper with charge per unit area +. + + + + + + + + + + + + + + A B 2. ... +2. y = uyT y= (u cos. . . At ymax. . . y and z EA = Ex + Ey + Ez 0 0 0 2 2 2 2 . sin2. . . . . . and . .)T y u cos 2u msin qE .. y u2msin 2 qE . . . 2 0 max y 2u m q. fromeqn (1) . .5m 2. . The electric field intensity at point Adue to plate x. . y 2 0 2u m sin 2 q . . Find the ratio of the net electric field at that pointAto that at point B. . . . ... . .. .(1) A B +s +2s s x y z 2s 2Î 0 s 2Î 0 s Î 0 At point B B 0 0 0 E 2 2 2 2 . . . .(2) Fromeqn (1) and (2) A B E 0 E = . . . . B 0 E 4 2 .. .EA = 0 . Electric field intensitydue to nonconducting sphere 0 E x 3 .(1) (where x < R) Where . (C) decreases as r increases for R < r < . Find the frequencyofS. r > R 2 E 1 r . inside a cond uctor) and they fall normally on the surface of a metallic conductor (because whole surface is at same potential and lines are perpendicular to equipotential surface). Electric field lines never enter a metallic conductor (E = 0. of the particle if th e amplitude does not exceed R. both (a) and (c) are correct. Hence. Thema gnitude of electric field due to the sphere at a distance r from its centre : (A) increases as r increases for r < R (B) decreases as r increases for 0 < r < .r r < R 2 0 E Q 4. r < R E . (D) is discontinuous at r = R (E) both (a) and (c) are correct Sol.in 16 Example-18 : A metallic solid sphere is placed in a uniformelectric field. Sol. Example-20 :Aparticle ofmassmand charge qmoves along a diameter of a uniformlycha rged sphere of radius Rand carrying a total charge+Q. .. The lines of force follow the path(s) shown in figure as : (A) 1 (B) 2 1 2 3 4 4 3 2 1 (C) 3 (D) 4 Sol. The force onnegative charge is opposite direction of electric field.. .ELECTROSTATICS www. r . . 0 E r 3 . Example-19 : Anon-conducting solid sphere of radius Ris uniformly charged. is volume charge density. ..physicsashok.M. qE = ma Fromeqn (1) 0 .H. . (3) Fromeqn (2) and (3) 2 0 q 3 m . . . . . . . . . . 0 2 f q 3m . . . .M. .Apointmass having charge +q andmassmis fired towards the centre of the spherew ith velocityv froma point at distance r (r >R) fromthe centre of the sphere. {. . .H. Find theminimumvelocity vso that it canpenetrateR/ .. 3 Q 4 / 3 R . . .. . . Example-21 : Apositive chargeQis uniformlydistributed throughout the volume of a dielectric sphere of radius R. . a = . . . ..(2) x R q For S.. 3 0 1 qQ f 2 4 mR . .. . . . .q x ma 3 .f } 0 1 q f 2 3m . 0 a q x 3 m . . . . .2x . . . . 0 q 3 m . 2. .. .physicsashok. f 3 0 Q 12R R V 32.ELECTROSTATICS www. . Frommachenicalenergyconservation U i + T i = Uf + Tf qV i + ½mv2 = qVf + 0 . . 2 2 . .(1) i 0 V Q 4.. Charge on the smallmass remains constant throughout themotion. r . . where r R 2 = R O vmin 2 R r +q 2 2 f 3 0 Q 3R R 1 4 V 2 4. .in 17 2 distance of the sphere. f 0 V Q(11) 32. . R . . R . R . Neglect any resistance other than electric interaction . . Sol. . . . . . . R . 2 2 f 3 0 1 Q(3R r ) V 2 4. r . .. . R 8 r . . f 0 V 11Q 32. . .. ... . . The potential at the point P due C 4a a +Q B x g P to +Q isV. . R . . . r 2 32.. . . 2 0 v qQ 11 R 2m. R 2. . . . . . The velocitywithwhich the bead should projected fromthe point P so that it can complete a circle should be greater than . . Example-22 : The diagramshows a small bead ofmassmcarrying charge q. R . . Putting the values ofVi andVf ineqn. . . . . R 4. . r .. 2 0 0 mv 11qQ qQ 16.. .. 2 0 0 1 mv 11qQ qQ 2 32. In the same plane a charge +Qhas alos been fixed as shown. . . . . 8R r . 1 2kqQ r R 3 2 v mR r 8 . . . . . . . . . .. . v2 2kqQ r R 3 mR r 8 . . . v2 2kqQ 1 3 R mR 8 r . . . . . (1) 2 0 0 qQ 1 mv 11qQ 4. . The beamcan freelymove on the smooth fixed ring placed on a smooth horizontal plane. . . .. . . 2 0 mv qQ 11 1 2. . .. Ui + Ti = Uf + Tf 20 0 qV 1 mv qQ 0 2 4.(A) 6qV m (B) qV m (C) 3qV m (D) none Sol. (4a) . . . . fromeqn (1) . . According to conservationprincipalofmechanical energy. a .(1) where v0 is velocityof at point p the potential at the point p due to +Qis C 4a a +Q B x g P v0 0 v Q 4.. d to the right (C) 2 0 362qQ 361. Example-23 : Two spherical.. . . 0 v 6qV m = Ans.. . d to the left (D) 2 0 360qQ 361.. and verythin shells of uniformlydistr ibuted positive chargeQand radius d are located a distance 10d fromeach other..in 18 2 0 qv 1 mv 4qV 2 . on the line connecting the centers of the two shells.. What is the net force on the charge q ? (A) 2 0 qQ 361.ELECTROSTATICS www. . . d to the left (B) 2 0 qQ 361..physicsashok. Q Q B q F 19 d 2 . 20 v 6qV m . 20 1 mv 3qV 2 . as shown in th e figure.. Electric force on charge q due to sphreAis zero. d to the right Sol. . But electric force due to sphere B on charge q is 2 0 F qQ towards left 4 19 d 2 . nonconducting.A positive point charge q is placed inside one of the shells at a distance d/2 fro m d/2 10 d d Q Q the center. . = angle made by E2 with x-axis But 2 cos cos x 1 x .0 (C) (3. . Here r = 1+ x2 E1 = The magintude of electric field due to wire along y-axis = 2 0 3 2 1.. . . The net component of electric field along x-axis is x 2 2 0 0 E 3 cos cos 2 1 x 2 1 x . . 1.. ln 2)/. 1) to (0. . where . x . . Example-24 : The diagramshows three infinitely long uniformline charges placed o n the X.. . directed paerpendicular to y-axis E2 = The magnitude of electric field due to wire 1+ x2 E along z-axis = 2 1 x l + directed perpendicular to z-axis. .. 1... . d . . 2 .. = angle made by E1 with x-axis and . . The electric field due to wire along x-axis is directed perpendicular to x-axis. . .0 (B) (. . . . The work done in moving a unit positive charge from(1. ln 2) /2. . .. 0 E 4 x 2 1 x .0 (D) None Sol. . Yand Z axis. x . .A d/2 2 0 qQ towards left 361. ln 2) /2. 1) is equal to Y X Z l 3l 2l (A) (. . . . x 0 0 x 1 dz z . . .. . . . .. . . . . . . .. dV = Ex dx or . 2 1 v x 0 2 v 0 x 1 dV 4 x dx 2 1 x . . .in 19 . . . . . . x 0 2 x 1 0 dV 4 x dx 2 1 x . . . . . .. . . . . Put 1 + x2 = z or 2x dz dx = or dz = 2 xdx x 0 1 2 0 x 1 V V 2 x dz z 2x .physicsashok. x 0 1 2 2 0 x 1 V V 4 x dx 2 1 x . . .ELECTROSTATICS www. .. . . . Find the minimuminitial velocity of projection required if the particle just grazes the shall. Applying mechanical energy conservation principle. . d r 0. (A) 2 m/ s 3 (B) 2 2 m/ s 3 (C) 2 m/ s 3 (D) none of these Sol.x 0 x 1 0 lnz ..5 mm 2 . . ..x) .. . . . 0 1 0 .ln2.. .. . . . . . .ln1. . 2 ln(1.. ... Example-25 : A particle ofmass 1 kg & charge 1 µC 3 is projected towards a non conducint fixed spherical shell having the same charge uniformly V from 1 mm distributed on its surface. . 0 1 0 .. m vd = mv0 r . ln 2 . Apply conservation principle of angular momentum. ... . 0 v vd v r 2 . . . 2 1 0 v ... . . . v . Ui + Ti = Uf + Tf 2 2 2 0 0 1 q 1 0 mv mv 2 4 r 2 . or 2 2 2 0 0 1 q 1 mv mv 2 4 r 2 . .. . . . . 9 12 3 8 9 10 1 10 9 8 3 1 1 10 3 .. . . or 2 2 0 1 3 q mv 2 4 4 r . .physicsashok. . or 2 2 2 0 1 mv q mv 2 8 4 r . . . or 2 2 0 8q v 3 4 mr ..ELECTROSTATICS www.. .in 20 or 2 2 2 0 1 mv q 1 m v 2 4 r 2 2 .. . . . . . .. or 2 2 0 1 q mv 2 3 r .. . . . . v 8 2 2 m/ s 3 3 = = ELECTRIC POTENTIAL ENERGY Two like charges repel each other while the two unlike charges attract each othe .. . . . . . 2 Example-26 : Four charges q1 = 1µC. . . when the charges are moved awayfromeach other or they are brought near each other. 4 3 4 2 4 1 3 2 3 1 2 1 0 43 42 41 32 31 21 1 q q q q q q q q q q q q U 4 r r r r r r .r. . . q2. ..1. . Electric Potential Energy of a System of Charges : The electric potential energy of a systemof charges is given by i j i j 0 ij 1 q q U J 4 . . r ... either somework is obtained or somework is done. Here. . For example. 1 2 0 U 1 q q J 4 r . . q3 = 3µC and q4 = 4µC are kept on the vertices of a square of side 1m. This sumextends over allpairs of charges. electric potential energyof four point charges q1. q4 q3 q1 q2 1 m 1 m Sol. NOTE : Total number of pairs formed by n point charges are n.Thiswork is stored in the systemofcharges in formof electric potential e nergy. . So. all the charges are to be substitutedwith sign. . because that would b e an interaction ofa charge with itself andwe include onlytermswith i < j to make sure that we count each pa ir only once. Find the electric potential energy of th is systemof charges. In this problem. q2 = 2µC. . q3 and q4 wo uld be given by. It is represented byU. .n . . The electric potentialenergyof a systemof two point charges q1 and q2 invacuumat a separation r is given by. . .We do not let i=j. The electric potential energyof a systemof different point charges is equal to th ework done inbringing those charges frominfinityto formthe system. . .in 21 r41 = r43 = r32 = r21 = 1m and 2 2 r42 . 2 f i 0 q U U 3 2 4 a . .0 10 3 . . Sol. . (1) . . . . ext w . .. Ui =Electricalpotential energyinsquare arrangement 2 2 i 0 0 U 4q 2q 4 a 4 2a . . . .. (1) . . Find thework done byexternal agent in changing the +q +q +q +q a a fig (i) fig (ii) +q +q +q +q a configuration of the systemfromfigure (i) to fig (ii).62 × 10 2 J NOTE : Here negative sign of U implies that positive work has been done by elect rostatic force in assembling these charges at respective distances from infinity. . . 12 5 2 . . . we get U = (9. Example-27 : Consider the configurationof a systemof four charges each of value +q. Uf = Electric potential energyof circular arrangement of charge q a2 a a q q q 2 2 f 0 0 U 4q 2q 4 2a 4 (2a) .. = 7. .0 × 109) (10 6) (10 6) (4)( 3) (4)(1) (4)(1) ( 3)(2) ( 3)(1) (2)(1) 1 2 1 1 2 1 .U . . .physicsashok. . . . . .9. . r31 . . . . . . . 2 m Substituting the proper valueswith signin the relation givenabove in the theory. .ELECTROSTATICS www. . Ar dx q x . 0 dV Q2 rdx RL4 x . ext f i w . . . Example-28 : Aconemade of insulatingmaterialhas a total chargeQspread uniformly over its sloping surface. . . . .Calculate the energyrequired to take a test charge q frominfinity to apexAof cone. . B A AB = L Sol. . . U A q(V V ) . . U . . wext A A . . The slant length is L. . . ..(1) HereVA = Electric potential at pointA. .. q(V . The electric charge on the considered ring dq Q2 rdx RL . 0) . sin r R x L . .. . qV . . . .U . . Electric potentialdue to consideredring is 0 dV dq 4 x . . r. . . . .physicsashok.r. r Rx L . . . E V i V j V k . . x) y E V y . . . . ext A 0 w q V qQ 2 L . . x E V x .. .t. . .ELECTROSTATICS www. . . . 0 2 Q R x dx dV L RL4 x . (partial derivative ofVw. (partialderivative ofVw. . . L A 2 0 0 V dv Q dx 2 L . . E i . . Here. E j. .. . .r. (partialderivative ofVw. . y) z E V z .. fromeqn (1) RELATION BETWEEN ELECTRIC FIELD AND POTENTIAL In case of cartesian co-ordinates x y z E . 0 Q L 2 = p Î .t. . .t. . . .in 22 . E k . . z) . . . . . For example. . . . k . 3 y y .grad V . . . . . electric potential due to a point charge q at distance r is 0 V 1 . . This is alsowritten as. . . . .. . . . . Here. . E.2x 3y z. . . q 4 r .. . In polar co-ordinates r E V r . = 0 V 0 . E . q 4 r . . . 1 z z . . E V i V j V k x y z .. . . Example-29 : The electric potential in a region is represented as. . . . . . . 3 j. . . z...2 i . V . . V= 2x + 3y Obtain expression for electric field strength.V . .. . . . .x y z . . . V . .2x 3y z.. . . . and r 2 0 E 1 . . E . . . .. . . V r . . . Sol. . . . . and E 1 . . . . . . . . . . .. . . V 2x 3y z 2 x x . . . . . . .. . . . . . a r . .. the given field is uniform(constant). dy j. . . . .E r Here. . j. d. dV = -E r or . d . where . ..physicsashok. . . . is existing in x yplane as shown in figure.2 i 3 j 4 k . . . . 2 j.1. 2 2x 3y 4z . b a a V -V = .1.. electric field E dV dr . b r . .d. = 1 V Example-31 : Auniformelectric field having strengthE. . . 2.. or E = (slope of v r graph) NOTE : If electric field . Find the p. i .1. then electric potential can be determined from the relation as given below: . dV .1.ELECTROSTATICS www.. 2 i . Here. V . dx i dy j dz k . . . 2. . or a ab a b b V .1. 2.. 1. 1. . between originO &A(d. . . . dr . . and . . 0) . d b . 2k m .. . . 2 2 i 3 j 4k . 1.. . dr . . 2. N C .. E . . . . 2 2dx 3dy 4dz . . . dr .1. . . So using. V . . E is known. . . . k m . . dx i . 2. dz k . . .E . 2... .in 23 Thus. In uniform electric field V = Ed Example-30 : FindVab in an electric field E . . Sol. . .d.. + sin. .) 0 A ..) E A(d.) (B) Ed(sin. . Ed(cos. sin. Ed(cos. = 1 . . 1)&B (1. Ed(cos. Esin. .v . . Ed cos..(A) Ed(cos. Ed sin. Ecos. ) A 0 .0) o x y z (C) 2 Ed (D) none of these Sol. . . V . . Here E . . E . d j . cos.v . .. .. . j . sin. Find the potential difference between pointA(3. (V . 3) (A) 100 V (B) 200 2 V (C) 200 V (D) 0 Sol.V ) . i . y = 3 + x . sin. d i . .. ) Example-32 : Uniformelectric field ofmagnitude 100V/min space is directed along the line y= 3 + x. V . . tan. . OA . 2 2 . 2 i 2 j. .2 i . A D x E V V 2 5 30 15 v /m 0..2 0. = 45º E . . ...100sin. . . . . AB . j . ...100cos. i . . . 2 2 . . V 100 i 100 j . . . . . 3 j. .15 j .in 24 . E = . . The potent ials a these points are V (A) = 2 volt.. . . . .r . . .. .E .100 2 . . .2m (A) 10 Vm 1 along PQ (B) 15 2 Vm 1 along PA (C) 5 Vm 1 along PC (D) 5Vm 1 along PA Sol.100 2 . .r . B. . . .2 2 .. C.V = 0 Example-33 : A. . . P andQare points in a uniformelectric field. . . . .V . 100 100 V i j r 2 2 . D. . .physicsashok. . . . A B V .. ..2m A B P C Q D 0. V(C) = 8 volt. . . i .ELECTROSTATICS www. . . The electric field at P is 0. ... Also A B y E V V 2 5 30 15 v /m AB 0. j . V(P) =V(B) =V(D) = 5 volt. 2 j ..15 i . E 100 i 100 j 2 2 .. P y x 15 15 x A 45º . . . 3 i ..V . 15)2 = 15 2 v /m .we can say that the potential difference between anytwo points on anequipotential surface is zero. The negative slope v x graph give x component of electric field. PA E . The correct statement about electric field is (A) xcomponent at point Bismaximum (B) x component at pointAis towards positive x-axis x A B C v (C) x component at point C is along negative x-axis (D) x component at point C is along positive x-axis Sol. In the giv en graph. x component of electric field is positive. +q This spherical surfacewillbe the equipotential surface and the electrical lines of force emanating fromthe point chargewill be radial and normal to the sp herical surface.15 2 Hence (B) is correct.| E |= (. Fig. Hence. . So. shows the electric lines of force due to a point charge +q.. EQUIPOTENTIAL SURFACE Equipotential surface is an imaginarysurface joining the points of same potential inan electric field. slope at C is negative.15)2 + (. Example-34 : Variation of electrostatic potential along x-direction iswhosn in t he graph.The electric lines of force at each point ofan equipotential surface are normal to the surface. . (d) electric lines of force cross equipotentialsurface perpendicularly. E .2ax2 z .The equation of anequipotential surfacewill be of the form (A) z = constant/[x3y2] (B) z= constant/[xy2] (C) z = constant/[x4y2] (D) none Sol. (i) Equipotential surface due to line charge is cylindrical. (dx i .. E . dv . .. where a is a positive constant. r . (f) The surface ofa conductor in equilibriumis equipotential surface. . dy j. q +q Example-35 : The electric field in a region is givenby : E . (e) Work done to move a point charge q between two points on equipotential surfa ce is zero.But equipotential surface never be a point. dz k ) . = Constant .d r . dzk . (j) Equipotential surface due to an electric dipole is shown inthe figure.in 25 Regarding equipotentialsurface. . dy j. dr . At equipotential surface potential is constant. (c) Electric field is always perpendicular to equipotential surface... Therefore E v r = .ELECTROSTATICS www. . solid etc. .d d . So. . (g) equipotential surface due to isolated point charge is spherical.k . j. x i . z k . dr . constant . i . (h) Equipotential surface are planar in uniformelectric field. dx i . . E .4axy z . . (b) Equipotential surface is single valued. following points areworth noting : (a) Equipotential surfacemaybe planar.ax2y / z . y j. equipotential surfaces never cro ss each other.physicsashok. .in 26 . . . In option (D). . m1 = 2 . where . (2ax2 z) j.. = constant 2 4axy z dx 2ax2 z dy ax ydz z . ELECTRIC DIPOLE Electric dipole is a systeminwhich two equal and opposite point charges are placed at a smalldistance. thene . . The dipolemoment of such a dipole is given by p = q × 2l = 2ql Electric Potential and Field due to an Electric Dipole Let an electric dipole is placed along y-axis and its centre is at origin. m1m2 = 1 . dy j. tan 4 1 8 2 . t henthe electric field vector at (1. Example-36 : The equation of an equipotential line an electric field is y= 2x.ELECTROSTATICS www. dzk ) . 2) may be (A) 4 i +3 j (B) 4 i +8 j (C) 8i + 4 j (D) 8 i + 4 j Sol. y = 2 x or dy 2 dx = . . tan 1 2 . . = constant. 4 2 z constant x y . = constant 6 ax2 y z = constant 2 z constant 6ax y = .physicsashok. Let charges of an electric dipole are q and +q and are separated by a small dista nce 2l. . . . is anglewith x-axis. . . 2 m 1 2 = . . (ax2y / z)k . 2 4ax y z 2ax2y z 2ax2y z 2 .).(4axy z )i . (dx i . Electric field is perpendicular to equipotential line . . The line joining the two charges is called axis of dipole. The product of any of the charges and distance between two charge is called electric dipole 2l q P +q moment p. It is directed fromnegative charge to positive charge (fig. +q q z y l l x2 + z2 + (y+l)2 x2 + z2 + (y l)2 A(x.lectric potential at pointA(x. z) x . y. z) due to this dipole. y. l 2 2 0 V 1 p 4 (r ) . . . and z E V z . . ..ELECTROSTATICS www. l If y = r or axis 2 0 V 1 P 4 r . .. . . ) 2 2 2 0 1 2pr 4 (r ) . l . l ... . . .. . . . .. . . . . . .. . . .physicsashok. x E V x . Special Cases : (i) On the axis of dipole (axial position) : x = 0. l (alongp . . . y E V y . . . . If r > > l (b) Ex = 0 = Ez and y 2 2 2 0 E E 1 2py 4 (y ) . . .in 27 2 2 2 2 2 2 0 1 q q V 4 x (y ) z x (y ) z . z = 0 (a) 2 2 0 V 1 p 4 (y ) . . 3 0 1 . 3 0 1 psin ... .. . . . . .) : (a) 2 0 V 1 p cos 4 r . if r >> l (iii) In polar co-ordinates (r.. l (opposite to p . z = 0 (a) bisec tor V 0 . . l or bisec tor 3 0 E 1 p 4 r . . (ii) On the perpendicular bisector of dipole (equatorical position) : Say along x-axis (it may be along z-axis also) y = 0. . .. . 2pcos 4 r .. . Further at a distance r fromthe centre of dipole (x= r). negative sign implies that the electric field is along negative y-directio n or antiparalleltop . ) Here. . l If y = r axis 3 0 E 1 2p 4 r . p 4 (r ) . . V r . . . . E 1 . (b) Ex = 0. . . . . then 2 2 3/ 2 0 E 1 .. (b) r E V r .. . Ez = 0 and y 2 2 3/ 2 0 E 1 p 4 (x ) . .. . tan tan 2 . . 2 2 r E E E. .. 2 3 0 E p 1 3cos 4 r . E Er E P r O A(r. . . . . . .. ) . . (c) In Figure.4 r . . 9 . .P cos 45 x P = 2 2qa cos 45 . 2a .P cos 45 . Example-38 : Two point charges +3 µC and 3 µC are placed at a small distance 2 × 10 3 mfr omeach other. P . Here 1 P . Find: (a) electric field and potential at a distance 0. q .. 2 a 1 P = 2 2 qa 2 P . q .in 28 Example-37 : 4 charges are placed each at a distance a fromorigin. 2a . 2 q a j . 3 0 E 1 . The dipolemomen t of configurationis (A) 2qa j (B) 3qa j 3q 2q 2q q y x (C) 2aq[i + j] (D) none Sol. 2 qa 3 P . . 2 qa 3q 2q 2q q y x P1 45º 45º 45º P3 P2 2a 2a 2a .physicsashok.2qa cos 45 x P = 0 y 1 2 3 P = P sin 45 + P sin 45 . 2q . (b) electric field and potential at the same distance fromdipole as in (a) if th e dipole is rotated through 90º. Sol. x 1 2 3 P = P cos 45 . p 4 r .ELECTROSTATICS www..2qa cos 45 .= .P sin 45 y 2 2qa 2qa 2qa P 2qa 2 2 2 = + . Dipolemoment p = q × 2l = (3 × 10 6) (2 × 10 3) p = 6 × 10 9 C-m (a) Electric field in equatorialposition.6mfromdipole in its equatorialp osition. . 6) . the same point nowwill be in axial positio n. So. . E = 250 N/C Electric potential. 9 9 2 V 9 10 6 10 (0. . . .9 3 E 9 10 6 10 (0. . . . . 2p 500 N/C 4 r .6) . . and electric potential 2 0 V 1 p 4 r . . . . .. = 150 V . ..V= 0 (b) On rotating the dipole through 90º. . electric field 3 0 E 1 . . . .q. . 0 . Sol. Electric Dipole in Uniform Electric Field : (i) Torque :When a dipole is placed in a uniformelectric field as shown in Fig. It is found that the magnitudes of the electric intensity and electric potential due to the dipole are equal at a point distance r = 5 m fromorigin find the position ve ctor of the point. pEsin . Now |EP| = |VP| or P 2 P 3 2 K 1 3sin K sin r r . . . Let P be such a point at distance r and angle . or . q +q P qE qE l E ...in 29 Example-39 : Ashort electric dipole is situated at the origin of coordinate axis with its axis along x-axis and equator along y-axis.physicsashok. . the net force on it. . . . . .ELECTROSTATICS www. = 5 sin2. .= 45º Positive vector of r . point P is r 5 ( i j) 2 = + . . or 1 3sin2 sin 5 .. E.. F .. or .. or sin 1 2 . = qE × 2l sin . while the torque. . . Squaring both nets y x r P q 1 + 3 sin2. E . . . ..qE . . p . fromequator. NOTE :If the electric field is not uniform the dipole will experience both a res ultant force and a torque so its motion will be combined translatory and rotatory (if . dW= . i..d. .1 to . 2 1 1 2 W pEsin d pE [cos cos ] . W= pE[1 cos.d.1 = 0 to position ..2 = . . 0 or 180º) (ii) Work :Work done in rotating a dipole in a uniformfield througha small angle d.. . . .e. if a dipole is rotated fromfield direction. .2with respect to f ield.. So.willbe.= pE sin . i. So. . . ..e. .work done in rotating a dipole fromangular position . . .] .Fromthis expression it is clear that torque acting on a dipole ismaximum(= pE)wh en dipole is perpendicular to the field andminimum(=0) when dipole is parallel or antiparallel to the field . . . . PotentialEnergyU 1. . force. with 2 pE I . i.) pE (1 cos 90º) U . . TABLE : Dipole-Dipole Interaction S.. . .e. or 2 2 2 d dt . torque and potential energyare different.) or 2 2 d I pE dt . sin . E . .p . This is standard equation of angular SHMwith period T 2 .. . . . So. Interaction of Two Dipoles : If a dipole is placed in the field ofother dipole. . potential ener gy of dipole is defined aswork done in rotating a dipole froma directionperpendicular to the field to the given dire ction. about its equilibriumposition. . . . Relative position of Dipole Force F Torque .No. the restoring torque produced in itwillbe.. If it is given a small angular displacement .. .ELECTROSTATICS www. align itself para llel to the field. r .in 30 q +q P = 0 E Wmin = 0 (a) +q q P = 180º E (b) Wmax = 2pE (iii) Potential energy : In case of a dipole (in a uniformfield).pE cos. . U = (W. . (. (iv) AngularSHM:Adipolewhen placed in a uniformelectric field.physicsashok. . dipolewill execute angular SHMwith time period T 2 I PE . then depending onthe position s of dipoles relative to eachother. . . = pE sin . = pE. W90º) U = pE (1 cos . . . 4 × 10 10m. (b) What is thework done inrotating the dipole through 180º ? .. r (along r) zero 1 2 3 0 1 p p 4. r zero (perpendicular to r) on 1 2 2 3 0 p .p1 F F p2 1 2 0 4 1 6p p (along r) 4 r . (a) Find the necessarytorque required to rotate the dipole through 180º fromits eq uilibriumposition. This dipole is placed in a uniformelectric field of 4. 2... p r 1 p2 F F 1 2 4 0 1 3p p 4.. 1 p p 4. on 1 2 1 3 0 p . .2 × 10 19 C are placed at a distance 2 . r 3.. r F p F p 2 1 1 2 4 0 1 3p p 4 r .0 × 105V/m. . 1 2p p 4. r Example-40 : Two point charges +3...2 ×10 19 C and 3. zero 1 2 3 0 1 2p p 4 r . . from each other. U0 = pE or U0 = (7. W = 7. p = (3.14 × 10 23 J (c) The potential energyof dipole in this positionis given by U.4 × 10 10 m .2 × 10 19) (2. The torque of electric force about centre is balanced bytorque due to frict ion about the centre.2 = 180º. . . In equilibriumposition. where . Electric dipolemoment p = q × 2l Here.1 cos.2 × 10 19 C 2l = 2.68 × 10 29 × 4 × 105 × sin 180º . = PE sin.in 31 (c) What is the potential energy of dipole in this position ? Sol.68 × 10 29 C-m and E = 4.68 × 10 29 × 4 × 105 × (cos 0º cos 180º) W = 7.0 × 105 V/m . mgr sin. = pE sin .1 = 0.is the angle between the axis of dipole and electric field.07 × 10 23 J Example-41 : Awheelhavingmassmhas charges +q and q on diametrically opposite points.. .0 × 105) U0 = 3. .. .68 × 10 29) (4. = pE cos .. .68 × 10 29 × 4 × 105 × (1 + 1) W = 6.= 0 ( . p = 7.ELECTROSTATICS www.4 × 10 10) p = 7. =PE sin..=0 . (for equilibrium) . But f=mg sin. +q -q mgsin 9090E f P E mgr mgr P q 2r . It remains in equilibriumon a rough inclined plane in the presence of un iform vertical electric field E = +q q E (A) mg q (B) mg 2q (C) mg tan 2q q (D) none Sol.68 × 10 29 C-m (a) necessary torque . q = 3.2) Here.physicsashok.= 7. r f= PE sin. sin 180º = 0) (b) Work done inrotating the dipole through 180º is given by W= pE (cos. .. . . P = q × 2r (dipolemoment) E mg 2q . . in 32 Example-42 : Anonconducting ring ofmassmand radius Ris charged as shown.At time t = 0. . It is then placed on a roug h nonconducting horizontalsurface plane.physicsashok. .. 2RE . .E f = mR.Rd. / 2 2 E 0 . 2R . . 2R E . f = macm . . . Where f is friction surface. . d. dp = (2Rdq)} d.. .e. {.ELECTROSTATICS www.E 2R2 .E fR = mR2. whenit startsmoving. . .q 2R. element on ring then dq ..(1) Let d.d. The charged densityi. Sol.. . . dq . . 2 E . a uniformelectric field E = E0i . { d d R . . charge per unit length is . . . ... } y x + + + + p f rough non-conducting surface d dP E . y x + + + + Determine the frictionforce (magnitude and direction) acting on the ring. is swithced on and the ring start rollingwithout sliding. 2RdqEsin . d.. . sin d .Rsin. dq f dqq q dq E d p p. .. . = x E dq × 2 = 2x Edq / 2 2 /2 0 0 d 2x E dx 2E x 2 . . 2E 2 E 2 8 4 . . . . . . . . . 2R. .. . . Find the angular acceleration of the rod just after it is released. Sol.E = macm + mR. . . d. . . . acm = R. . ma . . . . (fromeqn (1)) for pure rolling. . . . . . 2R.RE cm 0 f . . . .2R. E 2 I 4 .Arod Rof length l andmass m is parallel to the sheet and hinged at itsmid point. The linear charge densities on the upper and lower half of the rod are shown in the x l . Electric field due to non-conducting sheet is 0 . . Edq S s Edq dq +dq x x or m 2 E 2 12 4 .E = 2 macm macm = . . .RE Example-43 : In the figure shown S is a large nonconducting sheet of uniformchar ge density . .l R S s figure.E macm = mR. 2 . . (a)] while inwa rd negative [Fig. the net electric flux through any closed surface is equal to the net charge inside the surface divided by . i. dS . . q .. 0 3 2 m . = 90º. . i. . E .in 33 . . dS .e.E = EdS cos .. n E i. . .physicsashok. According to this law. i. .. d. E n (d ) = 0 E min (c) For a closed body outward flux is taken to be positive [Fig.0.. in E S 0 . 2 2 E 12 3E 4 m m . .ELECTROSTATICS www. n dS .. electric field i s normal to the area with (d. . E dS .E)max = EdS.. = 1.e. ... .e. . . i. ELECTRIC FLUX Electric flux throughan elementaryarea dS is defined as the scalar product of ar ea and field. . d. . It represents the total lines offorce passing through the given area. Regarding electric flux following points areworthnoting : (a) Electric fluxwillbemaximumwhen cos. .. E n (d ) = EdS E max (b) Electric fluxwillbeminimumwhen cos. In symbols it can bewritten as. .e. i. E . .e. . field is parallel to the surfacewith (d. . . . = 0. = 0. .E)min = 0.(i) ...e. E . (b)] n (a) Positive flux (b) Negative flux GAUSS S LAW This lawgives a relation between the net electric flux through a closed surface and the charge enclosed by the surface.E . (i) or (ii) can be applied easily. Let us discuss fewsimple cases. Applications of Gauss s Law : To calculate Ewe choose animaginaryclosed surface (calledGaussian surface) inwhi chEqs. Inmost of the caseswewill use Eq. Here. (ii).but this formofGauss s lawis applicable only under following two conditions : (i) The electric field at every point onthe surface is either perpendicular or t angential. S is the areawhere electric field is perpendicular to the surface. (ii) Magnitude ofelectric field at everypoint where it is perpendicular to the s urface has a constant value (sayE). . It is nothing but Coulomb s law.rl) and qin = net charge enclosing this cylinder = . . .ELECTROSTATICS www. in 0 ES .We have to calculate the electric field at a point. .r2 and qin = charge enclosing theGaussian surface = q q r E . 2 0 E 1 q 4 r . a cylinder of any arbitrary length l of radius r and its axis coinciding with the axis of the l r E ++++ + + + + E line charge.r . q . l l . q .physicsashok. we can apply theGauss s lawas. S = are of sphere = 4. E Curved Surface E Plane Surface Here. (ii). a distance r fromthe line charge. . 2 .in 34 (i) Electric field due to a point charge : FromEq. S = area of curved surface = (2. 0 E(2 ) . Here... . 0 E 2 r . . q .l . (ii) Electric field due to a linear charge distribution : Consider a long line chargewith a linear charge density(charge per unit length) . 0 E 4. .. . Hence.We construct aGaussian surface. in 0 ES . E 1 r . q . (iii) Electric field due to a plane sheet of charge : FromGauss s Law in 0 ES . i.. 0 0 0 E(2S ) ( ) (S ) .Hence.e. . . q . . .. 0 E 2 .. + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + E E S0 . applying in 0 ES . . . (iv) Electric field near a charged conducting surface : This is similar to the previous one the onlydifference is that this time charges are on bothsides. . But it is equal to net charge enclosed within the surface only.in 35 Here.) (2S0) + E S0 E + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + . net flux through a closed surface depends upon . . . (D) None of these Sol. The surfacemay have any shape. S = 2S0 and qin = (.ELECTROSTATICS www. It me ans. which are enclosed inside the Gaussian surface. Hence (A) is wrong. electric field at a point is equal to electric flux passing per unit area. E.q. it is a general law. Obviously (D) is wrong. (C) Electric flux through a closed surface is equal to total flux due to all the charges enclosed within that surface only. Hence (C) is correct. Hence (B) is wrong. . It means. . + 0 0 0 E(2S ) ( ) (2S ) .physicsashok. Since. . Gauss lawis applicable to a closed surface. Example-44 : Which of the following statements is/are correct ? (A) Electric field calculated byGauss lawis the field due to only those charges. . . . . dS q .. (B) Gauss law is applicable onlywhen there is symmetrical distribution of charge . . 0 E . . dS . therefore. .. . . is the net flux emanating from a closed surface. Gauss law is 0 E. Though net flux through the closed surface depends upon the charges enclosed in that surface only but electric field E at a point depends no t onlyupon charges enclosed but it depends upon charges lying outside the surface also. . (b) Nowas cube is symmetrical bodywith 6-faces and the point charge is at its ce ntre. . . a hemispherical body is placed in such away that field is parallel to its base (as shown in figure).pb = E × . in total 0 0 . (a)According to Gauss s Law.curved surface = 0 Example-46 : If a point charge q is placed at the centre of a cubewhat is the fl ux linked (a) with the cube (b) with each face of the cube ? Sol.R2E (C) .R2 cos 90º = 0 E n . so electric flux linkedwith each facewill be total 0 ´ q 6 6 . E O C The flux linked with the curved surface is : (A) zero (B) . . .R2E (D) R2 E 2 . . . . the flux linked with base . The flux linked with the body is zero as it does not enclose any charge.pb = 0 As field is parallel to base. Sol.cs + .paralled to base = 0 . . . q . q . = .Example-45 : In a region of uniformelectric field E. . . Example-48 : Calculate flux through the cube and flux through the each surface o f cube when q is placed at one of its corner. Flux throughthe cube 0 q 8 . . . Flux through 0 ABCD q 3 24 . . Flux through 0 ABFE q 3 24 . To cover charged particle completlywe have to use 7 extra cubes around the charged particle. . t he chargewill be in the centre. Flux through EFGH = 0 .ELECTROSTATICS www. .0). Flux Through cube 0 ( ) q 8 . By placing three cubes at three sides of given cube and four cubes above. . .in 36 NOTE : If charge is not at the centre of cube (but any where inside it). . total f lux will not change but the flux linked with different faces will be different. . Example-47 : If a point charge q is placed at one corner of a cube. . the fluxwith each facewillbe one-eight of the flux (q/. So. . Sol.physicsashok. D C A E F B H G q Flux through 0 ADEH q 3 24 . what is the flux linkedwith the cube ? Sol. So. 0 E . . . . . Totalflux 0 . . . q .12 C . 0 EA . q . .10. . q 5 103 8. . . . q .3 12 2 3 2 5 10 8. 3 0 0 E a q .. x = a. .2. . E x i . . . z = 0 and z = a. . l = 2cm and a = 1cm. .Flux through FGCB = 0 Flux through GCDH = 0 Example-49 : The electric field in a region is given by 0 E = E x i . . Sol.2.. 3 0 0 E a q . y = a.85 10 1 10 2 10 . . . l . . Take E0 = 5 × 103N/C. a . Find the charge contained inside a cubical volume bounded by the surfaces x = 0. .85 10 4 2 . q . q. y = 0. . . whereAis area of surface 0 2 0 E a . . Thus. . .. FromGauss s Law C 0 E . . the electric field at anyexternalpoint is the same as if the total charge is concentracted at centre.We can construct a Gaussian surface (a sphere) of radius r > R. .r ) . . q . 2 0 E 1 q 4 R .. we canwrite.. . r R Gaussian surface E + q + + + + + + + + + + + Hence.physicsashok. . 2 0 E(4. . outside 2 0 E 1 q 4 r .ELECTROSTATICS www.. ds .in 37 ELECTRIC FIELD AND POTENTIAL DUE TO CHARGED SPHERICAL SHELL OR SOLID CONDUCTING SPHERE Electric Field The allpoints inside the charged spherical conductor or hollowsphericalshell. . Einside = 0 surface 2 0 E 1 q 4 R . as there is no charge inside such a sphere. At the surface of sphere r = R.. q . el ectric field E . 0. 2 0 E 1 q 4 r . . . 0 V 1 q 4 R . . .. . outside 2 0 E 1 q 4 r . 0 V 1 q or V 1 4 r r . . r . . 2 surface 2 0 0 0 E = 1 q = q/4pR = s 4pe R e e Potential : As we have seen. . . Thus.. . . the potentia l is same at allpointswhich is equal to the potential at surface... .... The variation ofelectric field (E) with the distance fromthe E O R r 1 q R2 = E r2 1 centre (r) is as shown in fig. .. Thus. At some internalpoint electric field is zero everywhere. (V. . . . V r 0 outside 2 0 dV q dr 4 . . . at externalpoints the potentialat anypoint isthe samewhenthewhole charge i s assumedto the concentrated at the centre. . E dV dr .At the surface of the sphere.. . . . . . r = R. therefore. outside 2 0 dV 1 q dr 4 r .we canwrite. = 0) . NOTE : At the surface graph is discontinuous. . . ..inside surface 0 V V 1 q 4 R . . .r2 and 3 in q ( ) 4 r 3 .30 . . .. . .. 3 . Sol. q = 500 µC = 500 × 10 6 C and R = 30 cm= 0. .0 10 0. In side hollowconducting sphere. . . intensityof electric field E = 0 and potential 0 V 1 q 4 R .ELECTROSTATICS www. . q .We have to find the intensityof electric field due to this charged sphere at point P distan ce r fromcentreO. = charge per unit volume = q 4/3.. Here.(i) Here.5 × 107 V ELECTRIC FIELD AND POTENTIAL DUE TO A UNIFORMALY NON-CONDUCTING SPHERE Electric field : Let positive charge q is uniformlydistributed throughout the volume of a solid s phere of radius R. S = 4. . Here. V O R r 1 q R = V r 1 R The potential (V) varieswith the distance fromthe centre (r) as shown in Fig.in 38 and outside 0 V 1 q 4 r . The point at 10 cmdistance fromcentre of spherewillbe inside the sphere. . Hence. Example-50 : The diameter of hollowmetallic sphere is 60 cmand charge on sphere is 500 µC.. . electric field is zero everywhere and potential is uniform(same as at the surfac e).. ApplyingGauss s law in 0 ES . . .R . Find the electric field and potential at a distance 10 cmfromcentre of sphere. ..30 m . . 6 9 500 10 V 9.physicsashok. . V = 1. r 4 R . so qin = q. and Gauss s law gives. 2 0 E 1 q 4 R .r ) .we have 3 0 E 1 . 2 0 E(4. (i). inside 3 0 E 1 . E O R r 1 q R E r E r2 2 1 .. or 2 E 1 r .we have the following formulae formag nitude of electric field.. E = 0 At surface r = R. r At the centre r = 0. q .. . q . r 4 R . Thus.. . . q . so. q 4 R . or 2 0 E 1 q 4 r . . To find the electric field outside the charged sphere.This surface encloses the entire charged sphere. for a uniformly charged solid sphere. . surface 2 0 E 1 .Substituting these values inEq. . outside 2 0 E 1 .. q 4 r . or E ..we use a sphericalGaussian surface ofradius r(>R). so. .. . . . At r = R. . . .. . .. 4 R . . or V 1 r . The electric intensityinside the sphere. . V r outside 2 0 dV 1 . 4 r .in 39 The variation of electric field (E)with the distance fromthe centre of the spher e (r) is shownin fig.physicsashok.. i. q dr . q . q 4 r . inside 3 0 E 1 .. 0 V 1 q as V 0 4 r . . outside 2 0 E 1 . . at the surface of the sphere potential is S 0 V 1 q . . Potential : The field intensityoutside the sphere is. . inside inside dV E dr .ELECTROSTATICS www. outside outside dV E dr . r 4 R . . 0 V 1 q 4 R . ..e. . . . . . . . . i. . . 0 V 3 1 q 3 V 2 4 R 2 . Substituting S 0 V 1 . we get 2 2 0 V 1 . . . . .e. potential at the centre is 1. 2 r S 3 0 R V V 1 q r 4 R 2 . . . q . .S V r V inside 3 R 0 dV 1 q r dr 4 R .. surface 0 V 1 q 4 R . .. Thus. . . 4 R . . . .5 times the potential at surface. . q 3 1 r 4 R 2 2 R . . .. . . . . . . . . ... . At the centre r = 0 and C S. . 1 q . . .. .. and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . . .. . outside 0 V 1 q 4 r . for a uniformlycharged solid spherewe have the following formulae for pote ntial. . R 1 q R 32 V The variationof potential (V)withdistance formthe centre (r) is as shown in figure. . .. inside 3 0 E 1 . and 2 inside 2 0 V 1 q 3 1 r 4 R 2 2 R . r = 10 cm = 0. .6 m .60) 2 2 (0. Sol. . 7.. ..0R3 and (b) the electric field inside the sphere has amagnitude given by. Showthat: (a) the total charge on the sphere isQ = . .. The point 10 cmfromcentre of spherewill be inside the sphere.ELECTROSTATICS www. . .60) . . . q = 500 × 10 6 C. .60) . . . . qr 4 R . . . . . .1 × 107 V Example-52 : Asolid non conducting sphere ofradiusRhas a non-uniformcharge distr ibution ofvolume charge density.5 106 3 1 2 72 . R . .0 is a constant and r is the distance fromthe centre of the sphere. . where . . Hence. = 1. . Here. .5 106 107 72 . . . . .0 10 500 10 0. . 2 4 KQr E . 6 2 9 2 9 10 (500 10 ) 3 1 (0. . 7. . . .10 2 10 N/C (0.physicsashok.. . 0 r .10 m R = 60 cm= 0. . . . . Find the electric field and potential at a distance 10 cmfromcentre of sphere. . . . .in 40 Example-51 : The radius of a solidmetallic non-conducting sphere is 60 cmand cha rge on the sphere is 500 µC. R . .10) (0. . 6 9 6 inside 3 E 9. . r dr R 4 0 3 0 4 R Q Q R R 4 . . . . . in 2 0 E q 4 r . .r2dr 0 2 dq dV r 4 r dr R . ... . .. .Sol. . .. .. . 0 3 4 r dr R .. . . (a) dv = 4.. . 4 0 2 0 4 r E 4R 4 r . (b) r 0 3 in 0 4 q r dr R . . .. . . . . . . 4 4 0 0 4 r 4 r R 4 4R . R 0 3 0 4 Q r dr R . .0R3 .But Q = . . B and C. .. . a . . . .. . . . and 0 K 1 4 . b and c (a < b < c) have surface charge densities . (ii) If the shellsAand C are at the same potential. . . . . . . (i) Find the potential of three shellsA. .physicsashok.on C) 4 a2 4 b2 4 c2 K a b c . . . b .ELECTROSTATICS www. . . 0 . . Example-53 : Three concentric metallic shellsA. . 2 4 E KQr R . .onA) + (Potential ofAdue to . 2 0 a b c b .on B) + (Potential ofAdue to + .on B) + (Potential due to + . The three shells are shown in figure. . . .. respectively. B and C of radii a.. . . . . .on C) 4 a2 4 b2 4 c2 K a b c .onA) + (Potential due to . obtain the relation between the radii a.in 41 0 3 Q R . .s Potential of B = (Potential due to + . c . . . c b C B A a +s +s . .. . . .. Sol.. . . (i) Potential ofA= (Potential ofAdue to + . . and . . . .. . . b and c. . . Example-54 : An electric field converges at the originwhosemagnitude is givenby the expression E = 100N/C. .on B) + (Poten tial due to + . . . .. . . .. . (A) total charge contained in any sphericalvolumewith its centre at origin is ne gative. . (D) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10 9 C. . . . c = (a + b). (C) total charge contained in a spherical volume of radius 3 cm with its centre at origin has magnitude 3×10 13 C.onA) + (Potential due to . . . (B) total charge contained at anysphericalvolume. .on C) 4 a2 4 b2 4 c2 K a b c . Solving we get.Potential of C = (Potential due to + . where r is the distancemeasured fromthe origin. . . . . . . 2 2 0 a b c b c . . . . . . . . (ii) Given that VA =VC or .. . . . . . 2 2 0 0 a b c a b c b c . . . . . . or a2 b2 a b c c b c . . . . . . is negative. . irrespective ofthe location of its centre. So. . . Example-55 : Figure shows a section through two long thinconcentric cylinders of radii a&bwith a < b. . . . Since. .ELECTROSTATICS www.physicsashok. Find the electric field at a distance r fromthe axis for b a (A) r < a (B) a < r < b (C) r > b Sol. 2 0 E q 4 r .. The cylinders have equal and opposite charges per unit length .. electric lines of forces are terminated into the sphere. . . . . O . total charge contained by should the sphere should be negative. q = 3 × 10 13 m . .. Hence. ds . . . options (A) and (B) are correct E E E E . . 2 r . . 0 E 2 r . . 2 r . r b E a 0 E . .. Hence. ds . . . a l b . option (C) is correct. E = 0 (B) a < r < b C 0 q E . Here r = 3 × 10 2C . r b a 0 E .l (A) r < a C 0 0 E .in 42 Sol. . . . Howis the Colulomb force between two charges affected bythe presence of a thi rd charge ? 6. 0 THINKINGTYPEPROBLEMS 1. .Why? 5. What are insulators and conductors? 3. . Atruck carrying explosives has ametal chaing touching the ground. E . after touching the rod. (C) r > b C 0 E ds 0 . ... Aperson standing on an insulating stool touches a charged insulated conductor . . Electric lines of fore never cross. Acharged rod attracts bits of dry cork dust which. The electric charge ofmacroscopic bodies is actually a surplus or deficit of electrons. 0 a b r E . . Explain.r. 2. .Why not protons? 2. . of ten jump away fromit violently. 4.Why ? . . Is the conductor completely discharged ? 7. . Of the po ints A. Will the leaves of the electrometer be deflected ? What will happen to the leaves if the wire is removed and the rod is then charged ? 16. Will th e leaves of the electrometer be deflected in this case? Will the deflection change if the rod is earthed ?There is no connecion between the rod and the housing. Is this true or fa lse ? 9. Theelectricpotentialinsideahollowchargedsphericalconductor isthesameat allpoints andisequalto the potential of the surface. The electric field inside a hollow charged conductor is zero.. AchargeAsituatedoutsideanunchargedhollowconductorexperiencesaforceifanother char geBisplaced insidetheconductor. If only one charge is available. But if the wire is removed and the charge is transferred by the ball to the ball of the electrometer. In the previous question. Can a small spherical body of radius 1 cm have a static charge of 1 C? 15. Can two balls with like charges be attracted to each other? 20. Why ? . The electric field inside a hollow charged sphericalconductor is the same at all points and is equalto the field at the surface. Is this true of false? Briefly explain your answer. the deflection depends on which surface of the bucket (external or internal) is touched before that. The rod and housing of the electrometer are connected by a wire. B and C. Is this true or false ? 10. The work done in carrying a point charge from one point to another in an ele ctrostatic field depends on the path along which it is taken. The housing of the electrometer of question no. but Bdoesnot experienceanyforce. Abroad metalplate is connected to the earth through a galvanometer. Draw an approximate diagram showing how the current flowing through the galvanometer depends on time. 18. The values of the ptentials are written in brackets. does it make a difference which face of B is touch ed in order to remove the free charge (include positive charge) from B? 14. and then a certain charge is imparted to the housing.15 is given a charge. The figure shows lines of constant potential in a region in which on electri c field is present. 17. By touching different points of a metal bucket with a test ball B connected by a wire to an earthed electrometer it can be observed that deflection ofthe leaves of the electrometer is the same for any position of the ball. G ive reasons for your answer in brief. 19...ELECTROSTATICS ELECTROSTATICS 8.. 21. Is this true or false ? 11. can it be used to obtain a charge many time s greater than it in magnitude? 13. Apositiv ely charged ball flies along a straight line above the plate at a distance much less than the linear dimensions of the plate.. the magnitude of the electric field is greatest at the point.Why?Doesitnotviolatethethird lawofmotion? 12. A metal leaf is attached to the internal wall of an electrometer insulated f rom the earth. The gravitationalfield strength is zero inside a sphericalshellof matter. Th e electric field strength is zero not only inside anisolated charged spherical coonductor but inside an isolated conductor of anyshape.say.22.What is the magnitude of the electric field for points inside the sphere ? Will your ans wer change if the sphere carries a charge? 26. Is the gravitational fieldinside. Cantwo equipotentialsurfacesintersect? 24. (b) equal to and (c) greater than Q. conducting spherical shell of radius R carries a negative charge Q. Areequipotentialsurfaceswhichariseduetoapointchargeandwhosepotentialsdiffer byac onstant amount (say 1 volt) evenly spaced in radius ? 28. An isolated.in 43 . Sket ch the lines of force and www. acubicalshellofmatter zero ?Ifnot. 25. positivelychargedmetalobject isplacedinsideandconnectedbyawire?Assume that thepo sitivechargeis (a) less than.physicsashok. Is charge uniformlydistributed over the surface of an insulated conductor of any shape ? If not. what is the rule for the distribution of charge over a conductor? 27. What will happen if a small. Anunchargedmetalspheresuspendedbyasilkthreadisplacedinauniformexternalelectricfi eldE. Two point charges Q and 4Q are fixed at a distance of 12cm from each other. inwhatrespect istheanalogyno t complete? 23. ELECTROSTATICS www. The kinetic energyof the electrons inthe beamincreaseswhile the batteryproducing the potentialdifferenceV performs nowork. Is this true or false ? 34. Indicate the surface distribution of charge on a squaremetal plate byusing d ots in such away that the greater the surface densityof charge. But the special rubber tyres of aircrafts a re made conducting.As the balloon is blown up. Conductors arematerials inwhichthere are a fewfree electrons per atomof thema tter. An electric dipole is placed in a non-uniformelectric field. and (c) if a second charge is placed inside the gaussion surface? 40. Zero work is donewhen a charged particle is transferred fromone equipotentia l surface to another. Ordinary rubber is an insulator. unless conducte . however. There is an electric field near the surface of a conductor carrying direct c urrent. and (c) outside the balloon? THINKINGPROBLEMS SOLUTION 1. (b) on the surface of the balloon. Is this true or false? 36. the farther awaythe dot s are fromthe plate (shown in fig. Asmall sphere is charged to a potentialof 50Vand a big hollowsphere is charg ed to a potentialof 100V. 3. This charge. To conduct away the charge produced by friction. (b) if a second charge is placed near. Howcan his fact be reconciledwith the principle of conservation of energy? 35. Is an electric field of the type shown by the electric lines in the figure physicallypossible? 33. 2. D oes any charge develop on its surfaceswhich are perpendicular to itsmotion? 32. Does electric flux e . How can youmake electricityflowfromthe smaller sphere to the bigger one? 38. 4.physicsashok. if any. Is there a net force on it? 39. Insulators arematerials inwhich the electrons are not as free as in conductors. Aspherical rubber balloon carries a charge that is uniformlydistributed over its surface. the original sphere. 30. ApotentialdifferenceVis set up betweena filament emitting electrons in avacc umtube and a thinmetallic ring. it acquires like charge and so is r epelled strongly bythe charge on the rod. change (a) if the sphere is replaced by a cube of the same volume.They cnnot be removed fromthe re easily.). Draw the lines of force indicating the direction clearly.in 44 locate the neutralpoint. Apoint charge is placed at the centre of a spherical gaussion surface. The electron beampasses past the ring through its central regionwithout spreadin g. Acharged rod first attracts the dust by producing unlike charge at the near e nd and like charge at the far end. Because protons are tightlybound in the nucleus. since no current flows through the circuit. When the cork dust touches the rod. Ametal plate ismoved with a constant acceleration a parallel to its plane.Why is this necessary? 37. 31. and outside. how does E vary for points (a) inside the balloon. Acharge +Qis fixed at a distance d in front of an infinite metal plate. 29. theman acts as a conducting body. 7.d away to the earth. 5. 6. 8. It is true. Because if theycross. may produce sparks causing the explosives in the truck to catch fire and explode.which is physicallyimpossible. No. that is. the electric field at the point ofintersectionwillhave two directions simultaneously. This follows fromthe conditionthat all points on conductor have the same potential. The electric field inside a hollowcharged conductor is zero. He shares chargewith the charged body. The force on the charge due to another does not depend on the presence of oth er charges. electrical forces are physically independent. . in 45 9.physicsashok. 10. Place an insulated conductor B close to the charged body. Next.An opposite current is producedwhen the cha rgemoves away from the plate. 15. But the unlike charge at the near endwill remain bound due to the force of attraction. deliver this charge to a big b ody by conduction. It is true. the stronger the field. E 9 109 1 9 1013 NC 1 0. 12. No current is produced when the chargemoves above the plate. equal inmagnitude to the inducing charge. the equipotential surfaces are closest in the neighbourhood ofB. an equal unlike charge is induced at the near end ofBand equal like charge at the far end. It is greatest at the pointB. 16. Since the electric field is the rate of fallof potential. 11.A.Air cannot sustain sucha strong electric fiel d. The unlike charge.No. . It is a fact that positive charge always flows fromhigher to lower potential. the charge B is simplyan internal part of the sphere. that is. so a small bodycannot have a charge of 1 coulomb. the closer the equipotentialsurfaces. 20. irrespective ofwhich part ofB is touched. It is false. Repeat the process after completely dischargingB. To ob tain a similar charge in largemagnitude. When the housing of the electrometer is given some charge. The field inside is zero but on the surface. so th e leaveswillnot be deflected. In the figure. the potentialof t he rod cannot be the same as that . Yes. .Apositive current pulse passes through the galvanometer. first deliver unlike charge to a bodyC and thenrepeat the induct ion process betweenC and the body inwhich like charge is to be stored. 19. The free chargewillgo to the earth via the finger. No. 18.while the induced negative charge accumulates on the upper surface of the plate. electrostatic induction causes the induced positive charge to pass into the earth. thework done in carrying a point charge fromone point to anothe r does not depend on the path alongwhich it is carried because electric field is conservative. both the leaves will be deflect ed because of the potential difference between the rod and the housing. so the field at Bis the greatest. In fact. yinduction. irrespective ofthe path.ELECTROSTATICS www. the field is finite . It is false. .The field is everylarge. if the charge of one ball ismuch greater than that of the other.01 . force a rises betweenAand the hollow sphere. because the free positive charge inB exists at a higher potential due to the positive charge onA. As the charge approaches the plate.The like charge at the far end will flowto the earth. this is not in violationof the third lawofmotion. . Theycan. Nowremove the bodyB to a distant place. After thewire is removed and the rod is charged. by touching this bodywithB. The housing and the rod connected togetherwillhave the same potential.The forces of attraction caused by the induced chargesmay exceed the forces of repulsion. . 13.will spread over the surface ofB as there is no force of attraction to keep it on one side. by repeating the induction process. 14. This is so because there is an electric field overAbut there is no electric field over B as it lies inside a hollow conductor. 17.Touch the body B with your finger tip in the presence ofA. But when the ballB is disconnected fromA. 21. so the deflectionwillbe greater. the potentialof the leaves (and also of the rod) is raised above the potential of the housing. the leaveswill be deflected. the potentialdifference between the rod and the housing is increased further. so there is deflection of the leav es.On account of thepotentialdifference between the rod and the housing. When the rod is earthed. Since the surface of the bucket is equipotential.When it delivers this charge toA.When B touches the inner . the leaves showthe same deflectionwherever the testing ballBmaytouch the surface of the bucket.of the interior of the housing because the rod lies partlyoutside. The electrometermeasures the potentialdiference between the bodyand the housing of the electrometer. the ballBcollects charge by conduction. .Vis negative. V.. But when it touches the outer surface.ELECTROSTATICS www. . . . .e. . The field inside themetal sphere at all points is zero because r . it collects no charge because there is no potential difference between it and the inner surface of the bucket. the answer does not change if the sphere carries a charge. . It ismore concentrated at the edges and corners. . qmust be reduced to zero. where R= radius of the shell and q is the charge of themetal body. VA (potentialof the surface of the shell) 0 1 Q q 4 R . . No. The analozyis not complete in respect of distribution of mass and charge. so there is deflection. . E . (b) when q =Q.. the entire charge on themetalbodymust flowto the outer shell. there can be no flowof charge fro mone body to another. Vis positive.Mass is uniformlydistributed but charge is not distributed unifo rmly. 22. Unless there is a potentialdifference. the field inside a cubical shellofmatter is not zero. . . . V So chargewillflowfromthemetalbodyto the shell. . . 23..whichis physicallyimpossible. . .physicsashok. the electric field at that point can have two directions simultaneously.. .. 24.V= 0. . . . B A 0 V V q 1 1 4 r R . (a) when q <Q.. VB (potential of themetal body) 0 1 q Q 4 r R .. . the common potential 0 q Q 4 R . So no deflection takes placewhen the ballB touches the inner surface and then touchesA . .. Thus. No. . So E = 0. i. it collects a certain amount of charge.. 25.. For the sake of simplicity. is infini tyformetals and 0 r E . B A V . (c) when q > Q..in 46 surface. . . No. two equipotentialsurfaces can never intersect because if they intersect at a point. SinceVB =VA at equilibrium. let us consider themetal bodyto be a sphere plac ed at the centre of the shell. . . . . The lines of force are shown in the figure. The lines of force are shown in the figure.At the neutral point N. ... No. Note : One everyimportant result follows fromthismap of electric flued due to a point charge and the induced charges on ametalplate. or . . . .. . . . the greater the surfa ce density of charge. . . ... r is not constant. . . . Charge is distributed according to the curvature of the surface. 27. Let it be at a distance x from4Q. Then .26. .r Obviously. Charge is not distributed uniformlyover the surface ofa conductor ofanyshape ..r . . the total field is zero. its surface is an equipotential surface. . .The greater the curvature.. .. Explanation Since it is ametalplate. 0 1 1 1 q r 4 r r r . . So the equipotential surfaces differing by 1Vare not equispaced. So lines of forcemust terminate normallyon it. . . . . or x = 8 cm 29. 2 1 1 2 0 1 2 0 1 2 q 1 1 q r r V V 4 r r 4 r r . the centre ofgravity of the charges.. 28. Here. where 2 1 . all the lines near the edge of the figurewill appear to radiate uniformlyfromthe point P. r ..2 2 0 0 1 4Q 1 Q 4 x 4 12 x ... If a charge Qis placed as far behind the plate as +Q . This is called the electrical image of +Q.Aforcemay be applied on an electron. it experiences a force towards the ring and so it is accelerated. Hence. Now. E .0.0 . The potential of the smaller conductor is now 100+50 =150V. an electric field of the type represented is physically impossible .This is equal to V/l. 38. Connecte the two conductors byawire. so that there develops an elec tric field inside themetal plate. Eoutside Hence.physicsashok. consider a closed path abcda as shown in the figure. 37.Nowmove a charge along this path.E . Yes. there is a field near the surface of the conductor.0. is converted into its kinetic energy. There is an electric field between the filament and the ring. the electron can gain velocityat the cost of its ownpotential energy. the electrons inside are also acceleratedwit h the same acceleration.As potential inside a hollowcond uctor is the same as that of its surface. Since E = . Thus.ab. So for interaction between +Qand the plate or for electric fields between them.wemayuse the formali smof replacing the plate by Q. There is an electric field inside a current-carrying conductor.Work is always done byanexternal agent when a charged particle is tran sferred fromone equipotential surface to another. 36. there is a net force on the dipole given by F p E . the surface density of charge developed is given by . 33. No. = . It electricalpotential energy. To conduct awayelectricityproduced byfriction.work done is essential lyzero as it is a conservative field. The distributionof charge is shownin the figure. dots are equidistant fromthe s traight portion and far away fromthe corners. Explanation Since the surface density of charge is proportional to curvature and corners have the greatest curvature and a plane surface has zero curvature.The intens ityof the electric field is given by eE = ma. inside outside inside E . So theremust d evelop charges on the faces that are perpendicular to the direction ofmotion. Thework done per unit charge = potential difference. the potential inside the hollow conductor is 100V./. The front facewill be charged positively. Imagine a rectangular pathwith its two sides perpendicular to the electric lines . .0.Work done byan external agent along abcda is zero. 31.But inanelectric field.cd. 35.in 47 is infront and the plate is removed. . Since the plate is accelerated. Place the smaller one inside the bigger one.0 ma/e 32. whereVis the voltage across the conductor and l is the length of the conductor.As the electron is emitted fromthe filament.ELECTROSTATICS www. though the batterydoes not supply any energy. and the rear one negatively. Chargewill flowfromthe smalle r one to the bigger one as the smaller one is at 150Vand the bigger one at 100V.0E = . 30.AB(positionof the plate) is still an equipot ential surface. Some networkwillbe performed. True. The concentration of lines at the botomindicates a field which is strong er at the botoomthan on top. 34.which it acquires on emerging fromthe filament. onlywhen an electric field is created. False. . . q / 4. it remains the same as the total fluxis determi ned solely bythe charge inside the surface and not the charge outside.. (b)E decreases as the surface densi tyof charge decreases . E . .l 39. 1 q . 40. . where r is the distance of the point fromthe centre of the balloon..E = 0.. (a) For points inside the balloon. . 0 E .. . flux will change as the total c harge inside the surface has changed. . . (a) No.s. the fluxwill not change as it depends onlyon the total charge inside the surface and not on the extent and shape of the surface. (c) E remains constant because 2 0 E ./ .. r . byGauss s law. . (c) Yes. (b)No. Total flux E 0 . . (E) If A is fals e but R is true. When a positively charged glass rod is brought close to the metal sphere. 2. Assertion (A) : A bird perches on a high power line and nothing happens to t he bird. but there does occur a loss of energy. both of them may not be charged. 3. Two copper spheres of same radii one hollow and other solid are charged to th . Assertion (A) : Electric current will not flow between two charged bodies whe n connected if their charges are same. 10. Reason (R) : In case of sharing of charges. the electric field inside it is zero a t every point. Reason (R) : Electromagnetic radiations exert pressure. But if the sphere touches the rod. Reason (R) : Surface charged density = charge/area. Assertion (A) : Dielectric breakdown occurs under the influence of an intense light beam. Reason (R) : Current is rate of flow of charge. the sphere is drawn towa rd the rod. Initially. the energy of conservation fails. Objective Type Question 1. Reason (R) : X-rays emit photoelectrons and metal becomes negatively charged. 9. Reason (R) : If a conductor is connected to ground. can be built t o block an electric field. Reason (R) : The potential of a conductor depends upon the charge given to it. it suddenly flies away from the rod. then repelled.in 48 Reasioning Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING INSTRUCTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R ) is given just below it of the statements. Ametal sphere is suspended from a nylon thread. Explain why the sphere is first attracted. 11. If high energy X-ray beam falls on the ball. Reason (R) : Due to induction effects a charged body can attract a neutral body. 1.ELECTROSTATICS www. 8. 4. Then the electric field intensities near their surfaces are also equal. Assertion (A) : Metallic shield in the form of a hollow shell. Assertion (A) : A small metal ball is suspended in a uniform electric field w ith an insulated thread. Level # 1. Assertion (A) : The surface densities of two spherical conductors of radii r1 and r2 are equal. mark the correct answer as (A) If both A and R are true and R is the correct explanation of A. the ball will be deflected in the electric field. 5.physicsashok. Assertion (A) : Two adjacent conductors. Reason (R) : The level of bird is very high from the ground. Reason (R) : In a hollow spherical shell. (C) If A is true but R is false. Assertion (A) : If there exists colomb attraction between two bodies. 6. the metal sphere is uncharged . carrying the same positive charge ha ve a potential difference between them. (B) If both A and R are true but R is not the correct explanation of A. there occurs no l oss sof charge. the extra charge induced on conductor will flow to ground. Assertion (A) : When charges are shared between two bodies. Assertion (A) : The tyres of aircrafts are slightly conducting. (D) If both A and R are false. 7. moving on a line perpendicular to the internuclear axis. -particle is (A) 2 2 0 4 3 3.. B and C satisfy (A) VA > VB (B) VA < VC (C) VA < VB (D) VA > VC . The distance between the two hydrogen nuclei is b The maximum force experienced by the . Let A be the origin.. e b (B) 2 2 0 8 3. e b (D) 2 2 0 4 3. Through the exact centre of a hydrogenmolecule. an . -particle passes rapidly .. e b 3. A uniform electric field pointing in positive x-direction exists in a region.e same potential then (A) both will hold same charge (B) solid will hold more charge (C) hollow will hold more charge (D) hollow cannot be charged 2. e b (C) 2 2 0 8 3 3. Then the potential at the points A. . B be the point on the x-axis at x = +1 cm and C be the point on the y-axis at y = +1 cm. 2 0 1 3 2 4 . 2 0 1 3 4 4 q . .physicsashok. . Two circular rings A and B. 2C and 3C. Three identical spheres each having a charge q and radius R. .45 x 1010 J 5..C . . . .. 7ms. A particle of mass m = 100 gm and carrying a po sitive charge q . .1 7.8 x 1010 J (B) 39.. (A) (B) (C) (D) 6. (D) . .10. each of radius a = 130 cm are placed coaxially wi th their axes horizontal in a uniform electric field E = 105 NC 1 directed vertically upwards as shown in Figure.C while ring B has a negative charge of magnitude 2 q . 2 0 1 2 4 4 q . R .. q .in 49 4.y plane as shown in figure. Distance between centers of these rings A and B is h = 40 cm.. R . Calculate its velocity when i t has moved a distance of 40 cm.1 (C) v . . 10.1 (D) v . 32ms.ELECTROSTATICS www. Correspond ence electric field lines in x. The potential field depends on x and y coordinates as V = (x2 y2). 20.. are kept in such a way that each touches the other two. . (C) . 6 2ms. . . . . (A) v . 4 2ms.. the magnitude of the electric force on any sphere due to other two is (A) .C is released from rest at the centre of the ring A.5 m is (A) 19. 2 0 1 2 4 4 R .6 x 1010 J (C) 9. (B) .1 (B) v . On an equilateral triangle of side 1 m there are three point charges placed a t its corners of 1C. . ring A has a positive charge q1 . The work required to move these charges to the corners of a smaller equilateral tria ngle of side 0.9 x 1010 J (D) 4. are placed at the vertices of an equilateral triang le. is a positive constant and r is the distance from the centre of the sphere. 3q 10.. A hemisphere of radius R is charged uniformly with surface density of charge . . (B) 0 2 .. . A system consists of a uniformly charged sphere of radius R and a surrounding medium filled by a charge with the volume density . 8. .q (B) . (D) 0 4 3 . r . . The charge needed at the centre of the triangle for the charges to be in equilibrium is (A) 3 . ..25 x 106 N/C (D) 4.q . Hollow spherical conductor with a charge of 500.5 N.C is acted upon by a force 5 62. Find the charge of the sphere for which the electric field intensity E o utside is independent of r.125 x 106 N/C (C) 2. . What is electric intensity at its surface? (A) zero (B) 1. R . (C) 2. 3q (C) 3q (D) . .5 x 106 N/C 11. (A) 0 . (C) 0 2 . . where . (B) 0 4 . R2 (D) None of the above 9. R . Three charges each of +q. R . What will be the potential at centre? (A) 0 2 .. 2 .. (A) Periodic. It is then disconnected from the source of potential.ELECTROSTATICS www. x = 5x0 . R (C) Such that P crosses O and continues to move along the negative z-axis toward s z = . T hen the motion of P is. lying along x-axis. A positive charge q is placed A B at the centre (A) The electric field at A and B are equal (B) The electric charge density at A = the electric charge density at B (C) Potential at A and B are equal (D) All the above. (B) Simple harmonic. Take the electric potential at a point due to a charge Q at a distance r from it to be Q/ ( 4. z0) where z0 > 0. x = 6x0 . 0. for all values of z0 satisfying 0 < z0 < . Apositively charged thinmetal ring of radius R is fixed in the xy-plane with its centre at the originO.physicsashok.0 0 e (C) . the work done in moving charge from C to A arc CA is. (A) kW0 (B) 0 W k (C) 0 2 W k (D) W0. Anegatively charged particle P is released from rest at the point (0. (B) log 2 4 q e .. with dielectric constant k. (D) None of these 16..0 . and a charge q is fixed at each of the points x = 2x0. its charge being left unchanged and i s immersed in a large volume of dielectric... A charge +q is fixed at each of the points x = x0..in 50 12.. The electrostatic energy will be. on the x-axis. Then.r ).0 .. (D) 0 0 e 4 x q log 2 . for all values of z0 satisfying 0 < z0 . For an infinite line of charge having charge density .. . x = 3x0. 13... x = 4x0. ad inf. 14. (A) log 2 2 q e . (A) Zero (B) 8 x log 2 q .. An isolated metallic object is charged in vacuum to potential v0. ad inf. 15. the potential at the origin due to the above system of charges is. A circular cavity is made in a conductor. Here x0 is a positive constant. its electr ostatic energy being W0.. where k is a constant. 3 : 4 (C) 4 : 3 : 1 (D) 4 : 3 : 2 (A) 1 : 2 : 3 (B) 2 : 19..0 .0 . a). The algebraic ratio of charges q1. the ratio of their speeds vA : vB will be (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4 18.(C) log 2 4 q e . q2 and q3 are taken such that when q1 and q2 are pla ced close together to form a single point charge. 0) to (a. An electric potential is given by V = k(xy). then directly from (0. Three point charges q1. A partic le of charge q0 is first taken from (0. Also q3 and q1 when combined exert an attractive force on q2 of magnitude 18 units at same distance L.. when q2 and q3 are so combined the force on q1 at distance L is an attractive fo rce of magnitude 4 units. a) and lastly from (0. q2 and q3 is. 17. If W1. a) to (a. (D) 2 log 1 2 q e . A particle A has charge +q and particle B has charge + 4q with each of them having the same mass m. a). W2 and W3 be the work done for the individual paths respectively then (A) W1 = W2 = W3 = q0ka2 (B) W1 = W3 = 2q0ka2 (C) W1 = W3 > W2 (D) W1 > W2 > W3 . 0) to (0. 0) to (a. 0) to (a. When allowed to fall from rest through same electrical potential difference. the force on q3 at distance L from this combination is a re pulsion of 2 units inmagnitude. What is the new force of repulsion between A and B ? (A) F (B) 8 3F (C) 2 F (D) 4 F 22. A third uncharged sphere of the same size is brought in contact with sphere. Eight dipoles of charges of magnitude are placed inside a cube. The kinetic energy attained by the particle after moving a distance x is. The net electrostatic a Q +q +q energy of the configuration is zero if Q is equal to [2000] (A) 1 2 . Four charges 2C. (A) qEx2 (B) qEx2 (C) qEx (D) q2Ex 25. Neglect the effect of gravity. (D) The dipole momen t is 2qd along the x-axis. A and removed. the ratio t2 /t1 is nearly equal to (A) 1 (B) (mp /me)1/2 (C) (me /mp)1/2 (D) 1836 23. It is then brought in contact with sphere B and removed. An electron of mass me. A particle of mass m and charge q is released from rest in a uniform electri c field E. 26. Two metallic identical spheres A and B carrying equal positive charge + q ar e a certain distance apart. 3C. The force of repulsion between them is F. A proton of mass mp. Which of the following statements is true for the point of intersection of the diagonals ? (A) Electric field is zero but electric potential is non-zero (B) Electric field is non-zero but electric potential is zero (C) Both electric field and electric potential are zero (D) Neither electric fie ld nor electric potential is zero 21. 0) and (d. moves through a certain distance in a uniform electric field in time t1. The total el ectric flux coming out of the cube will be (A) 0 8e . Three charges Q. 0) respectively of a x-y coordinate system. 4C and 5C respectively are placed at all the corners of a square. (D) Zero 24.in 51 20. (B) Work has to be done in bringing a test charge from . also initially at rest. (C) 0 e . takes time t2 to move through a n equal distance in this uniform electric field. to the origin [1995] (C) Electric field at all points on y-axis is along x-axis. (B) 0 16 e . Two point charges +q and q are held fixed at ( d. +q and +q are placed at the vertices of a ring-angled isosceles triangle as shown.ELECTROSTATICS www. Then (A) The electric field E at all points on the x-axis has the same direction. initially at rest.physicsashok. . q . Three positive charges of equal value q are placed at the vertices of an equ ilateral triangle. Two equal point charges are fixed at x = -a and x = +a on the x-axis. (B) 2 2 2 . Anothe r point charge Q is placed at the origin. is approximately proportional to [2002] (A) x (B) x2 (D) x3 (D) 1 x . (C) 2q (D) +q 27. when it is displaced by a small distance x along the x-axis.q . The resulting lines of force should be sketches as in [2001] (A) (B) (C) (D) 28. The change in the electrical potential energy of Q. +. a) on y-axis. A positive charge Q is released from rest at the point (2a. (C) 0 2 k . +. (B) 0 4 k .in 52 29. [2004] (A) +. . . (D) 0 4 k . . (B) move to the origin rema in at rest . +. 0) on the x-axis. . (B) . q1 q1 q2 The electric field on the surface will be [2004] (A) due to q1 and q2 only (B) due to q2 only (C) Zero (D) due to all 32. Z Z=a Z= a Z= 2a P x 2. such that field at the centre is double that of what is would have been if only P Q R T S U O one +ve charge is placed at R. +. +. 31. The electric field at point P is [2005] (A) 0 2 k . . Which one of the following diagrams correctly represents the electric lines of forces? [2003] (A) (B) (C) (D) 30. Two equal negative charges q are fixed at points (0.ELECTROSTATICS www. 3 positive and 3 negative are to be placed on PQRSTU corners of a regular hexagon. . The charge Q will [1984] (A) execute simple harmonic motion about the origin. (C) . Three infinitely long charge sheets are placed as shown in figure. (D) +.physicsashok. . A metallic shell has a point charge q kept inside its cavity. . A Gaussian surface in the figure is shown by dotted lime. +. +. . . . Six charges of equal magnitude. . . +. +. 33. +. . Q 35. A solid conducting sphere having a charge Q is surrounded by an uncharged co ncentric conducting hollow spherical shell. the n ew potential difference between the same two surfaces is: [1989] (A) V (B) 2V (C) 4V (D) 2V 36. A charge q is placed at the centre of the line joining two equal charges Q. Q (C) 4 . A metallic solid sphere is placed in a uniform electric field. The lines of force follow the path(s) shown in Figure as [1996] (A) a (B) 2 (C) 3 (D) 4 . Let the potential difference between the surface of the solid s phere and that of the outer surface of the hollow shell be V. Q (B) 4 . If the shell is now given a charge of 3Q.(C) move to infinity (D) execute oscillatory but not simple harmonic motion 34. Q (D) 2 . The system of the three charges will be in equilibrium if q is equal to: [1987] (A) 2 . When positively charged spheer is brought near a metallic sphere. and . . respectively. More Than One Choice Questions: 39. . . . . . . . . . Three concentric spherical metallic shells A. . The points A A B q and B are on the cavity surface as shown in the figure. . . . . (b) 2 0 1 B V a b c c . It means: . for r < R (B) decreases as r increases. . 0 1 A B C V V V a b c . . . 40. .. An ellipsoidal cavity is carved within a perfect conductor Figure. . B and C are: (a) . A non-conducting solid sphere of radius R is uniformly charged. . . Then [1999] (A) electric field near A in the cavity = electric field near B in the cavity (B) charge density at A = charge density at B (C) potential at A = potential at B (D) total electric field flux through the surface of the cavity is 0 q . B and C of radii a. . . (D) None of these 38. . for 0 < r < . . . b and c (a < b < c) have charge densities of .in 53 37. The magnitud e of the electric field due to the sphere at a distance r from its center [1998] (A) increases as r increases. a positive charge q is placed at the centre of the cavity.ELECTROSTATICS www. . . (c) 2 2 0 1 C V a b c c c . . . .physicsashok. . The potentials of A. . (C) is discontinuous at r = R. it is obse rved that a force of attraction exists between two. 0 1 A V a b c . (d) . . . (c) electric field is zero at a point on the axis which is at a distance x on th e right side of the charge Q/4. 44. (b) tension in the thread may increase to twice of its original value if another charged sphere is placed vertically below it. where electric field is zero. (c) electric potential is zero at every point of one circle only. Then: (a) the charge is uniformly distributed over its surface. (d) net electric flux through the surafce is zero. (b) metallic sphere may b e electrically neutral. (c) electric field strength inside the sphere will be equal to zero only when no exxternal electric field exists. Then: (a) potential is zero at a point on the axis which is x/3 on the right side of t he charge Q/4. Two equal and oppositely charged particles are kept some distance apart from each other. A conducting sphere of radius R has a charge. then: (a) tension in the thread may reduce to zero if anohter charged sphere is placed vertically below it. (b) potential is zero at a point on the axis which is x/5 on the left side of th e charge Q/4. (c) tension in the thread is greater than mg if another charged sphere is held i n the same horizontal line in which first sphere stays in equilibrium.(a) metallic sphere is necessarily negatively charged. (d) there exist two points on the axis. if there is no externa l electric field. (d) potential at every point of the sphere must be same. A spherical surface having radius equal to separation between the particles and concentric with thei r midpoint is considered. Two point charge: Q and Q/4 are separated by a distance x. A small sphere of amss m and having charge q is suspended by a light thread. 42. if an external electric field exists in the space. 41. 43. (b) electric field is zero at no point. (c) metallic sphere may be negatively charged. (b) distribution of charge over its surface will be non-uniform. . (d) mothing can be said about cha rge of metallic sphere. Then: (a) electric field is normal to the surface at two points only. ... [1992] 5.... B and C... The whole set up is take n in a satellite into space where there is no gravity (state of weightlessness).. 46.... (c) Charge is independent of state of rest or motion.... R and S are (a. O) (a.in 54 45.... b. the magnitude of the force on the point charge of q coul. Select the correct statement(s): (a) Charge cannot exist without mass but mass can exist without charge. are placed on five vertices of a regular hexagon of side L metres.physicsashok. The angle between the two s tring is . (b) Potential difference between two points.. (d) Change in potential energy of system of two charges... [1989] 4. [1988] 3.. 0.ELECTROSTATICS www. O.... is perpendicular to the line joining the charges. The coordinates of the points X Y P Q R S E P. A point charge q moves from point P to point S along the path PQRS (figure) in a uniform electric field E pointing parallel to the positive direction of the X-axis... Which of the following quantities do not depend on the choice of zero potent ial or zero potential energy? (a) Potential at a point.... newtons. Five point charges. y... b. (b) Charge is conserved but mass is not... The work done by the field in the above process is given by the expression .. O. [1984] 2. Two identical charges +Q are kept fixed some distane apart. If P is given a small displacement .... the magnitude of the electric A B C (50 V) (40 V) (30 V) (20 V) (10 V) field is greatest at the point . (c) Potential energy of a system of two charges. Of the points A.. . The value of the potential are written in brackets. O).. each of value +q coul. (d) None of these 47. A small particle P with charge q is placed midway between them.. z (all in metres) in space is giv en by V = 4x2 volts.. it will undergo simple harm onic motion if: (a) q is positive and .2 m) is ... and the tension in each string is .... O) respectively... (c) q is negative and .... Figure shows line of constant potential in a region in which an electric field is present... (b) q is positive and . V/m. (2a. Q. (d) q is positive and . Two small balls having equal positive charges Q (coulomb0 on each are suspend ed by two insulating strings of equal length L (metre) from a hook fixed to a stand.. O) and (O..... is perpendicular to the line joining the charges. is along the line joining the charges.. Placed at he centre of the hexagon 1 . is along the line joining the charges.. Fill in the blanks: 1.. The electric field at the point (1m.... The electric potential V at any point x. the ball will be deflected in the direction of the field. Two identical metallic spheres of exactly equal masses are taken.. One is give n a positive charge Q coulombs and the other an equal negative charge. [1983] 8.. newton... A small metal ball is suspended in a uniform electric field with the help of an insulated thread. Their masses after charging are differen t.2 3 5 4 q q q q q q L is . If high energy X-ray beam falls on the ball. The work done in carrying a point charge from one point to another in an elec trostatic field depends on the path along which the point charge is carried. [1983] . [1992] 6 True / False : 6..... [1981] 7.. .in 55 9. . The point is on the line passing through the centre of the ring and perpendicular to the plane of the disc. . .ELECTROSTATICS www. . . V is the voltage at a given point in the field E. IV-A (D) I-B. At a distance x from an infinite line of charge D. [1988] Table Match 10.physicsashok. 2 2 0 1 2 E x x R . At a distance x from an infinite sheet of C. III-C. The point is on the line passing through the centre of the ring and perpendicular to the plane of the ring II. IV-D (B) I-D. At a distance x from the centre of a A. At a distance x from the centre of a uniformly B. . . 2 2 3 2 0 1 4 E qx . IV-B (C) I-C. II-A. . . is the surface charge density. is the linear charge density) List I List II I. . . II-D. Assume that the . The particl e executes a simple harmonic motion along the axis of the ring. 0 2 E x . R x . . III-D. II-B. . . . uniform distribution of charge IV. . . Match List I and List II and select the correct answer using the codes given below the lists: The electric fields due to various charge distribution are (q is the total charg e on the body. 0 2 E . . charged ring of radius R. . . uniformly charged ring of radius R. . . III. II-C. III-A. III-B. IV-C Passage Type Questions THE NEXT QUESTIONS REFER TO THE FOLLOWING PASSAGE E is the electric field created by q1. . A point charge q is placed in the axis of the ring at a distance 2R from the center of the ring and released from rest. A ring of radius R carries a uniformly distributed charge +Q. (A) I-A. k is Coulomb s Law const ant. What is the work done on q2 when it is moved at constant velocity along the d istance d? (A) Zero (B) 1 Vq (C) 2 Vq (D) 2 Edq 3.electric field created by q2 is negligible compared to E. 1. then when q2 is moved from its present lo cation to a distance r from q1. m1 si the mass of q1. and m2 is the mass of q2. Which of the following represents the work done on q2 when moved from its pre sent position to a distance r from q1? (A) 1 2 1 2 kq q r (B) 1 2 kq q r (C) 1 2 2kq q r (D) 2 1 2 kq q r 4. the force on q2 due to q1 (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 d r 2r q1 q2 . If q1 and q2 have opposite charges. E can best be described as (A) constant (B) decreasing as r increases (C) increasing as r increases (D) increasing as q2 increases 2. with the wire at lower potential. producing a region of ionized air near the wire.0 cm.m. One form of precipitator consists of a vertical hollowmetal cylinder with a thin wire. in particular in the smokestacks of coal-burning power plants.g ash particle have if the electric f ield computed in part (a) is to exert a force ten times the particle s weight? 9. Electrostatic precipitators use electric forces to remove pollutant particles from smoke. This sets up a strong radial electric field directed inward. (a) What is the electric-field magnitude midway between the wire and the cylinde r wall? (b) What magnitude of charge must a 30. Assume that the wire and cylinder are both very long in comparison to the cylinder radius.0 . the radius of the cylinder is 12. and the charged pollutants are accelerated toward the outer cylinder wall by the electric field. If q1 is positive then when q2 is moved from its present location to a distan ce r from q1. The device consists of a thin wire on the axis of a hollow metal cylinder and insulated from it.0 kV. Suppose the radius of the central wir e is 80.in 56 5. A Geiger counter detects radiation such as alpha particles by using the fact that the radiation ionizes the air along its path. running along its axis. insulated from the cylinder. If q1 and q2 are both positively charged and q2 is released. and the potential difference between the wire and the cylin der is 60. what is the maxi mum velocity that can be achieved by q2? (A) 1 2 2 kq q m r (B) 1 2 2 2 kq q m r (C) 1 2 2 2 2kq q m r (D) 1 2 2 2kq q m r 8. The strength of the electric field E at r is (A) half the field strength at 2r (B) the same as the field strength at 2r (C) twice the field strength at 2r (D) four times the field strength at 2r 7..physicsashok.0. Smoke en ters the precipitator at the bottom. . A large p otential difference is established between the wire and the outer cylinder. the magnitude of the voltage experienced by q2 due to E (A) decreases by a factor of 4 (B) remains the same (C) increases by a factor of 2 (D) increases by a factor of 4 6. ash and dust in the smoke pick up electrons.ELECTROSTATICS www. ionize many more air molecules. When io nizing radiation enters the device. this sets up a strong radial electric field directed outward.00 cm.A large potential difference is established between the wire and the outer cylin der. on the way there. S uppose the radius of the central wire is 50.00 x 104 N/C at a distance of 1. with the wire at higher potential. The free electrons produced ar accelerat ed by the electric field toward the wire and. it ionizes a few air molecules. Thus a current p ulse is produced that can be detected by appropriate electronic circuitry and converted to an audiable click.50 cm from the wire? (Assume that the wire and cylinder are both very long in comparison to the cylin der radius. .0. What potential difference between the wire and the cylinder is required to produ ce an electric field of 6.m and the radius of the hollow cylinder is 2. Five thousand lines of forces enter a certain volume of space. enter the charge q q1 q2 2 ? 6. Find the potential at a point P. Particle 1 located far from particle 2 and possessing the kinetic energy T0 and mass m1 strikes particle 2 of mass m2 through the aiming parameter . An electron describes a circular path due to the attraction of the charges in a plane biseA B d 5e 5e O d P R cting perpendicularly the line joining the two point charges.physicsashok. . which is at a distance x from C. Two point charges each carrying a positive charge of 5e (e being the magnitude of the electronic charge) are separated by a distance 2d. The drop is split into two identical spherical droplets (each of which has charge Q/2 spread uniformly over its face). 3.in 57 Level # 2 1. (a) Find the net force acting on the dipole. will the field + . (a) Compute the change in the electrostatic potential energy caused by the split ting. As shown a solid spherical region having a spherical cavity whose diameter R is equal to the radius of spherical region P C Q R that has a total charge Q. in the figure. find the t ime period of oscillation. An electric dipole is placed at distance x from centre O on the axis of a charged ring of radius R and charge Q uniformly distributed over it. to the straight line connecting it to a negative point charge q2. . (b) What is the work done in rotating the dipole through 180°..ELECTROSTATICS www. (b) Repeat the above calculation for the case in which the charge is uniformly d istributed in the drop volume. 4. determine the orbital speed of the electron. Find the smallest distance between the particles when m1 < < m2. and three thou sand lines emerge from it. which are kept very far from one another. 7. Assume that the dipole is linearly restrained. 2. before and after the splitting. An electric field line emerges from a positive point charge +q1 at an angle . the arm of the momentum vector relative to particle 2 as. 1 2 r P P0 . If the radius of the circular path described by the electron is R. A spherical water drop of radius a has a charge Q spread uniformly over its f ace. (c) If the dipole is slightly rotated about its equilibrium position. 5. Each particle carries a charge +q. At what angle . find its time period of oscillation (iii) What is the work done by an external agency to rotate it through an angle . hinged at its centre is placed in a uniform electric field as shown in the figure. One ball (ball 2) is fixed and the other (ball 1 is free. = 2 times.. Two small identical balls lying on a horizontal plane are connected by a weight-less-spring. A semi-circular ring of mass M and radius R with linear charge density .What is the total charge in coulombs with in the volume ? 8. Determine the strength E of the electric field at the centre of a hemisphere produced by charges uniformly distributed with a density . (ii) It the ring is slightly rotated about O and released. 9. Determine the change in frequency. (i) Find the net force acting on the ring. . The balls are charged identically as a result of which the spring length + 1 2 increases . O 10. over the surface of this hemisphere. . y0) and projected with velocity v = v0 . where . . . r is the distance from centre of the system. B and C have radii a = 10 c m. i and assuming o x y v A 0 electric field strength to be zero at y = 0. Suppose in an insulating medium. having di-electric constant k = 1. = ay. What should be this charge Qmax on the ball and the charge Q on t he Earth (assuming it to be uniform sphere of mass m and radius R) such that it may be launched from the Earth s surface with zero launch velocity. Two identical balls are suspended from the same point by two threads. A charge q is placed on the surface of an originally uncharged soap bubble of radius R0. 1 and t he other with a surface density . 18. The ba lls are given equal charges and immersed in kerosene. 16. b = 20 cm and c = 30 cm respectively. The density of kerosene . the radius is increased to a somewhat large va lue R.physicsashok. Show that q = 1/ 2 2 0 0 2 0 0 . . A metal ball of radius r and density .0 = 0. Three concentric.0 = 2. 14. A particle of mass m having negative charge q moves along an ellipse around a fixed positive charge Q so that its maximum and minimum distances from fixed charge are equal to r1 and r2 respectively. conducting spherical shells A. The innermost shell A is earthed and charge q2 = 4 . Calculate angular momentum L of this particle. B A 17.0 and . Find the magnitude of the electric field strength vector as a function of r. volume density of positive charge varies with y-coordinate to law . Find the potential difference between points A and B. calculate slope of trajectory of the particle as a function of y. 2. . Determine the density of the material of the balls if the threads deflect equally in vacuum and kerosene. The system consists of a hemispherical dielectric with volume charge density . C are given to shells B and C respectively. 12. are positive constants. is charged by direct contact from the Earth s surface till it acquires its maximum value. .C and q3 = 3 . Determine the force F of interaction between two hemispheres of radius R tou ching each other along the equator if one hemisphere is uniformly charged with a surface density. Calculate charge q1 induced on shell A and energy stored in the system. 13. 19. A space is filled up with volume density of charge r 3 0 e .in 58 11. .ELECTROSTATICS www.8 g/cm3 and its relative permittivity . 15. Due to the mutual repulsion of the charged surface. A particle of mass m having positive charge q is placed in the medium at point A(0. PR R(R RR R ) 3 32 . .0 are respectively. (c) If a particle of charge +q starts form rest at the centre of the circle. sho w by a short quantitative argument that the particle eventually crosses the circle. [IIT 91] 2. the electric permittivity and magnetic permea bility of free space. the corresponding quantities in a medium.. .0 and .. . . (b) Given the expression V(x) at a general point on the x-axis and sketch the fu nction V(x) on the whole xaxis. the index of refraction of the mediu m in terms of the above parameters is . Level # 3 1. (a) Show that all points in the x-y plane where the electric potential due to th e two charges is zero. 0) respectively in the x-y plane.. . in which P is the atmospheric pressure. Two fixed charges 2Q and Q are located at the points with coordinates ( 3a. In you r answer. 0) a nd (+3a. the sequence of the sub-questions should be the same as given in the question paper. and . . Find its radius and the location of its centre. . .2. (i) If . lie on a circle. Write in yo ur answer book the subquestion number and write down against it your answer corresponding to each blank. Find its speed when it does so. ... A blank to be filled appears in each of the following statements.. a long the axis of the disc. 0. each of value +qC. Find the least value of v0 fo r which the particle will cross the origin. y... An electron of charge ( e) at a radial distance 1 Å moves insid e the sphere.. Calculate (a) Distance travelled by the particle till it comes to rest and (b) Acceleratio n at that moment. The electric field at the point (1m. from a height H with zero initial velocity. 8.. [IIT 92] 3. A particle of mass m and positive charge q is dropped. 0. (iii) Five point charges. . 0. 3/ 2m and . Find the force attracting the electron to the centre of the sphere.ELECTROSTATICS www. newton. is v0. The particle has q/m = 4. V/m.. If the x-component of the electric field at (3.1 .C. Find t he smallest (non-zero) value of the speed v such that the particle does not return to P. per unit l ength is located in the y-z plane with its centre at the origin O. 1) ? [REE 95] 6. 4. Find also the kinetic energy of the particle at the origin. 3 / 2 m.physicsashok. Consider the classical-model of an atom such that a nucleus of charge +e is u niformly distributed within a sphere of radius 2 Å. . 1 . A non-conducting disc of radius a and uniform positive surface charge density . Its speed at x = + . 2m) is . (b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position. Calculate the frequen cy with which the electron would oscillate about the centre of the sphere. 27 / 2 m respectively on the y-axis.. Assume that space is gravity free.C are fixed at the points 27 / 2m. 1 . is placed on the ground. 1) is zero calculate the value of Q. are placed on five vertices of a regular hexagon of side L metre. z (all in meters) in space is g iven by V = 4x2 volts. with an initial speed v. 1. 0) on the positive x-axis directly towards O. 0).. A radioactive source in the form of ametal sphere of radius 10 2m. 1.C moves along the x direction. Four point charges + 8 . A particle of charge q and mass m moves rectilinearly under the action of an electric field E = A Bx where B is a + ve constant and x is a distance from the point where the particle was i nitially at rest..0g/. A circular ring of radius R with uniform positive charge density . [IIT 99] (a) Find the value of H if the particle just reaches the disc. 9. How long will it take for its potential to be raised by 2 volt assuming that 40% of the emitted beta particles escape the source ? [REE 97 ] 7. Is the y-component zero at (3. A particle of mass 6 × 10 4 kg and of cha rge + 0. A particle of mass m and positive charge q is p rojected from the point P(R 3. The magnitude q q q q q of the force on the point charge of value qC placed at the centre of the hexagon is . A charge 10 9 coulomb is located at origin in free space and another charge Q a t (2.C. Given .. [REE 95 5.C and + 8 .. with its axis vertical. emits beta p articles at the rate of 5 × 1010 particles per s. The source is electrically insulated.in 59 (ii) The electric potential V at any point x. 0 ) = 9 × 109 Nm2/C2. Eight point charges are placed at the corners of a cube of edge a as shown in figure. [IIT 2000] 10. [JEE 2003] . Find the work done in disassembling this system of charges.1/(4.. .15 x 10 11 C Level # 2 1. . (a) The fractional change = .. T 8.37 x 105 volt/m (b) 2.. 2 2 0 1 4 4 Q . . True-False & Match the column 1. . A C A C A A A C A B Q.ELECTROSTATICS www. F 7. 1 2 3 4 5 6 7 Ans. 3. 6. B 2. A C D B C B B A A C Level # 1 Q. qEa 4. . . ABD ABC ABC ABCD AC ABC BD Fill in the blanks.. . B Que. . 31 32 33 34 35 36 37 38 39 40 Ans. .physicsashok. 21 22 23 24 25 26 27 28 29 30 Ans. 2 0 1 4 q q . . B B D C C B C B C C Q. 11 12 13 14 15 16 17 18 19 20 Ans. .. . 1 2 3 4 5 6 7 8 9 10 Ans. (a) 1. A A B D A A B D A B Q. T 9.in 60 Answer Key Assertion-Reason Que. L . B A A D C D D Passage Type 8. . L . . F 10. 8 5. 0 2 R2 4x2 1 x 2 7 V Q 2. D C D B A D A C BC BC Q. 1 2 3 4 5 6 7 8 9 10 Ans.. . 180°. 41 42 43 44 45 46 47 Ans. 4 0 1 .. 22 / 3 1 1 (b) The fractional change in this case is also the same. (R ... 3. . . x ) 4. . . . . . . . . . . (b) 2 2 3/ 2 0 aqQx . .. . (a) 2 2 2 2 5/2 0 aqQ R 2x 2 (R x ) . . 2 2 0 0 2 min (1 1 (2 T / q ) 2T r q . . .. . . 2 (r r ) mr r Qq 0 1 2 1 2 3 . sin / 2 q q 2 sin 2 1 . 17. . .ELECTROSTATICS www. . . (i) . Energy = 9..45 J 16. . 15.. n 3n . 3 0Q 4 r gR 3 . . . 2 2 3 / 2 2 0 (R d ) V Re 5 .RE (ii) 2 N 2 E .ER2 11. 2 times 10. 7. . 2 0 max Q 4 r Rg 3 . . . ..7 nc 8.. . 12.. . 14.0 . . q1 = 0 3 . . . (y y ) 3m v a q 30 3 20 0 ...C. (iii) 4.in 61 5. 4. 13. 9. . .physicsashok. . . 6. . 1 2 0 2 2 F R . . . 18. Q = 4. . 1. . . r3 .. 0) (b) . centre dt (5a.10 q (N) 3. 4. . . . . 0 2 0 1 e 3 r .. . . (i) ... 0 4 11 11 E 2 10 0 9 y .0 2 3 V R 2 2 R . .0 . . 8 ma V 9q .0. the charged particle eventually crosses the circle. . .. . 2 m V q ..0 .27 × 10 10 C. . . . 2.. | x a | 2 | x 3a | 1 4 V Q 0 (c) At x = 9a where V = 0.6 gm/cm3 Level # 3 1. 8 V/m (iii) 2 (ii) 9 2 L 9 . . 17.0 . (a) radius = 4a. 9 × 1014 Hz 5. . x B 2A . . . qA 9. 6. W = 5.E at origin = 2.5 × 10 4 J 10. .. Vmin = .824 . (a) H = 3 (b) 3 = 0.. ..0 . 8. .F 4a and H = 0 a . .. . a = n . a . q 4 1 2 0 X X X X 7. 700 . . x = 3 m/s. K. . PH NO. TIRUPATI.CURRENT ELECTRICITY DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. 9440025125 . (d) By convention. and instantaneous current t 0 i lim q dq .TheCGS unit of current is emu of current and is called biot (Bi). .. (c) Though conventionallya direction is associatedwith current (opposite to the motion of electrons) it is not a vector as the directionmerely represents the sense of charge flow and not a true direction. . Further current does not obey i1 i2 i = i1 + i2 the lawof parallelogramof vectors. . . t dt .e.. i. . Example 1. iftwo currents i1 and i2 reach a point we always have i = i1 + i2 whatever be the angle between i1 and i2. (f) Current indifferent situations is due to motion of different charge carries. the direction of current is taken to be that inwhich positive chargemoves and opposite to the direction of flowof negative charge. . Regarding electric current following points areworth noting : (a) Current is assumed to be a fundamentalquantityin physicswith unit ampere and dimension [A].0A) 2.physicsashok. if through a crosssection. . (e) As charge is conserved and current is the rate of flowof charge.ELECTRIC CURRENT www. ..q charge passes in time . i. inelectrolytes due tomotion ofbothpositive and negative ions. so that charge rate remains unchanged. .t. . . The refore. Similar towater flow. Current inconductors and vacuum tubes isdue tomotion ofelectrons. charge flows fasterwhere the conductor is smaller in cross-section and slowerwhe re the conductor is larger in cross-section. (a) How many coulombs of charge pass a cross-section of the wire in the time int . 1A = 1C = (1/10)emu of charge s s 1A = 1 Bi 10 (b) The current is same for allcross-sections of a conductor of non-uniformcross -section.e. The current in awire varieswith time according to the relation i (3. .in 1 THEORY OF ELECTRICCURRENT CURRENT ELECTRICITY The time rate of flow of charge through any cross-section is called current. the charge entering at one end per second of a conductor is equal to the charge leaving the other end per second..0 A t s . . the average electric current through that area av i q t . indischarge tube due tomotion of positive ions and negative electrons and in semiconductors due tomotion of electrons and holes. 16 . 4 0 q ..3t . . . 2 4 0 q .0 s? (b)What constant current would transport the same charge in the same time interv al ? Sol. 28 C . t . .. dq . 2t)dt . . idt . (a) i dq dt . (3. q 4 0 0 . 12 . .erval between t = 0 and t = 4. . J di dScos . Cell.ACdynamo is the source of it .in 2 (b) i q 28 7 A t 4 . . Rectifier Inverter DC AC Current Density Current densityJ . ..S as shown in Fig. the cross-sectiona l area normal to the current in accordancewith Fig. . if at point P current . di = JdS cos . . S . . So.e. Types of Electric Current According to itsmagnitude and direction current is usuallydivided into two types : (i) Direct current : If themagnitude and direction of current does not varywith time. J di n dS .. . .physicsashok. it is said to be direct current (DC).. dS cos .I passes normally through area . i J dS i.e. . + dS i P n J i. at P will be given by S 0 J lim i n . it is said to be alternating current (AC). . (ii) Alternating current : If a current is periodic (with constant amplitude) an d has half cycle positive and half negative. . will be dS . at at a point is defined as a vector havingmagnitude equalto current per unit ar ea surrounding that point and normal to the directionof charge flowand directioninwhich current passes through that point. Regarding current densityfollowing points areworth noting : (a) If the cross-sectional area is not normal to the current. c urrent densityJ .. NOTE : It is worthy to note here that rectifier converts AC into DC while inverter DC into AC. battery or DC dynamo are its sources.ELECTRIC CURRENT www. dS cos . . . i. J . . . . i. . 1 . or J . i L S . .or di . . . i . . dS .. . E . dS .EL. . J . E V and R V L S . . . . . J i 1 E S . . . . .e. .e. So.. . . (b) In case of conductors asV= iRand by definitions. .8mAwhen the voltage is applied across the length of the sample. . The current flowing throughconductor d d i neAv or v i neA . . . then find howmuch time does it take for the electrons to travel the full length of the sample ?Given that charge on an electron e = 1. [(. . / .ELECTRIC CURRENT www. . . . . . .8 10 A/m 25 10 4 10 . .physicsashok. d a = e E m . . . . Sol. . . nq v . where n= number ofmoving electrons per unit volume A= area of cross-section NOTE : Some books have taken average drift velocity as half of initial and final drift velocities of an electron giving vd = (eE.6 × 10 19C. This is wrong as vd is the avera ge of drift velocities of large number of electrons at same instant and as for each electron v . J i i S b d . [as s = (b × d)] So. / . Example 2. It is repr esented by vd . thickness 25 × 10 5 mand length 6 × 10 2 mcarries a current of 4.)charge = nq] Drift Velocity The average uniformvelocity acquired by free electrons inside ametalby the appli cation of an electric field which is responsible for current through it is called drift velocity. . . .8 10 4.What is the current density ? If the free electron density is 1022 m 3. 3 3 2 5 3 J 4. An n-type silicon sample ofwidth 4 × 10 3m. a= e E m constant . . ..) ./2m). d = a f with . v (.. . . J i n (nqv) n S . ..in 3 (c) In case of uniformcharge flowthrough a cross-section normal to it as i = nqvS So. or charge J . Bydefinition. v v . Also find the drift velocityof electrons in themetalwhen a cur rent of 16Apasses through it.6 × 10 19C . 3 d 22 19 4. . . . . . . . The area of cross-section. 1mand5 × 103 kg/m3 respectively. . .8 10 v 3 m/ s 10 1. . length and density of a piece of a metal o f atomicweight 60 are 10 6m2. Example 3. Hence. Given thatAvogadro s number NA = 6 × 1023/mol and charge on an electron e = 1. .02 s. .6 10. So. and as in case of electric conduction inmetals J = nevd or d v J ne . . . v 3 . . . . Find the number of free electrons per unit volume ifeveryatomcontributes one free electron. . time taken by electron to travel the length L (= 6 × 10 2 m) of the conductor 2 d L 6 10 t 0.. . . . i. and the ratio ofthemagnitudesE and J is constant. . and as J = nevd .. . .physicsashok. Thus. .) and this relationship is called the Ohm s law. . is nearlydirectly proportional to E. . A A A N m so n N N m N d as d m N M V VM M V . . . . resistivity E J . the electric field inside the me tal E = . is taken to 2 × 10 8 ohm m. .6 10 .) to travel 1 m length of wire if it can. .3 min. . . RESISTANCE AND OHM S LAW For somematerials. . . m(ohm metre).The reciprocalof resistivityis conductivit y(. . . .. . . 500 s = (8. the number of electrons per unit volume ne = 1 × n = 1 × 5 × 1028 = 5 × 1028/m3 Further as here. As according toAvogadro s hypothesis. . .in 4 Sol. . 6 3 d 28 19 v J 16 10 2 10 m/ s ne 5 10 1. especiallymetals. . SI units ofresistivity are . at a given temperature.J = (2 × 10 8) × (16 × 106) = 0. . Nowas each atomcontributes one electron. J .e. . If resistivity of metal . . . .This ratio is called the resistivity(.32 V/m and not zero as in electrostatics.. . 6 2 6 J i 16 16 10 A/m A 10. . .).ELECTRIC CURRENT www. NOTE : From this example it is clear that : An electron will take 1/(2 × 10 3). 23 3 28 3 3 6 10 5 10 n 5 10 /m 60 10. . . . . l . . Resistance Suppose a conductingwire has a uniformcross-sectional areaAand length l as shown in Fig. . . l Here.Materials which show substantial departures fromOhm slaw arecalled non-ohmic or non-linear. .. If themagnitudes of the current densityJ . l is constant for ohmicmaterials. Amaterial that obeysOhm s lawreasonablywell is called an ohmic conductor or a line ar conductor. 1 . This is called the resistance R.Thus. and the electric field E. . the total current i is given by i= JAand the potential differenceVbetween the ends isV= El. V E i JA A . are J E A + V l i uniformthroughout the conductor. A . Let Vbe the potentialdifference between the ends of the wire. G = 1 = i r V SI unit of G is ohm 1 which is called mho. i. The slope of this line is equal to the resistance of the conductor. Vand .e. l . Thus.e. (b) Specific resistance (. R V i . (i. both thewireswillhave same specific resistance.e. Awire has a resistance R. .we have 2 R V . l i.What will be its resistance if it is stretch ed to double its length ? Sol. whi chwire has : (a) greater resistance and (b) greater specific resistance ? Sol. V l Substituting this in R A . V = iR is often calledOhm s law. So. . but only when R is some constant we can correctly call this relationship Ohm s law. It does not depend on l orA. R = V = tan i . Reciprocal of resistance is called conductance (G). l So.in 5 Thus. . R 1 A ..) is amaterial property.ELECTRIC CURRENT www. for ohmic conductors V i group is a straight line possing through origin. . are constant) . Two copper wires of the same length have got different diameters. So. for given volume andmaterial.physicsashok. then V =Al A . the thinner wirewill have greater resistance. whether or not it obeys ohm s law. Example 5. Example 4. The resistanceR of a particular conductor is related to the resistivity . Let Vbe the volume ofwire. ofitsm aterial by R A . l The equation. NOTE : The equation R= V i defines resistance R for any conductor. (a) For a givenwire. V i . R A . l2 When l is doubled. . or the newresistancewill be 4R.R . resistancewill become four times. .2 Solving this. . . (a) The current density across a cylindrical conductor of radius R va ries according to the equation 0 J J 1 r R .physicsashok.2 .eff .(v) Putting the value of (i). is temperature coeffi cient of resistivity. Mathematically the dependence of resistance R on temperature is expressed as R0 0 T0 T R Slope = R R (T) = R O 0[1 + .8 × 10 3 perº C. . is connected in parallelwith a resistance = 3R.t) . . ..(ii) But R1 = R10 (1 + . (iii).. Find the value of . having thermal coefficient of resistivity= 2.efft) .What is the temperature of the bath ? ... (v) in eqn (ii). eff 5 4 .6º C Example 7. Sol.ELECTRIC CURRENT www.we have 1.at 20ºC. (iv). R (T) = R0[1 + ..2 = 1.(iv) and R = R0(1 + .. Example 8. where r is the distance fromthe axis.. Thus the current density i .. Thewire is placed in a liquid bath and its resistance rises to 1.in 6 VARIATION OF RESISTANCE WITH TEMPERATURE The resistance of a conductor varieswith temperature.. .. The factor . . R0 = 1.(T T0)] In this equationR(T) is the resistance at temperature T and R0 is the resistance at temperature T0. . . Example 6.0. .. The graph of variation of resistance of puremetalwith temperature is shown.0 . .. The resistance of thin silver wire is 1. . Sol. . R (T) = 1..(i) The equivalent resistance at tºC is 1 2 1 2 R R R R R .2 .for silver is 3. The equivalent resistance at 0ºC is 10 20 0 10 20 R R R R R .8 × 10 3 (T 20) = 0.t) .= 3.(T T0)] Here. we get T = 72. often taken to be 0ºC or 20ºC...8 × 10 3(T 20)] or 3.(iii) R2 = R20(1 + 2..8 × 10 3 perº C and T0 = 20ºC Substituting thevalues. Aresistance Rof thermal coefficient of resistivity= .0[1 + 3.. (b) Suppose that instead the current densityis amaximumJ0 at the surace and decr eases linearlyto zero at the axis so that 0 J J r .R2.s a maximumJ0 at the axis r = 0 and decreases linearlyto zero at the surace r = R.Calculate the current in terms of J0 and the conductor s cross sectional area isA= . R . . Calculate the current. rdr The current in considerd element is 0 dI JdA J 1 r 2 rdr R ..Anetwork of resistance is constructedwith R1&R2 as shown in the figure. . V3. The cross-sectional area of considered element is dA= 2. . .in 7 Sol. .ELECTRIC CURRENT www. . . .. . . 3.. . . . .. .. N are V1.. .physicsashok... . or 0 dI 2 J 1 r rdr R . VN respectivelyeach having a potential k time smaller than R2 R2 R2 R3 V0 R1 V VN N 1 R1 R1 V1= V0 k . Example 9.... . .. 3 R 0 0 I 2 J r R 3 . .. .. (a)We concider a hollowcylinder of radius r and thickness dr. . . .. 3 0 0 2 J R 2A I J R 3 3 . ... .. . The potential at the points 1. . 2.. R 0 0 I 2 J 1 r rdr R . V2. . . 2 0 0 R A I J J 3 3 . . . .. (b) R 2 0 0 r I 2 J dr R . .. .. or . . k&R3. . . . Sol. . Find: (i) 1 2 R R and 2 3 R R in terms of k (ii) current that passes through the resistance R2 nearest to theV0 in termsV0. .I I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 I2 previous one. . . I = I1 + I2 or 0 0 0 0 0 2 1 2 1 V V V 0 V V k k k k R R R . R2 R2 R2 R3 V0 R1 V VN N 1 R1 R1 V1= V0 k I I1 V2= V0 k2 VN= V0 kN I´ I´1 I´2 (i) According to kcL. . 0 0 0 2 1 2 1 k 1 V V k 1 V kR kR k R . . . physicsashok. . . 0 1 1 2 2 V 0 V 0 k I R R . .Thematerial obeysOhm s lawand its resistivityvaries along the rod according to . Also. . . . (b) Find the electric potential in the rod as a function of x. . . .2 1 2 R k 1 R k . (a) Find the total resistance of the rod and the current in thewire. .. . ..0 e x/L. I´ = I´1 + I´2 or 0 0 0 0 N 2 N 1 N 1 N 1 1 2 1 3 V V V 0 V 0 k k k k R R R R . . . . Arod of length Land cross-sectionareaAlies along the x-axis between x= 0 and x= L. Example 10. 0 0 0 1 2 2 3 3 V V k 1 V I kR k k R k R k 1 . The end of the rod at x = 0 is at a potentialV0 and it is zero at x = L. Sol. . 2 3 R k R k 1 . . .in 8 . . . (ii)Here. (a) The resistance of considered element is 0 x / L dR dx e dx . .(x) = . . . .ELECTRIC CURRENT www. After solving. . . . e. . . Vx = Ex . . 0 L x / L 0 x /L L 0 0 R e dx L e A A . 0 x E V V x . . . .. . . . . . . (b) . .A A . . . . . V0 . . . . . x dx 0 1 0 L L 1 R e 1 1 A A e . . . . . E = . . . J E = 0 x /L I Ie A A . . 0L e 1 R A e .. . . . . . . x / L 0 0 0 e AeV E A L e 1 . . . . . . 0 0 0 0 I V 0 V AeV R R L e 1 . . . . . . x /L 0 1 E e V L 1 e . .J or x /L 0 E . . . .. . . . . . . . . . . . . . . . ELECTRIC CELL An electric cell is a devicewhichmaintains a continuous flow of charge (or elect ric current) in a circuit by a chemicalreaction. a cellconverts the chemicalenergyinto electricalenergy. The emf of a cell in a circui t is taken to be positive. ifWis thework done bya cell inmoving a charge q once ar ound a circuit including the cell. (i.physicsashok. Inside the cell. . x /L 1 0 x 0 1 V e e V V Ex 1 e . These electrodes are kept in a solution called electrolyte..e.e. there are two rods of differentmetals called as electrodes. . E1 E2 + i i + E = E1 + E2 (a) E1 E2 i + + i E = E1 E2 (b) E1 E2 i + + i E = E2 E1 (c) NOTE : The term electromotive force is misleading introduced by Volta who though t it to be . .in 9 or . . i. Therefore. . Inthe light ofmodernviews inreference to a cell following terms need to be reviewed. emf E W q . Electromotive Force (EMF) The emf ofa cell is defined aswork done bythe cell inmoving unit positive charge in thewhole circuit including the cellonce. current flows inthewire. cell is disc harging) otherwise negative as shown inFig. is fromnegative to positive terminal. On joining two electrodes by a wire the charge begi ns to flow in the wire. . . SI unit of emf is joule/coulomband is called volt. . if circuit current inside a cell. Thus. a chemical reaction takes place in the electrolytewhichmaintains the charge onthe electrodes and the flowofcharge inthewire is continuouslymaintained . .ELECTRIC CURRENT www. In an electric cell.. V = E ir This can also be shown as below: E r i A B VA E + ir =VB or VA VB = E ir Following three special cases are possible : (i) If the current flows in opposite direction (as in case of charging of a batt ery). denoted byr. Internal Resistance (r) and Terminal Potential Difference The potential difference across a real source in a circuit is not equal to the e mf of the cell.whena current is drawnthrougha source. the potentialdifference between the terminalofthe source is. Actually emf is not a force but work requ ired to carry unit charge from lower potential to higher potential inside the cell. thenV= E + ir (ii)V= E.As the currentmoves through r.We cellthis the internalresistance ofthe source. The reason is that chargemoving throughthe electrolyte ofthe cell encounters resistance.force that causes the current to flow.Thus. if the current through the cellis zero . it experiences an associated drops in potential equal to ir . This is because current in the circuit i E r .. represents a circuit consisting of a source of emf and two resistors conne cted in paralle.. Then R = R1 + R2 This result can be readily extended to a network consisting of n resistors in seri es. + Rn In Parallel Fig. A B V R1 A B V R i i1 i2 R2 i IfR be the equivalent resistance.. then 1 2 . A B R1 R2 V V1 V2 A B V R i Let equivalent resistor is R as shown. if the cellis short circuited.we can summarise it as follows : E r i V = E ir or V < E E r i V = E + ir or V > E E r V = E if i = 0 E r i =Er V = 0 if short circuited COMBINATION OF RESISTANCES In Series Fig...physicsashok. E ir = 0 or V = 0 Thus.ELECTRIC CURRENT www..in 10 (iii) V= 0. R = R1 + R2 + .. represent a circuit consisting ofa source of emf and two resistors connect ed in series. . or E = ir Short Circuited E r .. . This result can be extended to a network consisting of n resistors in parallel. The result is .1 1 1 R R R . (b) (Star)Y.. . (Delta). to be equivalent to Fig. 1 R R R R .. .. Nowthis 2 . Example 11.. R R R R R R R R R . i 18V 6 Current drawn fromthe batteryis.physicsashok.... the same current flows through each resistance whi le in parallel combination. i = net emf = 18 = 3A net resistance 6 Star-Delta (. resistances are in series and their equivalent resistance is 4 + 2 = 6 .. the voltage drop across each resistor is equal to the source voltag e V. (a) . R2 A B C R1 R3 R1 R3 R2 RA A B C RB RC A B C RA RB RC (a) (b) (c) 1 3 1 2 A B 1 2 3 1 2 3 R R R . and 3 .ELECTRIC CURRENT www. Compute the equivalent resistance of the network shown in Fig. ... The 6 . . . equivalent resistance of the network = 6 .in 11 1 2 n 1 1 1 .. Their equivalent resistance is .) Conversion For Fig. . resistances are in parallel.. 18V i i 4 6 3 Sol. . Therefore. NOTE : In series combination. . and 4 . . . . 18V i 4 2 1 1 1 or R 2 R 6 3 . and find the current idrawn fromthe battery. . . . the equivalent circuit can be drawn as shown in f ig. . .5 5 9 5 A C D O B A O B A B CB CD C CB CD DB R R R 20 10 5 R R R 20 10 10 . . . . . .5 R R R 10 10 20 . . Example 12. Find the equivalent resistance betweenAand B in Fig. using Star-Delt a Theorem. Using Star-Delta theorem. . . . . . R´ 15 22. . . . . 10 10 10 RC = 5 RB = 5 RD = 2. . .ELECTRIC CURRENT www. . . . .5 15 22. DC DB D DC DB CB R R R 10 10 2.5 15 22. BC BD B BC BD DC R R R 20 10 5 R R R 20 10 10 . . 10 A B C D 10 10 10 10 10 10 Sol. .physicsashok. . .5 .5 9 15 22.5 37.in 12 and 2 3 C 1 2 3 R R R R R R . . . Sol. . . RAB = R´ + 5 = 9 + 5 = 14 . .What willbe the change in the resistance of a circuit consisting of five identical conductors if two similar conductors are added as shown by the dashed line in figure.. Example 13. . . Hence. For example. For example.DCEF DandABEFAare loops.ELECTRIC CURRENT www. First here are two terms that wewill use often. Loop : Aloop is any closed conducting path. For example. Similarly. . . J unctions are also called nodes or branch points.in 13 r r r r r r r B A r r r r r r B A r 2r 2r A r B A r r r B R2 = Req = 3r But before added the conductors. KIRCHHOFF S LAW Manyelectric circuits cannot be reduced to simple series-parallel combinations. 5r . de vised byKirchhoff. the equivalent resistance is 1 eq R . Junction : Ajunction ina circuit is a point where three ormore conductorsmeet. 2 1 R 3 R 5 . in figure (a)ABCDA. in figure (b) p oint B and F are junctions.physicsashok. it is always possible to analyze such circuits byapplying two rules. R´ . in figure (a) pointsDand Care junctions. R1 R2 R3 E1 E2 A B D C F E (a) (b) E1 E2 E3 R1 R2 R3 R4 A B D E F G I H R5 C However. two circuits that cannot be so broken down are shown in fig. in figure (b). CBFEC. (i) Junction rule : The algebraic sumof the currents at any junction is zero. junction . the sumof all the curents directed towards a i4 point incircuit is equal to the sumof all the currents directed awayfromthat poi nt. i .Similarly. BDGFBare loops. 0 i2 i2 i3 This lawcan also bewritten as. . That is. Kirchhoff s rules consists of the following two statements. 2 A B C D 4 4 2 F E 2V 4V 6V Sol. In applying the loop rule..(iii) Solving Eqs.. . 2i3 6 4i3 4 = 0 . the emf is considered to be positive.physicsashok. E A B Path (a) (b) Whenwe travel form+ to . the iR termis positive because this represents a rise of potential. (ii) Loop rule : The angebraic sumof the potentialdifferences in anyloop includi ng those associated emf s and those of resistive elements. closed loop .V . 4i1 + 4 2i1 + 2 = 0 . ApplyingKirchhoff s first law(junction law) at junctionB..must equal zero.(i) ApplyingKirchhoff s second lawin loop 1 (ABEFA). the emf is considered to be negative.(ii) ApplyingKirchhoff s second lawin loop 2 (BCDEB). 0 The loop rule is based on the fact that the electrostatic field is conservative in nature. (ii) and (iii).we get i1 = 1A C D 2V 1 4V 2 6V . Find current in different branches ofthe electric circuit shown in f igure. we need sign convention as discussed below: (a) Whenwe travel through a source in the direction from to +. (i)...ELECTRIC CURRENT www.. That is. (d) Whenwe travel through a resistor in the direction opposite to the assumed current. R A B Path (d) i Example 14.in 14 Thus. the iR termis negative because the current goes in the direction R A B Path (c) i of decreasing potential. infigure i1 + i2 = i3 + i4 The junction rule is based on conservation of electric charge. i1 = i2 + i3 . E A B Path (b) (c) Whenwe travel through a resistor in the same direction as the assumed current. negative sign of i3 implies that current i3 is in opposite direction ofwha t we have assumed. .A B F E 2 4 4 2 i3 i2 i1 i3 i1 2 i 8 A 3 . . Here. 3 i 5 A 3 . .. . . Distribution of potential in series connections :Whenmore than one resistance s are connected in series. . . . 1 V 1 V V 1 2 3 6 . . Distribution of current in parallel connections : When more than one resistances are connected in parallel. inthe figure.. the potential difference across themis equal and the current is i i2 3 i1 i R 2R distributed among themin inverse ratio oftheir resistance. inthe figure. . . . .. . ... ... ..g. and 3 i 2 i 2 i 6 3 2 11 . 1 2 3 i : i : i 1 : 1 : 1 6 : 3 : 2 R 2R 3R . 2. . . . . or i 1 R .g.. . 2 V 2 V V 1 2 3 3 . . . e. . R for same value of i. . . for same value ofV e. ... . . V1 V2 V3 R 2R 3R i V1 : V2 : V3 = R : 2R : 3R = 1 : 2 : 3 . . . . . . . COMBINATION OF CELLS .ELECTRIC CURRENT www. 2 i 3 i 3 i 6 3 2 11 . the current through themis same and the potential distributed in the ratio of th eir resistance. . . . .in 15 IMPORTANT FEATURES 1. and 3 V 3 V V 1 2 3 2 . . 1 i 6 i 6 i 6 3 2 11 . . . ..physicsashok... . as V = iR or V . as 3R i V R . . Then .Cells are usuallygrouped in following threeways : In Series Suppose n cells each of emfE and internal resistancer are connected in series as s hown in figure. . E/r = 1/ r . Net emf = Eeq . . R 1 1/ r .n . . . i2 i3 i1 i R E1 2m)E .in 16 r i R E E r E r Net emf = nE Total resistance = nr + R . i R i E r E r E r . then equivalent emf = (n while total resistance is still nr + R . Net emf = E Total resistance r R n . Current inthe circuit i = net emf total resistance or i = E R + r/n IInd case : Let n cells have different E and r . . Current in the circuit = Net emf Total resistance or i = nE nr +R NOTE : If palarity of m cells is reversed. . . . Total resistance = Req . E i= nr+R In Parallel : Consider the following three cases : Ist case : Let n cells each of emfE and internal resistance r are connected in parallel.2m.ELECTRIC CURRENT www.physicsashok. . Net emf = Eeq . IIIrd case : This ismost general case of parallel grouping inwhich E and r of diffe rent cells are different and the positive terminals of fewcells are connected to the negative terminals of the ot hers as shown. or .E2 E3 r1 r2 r3 i A B F E Hence. . . . E / r i 1 R 1/ r . C D eq eq E i R . . . . . . . . . r i E R i . . . 1 1 2 2 3 3 1 2 3 E / r E / r E / r 1 1 1 r r r . . . . . . . . . . . . . . . . . . . . . .in 17 . i i2 3 i1 i R E1 E2 E3 r1 r2 r3 i Total resistance = Req 1 2 3 R 1 1 1 1 r r r . eq eq E i R . . . . . . . .ELECTRIC CURRENT www. . . .physicsashok. 1 1 2 2 3 3 1 2 3 E / r E / r E / r i 1 R 1 1 1 r r r . Net emf = nE Total resistance = Req R nr m . . . . Hence. . In Mixed Grouping There are n identical cells in a rowand number of rows are m . Further. 1 2 1 2 eq 2 2 E E 10 4 E r r 2 2 3V 1 1 1 1 r r 2 2 .Hence. .. . For parallel combinationwe can apply. i. Nowthis is in serieswith the third one. req = 1 . . . . . Find the emf and internal resistance of a single batterywhich is equ ivalent to a combination of three batteries as shown in figure. i nER nr m . . eq 1 2 1 1 1 1 1 1 r r r 2 2 . . 4V 10V 2 2 1 6V Sol.e. . The givencombination consists of two batteries in parallelandresultant ofth ese two inserieswiththe third one. . . . . Example 15. (1 + 1) or 2 . E n r I E nE r r n = = I nE r = I . Abatteryconsists of a variable number n of identical cells having in ternal resistance connected in series.ELECTRIC CURRENT www. In previous problem. Which one of the graph belowshows the relationship between I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol.in 18 1 1 6V 3V The equivalent emf of these two is (6 3)Vor 3Vand the internal resistancewill be . r = 2 E=3V Example 16. if the cell had been connected in parallel (ins tead of in series) which of the above graphswould have shown the relationship between total current I and n ? (A) I/A O n (B) I/A O n (C) I/A O n (D) I/A O n (E) I/A O n Sol.. nE nr I nE E nr r = = I n Example 17.physicsashok. The terminals of the batteryare short circuited and the current Imeasure d. Underwhat conditioncurrent passing through the resistanceR can be increased by short circuiting the battery of emfE2. E1 r1 E2 r2 R (A) E2r1 > E1(R + r2) (B) E1r2 > E2(R + r1) (C) E2r2 > E1(R + r2) (D) E1r1 > E2(R + r1) Sol.n Example 18. The current throughR before short circuit . The internal resistances of the two batteries are r1 and r2 respectively. e..ELECTRIC CURRENT www. i.physicsashok. ig = 0 and hence. The bridge is said to be balancedwhen deflection in galvanometer is zero.in 19 1 2 1 2 I E E r r R = + + + After short circuit : 1 1 I' E r R = + E2 E1 r1 r2 I > I R 1 1 2 1 1 2 E E E r R r r R > + + + + E1r1 + E1r2 + E1R > E1r1 + E1R + E2r1 + E2R . G R S P Q B A C D G R S P Q B A C D . NOTE : In Wheatstone s bridge. 1 2 2 1 2 E r > E r + E R WHEATSTONE S BRIDGE This is an arrangement of four resitances inwhich one resistance is unknown and rest known.TheWheatstone s bridge is shown in fig. cell and galvanometer arms are interchangeable. the conditionof deflection is G R S P Q B A C E D i2 ig = 0 i1 i2 i1 i P R Q S . r 2 2 r 2 2 r3 r 2 1 r 2 1 B C P O and r2 = ROB = (a)r . . 1 PB PD a r r R R 2 . 1 2 PO R r r 4 . condition of balance bridge is.In both the cases.Ahemisphere network of radius a ismade by using a conductingwire of resistance per unit length r.And. Hence. This is a balancedWheatstone bridge between P andO Hence. P = R Q S Example 19. . Find the equavalent resistance acrossOP. r3 can be removed. the circuit can be drawn as shown in figure. . P B O D A C Sol. . . Point (AandC) and (DandB) are symmetricallylocatedwith respect to points Oand P. Here . . 3 I IR + 15 2I = 0 I 18 3 R = + . P Q V . PO 2 ar R 8 . . In the circuit shown. Two batteries one of the emf 3V. internal resistance 2 ohmare connected in series with a resistance R as shown. internal resistance 1 ohmand the other of emf 15 V.physicsashok.ELECTRIC CURRENT www. . Ans. If the potential difference between a and b is zero a b R the resistance of R in ohmis (A) 5 (B) 7 (C) 3 (D) 1 Sol.According to loop rule. Example 20.V = 2 V Example 21. what is the potential difference VPQ? (A) +3V (B) +2V 4V 2V Q P 1V 2 1 2 3 (C) 2V (D) none Sol. Va Vb = ( 3 + I) a b .in 20 . In ABCPA 2I + 4 I + 2 = 0 I 6 2A 3 = = VP VQ = {algebriac sumof rise up and drop up of voltage} 2V 4V Q P 1V I A B C D I I (I I ) 1 I1 I1 I1 2 1 2 3 = (2 2I) = (2 4) . in figureAvoltage V is applied between the points AandB.R 15V 3V I or 0 = 3 I . or 1 2 1 v v v 2R 2R 4R = + . R 9 3 3 . This applied velue of voltage is halved after each R1 R1 R1 R1 R1 R2 R2 R2 R2 R2 A B section. I = I1 + I2 1 2 1 0 v v v v v 2 2 2 4 R R R . I = 3 A . (A) R1/R2 = 1 (B) R1/R2 = 1/2 (C) R1/R2 = 2 (D) R1/R2 = 3 Sol. Consider an infinite ladder network shown . Example 22. . . 3 18 3 R = + or 9 + 3R = 18 . . VAB = VA VB = 0.ELECTRIC CURRENT www. calculate the following : .5I 0. Fromthese points of view.Aswe know. resistance decreases. 9 6 6 3 1 A 7 A 24 V I I I 0. in series resistance increases. Forminimumcurrent through battery. 6 3 6 S1 1 1 1 S3 S2 9 A B Sol.5 × I = 0. all switches should be open. but in parallel.5I = 0 24 . 7I 4.5 1 1 24 V According to loop rule.physicsashok. equivalent resistance 24V across battery should bemaximum.5 B 4. In the circuit shown in figure. S2 and S3 in the figure are arranged such that current through the batteryisminimum.5 × 2 = 1 V Example 24. find the voltage across pointsAand B. I = 2A .in 21 or 1 2 1 1 1 v R R 2R = + I2 v v = 0 v/2 I I1 v/4 R2 R2 R2 R2 R2 v = 0 v = 0 v = 0 v = 0 v = 0 or 1 2 1 1 2R R = or 1 2 R 1 R 2 = Example 23. If the switches S1. Also. 36 Va = I1 × 6 = 6 I1 or 36 Va = 6 × 4 = 24 V . Va Vb = 12 24 = 12 V (ii) In loopABCIHA. . 3 6 6 3 a S b 36V Sol. Va = 36 24 = 12 V 3 6 6 3 a b V = 0 V0 = 36V I1 I2 36 Vb = 3 × 4 = 12 V and Vb = 36 12 = 24 V .(i) Potentialdifference between points a and b when switch (ii) Current through S in the circuit when S is closed. . Also 2 I 36 0 4 A 9 . 6(I I1) + 3I1 = 0 or 6I + 6I1 + 3I1 = 0 or 9I1 = 6I S is open. . . . (i) Here 1 I 36 0 4 A 9 . S.(ii) In loop EFGHIJE.. .. . 1 1 I 9 I 3 I 6 2 . . Let the equivalent resistance of circuit is R...When he joins a second identical cell is serieswith the first. . L. 36 3I1 6(I1 I2) = 0 or 9I1 6I2 = 36 .. 2 3 2 6E 3I I .. . = 3 2 . . the current becomes I2.An enquiring physics student connects a cell to a circuit andmeasures the current drawnfromthe cell to I1. (ii) and (iii) I2 = 3Afrom b to a . and 2 I 2E 2r R . . and 3 I E r / 2 R .. .ELECTRIC CURRENT www. 3 6 3 6 D C B A H I E J F I G (I I1) I1 I2 (I I1 + I2) 3(I I (I1 I2) 1 + I2) + 6(I1 I2) = 0 or 3I + 3I1 3I2 + 6I1 6I2 = 0 or 3I + 9I1 9I2 = 0 . 3I I 3E 2E r / 2 R 2r R . 1 I E r R . .H.. Example 25.When the cells are connected are in parallel. Show that rela tion between the current is 3 I3 I2 = 2I1(I2 + I3) Sol.(i) In loop CDEJIC. . .in 22 . .(iii) After solving eqn(i). The emf and internal resista nce of cell is E and r respectively. . . .physicsashok. the current through the circuit is I3.. . Example 26. .H.S. .. . . Hence. 1 1 1 1 1 1 1 1 1 1V 1V 1V 1V 1V 1V B 1V 1V A Sol. L. i1 + i2 = i 4i + i1 = 0 VB for the circuit shown in the figur . . = 2I1 (I2 + I3) 2E 2E E r R 2r R r / 2 R . . R. Find the potential differenceVA e.S. . . . . . . 6E2 r / 2 R 2r R .H. . . .r / 2 R 2r R .S.. .H. . = R.... . . . Then equivalent resistance is 0 2 r r R n n = = . r = the resistance of each piece R nA n = r . as shown i s : (A) 9 (B) 27/4 . (i) Example 28... Power generated across a uniformwire connected across a supply is H.ELECTRIC CURRENT www. . The ratio of powers dissipatted respectively in R and 3R.. A B V V 2 4 22 V 9 9 . . the totalpower generated in the wire is : (A) 2 H n (B) n2H (C) nH (D) H n Sol. = v2 H t R = . . 4 B A 3 2 1 0 4i x i i2 i1 3i 2i x+4 x+3 x+2 x+1 i1 . .(i) when all pieces are corrected in parallel. Let the resistance of wire is R A = r . . .. Example 27.physicsashok. 2 2 2 1 0 v n v H t t r R = = .. . x 4 9 . If the wire is cut into n equal parts and all the parts are connected in parallel across the same supply. n2H fromeqn.in 23 . . ..3R 2R R (C) 4/9 (D) 4/27 Sol. P = I2y R y 10V I I I or 2 P 100 y (2 y R) = + + For maximumpower dissipatted. (C) 5 .. Sol.. . (D) 3 . .. 2R R y 10V.. .... (A) 2 . .. and P2 = Power dissipatted in 3R = I2(3R) = 3I2R 3R 2R R I/3 2I/3 I I .. . According to KOL : 10 2 I I y I R = 0 . . . In the figure shown the power generated in y is maximumwhen y = 5. 2 1 2 2 P 4I R P 9 3I R . 1 2 P 4 P 27 = Example 29. P1 = Power dissiatted in resistance R 2 2I R 4 I2R 3 9 . . (B) 6 .. I 10 2 y R = + + . . . . The rod is connected in series with a resistance to a 6Vbattery of negligible internal resistance. Example 30. 10 5 11 5 10V 5V 20V 30V 25V 25V 25V 25V 25V 25V V=0 V=0 V=0 V=0 V=0 15V I4 I3 I2 I1 I=I +I +I +I4 1 2 3 4 30V 5V 55V 3 I 30 0 3A 10 .44 (the value of (2. R . . Find the current through 25Vcell&power supplied by20Vcell in the fig ure shown. By solving. . 4 I 15 0 3A 5 . . . 3.in 24 dP 0 dy . 10 5 11 5 10V 5V 20V 30V 25V Sol. .What value should the series resistance have so that : (i) the current in the circuit is 0.37) . .physicsashok. where V is the potential difference across it. Fromfigure. 1 I 55 0 5A 11 . The current I through a rod of a certain metallic oxide is given by I = 0.2)2/5 = 1. 2 I 5 0 1A 5 . .ELECTRIC CURRENT www. y = 2 + R or 5 = 2 + R . Electric current through 25Vcell = I1 + I2 + I3 + I4 = 5 + 1 + 3 + 3 = 12A The power supplied by 20Vcell is 20I2= 20 × 1 = 20 W P = Example 31.2 V5/2. 44 is 2 0. 0. . . 2. . .44R . R = 13.44 5 V´ 0. .2 6 0.44R or .52 .(ii) the power dissipated in the rod is twice that dissipated in the resistance.2 5 R 0. . Sol.12 R = 10. . .44 .2 5 .2 .2 5 The potential difference across connected resistor isV´´= 0.44 R E = V´ + V´´ or .2 V . 2. (i) the potentialdifference across rod for current 0.44 0.. .64 3. 6 . . (ii)Total power supplied by battery is used byrod and resistor . 2.2 0. . E I = V I + I2 R .2 5 .2 6 .. 3125 6. .. . Remarks : In the case of rod. . The bath tub is located at a height of 10mfromthe ground&it holds 200 litres ofwater ... .4A Fromeqn (i). . .. V´ + V´´ = E 2 I 5 IR 6 0.ELECTRIC CURRENT www.. EI = 2I2R + I2R = 3I2R or E = 3IR or 6 = 3IR . or 2 5 5 I 4 2 32 0. .. dm 40 40 4 kg / sec. . .. or 2 I 5 2 6 0. or I = 32 × 0. R 2 20 0. if the bath tubwas full initially. . . . Power P gh dm dt .2 . .. . . Aperson decides to use his bath tubwater to generate electric power to run a 40watt bulb.4 64 . So. or 40 0. IR = 2 .2 . .2 .. . IR = 2 . or 2 I 5 4 0.2 . . .physicsashok.in 25 But VI = power dissipated in rod V I = 2I2R . . ohm s lawis not ap plicable in the case of the givenrod. The efficiency of gen erator is 90%.(i) Also.V I graph is not straight line. . at what rate should thewater drain fromthe bathtub to light bulb?Howlong canwe keep the bulb on. or 40 gh dm 90 dt 100 .. . . Example 32.2 = 6. (g = 10m/s 2) Sol. . . .9 gh dm dt . . Ifwe installawater driven wheelgenerator on the ground. ..dt 0. dm 4 dt 9 . . 0 m dm t dt .9 gh 9 10 9 . 0 t m 200 450 sec. . . . .. .. .. . . . . . . . . . Ans. 3 4 .ELECTRIC CURRENT www. 1234 isasquare. then in follows fromsymmetry that there is no c urrent through thiswire. The circuit shown in figure ismade of a homogeneouswire of uniform cross-section. . .physicsashok.. Let Vbe the voltage between 1 and 2. . R12 = R13 = R34 = R24 = r and 15 25 36 46 R R R R r 2 . .. Find the ratio Q12/Q34of the amounts of heat liberated per unit time in conductor 1 2 and 3 4. . . 3 4 1 2 Sol. 2 12 V Q r .(i) Current through3 4.2 1 2 3 4 Q 2 3 11 6 2 Q . G l (100 l) D A C . . . Metre Bridge Themetre bridge is the practicalapplication of theWheatstone s network principle. . . . 3 4 1 2 6 5 3 4 1 2 6 5 Then the amount of heat liberated in conductor 1 2. . Therefore. . Let us represent the central junction ofwires in the formof two junctions c onnected by thewire 5 6 as shown in figure. . Further. the central junction can be removed fromthe initialcircuit. . . . 2 2 3 4 3 4 2 Q i r V r 2 3 . . . . i V r 2 3 .in 26 Example 33. . resistance . .AC is a 1mlong uniformwire. thenDC = (100 l) cm Since. Let AD = l cm. In Fig. l l If P is known. P AD Q DC 100 . . length . can be determined from the ratio of their balancing lengths.. . Example 34. The potentiometer wire AB is 100 cm long. thenQcan be determined. When AC = 40 cm. no deflec tion occurs in the galvanometer.P Q B R J S In such a bridge. the ratio of two resistances sayRand S. findR. . = 10 x where xis inmetre. . R 10 CB (10) 100 40 10 60 15 AC 40 40 . ./m. . . . . . . . G R1 P R2 I 10V I I A B E = r E1 r1 E2 r2 + = 56 21 4 + 5 = 14 6 V r = 1 × 5 1 + 5 = 56 But VA VP = E = IR1 . . . .ELECTRIC CURRENT www. . . . 14 1 10x .0 = 10Vis connected across a 1 mlong uniform wire having resistance 10. Abatteryof emf . If a galvanometer shows no deflection at the point P.. . The resistance inAP is R1 = x. find the G 5 1 10 P A B 0 =10V 1=2V 2=4V distance ofpoint P fromthe pointA. x)10 The resistance in PB is R2 = (1 The equivalent circuit is According to loop rule 10 IR1 IR2 10I = 0 . . . Example 35. . 10 AC R CB . Two cells of emf .2 = 4Vhaving internal resistances 1 . .and 5 .respectively are connected as shown in the figure.physicsashok. Sol. . . . ..1 = 2Vand . . . .in 27 G A B 10 R Sol. . 1 2 I 10 10 1 R R 10 10 10 2 . = R2 x. balance shifts to 50 cm. (AB = 1m) : G R1 A B R2 Sol. or R1(1 x) = R2x . Find R1 and R2.. AB = 1 m) According to balance condition ofwheatstone bridge. .(i) whenR2 is shunted. R1R4 = R2R3 or R1(1 x).. Example 36.Assume that P is contact point of potentiometer.4667 m 46.6 2 . In the figure shown for which values ofR1 and R2 the balance point for Jockey is at 40 cmfromA.WhenR2 is shunted by a resistance of 10 . .67 cm 6 10 . . The resistance inAP is R3 = x. .. . Let resistance perunit lengthof potentiometer is . . and the resistance in PB is R4 = (1 x). x 14 2 0. ( . .. in 28 Then 2 2 2 2 2 R R 10 10R R 10 R 10 . It consists of a long resistancewireAB of uniformcross-section inwhich a steadydirect current is set u p bymeans of a battery. . the balance position remains undisturbedwhile sensitivityof bridge changes. .. . . (i) and (ii). 2 1 2 R 1 x 10R x R 10 . .(ii) Fromeqn.. = R´2 x. 1 R 10 3 . Two other common forms of balancedWheatstone s bridge are shown. . IMPORTANT FEATURES 1. R1(1 x).. . . . r i i Potentialgradient k Potential difference across AB Total length . . and R2 = 5. .ELECTRIC CURRENT www. When batteryand galvanometer arms of aWheatstone s bridge are interchanged. I I G 1 R P I1 IG (I I1 + IG) S I A B I (I1 IG) Q 2.physicsashok. . L E1 i2 = 0 A B i C G E2 . A B G P Q R S POTENTIOMETER Potentiometer is an ideal device to measure the potential difference betweentwo points. . .AB k V L . AB k iR i L . . Then. EK = i(.. l2) Here. l l So. i R .Aknown resistance R is then connected to the cell as shown.The terminalvoltageVis nowbalanced against a smaller lengthAD´ = l2. = resistance per unit lengthof potentiometerwire The emf of source balanced between pointsB and C CB 2 E . l l or 1 U K 2 E E . . . For this the switch S´ is left opened and S is closed. (ii) To find the internal resistance of a call : Firstly the emfE of the cell is balanced against a lengthAD= l1. no switch S Is opened and S´ is closed. K 1 U 2 E E . Let the null points are obtained at lengths l1 and l2. . l1) and EU = i(..Here. Then. kl . . we can find the emf of an unknown battery . .physicsashok. .= resistance ofwireAB per unit length. .ELECTRIC CURRENT www.in 29 where AB R L . bymeasuring the lengths l1 and l2. l l E2 = iRCB Applications (i) To find emf of an unknown battery : l E1 i2 = 0 A B i C G EK 1 E1 i2 = 0 A B i C G EU 2 1 l2 i We calibrate the device by replacing E2 by a source of known emfEK and then by u nknown emfEU.. l l Example 37. . r) D´ D A S´ R E l2 S l1 Since. . l l G (E. . . {. .1 2 E V . E = i(R + r) and V = iR} or 1 2 R r R . Apotentiometer wire of length 100 cm has a resistance of 10 .What is the value of external resista nce ? .. B E R r V R . . It is connected in series with a resistance and a cell of emf2Vand of negligible internal resistance. l l or 1 2 r 1 R . . .Asource of e mf 10mVis balanced against a length of 40 cmof the potentiometer wire. . . . CB R 40 10 4 100 . and E = 10 × 10 3 V Substituting in above. . ...With suitable modifications. . to convert a galvanometer into an ammeter. . Hence. to convert a galvanometer into ammeter a low resistance. VCB = E. Conversion of galvanometer into an Ammeter An ammeter is an instrument whichis used tomeasure current in a circuit in amper e (ormilli-ampere ormicroampere). and . if no current is drawn fromthe battery or 1 CB AB E R E R R .ELECTRIC CURRENT www. it is always connected in series in the circuit. . A . g . . it can be used to measure current and potentialdifference. . . . RAB = 10 . its resistance is to be decreased so. Fromthe theoryof potentiometer.physicsashok. i S i S G . . MOVING COIL GALVANOMETER Moving coilgalvanometer is a device used to detect small current flowing in an e lectric circuit. . . E1 = 2V. .in 30 Sol. Since. therefore. the galvanometer coilhas some resistance of its own. . .we get R = 790 . . called shunt (S) is connected is S a G b i Ammeter in parallelto the galvanometer as shown in figure. Here. . . where RA = resistance of ammeter S = resistance of shunt G= resistance of galvanometer . S G R G S . E1 E A B C G R i Here. . Here. where R +G= RV = resistance of voltmeter Example 38. Amoving coil galvanometer of resistance 10 . hence. high resistance Ris series is connectedwith the galvanometer. to convert a galvanometer into a voltmeter. produces full scale def lection. betweenwhich the potentialdifference ig G R is to be measured. when a current of 25 .Conversion of Galvanometer into Voltmeter AVoltmeter is an instrument which is used tomeasure the potential difference between two points of an electric circuit directly in volt (ormilli-volt ormicro-volt). g i V R G . the resistance of coilofgalvanometer of its ownis low. . Voltmeter Since.When it is connected. Hence. it is connected in parallel across those two points of the circuit. 1 A Example 40.. I = 1000 mA . R . . . reading of voltmeter is V2 when only S2 is closed. . 10 mA at B (C is left isolated). howwill you convert the galvanometer into a voltmeter reading upto 120V. . is joined in a circuit as shown. ig = 25 mA= 25 × 10 3A To convert the galvanometer into voltmeter reading upto 120V: To convert a galva nometer into voltmeter of rangeV.1 × (I 10) = 0 or I 10 90 9 990 0. .. 3 R 120 10 4790 25 10. Here.physicsashok. The value of I is A B C (A) 100 mA (B) 900 mA (C) 1 A (D) 1. I 4R . .. Describe showing full calculations. a large resistance Rhas to be connected in series to it.= + = 10 mA I 10 mA I 10 I . . Example 39. 9 × 10 0. The reading of voltmeter is V3 when both S1 and S2 are closed then V 6R 3R R E S2 S1 (A) V2 > V1 > V3 (B) V3 > V2 > V1 (C) V3 > V1 > V2 (D) V1 > V2 > V3 Sol. V = 120 V . . 1 v I 3R 3R 4R . Sol. . Step-I When S1 is closed. According to loop rule.The value ofR is given by g R V G i . Themetre gives full-scale deflection for current I whenAand B are used as its terminals. . Amilliammeter of range 10 mAand resistance 9. Here.1 . current enters at Aand leaves . . . In the circuit shown in figure reading of voltmeter is V1 when only S1 is closed. i.1 A Sol.in 31 mAis passed through it.e.ELECTRIC CURRENT www. G = 10 .9 × 10 + 0. .3R 1 v 3 4 . Step-III When S1 and S2 both are closed. . Step-II When S2 is closed I 7R . 2 v . I 3R . . . . 6R R 6R 2 v 6R 6 7R 7 . . v3 = I × 2R R 2R . . I . I 1.5V R 100 50 A no current I1 I1 .5 V In loop BCGFB 100I + (I1 I) R = 0 or (I1 I)R = 100 I I1R = (100 + R)I . In the circuit shown in figure the reading of ammeter is the samewith both switches openaswith both closed.in 32 3 2 v 2R 3R 3 . .. I 300 IR 1. (ammeter is ideal) + 300 1. In loopABCDEA IR + 1..5 300 I 100 I 50 I = 0 . Step-I : Discuss the circuit when both switches open : According to loop rule : 1.5V R 1000 50 A Sol. . then find the resistance R.5 R .100 R. . .ELECTRIC CURRENT www..(1) A 50 I I I I 100 300 1.(2) Fromeqn (1) and (2) . Example 41. Step-II :Discuss the circuit after closing the switch. . .physicsashok. .. 1 100 R I I R .5 300I1 = 0 or 300I1 + IR = 1. . 300 1. .5 15 1 A 450 4500 300 . . .5 . .0 . It is observed that when the shunt resistance are 10.and a battery of internal resis tance 10 . VA VB = {algebraic sumof rise up and drop up of voltage} 120 = { IR 1 × I 100} 120 = IR + I + 100 100V I B A R I or 20 = 10R + 10 . R = 600 .. In order to charge the battery at 10 amperes charging current.. respectivelythe deflection are respectiv ely9&30 divisions. the resistance R should be set at (A) 0. The battery in the diagramis to be charged by the generator G. (B) 0.0 .What is the resistance of the galvanometer ? .I1 A F B I C D E (I1 I) G or . Example 42. The battery has an emf of 100 volts and an internal resistance of G R + + 1 ohm. . Agalvanometer having 50 divisions providedwitha variable shunt s is us ed tomeasure the current when connected inserieswith a resistance of 90 .5 R 300 300 .1 . 1 .100 R. R . . 50. Sol.. The generator has a terminal voltage of 120 volts when the charging current is 10 amperes.. 1 R 300 1. Example 43.5 . (D) 5. (C) 1. g S g S I ER R 100 R R .physicsashok. g g I 9 3 I´ 30 10 . . But . S g S g S g g S g S g S g S g S R R R 100 R R I R R I´ R´ R R´ 100 R R´ .in 33 Sol. . . . . .ELECTRIC CURRENT www. . S g g S IR I R R . . . . . S g g S g S g S ER I R R 100 R R R R . The electric current through galvanometer is proportionalto its deflection. . . . . or . . . . Also. I × . . . . . . ... . . actual current throughRis. . g g g g g g 10 10R 100 R 10 3 R 10 10 50 R 50 50R 100 R 50 . .. .3 . . i´ E R A . i E R . . . . in serieswithRis. . . . . . . . . or . . . . . . (a) The reading of an ammeter is always lesser than actual current in the cir cuit. IMPORTANT FEATURES 1. . . . .. . . . . . .(i) while the current after connecting an ammeter of resistance A GS G S . . . . . . . in Fig. (a). Rg = 233. . . . .R R´ . .. . . E R i (a) E R i´ (b) A G i´ i´ S (c) For example. . . .. . .. . e. (i) and (ii). i. .. resistance of an ideal ammeter should be zero. we see that i´< i and i´= i .WhenA= 0.(ii) FromEqs. ... .. . . . . .(i) Nowif a voltmeter of resistance RV(=G+R) is connected across newvaluewill be . .. (a) The reading of a voltmeter is always lesser than true V G R RV r i i i r For example. . . . .. . . . . . .. or V V´ ir 1 r R . . . . . . . .ELECTRIC CURRENT www. . . the resistance voltmeter is. . the actualvalue is. . . . resistance of an idealvoltmeter should be infinite. . . . . . . or % error A 100 R A . . . . r . .(ii) FromEqs. ..physicsashok. . RC CIRCUITS is value. .. . 2. V V i rR V´ r R . . . .. . .. Thus. . we can see that V´ < V and V´ = V if RV = . the . . (b) Percentage error inmeasuring the potentialdifference bya V V V´ 100 1 100 V 1 r R . . . if a current i is passing through a resistance V = ir .. r. . . (i) and (ii). . . . .in 34 (b) Percentage error inmeasuring a current throughan ammeter 1 1 i i´ 100 R R A 100 i 1 R . . . or V %error 1 100 1 r R . . . were discuss about time-varying currents.Acharge q0 = CVcomes in the capacitor as soon as switch is closed and q t graph in this case is a straight line q0 q r parallel to t=axis.In precedingsectionswe dealt onlywith circuits inwhichthe current did not varywi th time.Here. there is no time tag between connecting and charging.) . The q t equation in this case. Charging of a Capacitor First we consider the charging of a capacitor without resistance. charging takes some time Fig. q = q0(1 e t/. Ifwe employed a resistance in the circuit. On closing the switch. + S V C + V q = CV + 0 Consider a capacitor connected to a battery of emfVthrough a switch S. 2%charging is over in a C R circuit . The charge q increases expponentially from0 to q0. q0 = CV S C C and .368q0 Hence. q = 0 q0 q t 0. current decreases exponentiallywith time.e. as shown in Fig. q = q0 q0 q t 0.This current can be found bydifferentiating . in this case . Current flows ina C Rcircuit during charging of capacitor. The positive plate has a charge +q0 and neg ative plate q0. i0 i t Charging current is. q e.632 q0 Here. is charged. q = 0 In case of discharging definition of .in 35 Here. . S C R q0 + In case ofC Rcircuit discharging also takes time.. = CR= time constant V q t graph is anexponentially increasing graph. Thus. Suppose a capacitor has a charge q0. So. During discharging current flows inthe circuit till q becomes zero.8%of i tsmaximumvalue q0. At time t = . q = q0 At t = . The current at any time t can be calculated bay differentiating qwith respect to t..ELECTRIC CURRENT www. i.368q0 At t = . At t = 0. can be defined as the time inwhich 63. Discharging of a Capacitor Againwe consider the discharging of capacitor with resistance. q decreases exponentially fromq0 to zero.. t / c 0 q . The q t equation in this case is. q = q0 e 1 = 0.physicsashok. i = i0e t/. S q = 0 + q0 When the switch is closed.632q0 and at t = . . Fromthe graph and equationwe see that. At t = 0. q = q0(1 e 1) = 0..Once charging as over or the steady state condition is reached the current becomes zero. the extra electrons on negative plate immediately comes to the positive plate and net charge on both plates become zero.. can be defined as the timewhen charge reduces to 36. we cansay that discharging takes place immediately. Time i t graph is as shown in fig. Hence. This is an exponentially decreasing equation.Hence. i0 i t . discharging current is. The i t graph is as shown in Fig. i = i0e t/. Thus.q with respect to t butwithnegative signbecause charge is decreasingwith time. i t graph decreases exponentiallywith time fromi0 to 0. 1k 1k at t = 0 C R = 0. 1 1 2 2 t RC t 1 1 1 R C C t . the capacitor discharges with 1k 1k 1k C 9V S time constant (A) 33 ms (B) 5 ms (C) 3. Let i1 and i2 be the currents flowing throughC1 and C2 at any time t .When switch S is opened. = RC = 0. In the circuit shown in figure C1 = 2C2. Switch S is closed at time t = 0. and 2 t RC 2 I V e R .5 × 103 × 100 × 10 6 = 50 × 10 3 S = 50 m/s Example 45.3 ms (D) 50 ms Sol.physicsashok. .5 k . and a battery of emf 9V. then the ratio i1/i2 (A) is constant (B) increases with increase in time t C2 C1 V S R R (C) decreases with increase in time t (D) first increases then decreases Sol.in 36 Example 44. .The switch S has been closed for long time so as to charge the capacitor. Acapacitor C= 100 µF is connected to three resistor each of resistance 1 k. Time constant . .ELECTRIC CURRENT www. Here 1 t RC 1 I V e R . . . . . resistorwhen the switch S is closed.2 RC I e e I e . .6 × 10 3 J is dissipated in the 10 . . . Hence option (B) is correct. . In the R C circuit shown in the figure the total energy of 3. . . 1 2 t C C 1 RC C 2 I e I . . . . 1 2 . Example 46. . The initial charge o n the 2µF 10 S capacitor is . . . Find the charge on the capacitor at t = 2 R1C + R2C.in 37 (A) 60 µC (B) 120 µC (C) 60 2 µC (D) 60 2 µC Sol.6 10 2C . . . According to conservationprinciple of energy: Total energy stored on capacitor appears as heat in resistor.ELECTRIC CURRENT www. so that it carries a charge CVwith both the switches S1 and S2 remaining open. .6 × 10 3 q0 2 = 14. . or (R1 + R2)CI = EC q or . q0 = 12 × 10 5 q0 = 120 µC Example 47. . When t < R1C 1 t q CV e R C . 2 0 3 q H 3. or q0 2 = 2C × 3. .At t = R1C switch S1 E S2 C + R1 R2 S1 is opened and S2 is closed. . At t0 = R1C. q R1 R2 . 1 2 E q I R R 0 C . .6 × 10 3 q0 2 = 2 × 2 × 10 6 × 3. . . when S2 is closed. .4 × 10 9 . . 1 0 q CV e CV e .physicsashok. 1 2 R R C dq EC q dt . q0 q0 R1 R2 t = t E 0 At instant t (t > t0) . . . Sol. The capacitors shown in figure has been charged to a potential difference ofVvolts. Switch S1 is closed at t = 0. . . . . . . . . . 0 1 q t 1 2 CV t R C e R R C dq dt EC q . . . or . . . . . 1 2 1 EC CV R R C ln e t R C EC q . 1 q t 1 2 CV R C e R . . . . . . . But t = 2R1C + R2C Putting the value. t or . . .E + +q I I or .. . . . R C. . . . . . . . . . .. .. . . . or . . .ln EC. . . . 1 2 1 CV R R C ln EC ln EC q t R C e . . . q . . . . . . Inthe figure shown initiallyswitch is open for a long time. Example 48. Also. . V R R V2 R2 q q q1 q1 q +q q0 1 (q +q q0 1) or 0 V I R q q V IR 0 2 2 C C 2 2 . 0 q CIR CV 2 . . . Here I V 2R . . . . .physicsashok. . . Step-I :Discuss the circuit. V R R C C S Sol.. . . 0 q q IR V C .ELECTRIC CURRENT www. Also. . .. . . I I I +q q0 Step-II : Discuss the circuit after closing the switch : 1 V I R q 0 2 2 C . . or 0 1 V I R q q q 0 2 2 C C C . . . Nowthe switch is closed at t = 0. . . Find the charge on the rightmost capacitor as a function of time given that it was initiallyunchanged. . . 0 1 V R q q q I 0 2 2 C .when switch is open.in 38 2 q CE 1 1 CV e e . . or . . or 0 V IR q q 0 C C . . . or 0 t RC 0 V q q C e V q C . . . . . . . . . . . . .. . . . . . .. . . . . . . .. . . or 0 dq q q R V dt C . . . . . . . . . . . . . . . .. . . . . . . . . . . . or q 0 0 q q RC ln V t C . . . or 0 0 V q q RC ln C t V q C . . . .. . . or q t 0 0 0 Rdq dt V q q C . . . . .. . . In a hollow nonconducting pipe. . . (b) doubling V. 0 1 q q q 2 .What is the effect on the electron drift velocity of (a) doubling l.. . . A potential difference V is applied to a copper wire of diameter d and length l. . .physicsashok. the bulbs regain their original brilliance. . . Is a current-carrying conductor electrically charged? 2. . . Account for the increase in the resistance ofmetals with rise in temperature.(i) But potentialdiffer accross each capacitors are sume. be connected to produce an effective resistance of 4..Why? 11. . . two rheostats in parallel ar e preferable to a single rheostat. and 6. Acurrent is passed through a steelwire which gets heated to a dull red. 10. THINKING PROBLEMS 1. . Is there an electric field near the surface of a conductor carrying direct cu rrent? 3. A current i enters the top of a copper sphere of radius R and leaves through the diametrically opposite point. . . Then half the wire is immersed in cold water. . Why? 12. . The portion out of the water becomes brighter. . . The ions of one streamare negatively charged and constitute a current of strength l and those of . and (c) doubling d? 5. ? 8. . 7. 3. Answer briefly how can three resistances of values 2.Are all parts equally effective in dissipating Joule heat? 6. t RC 1 q CV 1 1 e 2 2 . 0 1 1 q q q q C C .in 39 or t V q0 q V q0 eRC C C . .ELECTRIC CURRENT www. Explain. . or q0 + q = 2q1 . . Putting the value of q. The brilliance of lamps in a roomnoticeably drops as soon as a highpower elec tric iron is switched on and after a short interval. Is current a scallar or vector? 4. . How can an electric heater designed for 220V be adopted for 110V without chan ging the length of the coil and also without a change in the consumed power ? 9. there are two streams of ions in opposite di rections. For manual control of the current of a circuit. although the bulbmay be quite far fromthe switch? 14. The drift velocity of electrons is quite small. Why is it easier to start a car engine on a warmday than on a chilly day? 17.. why does one . How then does a bulb light u p as soon as the switch is turned on. If the resistance of our bo dy is so large.What is the tot al current through the pipe? 13. The resistance of the human body is about 10 k. So the ch arges on the plates should be neutralized at once as it happens when two charged conductors are join ed by a wire. The electrolyte between the plates is a conducting medium.Why are the charges not neutralized immediately? 15.the other stream are positively charged and constitute a current of the same strength. Does emf have electrostatic origin? 16. Consider a voltaic cell in the open circuit with the copper plate at a highe r potential with respect to the zinc plate. No. and 6. TheP = V2/R where R is the resistance of the coil. This is equal to the rate of fall of potentialalong the conductor. An ordinary cellwith a small emf can produce larger current than an electros taticmachine which generates thousands of volts? Explain. P1 = (V/2)2/(R/2)=P/2 Total power = P/2 + P/2 = P 9. resistors in parallel and the 2./ne) (V/l) (. where Q is the charge passing through the conductor and V is the potential difference.E). and 6. The cold resistance of the coil of the iron ismuch smaller thanwhen it is hot . There is continuity of this field outside the surface of the conductor.Why? 20. Connect the 3. 4. Then 3. Thus. 8.physicsashok. resistor in series wit h the combination of 3. metals have mobile electrons. . E = V/l). Join the ends of the coil and apply the supply voltage (110V) between this co mmon terminal and the midpoint of the coil as shown in the figure.. and 6. or u = (. 3. willwork out to 4. in parallel will sumup to 3 6 2 3 6 . what is the fact? 19. the amount of positive and negative charge in any elementary volume remai ns the same through the electrons are inmotion. So a current-carrying conductor is electrically neutral.in 40 experience a strong shock from a live wire of 220 V supply? 18. with 2. and this. This amounts to increase in resistance with temperature. When it is connected as shown in the figure. Obviously from this expression (a) u is halved if l is double d.. the probability of electrons striking ionic cores increases. Lay people have the notion that a person touching a high power line gets stuc k to the line. . . Let P be the power of the entire coil when the coltage applied is V volts. 7. 2.With increase of temperature. 5. in series. u = j/ne = . 6. Current is a scalar but current density is a vector. the amount of energy libera ted is VQ. When a direct current flows through a conductor.E/ne where s is the conductivity and E is the electric field (byO hm s law j = . the lattice vibrat ions increase in amplitude. No. There is an electric field inside the conductor. the resistance of elements at right angles to the diameter varies fromele ment to element but current is the same through all sections and so heating effect varies fromsection to sectio n. (b) u is doubled if V is doubled.ELECTRIC CURRENT www. Is this true? If not. THINKING PROBLEMS SOLUTIONS 1. Therefore. while an energy of VQ/2 is liberated when a capacitor is discharged. the two par ts are in parallel and the resistance of each part becomes R/2. and (c) u remains uncharged on doubling d. So when the iron is switched . 13. Thusmore length of eachwire is r equired to produce the same current. 14. Then i = E/(l r/2) = 2E/l r. it is not necessary for the same electron to trave l from the switch to the bulb. Then i = E/lr where E is the emf of the cell and r is the resistance per unit le ngth of the wire. The current constituted by the negative ionsmoving opposite to the positive ions has the same sense as the current constituted by the positive ions. When the switch is turned on.As the coil gets heated. to produce a certain current i in a circuit. Hence the total current is 2I and not zero. every electron in the circuit begins to move simul taneoulsy. the length of the sing le rheostat wire required is l. there is a large drop of voltage and consequently the bulbs do not receive t he proper voltage.on. To produce the same current in the same circuit by using two rheostats in parallel. its resistance increases and so the voltage drop is made up wh en it is fully heated. let l be the length of each rheostat wire. . l = 2l. 10. This is why the portion outside the water becomes brighter. Suppose. 11. The charges are prevented frombeing neutralized because the electrostatic fi eld due to the charges on the plates is opposed by the charges of ions that migrate into the solution. 12. including those in the filament of the bulb. Obviously. For the bulb to light up. The resistance of the immersed portion decreases and so the current through the entire wire increases. This is definitely of advantage. This is why after some time the bulbs region their brilliance. 10. being heavier . Reason (R) : The drift velocity of electron in presence of electric field decrea ses. 11. 8.ELECTRIC CURRENT www. 2. Reason (R) : The number density of free ions in electrolyte is much smaller comp ared to number density of free electrons in metals. a high resistance is connected in series with the galvanometer. Further. Reason (R) : Temperature coefficient of these materials is very small. Reason (R) : If resistance is high. Assertion (A) : Heater wire must have high resistance and high melting point . Assertion (A) : The conductivity of an electrolyte is very low as compared to ametal at room temperature. Assertion (A) : When a wire is stretched so that its thickness is halved. ions drift much more slowly.physicsashok. Reason (R) : The emf of a cell is always greater than its terminal voltage. 5. the electrical conductivity will be less. Reason (R) : The current in a wire is due to flow of free electrons in a definit e direction. Reason (R) : Resistance of connected wires is very small and H . 3. its resistance would become 16 times. durin g charging of the cell. 4. Reason (R) : V 2 P R . Reason (R) : Illumination of the lamp is directly proportional tot he current th rough lamp. 7. (B) If both A and R are true but R is not the correct explanation of A. mark the correct answer as (A) If both A and R are true and R is the correct explanation of A. Assertion (A) : Terminal voltage of a cell is greater than emf of cell. the percentage de crease in the illumination of the lamp is 40%. Assertion (A) : A current flows in a conductor only when there is an electric field within the conductor. Reason (R) : As resistance increases current through the circuit increases. Assertion (A) : A current carrying wire should be changed.in 41 REASONINGTYPE QUESTIONS A statement of assertion (A) is given and a Corresponding statement of reason (R ) is given just below it of the statements. 1. Assertion (A) : Material used in the construction of a standard resistance is constantan or manganin. Assertion (A) : If the current of a lamp decreases by 20%. Assertion (A) : A 60 watt bulb has greater resistance than a 100 watt bulb. (E) If A is false but R is true. Assertion (A) : In meter bridge experiment. 6. Assertion (A) : The wires supplying current to an electric heater are not hea ted appreciably. . (D) If both A and R are false. Reason (R) : The data is insufficient to predict. (C) If A is true but R is false. 12. 9. R . Reason (R) : By adjusting the value of resistance in series with galvanometer th e range of voltameter can be adjusted. Assertion (A) : The range of given voltmeter can be increased. A potentiometer is to be calibrated with a standard cell using the circuit shown in the diagram. To improve accuracy the balance point should be nearer M. the drift velocity of the charge carriers. 3. for the shown RC circuit is (A) RC (B) 2 RC (C) 2 RC (D) not defined R R . the current through the 8. 7. The resistor in which the maximum heat is produced is given by (A) 2. (C) 1. (D) 1. The resistance of tungsten increases with increasing temperature. (B) 6.in 42 LEVEL # 1 1. between its ends is of the form. (C) n is increased but vd is decreased. (C) putting a shunt resistance in parallel with the galvanometer.5. The current in a copper wire is increased by increasing the potential differe nce between its ends. This may be achieved by (A) replacing the galvanometer with one of lower resistance. The balance point is found to be near L. Which one of the following statements regarding n. The time constant . the number of charge carriers per unit volume in the wire. (B) 3. Two cells of equal emf of 10 V but different internal resistances 3.ELECTRIC CURRENT www. resistance is 1 A. When the switch 1 is closed. a re connected in series to an external resistance R. resistance is 0. . (B) replacing the potentiometer wire one of higher resistance per unit length. flowing in the tungsten filament of an electric lamp and the potent ial difference. 4. (D) n is increased but vd is unaltered. and 2. and Vd. the current through the 2. V. the relation between the current. The value of R that makes the potential difference zero a cross the terminals of one of any cells is (A) 5. When the switch 2 is closed (only). The value of . 6. (A) I O V (B) I O V (C) I O V (D) I O V 2.75 A. (D) increasing the resistance R. .physicsashok. is (A) 5 V (B) 5 2 V (C) 10 V (D) 15 V 5. (C) 4. As a result . is correct? (A) n is unaltered but vd is decreased. (B) n is unaltered but vd is increased. (D) 12. . In steady state condition ratio of charge stored in the first and last capacitor is E R C R C R (A) n : 1 (B) (n 1) : R (C) (n2 + 1) : (n2 1) (D) 1 : 1 . n resistances each of resistance R are joined with capacitors of capacity C (each) and a battery of emf E as shown in the figure.8. What is the current it would supply to a 5. . + 3 . What is the heat developed in the resistance in the first four second. (B) 1 2 3 1 1 1 . resistor. are conductances of three conductors then their equival ent conductance when they are joined in series will be (A) 1 . .5 . What is the equivalent capacitance between A and B in the circuit shown. R = 12 . (D) None of these 15. (C) 1 2 3 1 2 3 . (C) drift velocity of conduction electrons is minimum at middle section. If 1 . Assume the voltmeter resistance be be 280. . . What is the internal resistance of the cell (A) 20. (B) 30. 9xy . (A) 44 A (B) 0.044 A (C) 4.4 A (D) None of the above. (C) 10.. . Then P Q (A) electric field intensity at P is greater than that at Q (B) rate of electric crossing per unit area of cross section at P is less than t hat at Q (C) the rate of generation of heat per unit length at P is greater than that at Q (D) mean kinetic energy if free electorn at P is greater than that at Q 12. . 14. .55 V. 2 .F (B) 1. . (A) 6.. A resistance R = 12. . . (Volt) 24 4 t(s) (A) 72 J (B) 64 J (C) 108 J (D) 100 J 11. and 3 . In the above question.in 43 9.40 V while a potentiometer reads its voltage to be 1. Its emf changes with time as shown in the graph. (D) at the ends of the conductor.F (C) Zero (D) 2. (D) 40. Suppose a voltmeter reads the voltage of a very old cell to be 1. A bat tery is connected across this conductor then (A) resistance of the conductor is equal to 4. Source .physicsashok. A source of constant potential difference is connected across a A B conductor having irregular cross-section as shown in the Figure.. Length of the conductor along the axis is equal to . . electric field intensity is same. 13. .) has rectangular cross-section. + 2 .ELECTRIC CURRENT www. is connected across a source of emf as shown in the fig ure.F 10.. A conductor is made of an isotropic material (resistivity . Horizontally dimension of the rectangle decreases linearly from 2x at one end to x at the other end and vertial dimension increases from y to 2y as shown in Figure. (B) rate of generation of heat per unit length is maximum at middle cross-sectio n. . 16. Find the current through 4. 1 m long metallic wire is broken into two unequal parts P and Q. resistor just after making the circuit (A) 0 A (B) 6 A (C) 12 A (D) 2 A 17. Find the ratio of the resistance of P and R (A) 1 : 4 (B) 1 : 3 (C) 1 : 2 (D) 1 : 1 . P part of t he wire is uniformly extended into another wire R. Length of R is twice the length of P and the resistance R is equ al to that of Q. The electron velocity is v and the charge density is . . If .physicsashok. (C) 2 nq 2m. If I is the equivalent current. It is required to make a suitable combination of these resistance s to produce a resistance of 5.5 V and internal resistance 0. and each c apable of carrying a maximum current of 1A. In the given circuit the ammeter reading is zero. . Which of the following statements is/are correct for potentiometer circuit (A) Sensitivity varies inversely with length of the potentiometer wire (B) Sensitivity is directly proportional to potential difference applied across the potentiometer wire. then (A) md v . which can carry a current of 4 A.5 A and 2 V (B) 1 A and 1 V (C) 1 A and 2 V (D) 2 A and 1. md (C) 2md .1. nq2 m (B) nq2 m. You are given several identical resistances each of value R = 10. 24. (B) 12. A cell of emf. then (A) . What is the value of resistance R ? (A) R = 100. The minimum number of resistances required is (A) 4 (B) 10 (C) 8 (D) 20 23. . 2eVx (B) 2eVxA 2 . . . -V graph is shown in Figure. 21. . each having a charge q is subjected to a potential V. . (D) None of these 20. is connected to a (non-lin ear) conductor whose .5 V 22. at any point between the electrodes at a distance x from the cathode. Assume that the electrons are emitted with zero velocity and they do not change the field between the electrodes.ELECTRIC CURRENT www.in 44 18. Find the current drawn from the cell and its terminal voltage (A) 1. On applying a potential V to the anode with respect to the cathode a current I f lows through the diode. 19. 1. . (C) 20. What is the equivalent resistance between A and B in the given circuit diagram. . (C) R = 0. A vacuum diode consists of plane parallel electrodes separated by a distance d and each having an area A. the number density of charge carriers in the conductor is n and the charge carriers (along with their random motion) are moving with a velocity v. A conductor or area of cross section A having charge carriers. is the conductivity of t he conductor and .5. (A) 2. (D) 10. m is the mass of ea ch charge carrier. (B) R = 10. (C) Accuracy of a potentiometer can be increased only by increasing length of th e wire (D) Range depends upon the potential difference applied across the potentiometer wire. is the average relaxation time. (D) 2 2nq m. . (C) 2 A 2 . . A long round conductor of cross-sectional area A is made of a material whose resistivity depends on the radial distance r from the axis of the conductor as 2 r . . 2md 25. . . . The total resistance per unit length of the conductor is R and the electric field strength in the conductor du e to which a current I flow in it is . (A) 2 A R 2 . is a constant. I .v . .I .. . eVx (D) eVxA 2 . . . (D) 2 A 4.. (B) 2 A R 4 . ... 55. B2 and B3 respectively. resistor is 2. The equivalent resistance between A and B is (A) .5 A (B) the current through the 4 . . . . B.5 A (D) the P. between the terminals of the 9 V battery is 7 V. Variation of current passing through a conductor as the voltage applied across its ends is varied as shown in Fig.D.physicsashok. A resistance R carries a current I . resistor is 3. The . 2 R 1 (C) R .ACB = . We will find that : (A) Resistance at C and D are equal (B) Resistance at B is higher than at A (C) Resistance at C is higher than at B (D) Resistance at A is lower than at B 28.. The total heat produced in R is (A) 6b a3R (B) 3b a3R (C) 2b a3R (D) b a3 R 31. W2 and W3 are the output powers of the bulbs B1. (2 ) 4 R 2 (B) . . . . between the points A and B in the circuit shown here is 16 V. NowW1. . . 9. 29. wher e C is the centre of the ring. If the resistance is determined at points A. . T . 2. Whic h is/are the correct statement(s) out of the following ? (A) the current through the 2 . (T T0) where . A and B are two points on a uniform ring of resistance R. 1. 4 2 30. 1. . D. The P. resistor is 1.. A 100 W bulb B1 and two 60 W bulbs B2 and B3 are connected to a 250 V source as shown in figure. A B 9V 3V (C) the current through the 3 . The charge flowing through a resistance R varies with time t as Q = at bt2. . . C and D.in 45 26. . The heat loss to the surroundings is .5 A 4. . . 3. Then (A) W1 > W2 = W3 (B) W1 > W2 > W3 (C) W1 < W2 = W3 (D) W1 < W2 < W3 [JEE 2002 (Scr)] 27. 2 (D) R . is a constant. ..ELECTRIC CURRENT www. (C) 3 calories/sec (D) calories/sec [JEE 1981] . 6.5 V (C) 3 V (D) 6 V 33. and T0 is the temperature of the atmospher e. I2 R (D) equal to . If the coefficient of linear expansion is .. . the strain in the resistance is (A) proportional to the length of the resistance wire (B) equal to . I2 R (C) equal to 2 1 . (A) 1 calorie/sec (B) 2 calories/sec 5. . The potential of the point O is (A) 3. . The heat generated in the 4 ohms resistor is 4.is the temperature of the resistance. In the circuit shown in Figure the heat produced in the 5 ohm resistor due to the current flowing through it is 10 calories per second. (IR) 32.5 V (B) 6. . . .T in the same time t. . (C) 2 r + 4 R (D) 5 2 2 R . 3 R R r R r . A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. (C) Q G .ELECTRIC CURRENT www. the temperature of the wire is raised by . . The three resistance of equal value are arranged in the different combinatio n shown below. Due to the current. the value of N is [JEE 2001] (A) 4 (B) 6 (C) 8 (D) 9 38. should be [JEE 1995] (A) 4 9 (B) 2 (C) 8 3 (D) 18 35. T he temperature of the wire is raised by the same amount . A steady current flows in a metallic conductor of non-uniform cross-section. (B) P G . . Express which of the following set ups can be used to verify Ohm s law? (A) A V (B) A V (C) A V (D) A V 40. . Arrange them in increasing order of power dissipation. (D) Q R . The effective resistance between points P and Q of the electrical circuit shown in the figure is [JEE 2002] (A) 2Rr R . r 39. 37. 36.physicsashok. the reading of the galvanometer is same with switch S open or closed. In the circuit P . A wire of length L and 3 identical cells of negligible internal resistances are connected in series. r (B) 8 . . In order that the maximum power can be delivered to the network. . the value of R in . electric field and drift speed. A battery of internal resistance 4. . R.T in a time t. The quantity/quantities constant along the length of the conductor is/are [JEE 1997] (A) current. Then [JEE 1999] (A) R G .in 46 34. (B) drift speed only (C) current and drift speed (D) current only. [JEE 2003] (I) a (II) b (III) c (IV) c (A) i (B) i (C) i (D) i R R R . is connected to the network of resistance as shown. R R 4R E 6R 4. P Q R S G v P Q r r 2R 2R 2R 2R 2R 2R . (B) Voltage drop in first half is twice of voltae drop in second half. Agalvanometer has resistance 100 .1 mA (B) 1.01 mA 46.1 mA (C) 10. Qand R as shown in the figure.. is connectedtomake it anammeter.1 mA (B) 1000. the smallest current required inthe circuit to produce the fullscale deflection is [JEE 2005 (Scr)] (A) 1000.in 47 41.1 mA (C) 10. Six equal resitances are connected between points P.Aresistor 0.ELECTRIC CURRENT www. Find the minimum current in the circuit so that the ammeter shows maximum deflection.and it requires current 100 µAfor full scal e deflection.n . . 4r I l/2 l/2 2r A B C (A) Power loss in second half is four times the power loss in first half. Consider a cylindricalelement as shown in the figure.A. In the givencircuit. (C) Current density in both halves are equal. Choose the correct option out the following.. is used as an ammeter using a resistance 0. For the post office box arrangement to determine the value of unknown resistance.01 mA (D) 1. In the figure shownthe current through 2. Themaximum deflection current in the galvanometer is 100. with respect to time. If the crosssectional diameter ofAB is doubled then for nullpoint ofgalvanometer the value ofACwould [JEE 2003 (Scr)] G A x (A) x (B) x/2 (C) 2x (D) None C B 42. If the resistance is changed to 2x. [JEE 2005] (A) 100.1 mA 48. Current flowing through the element is I and resistivityofmaterialof the cylinder is .1 mA (D) 100. the new graph will be [JEE 2004] (A) P (B) Q (C) R (D) S 45. the unknown resistance should be connected between [JEE 2004 (Scr)] B1 C1 B C D (A) B and C (B) C and D (C) Aand D (D) B1 and C1 44. A capacitor is charged using an external battery with a resistance x in seri es. Thenthe net resistancewillbemaximumbetween [JEE 2004 (Scr)] P Q R (A) P and Q (B) Q and R (C) P and R (D) any two points 43. A moving coil galvanometer of resistance 100. resistor is (A) 2 A (B) 0 A 10V 20V 10 2 5 (C) 4 A (D) 6 A [JEE 2005 (Scr)] 47.1. The dashed line shows the variation of . no current is passing through the galvanometer.physicsashok.1 .. [JEE 2006] P Q R S t ln I .(D) Electric field in both halves is equal. losses 50% of its initial charge sooner than C2 loses 50% of i ts initial charge. MORE THAN ONE CHOICE MAY BE CORRECT 50. is connected across one gap of a meter-bridge (the length of thewire is 100 cm) and an unknown resistance. (C) resistance R1 and R2 are interchanged. (D) Capacitor C1. (C) The currents in the two discharging circuits at t = 0 are unequal. (D) the battery and the galvanometer are interchanged. The heat developed is doubled if: (A) both the length and radius of the wire are halved. (B) There is an electric field inside a current carrying conductor. is connected across the other gap. resistor is 0. (B) both the length and r adius of the wire are doubled. resistor is 0. If resistance of connecting wires is equal to R. which of the following statements is/are correct? (A) At a particular value of r.ELECTRIC CURRENT www. 54. Some heat is produced in it. All the resistance in the given Wheatstone bridge have different values and the current through the galvanometer is zero. the balance point shifts by20 cm. (C) For manual control of the current of a circuit.50 A. (C) 5 . (C) the radius of the wire is doubled. When the se resistances are interchanged. (B) 4 . (B) the 3. [JEE 1989] (A) The current in each of the two discharging circuits is zero at t = 0. the unknown resistance is [JEE 2007] (A) 3 . In the circuit shown in Figure the current through (A) the 3. resistor is 0. The two capacitors are then separately allowed to dis charge through equal resistors at time t = 0.. Which of the following is/are wrong? (A) A current carrying conductor is electrically charged. (B) If R = 0. (D) the length of the wire is doubled and the radius of the wire is halved. (B) The currents in the two discharging circuits at t = 0 are equal but not zero . A constant voltage is applied between the two ensd of a uniform metallic wir e.25 A (C) the 4. Capacitor C1 of capacitance 1micro-farad and capacitor C2 of capacitance 2mi crofarad are separately charged fully by a common battery.physicsashok. negative terminal of second cell will be at higher potential than its positive terminal (C) Negative terminal of first cell can never be at higher potential than its po sitive terminal.50 A (D) the 4. (D) None of these 56. resistor is 0. their brightness will be: . 51. The current through the galvanometer will still be zero if. (D) None of these 53. greater than 2. Aresistance of 2 . (A) the emf of the battery is doubled. 55. potential difference across second cell can be e qual to zero.25 A 52. Neglecting any corrections. Two cells of equal emf and having different internal resistance R1 and R2 (R 2 > R1) are connected in series. (B) all resistance are doubled. (D) 6 . When electric bulbs of same power but with different marked voltages are con nected in series across a power line. two rheostats in series are preferable to a single rheostat.in 48 49. (B) The electric drift velocity is larger in the conductor of length 2L. (C) proportional to the squares of their marked voltages. (D) inversely proportional to the square of their voltages. 4.(A) proportional to their marked voltages. but have eq ual resistances. (D) The electric field in the second conductor is twice that in the first. 8. The two are connected in series in a circuit in which current is flowing. 57. 2. Which of the follo wing is/are correct? (A) The potential difference across the two conductors is the same. (B) inversely proportional to their m arked voltages. 3. 2. Two conductors made of the same material have lengths L and 2 L. R1 R2 R3 R4 . 2. 2. (C) The electric field n the first conductor is twice that in the second. 9V 8. 2. the current is 22 A and the voltag e is 12 V.physicsashok. resistor is 0. (C) 5 mA range with 1. (B) 10 V range with 200 k . the further reduction in the def lection will be (the main current remains the same). (D) the 4. then the ratio of curr ent passing through wires would be: (A) 54 : 64 : 75 (B) 9 : 16 : 25 (C) 4 : 9 : 25 (D) 3 : 6 : 10. When a galvanometer is shunted with a 4. (C) the 4. (D) two 2. It ca n be used as a voltmeter or as a higher range ammeter provided a resistance is added to it.ELECTRIC CURRENT www. A. resistors connected in parallel.5 mA (B) the potential difference across RL is 18 V (C) ratio of powers dissipated in R1 and R2 is 3 (D) if R1 and R2 are interchanged. resistors connected in series.in 49 58. internal resistance 1. (B) the 3. Mark correct statements: (A) Heat is always generated in a battery whether it charges or discharges. 2. A cell of emf 5 V. 4. 59. 2. heat generated in it is always less than electrical power developed in it. 9V 8. A mocrometer has a resistance of 100. 2. only. 64. magnitude of the power dissipated in RL will decrease by a factor of 9 . 8. wire. 60. The rate of change of internal energy is: (A) 240 J/s (B) 252 J/s (C) 264 J/s (D) 126 J/s.25 A. An electric current is passed through a circuit containing three wires arran ged in parallel. If the length and radius of the wires are in ratio 2 : 3 : 4 and 3 : 4 : 5. (B) 5/13 of the deflection when shunted with 4. and a full scale range of 50.50 A.5k 2k R1 R2 24V RL I [JEE 2009] (A) the current I through the battery is 7. the deflection is reduc ed to one-fifth. resistanc e in parallel. For the circuit shown in the figure 6k 1.25 A. 65. (A) 8/15 of the deflection when shunted with 4. The rate of heat generated in the battery is 12 W. 61. 63. the current through: (A) the 3. only. resistance in series.50 A. During the charging of a storage battery. resistor is 0. resistance in parallel. (D) 10 mA range with 1. will give maximum power output to: (A) a single resistor of 1. res istance in series. (D)None of these. 3. If the galvanometer is further shunted with a 2. 2. (A) 50 V range with 10 k . resistor is 0. (B) two 1. resistance. (C) Potential difference across terminals of a battery is always less than its e mf. 2. only. resistor is 0. (B) When a battery supplies current in an external circuit. (C) 3/4 of the deflection when shunted with 4. (C) two 1. Pick the correct ra nge and resistance combinations. (D) None of these 62. In the circuit shown in figure. resistors connected in pa rallel. Two bulbs A and B consume same power P when operated at voltage VA and VB res pectively. (1997) 2. the ratio of (P) RA/RB potential difference across A and B (B) In series connection..in 50 FILL IN THE BLANKS 1.physicsashok.. Column I Column II (A) Current (P) Mircoscopic quantity (B) Current Density (Q) Macroscopicquantity (C) Electric field (R) Parallel to the conductor boundaries (D) Resistance (S) Flux associated with current density 7.. A 2R 2R R B V Fig . [JEE 1982] 5.ELECTRIC CURRENT www. The equivalent resistance between points A and B of the circuit (Figure) give n below is ..c source then: Column I Column II (A) In series connection...2 ohm... so that the bulb delivers 500 Watts is . The resistance R that must be put in series with the bulb. the ratio of (S) V2 B/V2 A power consumed in A and B 8.. ohms.. Electrons in a conductor have no motion in the absence of a potential differe nce across it.. each battery is 5 V and has an internal resistance of 0. An electric bulb rated for 500 watts at 100 volts is used in a circuit having a 200 volts supply.. the ratio of (Q) V2 A/V2 B power consumed by A and B (C) In parallel connection... [JEE 1997] TRUE / FALSE 4. TABLE MATCH 6. The current-voltage graphs for a given metallic wire at two different tempera tures T1 and T2 are shown in the figure.2 The reading in the ideal voltmeter V is . 3. In the circuit (Figure) shown above.... the ratio of (R) RB/RA current in A and B (D) In parallel connection.. Bulbs are connected with a supply of d... ...1 Fig ... V... Column I Column II (A) Charging current at tiem t = 0 (P) 1 2 2 CV (B) Discharging current at t = 0 (Q) Maximum (C) While charging energy stored (R) Capacitor becomes short circuit (D) While charging energy dissipated as heat (S) Exponential law V T1 T2 I . [JEE 1985] The temperature T2 is greater than T1.... In a R-C circuit. The DC battery voltag e is 6 volts. Consider two identical cells each of emf E and internal resistance r connect ed to a load resistance R. . A variable resistor R is co nnected across the terminals of the battery. E . 6 V 3. Column I Column II (A) For maximum power transfer to load (P) 2 4 E r if cells are connected in series (B) For maximum power transfer to load if (Q) 2 2 E r cells are connected in parallel (C) For series combination of cells (R) eq E . only capacitor combinations. or capacitor-resistor combinations. each of 2 ohms resistance. One can have only resistor c ombinations. One capacitor and one resistor are connected in parallel. (B) Potential difference across the (Q) R = 0 terminals is maximum (C) Power delivered to the resistor is maximum (R) i E r .33 A. each of 1 microfarad capacitance. Column I Column II (A) Current in the circuit is maximum (P) R . (Ther e are also three resistors.0 V (D) 6.33 V (B) 1.in 51 9. 6 V (B) 0. The AC rms voltage is 120 volts (at 60 Hz.0 V (C) 2.33 A.ELECTRIC CURRENT www.) 1. The circuit is usually used fo r DC (direct current studies but can also be used for AC (alternating current) studies. There are also three capacitors.0 V 2. The ends of this co mbination are then connected t the 6-V DC battery. r PASSAGE TYPE QUESTIONS PASSAGE # I A physics instructor devises a simple electrical circuit setup in which one can easily insert various resistors and capacitors in series and parallel combinations. 2 eq r . respectively. Two of the 2-ohm resistors are connected in parallel and the 120-V AC voltage . All three resistors are connected in series and the combination is connected to the 6-volt DC battery. 3 V (C) 0. (D) Power delivered to the load is zero (S) r = R 10. what are the final current and voltage. E . r (D) For parallel connection of cells (S) 2 eq E .) The student inserts the resistors and /or capac itors as instructed and has available suitable ammeters and voltmeters for both DC and AC experiments. eq 2 r . A battery has an emf E and internal resistance r.physicsashok. across the 1 microfarad capacitor? (A) 0 A. 6 V (D) 6 A. What voltage drop occurs across each individual resistor as measured by the voltmeter ? (A) 0. 5 A (B) 2 A (C) 3 A (D) 6 A 5. What is the AC current through this series circuit? (A) 0. A.0 A . The 120 V.72 A (D) 40.045 A (B) 0. One capacitor and one resistor are connected is series. Two resistors are connected in parallel and the set of parallel resistors is then connected in series with the third resistor. If this series parallel combination is connected across the 6-vo lt battery.C .50 A (C) 0. What total DC current is drawn from the battery? (A) 0. What current will the AC ammeter measure if connected so i t measures only the current through one of the resistors? (A) 1 A (B) 2 A (C) 60 A (D) 120 A 4. voltage is then applied to this series RC circuit. 60 Hz.is applied to the ends of this parallel combination. in watts.9 W (B) 2.9 A (C) 1. and DC meters. interrupt .4°C PASSAGE # III In the laboratory. What power. What is the heat energy issipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no eat escapes).4°C (C) 34. The initial ELECTRIC CURRENT PASSAGE # II d h d h A set of experiments in the physics lab is designed to develop understanding of simple electrical circuit principles for direct current circuits. What is the current in the resistor? (A) 0.2°C (D) 56. What is the final temperature of the water at the end of the 10 minutes of heating? (A) 22. is consumed in the 8-ohm resistor in this case? 9.8 A (D) 3.9 W (B) 2. resistors. is consumed in the 8-ohm resistor? (D) 24 W (A) 0. A 12-volt battery is connected across a 4-ohm resistor and the heat energy dissi pated in the resistor in 10 temperature of the water is 20°C and the specific heat of the water is 4184 joule/ kg-°C. The student calcul ates expected current and voltage values using Ohm s law and Kirchoff s circuit rules and then checks the resu lts with the meters. What is the heat energy issipated in the resistor in 5 minutes? (A) 430 joules (B) 560 joules (C) 4300 joules (D) 5400 joules minutes is used to heat 2 kg of water (which is thermally insulated so that no eat escapes). 10.0 W (C) 4. and is directed to wire series and parallel combinations of resistors and batteries making measurements of the currents and voltage drops using the ammeters and voltmeters. A voltmeter has a very high resistance and should be connected in parallel to the circuit el ement whose voltage is being measured.0 W (C) 4. Two 4-ohm resistors are connected in series and this pair is connected in parall el with an 8-ohm resistor. How much power. The initial (A) 0.67 W (B) 2W (C) 12 W 8. 6. Resistorsof 4ohmsand8ohmsareconnectedinseries. The student is given a variety of batter ies. A 12 volt battery is connected across the ends of this parallel set.6 °C (B) 28. the voltmeter will affect the circuit.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor.Abatteryof 6voltsisconnectedacross theseries combination.8 A (B) 0.6 A 7. i n watts.4 W (D) 18 W A 6-votl battery is connected across a 2-ohm resistor.(A) 0. the voltage across a particular circuit element can be measur ed by a voltmeter. A student connects a 6-volt battery and a 12-volt battery in series and then con nects this combination across a 10-ohm resistor. Connected improperly. and in the third case across the battery (see figure 1). www. A light bulb is a resistor that. a curr ent will flow. After the current has flowed for a finite time. Vvoltmeter capacitor bulb battery Case 1 V voltmeter capacitor bulb battery Case 2 Vvoltmeter capacitor bulb battery Case 3 Voltage Time Fig. The bulb and the capacitor are connected in series w ith a battery and the voltmeter is placed in different position: in the first case across the capacito r in the second case across the light bulb.physicsashok. the capacitor will be ful ly charged.in 52 . 2 A capacitor consists of two conducting plates separated by a nonconducting mater ial. The ratio of the absolute amount of charge on one plate to the voltage across the plates is defin ed as the capacitance. When a battery is connected to a circuit containing a capacitor and a light bulb in series. causing positive charge to accumulate on one capacitor plate and an equal amount of negative char ge to accumulate on the other. becomes hot enough to emit energy in the form of light. An experiment is conducted in which a voltmeter is used to investigate voltages in a circuit containing a capacitor and a light bulb. The voltage for Case 1 as a function of time is shown in figure 2. when enough current flows through it. this is constant for a given capacitor. (Note : Assume that the battery has no internal resistance and that the resistance of the light bulb is constant). The voltmet er reading is recorded every 10 seconds.ing it and preventing current from flowing through the circuit element that it is meant to measure. The light bulb in the circuit stops shining. (D) It will increase. because as the capacitor plates fill with charge. because as the capacitor plates fill will charge. III. which will decrease the current and the voltage across the bulb. which of the following conditions would indica te that the capacitor was fully charged I. (C) It will increase. II. function of time (A) Time Voltage (B) Time Voltage (C) Time Voltage (D) Time Voltage 14. Which one of the following graphs could correctly represent the voltage across t he battery as a function of time during the experiment described in the passage Time (D) Time Voltage Voltage Voltage Voltage After the experiment described in the passage is completed. because as the capacitor plates fill with charge. and III 12. (B) It will remain the same. the battery is taken out of the circuit and the wires are reconnected.ELECTRIC CURRENT ELECTRIC CURRENT 11. Which of the following graphs. The voltage across the bulb equals the voltage across the battery (A) I Only (B) III Only (C) I and II only (D) I. which will create a greater voltage across the bulb. How will the voltage across the light bulb vary with time as the capacitor i s charging (A) It will decrease. the volt . represents the voltage acr oss the capacitor as a (A) (B) (C) Time Time 13. they wil l impede further charge. A voltmeter connected across the capacitor reads a constant voltage. II. In the circuit shows in figure 1. because as the capacitor plates fill with charge an d impede the current. the voltage output of the battery will increase to keep the current constant. they wil l induce further charge. in 53 . (D) the presence of resistors hinders the flow of charge. The total time taken for the capacitor to charge is m easured in both cases.age across the capacitor will decrease. (C) the more charge is absorbed by the resistors as the resistance of the circui t increases. 16. and then with the same two resistors in parallel. and found to be longer for the first case. which had been shining in the absence of the voltmeter immedi ately stop shining. It can be deduced that (A) When the resistance of the circuit si increased. The bulbs. The light bulb shown in figure 1 is replaced first with two identical resist or in series. (B) the presence of resistors affects the final voltage across the capacitor pla tes. and therefore the voltage across the light bulb will increase. 15. How might the circuit be modified in order to make the bulbs shine steadily again wi th their former brilliance voltmeter bulb V bulb battery (A) V (B) V (C) V (D) V www. withoutremovingthevoltmeter Inthediagram below. thus reducing the curr ent in the circuit. avoltmeterisconnectedinseriestoacircuit that includesabatter yandtwobulbsin series. the capacitance of the capa citor increases.physicsashok. 99 . S = 30 ohm and resistance of galvanometer is 4 ohm. Q = 2 ohm. 2. 8. 4. A B C D E F 50 V find the current passing through wire CD [in Ampere] 3. 1. 2. 3. 5. find the charge on 2 . It is required to send a current of 8 A through a circuit whose resistance is 5 . . Calculate the heat generated across 99 . .ELECTRIC CURRENT www.1 F 100 V. internal resistance by closing the switch. In the given circuit diaram. In the figure shown. What is the least number of cells which must be used for their purpose and how should they be connected? Emf of each cell is 2V and internal resistance is 0. Find how the voltage across the capacitor C varies with time t after capacitors C varies with time t after the shorting of the switch S at C S R R E the moment t = 0. In the circuit shown. 3. If an additional 6 . Find the value of the current through the galvanometer in that unbalanced condition of the bridge when P = 1 ohm.F 6V. 6.F 3 .5 . ( a < b) the length of each . resistance is connected across the battery find the value of x so that external power supplied by battery remains the same.5 . 0. [in Joule]. [in . In a Wheatstone s bridge a battery of 2 volt and internal resistance 2 ohm is u sed. the capacitor is charged by a battery of emf 100 V and 1.C ].in 54 LEVEL # 2 1. A homogeneous poorly conducting medium of resistivity . fills up the space be tween two thin coaxial ideally conducting cylinders. 2. 3.F each capacitor in steady state. is connected to a resist or of x ohms. resistance during the charging of 0. 2. s capacitor. The radii of the cylinders are equal to a and b. 1 . 7. 1. A 20 volt battery with an internal resistance of 6 . 4.physicsashok. Neglecting the edger effects. . Find the resistance of the medium between cylinders. .cylinder is . 11.F 10. 16. two resistors and a switch S. The meter deflects full scale for a current of 10 mA. Find resistance of the combination at a tempera ture of 50°C. and 300 . An ammeter and voltmeter are connected in series to a battery with an emf E = 6. In the circuit shown in the figure the current through 3 .physicsashok. .0 times whereas the reading of ammeter increases the some number of time. Initially S has been open for a longtime.0v.004 (°C) 1 respectively at 20°C. resistance is 2A. the reading of the latte r decreases . A circuit shown in figure has resistance R1 = 20. The meter behaves as an ammeter of three different ranges. E1 = 1. E3 = 3. what is the value E1 E2 E3 3. sec. 10. Find the volta meter reading after the connecting of the resistance. 1.5. . = 2. Calculate the resistance R1. range is 0 1 A between O and R.0 V. 14.0v... of E ? Internal resistance of each battery is 1 . R1 = 0. Find the current flowing through the resistance R in the circuit shown in figure.40 . At what value of the resistance Rx will the thermal power generated in it be practically independent of small variation of that resistance ? The A B R1 Rx R2 voltage between the points A and B is supposed to be constant in this case.. The internal resistance of the batteries are negligible.. and R2 = 30. .001 and 0. Find the current through the 10 ohm resistor at t = 75 . R2 = 0. When a certain resistance is connected in parallel with the voltmeter. 15. If E1 = 12V. The range is 0 10 A. In the given network switch S is closed at t = 0. The circuit shows a capacitor C two batteries. What is the effective temperature co-efficient of combination. . if the terminals R1 R2 R3 G O and P are taken. . R2 and R3. 13. 12. E2 = 14V.. 1. Two coils connected in series have resistance of 600 . The resistance of the galvanometer G in the circuit is 25. 1. 10.ELECTRIC CURRENT www. It is then closed for a long time by how much does the change on the capacitor change over this time period ? Assume C = 10 . . and tempera ture co-efficient of 0.in 55 9. 10V 20V S 2.20 . in 56 17.ELECTRIC CURRENT www. The internal resistance of the battery is negligible and the capacitance of the conductor C is 0.000 ohms resistance is u sed to measure the potential difference across the 400-ohms resistor. either by applying Kirchhoff s rules or otherwise. [JEE 1996] . as shown in figure.. is shown in the figure. OR Two resistors. (ii) What is the current that passes through the 2 ohm resistance nearest to the battery? 3. Calculate the steady current in the 2-ohm resistor shown in the circuit in t he figure. What will-be the reading in the voltmet er. what will be the reading in the ammeter? similarly. An infinite ladder network of resistances is constructed with a 1 ohm and 2 o hm resistances. Calculate the energy stored in the capacitor C (4.2 microfar ad.physicsashok. An ammeter of a 10 ohms resistance is used for this purpose. F) [JEE 1986] 2. Calculate the potential difference across the resistor of 400 ohm. An electrical circuit is shown in Figure. A part of circuit in a steady state along with the currents flowing in the br anches. and 800 ohms are connected in series with a 6-volt batt ery. [JEE 1987] The 6 volt battery between A and B has negligible internal resistance: (i) Show that the effective resistance between A and B is 2 ohms. It is desired to measure the current in the circuit. as will be measured by the voltmeter V os resistance 400 ohm. the values of resistances etc. 400 ohms. LEVEL # 3 1. If a voltmeter of 10. then calculate the leakage current at the instant t = 12 s.4 x 10 12 . R2 or R3. A leaky parallel plane capacitor is filled completely with a material having dielectric constant K = 5 and electrical conductivity . Find which of the above will give the most accurate reading and why? [JEE 2005] A B C G X R R = R1 or R2 or R3 . The capacitor is initially uncharged.ELECTRIC CURRENT www. Draw the circuit diagram to verify Ohm s Law with the help of amain resistance of 100. made to measure the unknown resistance X using the principle of Wheatstone bridge. [JEE 2004] 10. are connected by thick conducting strips. [JEE 2002] (A) Are there positive and negative terminals on the galvanometer? (B) Copy the figure in your answer book and show the battery and the galvanomete r (with jockey) connected at approxiate points. it is found that no deflection takes place in the galvanometer when the sliding jockey touches the wire at a distance of 60 cm from A. [JEE 1998] (A) Find the charge Q on the capacitor at time t. Find the emf (V) and internal resistance (r) of a single battery which is equivalent to a parallel combination of two batteries of emfs V1 and V2 and internal resistance r1 and r2 respectively. with emf V. and two galvanometers of resistances 106 . How a battery is to be connected so that the shown rheostat will behave like a potential divider? Also indicate the points out A B C R which output can be taken. 8.physicsashok. Obtain t he value of the resistance of X. A thin uniform wire AB of length 1m.. = 7. as shown in the figure. An unknown resistance is to be determined using resistances R1. Connections are to be X 12. 1 m 1. B and C. [JEE 1997] 6. (C) After appropriate connections are made. Th eir corresponding null points are A. [JEE 1997] 5. Show the correct posi tions of voltmeter and ammeter. In the circuit shown in Figure. (B) Find the current in AB at time t. : 7. Answer the following questions. with polarities as shown in figure. and 10 3 .. the battery is an ideal one.85 mC. If the charge on the plane at instant t = 0 is q = 8. and a source of varying emf. an unknown resistance X and a resistance of 12. The switch S is closed at time t = 0. [JEE 2003] 9. A battery and a galvanometer (with a sliding A B C D jockey connected to it) are also available.in 57 4. What is its limiting value as t . C A C B A C B D D A Que. D . D . 1 2 3 4 5 6 7 8 9 10 11 12 Ans. R and S.Q and R. D AC B ABD ABC C AC AB BC A Q. D . 41 42 43 44 45 46 47 48 49 50 Ans.P and Q.in 58 ANSWER KEY Reasoning Type Que. B . B B B D CD A A ABCD A C Q.Q. A . 51 52 53 54 55 56 57 58 59 60 Ans. F 5.R Que. 11 12 13 14 15 16 17 18 19 20 Ans. A . D . 61 62 63 64 65 Ans.P and S 9.S. C .R and S 8. R 2 3. B B AD D AD Fill in the Blanks / True False / Match Table 1.P and Q. C . R and S. 20 2. 21 22 23 24 25 26 27 28 29 30 Ans. B . C .Q and S. C A B A D A Passage Type .Q.physicsashok.Q.ELECTRIC CURRENT www.P. 31 32 33 34 35 36 37 38 39 40 Ans. T 6. 1 2 3 4 5 6 7 8 9 10 Ans. B . C C D A C A A C A D B A Level # 1 Q.Q.S.Q. B . D . C B D D A C C D D B Q. A A C B A B D A A BD Q. B C B AB C D A ACD A A Q.R and S. C . A .P and R. C . 1 2 3 4 5 6 7 8 9 10 Ans.P and R.Q 7. B A B B D A B A A A Q.P and S. A . P and R 10. 11 12 13 14 15 16 Ans. B . A . 0 4. 10. (b) 2 3 2 6 V V e t RC R R . 1 2 1 2 r r r . R1 = 0. 8. (ii) 1. 2t RC. decreases by 13. 2 A 3.0 V 12.9A or 4. . 0. 7.. 1 2 3 2 CV . 1. . E 1 e 2 . Rx = 1 2 1 2 R R R R .c 14.5 A 3. 1 9 Amp 6. r 5. V = . 0. I = . 1 E . . 2 .ELECTRIC CURRENT www.002 (°C) 1 16. . 10/3 V 4.95 V Level # 3 1. I = 20 mA 15. .199. . 954 .in 59 Level # 2 1. .. . a n b 2 . E = 7V 13.A 6. . . x = 7. 495 J 5. .physicsashok.025 .2275 . t RC . 72µC 11 2. . R3 = 2. 160 4. 0. R2 = 0. .96 x 10 3 A. . = 12.5275 . 2 3 2 3 2 3 0 3 R R R R R E R R E R . . . . 2. . 9. 17. = 2.5. . . 11. 1 2 2 1 1 2 V r V r r r . e.3 . . .88 x 10 4 J 2. . (a) . . . G J (c) 8. This is true for r1 = r2. X X X X . 8. Battery should be connected across A and B.V R 7. (a) No (b) A B C D X 12. Out put can be taken across the t erminals A and C or B and C. G1 G2 103 Ammeter Voltmeter 106 100 E 10. 9. So R2 given most accurate value. 9440025125 . PH NO. TIRUPATI.MAGNETIC FIELD DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. BIOT-SAVART S LAW According to this law. Applications of Biot-Savart s Law .in 1 THEORY OF MAGNETICEFFECTOFELECTRICCURRENT CONCEPT OF MAGNETIC FIELD The space around a current carrying conductor.l is given by 0 2 2 dB µ id sin Wb /m or tesla 4 r . inwhich itsmagnetic effect can be experienced. .One isBiot Savert s lawwhich gives themagnetic field due to an infinitesimallysmall current carrying wire at some point and the other isAmpere s law. then the rotationof the head oft he screwgives the direction ofmagnetic lines of force. l where µ0 is a constant and is called. permeabilityof free space. Inmagnetics. .MAGNETIC FIELD www. at the point P due to the small c urrent element of length d . × 10 7 Wb/A m Rules to Find the Direction of Magnetic Field (i) Right hand palm rule no. 1 : Ifwe spread our right hand in such a way that thumbis towards the directionof current and fingures are towards that point wherewe have to find the direction of field then the direction of field i Current carrying conductor P B will be perpendicular to the palm (ii) Maxwell s right handed screw rule : If a right handed cork screw is rotated s o that its tip moves in the direction offlowofcurrent throughthe conductor. and the direction perpendicular to the paper coming out is shown by . ..physicsashok. × P2 P1 Magnetic line of force i Current carrying conductor NOTE : By convention the direction of magnetic field . which is useful in calculating themagnetic field of a highly symmetric configuration carrying a steadycurrent..B perpendicular to paper going inwards is shown by . i r P µ0 = 4. is called magnetic field. the magnetic field dB. there are basicallytwomethods ofcalculatingmagnetic field at some p oint. d 1 2 p i (a) For aninfinitelylong straight wire. . .Let us consider fewapplications ofBiot-Savart s Law: (i) Magnetic field due to a straight thin conductor is 0 . . . 1 2 B µ i sin sin 4 d . .1 = .2 = 90º . . . i . . . . . . . x2 + R2 ..in 2 . 0 B µ Ni 2R . . . . 1 2 0 and 2 . 2 2 0 0 0 3 3 3 µ NiR µ 2Ni R µ 2M B 2x 4 x 4 x . . d 0 µ i B sin 0 sin 4 d 2 . .e. i. (b) For x > > R. . . .. . . .MAGNETIC FIELD www. . B d graph for aninfinitely long straight wire is a rectangular hyperbola as shown in figure. B (ii) Magnetic field on the axis ofa circular coil havingNturns is d . . . . . . B µ0i 2 d . (c) B 1 d . Here. . 0 µ i B 4 d . x2 . .physicsashok. . .R2 NOTE : This result was expected since. 2 0 2 2 3/ 2 B µ NiR 2 R x . . R = radius of the coil O x R P x = the distance of point P fromcentre and i = current in the coil (a)At the centre of the loop. . . . . . .. . . . . x = 0 . . Here. . M=magneticmoment of the loop M = NiA = Ni . . the magnetic field on the axis of dipole is 0 3 µ 2M .. . (b)Whenwire is semi-infinite. . The plane ofB is perpendicular to plane ofAand their centres coinci de.14 10 B 5 1. . . . . . . B1 = 8. . . . Find themagnetic field at the centre. Sol. .88 × 10 5 web/m2 . . Two circular coilsAand Bof radius 5 2 cmand 5 cmrespectively carry current 5Amp and 5 2 Amp respectively. . Themagnetic field due to first coil is 7 0 1 1 1 2 µ I 4 10 5 B 5 2r 2 10 2 .4p x Example 1. 7 1 2 20 3. . .441 10 . . each having equal radius R. B2 B . . .MAGNETIC FIELD www.71 . If current I is flowing through each ring then themagnitude ofthemagnetic field at the common centre is (A) 0 3 µ I 2R (B) zero (C) . . . Three rings.5weber /m2 B . . . .85 . . 2 2 B . 19. Themagnetic field due to ring inx yplane is 0 1 B µ I k 2R . (D) . 0 3 2 µ I 2R . . . . .. are placedmutuallyperpendicu lar to each other and each having its centre at the origin of co-ordinate system.. B1 .44 10 1.in 3 and 7 0 2 2 2 2 µ I 4 10 5 B 2r 2 5 10 2 . 9. .physicsashok.414 . (8. 78. themagnetic field due to ring iny z plane is 0 2 B µ I i 2R . . 0 2 1 µ I 2R .93 wb / m2 Example 2. Sol.14 10 B 4. .10. (4. and themagnetic field due to ring in x z plane is 0 3 B µ I j . . . . .44)2 .88)2 . 5 5 2 2 3. . . . the axis of a solenoid is .. . .. . . option (A) (c)Magnetic field 0 0 µ i µ i or B 2R 4 R . . . . . 0 . . . B µ i k i 2R . . . . .. B1 . . .2R .. . Hence. . .. .B2 . B= 2 O R Inwards i (iii) Field along 1 2 R x L O j is correct due to an arc of a circle at the centre is . B . ... . . 0 B 3µ I 2R . .B3 . .. . 0 A µ n 4 10 1000 .5m. i. . Total number ofturns in the solenoid. .52 × 10 3 T at its centre. n N 500 1000 m 1 0. . In a high tensionwire electric current runs fromeast to west.e. . Here. B = 2.52 10 i 2. (for L>>R) Example 3. . W N E S B i (a) (b) Example 4. Find themagnetic field at the centre of the polygon. .1 = 180º and . When the current flows fromeast to west. . Example 5. themagnetic field above thewire is towards north and belowthewire towards so uth. .5 mlong solenoid has 500 turns and has a flux density of 2. . l = 0. Sol. . . Find th e direction ofmagnetic field at point above and belowthewire.in 4 0 .A0. .physicsashok. µ0 = 4.N= 500 Therefore. number ofturns per unit lengthof the solenoid.5 .1 = 90º we get. 2 1 B µ Ni cos cos 2 . One side of the polygon is. × 10 7Hm 1. . . . × 10 7 Hm 1 Length of the solenoid.magnetic field lines are circular r ound it as shown infigure (a). µ0 = 4.And so. . 0 B 1 µ Ni 2 . .. (a) For a long solenoid (L >>R).2 = 0º.R which carries a current i .MAGNETIC FIELD www. then B=µ0ni or 3 7 0 B 2. Sol. .52 × 10 3 T. Arectangular polygon of n sides is formed by bending a wire of total len gth 2. l If i is the current throughthe solenoid. . Sol. Given.2 = 0º B = µ0Ni (b)At the ends of solenoid. Find the current in the solenoid. a 2 R n .. . . .. . 2 n 2 n . . .. . . . . d cot a / 2 . . . . . . .. . . . . . . . . . d a i . . . . . . . . . . . d a cot R cot 2 n n . . . Hence. . net magnetic field. . 1 2 3 4 B . . B . .. . . . . . . or 0 µ in B n tan 2sin 4 R n n . . ). .. . .. . B . k . B = (n) (magnetic field due to one side) 0 . .. .. . 3 a l (C) 0 µ I n 4 ( k ) 4 3 a . .. . . . .. . .. . . . . .MAGNETIC FIELD www.. B . . .physicsashok.. . Netmagntic field at point P is 30º 1 2 3 4 5 y x z a a P 30º (A) 0 µ I n 2 k 4.. or .. .. .. . . . . . . . Example 6. . . l (D) Zero Sol. 2 0 sin µ i n n B 2R cos / n . . 3 a l (B) 0 µ I n 4 k 4. 0 1 µ I 1 1 B 4 a cos30º 2 2 . . . . B . .. Here 0 . . . . . .. . ....in 5 All sides of the polygon produce the magnetic field at the centre in same direct ion (here . .Adjacent wires have current in opposite direction. . .. Infinite number of straight wires eachcarrying current I are equally placed as shown in the figure. . . . 1 B µ I sin 30º sin 30º 4 a . . . µ i B n sin sin 4 d . . .0 1 2µ I B 4 3 a . 0 2 1 µ I 2 2 R . . option (B) is correct. 0 2 1 µ I 4 2 R ..Magnetic field at point P. 4 3 a 2 3 4 .. 0 µ I B ln 4k 4 3 a ... 0 2 1 µ I 4 R . . Similarly. . . . .. . . . . . 0 2µ I 1 1 1 B 1 . ... . carrying current I. . . Hence. (B) . (C) . .. . and so on . . . .. . 0 2 1 µ I 4 R . . .. Example 7. ... Along straight wire. 0 2 2µ I B 8 3 a . 0 2µ I B ln 2k 4 3 a . distanceRfrompoint of bending is equal to : 45º I I P R (A) . (D) . is bent at itsmidpoint to formanangle of 45º. . . . B µ I 2 1 2 d . . 0 B µ I 1 1 2 d 2 2 . . .physicsashok.. . R 2 . B2 . . . . . Hence option (A) is correct Example 8. . . . . . . . Hence option (A) is correct Example 9.in 6 Sol. Find themagnetic field at P due to the arrangement shown (A) 0 µ i 1 1 2 d 2 . . . . . .. . . 90º 45º P 45º /4 d2 d20 2 µ I 2 1 B 2 d 2 . . 0 B µ I sin sin 2 4 d 4 2 2 . . . . 90º 45º d P (C) 0 µ i 2 d . . (B) 0 2µ i 2 d . . B . . 45º .. . 0 . . . . . . . B µ I sin 90º sin135º 4 .What is themagnitude ofmagnetic field at the centre O of loop of radius 2 mmade of uniformwirewhen a current of 1 amp enters in the loop and taken out of it by two longwires as shown in the figure. . . . (D) 0 µ i 1 1 2 d 2 . . . . ..MAGNETIC FIELD www. 0 B µ i 1 1 2 d 2 . . . . Sol. . 0 . . 0 . .. . . B1 . . . . .. . .. .. . B µ I 2 1 4 R . . . . .. .. . . O /4 d 2 B . . . . . . 1 2 I I 2 . .. .. . .. . . =magnetic field due to right wire 0 µ I sin sin ( k) 4 d 2 4 . . . . 1 B . . . . I1 I2 . . . . . . . .1 amp 90º 1 amp O Sol. 1 2 I I 2 . . . . . I 1 R . . . =magnetic field due to left wire is 0 1 µ I B sin sin k 4 d 2 4 . In circularwire. . d . 0 net C .in 7 . AMPERE S CIRCUITAL LAW It states that the line integral of B. 3 B . . (b)magnetic field has the samemagnitudeB at all places on the closed path. B1 . . 0 2 µ i B r 2 R . . 0 .. just as gauss s law is useful only for calculating the electric field s of highly symmetric charge distributions. . . each of raidus r . . . .. . B . . around any closed path or circuit is equ al to µ0 times the total current crossing the area bounded bythe closed path provided the electric field inside t he loop remains constant. . . . . B . .. . whose centers are separated by a distance d .. .l Its simplified formis Bl = µ0 inet This equation canbe used onlyunder following conditions : (a) at every point of the closed path B|| dI .e. B. .. Current of densityJ flows into the plane of the page along the shaded part of one conductor and an equal current flows out d Conductor . . Thus. 0 2 0 2 µ I µ (2 ) I 0 4 r 4 r(2 ) . .e.. (ii) Magnetic field of a solenoidwounded in the formof a helix is B = µ0Ni NOTE : Ampere s law is valid only for steady currents. . . Two long conductors are arranged as shown above to form overlapping cylinders..r (b) If r = R (i.. .. B3 .. it is useful only for calculating the magnetic fields of current configurations with high deg rees of symmetry. . µ i .physicsashok. . . . . i. Applications of Ampere s Circulatal Law (i) Magnetic field due to a longmetal rod of radiusR carrying a current i : (a) If r < R. Example 10. . B2 . . . . . at the surface) 0 B µ i 2 R .MAGNETIC FIELD www. Further more... =magnetic field due to circularwire 0 1 0 2 µ I µ I (2 ) 4 r 4 r . What are themagnitude and direction of themagnetic field at pointA? (A) (µ0/2.)4d2J/r. . . in the +y-direction (C) (µ0/2. inthe y-direction Sol. .. .. . . .)d2/r. . in the +y-direction (B) (µ0/2. . .. .y x Vacuum A of the plane of the page along the shaded portion of the other. . . 0 0 µ d µ d B j i j i 2 2 2 2 .dJ. B2 . .. B1 . . . . . as shown. . . . B . .)Jr2/d. in the y-direction (D) (µ0/2. . ..). . . . . 2 3 1 ..in 8 B µ0 jk d i µ0 jk d i 2 2 2 2 . .. . is along v r . . r v q p It is zero at . option (A) is correct. . = 90º (b) Direction ofB. MAGNETIC FIELD OF A MOVING POINT CHARGE The magnetic field vector B. r .MAGNETIC FIELD www. 0 2 B µ qvsin 4 r . .. . . at point P at position vector r . . 2 2 F . 0 1 . . . if q is positive and opposite to v . . relative to q1. (c) Suppose a charge q1 is moving with velocity and 1 v . 0 0 B µ jd j µ d j 2 2 .. . . . = 0º and 180º andmaximumat . . .. . along y-axis Hence....physicsashok. another charge q2 is moving with velocity 2 v . . . . then force on q2will be. .. . . . . . q v .. . B . . if q is negative. 0 0 B µ jd j µ jd j 4 4 . is found to be 0 . . . . . . . Note downthe following points regarding this equation. . . fromthe charge qmovingwith a velocityv . at position vector r .. . . (a)Magnitude ofBis. 3 B µ q v r 4 r . v r 4 r . This corresponds to Coulomb s electricalforce between the charges q1 and q2movingw ith velocities 1 v . Example 11. find the relation between electric andmagnetic fields. .µ q F v ... .. . . . . . For a charge q movingwith velocityv . . .. . . . . . . . .(ii) and 0 0 c 1 µ . . q q v v r 4 r . . . . . . .(i) 0 3 B µ qv r 4 r . 3 2 1 F µ . . or 2 0 0 c 1 µ . Sol. .. .. . . . .. . . . . . . . . respectivelyrelative to an observer at rest. 3 0 E q 4 r . 0 1 2 . and 2 v . . . . . . . . Case : (i) If . . we get B . . ( B i) . l Where B= intesityofmagnetic field i= current in the conductor l = lengthof the conductor and . (i) and (ii).= 1 then F = ilB(maximum) Therefore. B. This force is also called B F i Lorentz force. .. . l F . the forcewill be zero. j) .= 0º or sin .in 9 FromEqs. i k .. (A) P (B) Q (C) R (D) S Sol. . . .. i.Th e conductor is placed between two poles of two magnets. Then F = ilB × 0 = 0 Thus. . ...= 90º or sin . . i B(. . The direction ofmagnetic field on the conductor is along SR. forcewillbemaximumwhen the conductor carrying current is perpendicula r tomagnetic field.. E .when the current carrying conductor is parallel to t he field. But F .. .. .= 0. as shown.. Example 12. i .physicsashok. the conductor experiences a force in a directionperpendicular to both the direction ofmagnetic field and the direction of current flowing in the conductor.. . F . (ii) If . . = angle between the length of conductor and directionofmagnetic field. The S N Q R S P conductor willexperience a force inthe direction towards. l F . Astraight current carrying conductor is placed in such away that the current in the conductor flows in the direction out of the plane of the paper. FORCE ON A CURRENT CARRYING CONDUCTOR IN MAGNETIC FIELD When a current carrying conductor is placed in amagnetic field.MAGNETIC FIELD www. B .i B j . In vector form. 2 B v E c . i The direction of this force can be found out either by Fleming s left hand rule or by right handpalmrule. µ0. Themagnetic force is F = ilB sin .0 v . Hence option (B) is correct Example 13.The plane of the loop is perpendicular to themagnetic field.l l Hence. Find themagnitude of themagnetic force in both the cases (a) and (b). the direction ofmagnetic force on thewire is towards Q. Current i= 2Aflows inthe loop in the direction shown.0T.0m. The radius of t he loop is 1. i = 2A i = 2A × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × (a) (b) 1m . In the figure shown a semicircularwire loop is placed in uniformmagn etic fieldB = 1. net force on the loop in uniformfield should be zero. . × B 90º I × × × × × × × × × × × × × × × × × × × × × Sol. net force is not zero. . . B1 . Floop . Tesla. | Floop | . B1 . . . j) . FACD . . l 1 B .... | Floop | . .4 k . 2 IBR . . Example 15. l /2 R R (1) (2) 1 B . If the arc carries current I then find the force in the arc.. . 2i B sin .I B(. 1 2 B .Hence.B . . Themagnetic force on thewire is (in newton) y(m) A B .MAGNETIC FIELD www.. . Therefore.. . Refer figure (a) : It forms a closed loop and the current completes the loo p. 0 . × × × × × × . Aconducting wire bent in the formof a parabola y2 = 2x carries a cur rent i = 2Aas shown in figure. I B( 2R) j .physicsashok.in 10 Sol. Refer figure (b) : In this case although it forms a closed loop. thiswire is placed in a uniformmagnetic field B . Aswe know. but current doe s not complete the loop.. An arc of a circular loop of radius R is kept in the horizontal plane and a constantmagnetic field B is applied in the verticaldirection as shown in the figure. FACD . FAD . 2 FAD . I B j . FAD . .0 m) = (2) (2) (2) (1) sin 90º = 8 N Example 14. l (l = 2r = 2.magnetic force on a closed loop placed in uniformmagnetic field i s zero. 2| FAD | . .. × × C × × × × × × × × × × × × × × B A D . B2 . 16 i (B) 32 i (C) . . . Force on parabola + force on straight wireAB = 0 . Example 16..I j. The netmagnetic force on closed loop is zero. ..32 i (D) 16 i Sol.2 x(m) (A) .. . F .. Themagnetic force is F = I lB or F dq B dt .4k .32 i Hence option (C) is correct. The charge passed throughthe conductor is : (B is horizontal) (A) 1 Bmgh (B) 2gh Blh (C) gh Blh (D) m 2gh Bl Sol. A B . l . Suddenly a certain amount of charge is passed through it and it is found to jump to a heigh t h .4 j. The earth smagnetic induction isB. B. . Aconductor of length l and mass m is placed along the east-west line on a table. Force on the parabola = force on straight wireAB ..2. . . Fdt . l . B 20 10 0. l = 20 cm&h = 3 meters. l l . . 2gh . . l Hence option (D) is correct Example 17. . AU-shapedwire ofmassmand length l is immersedwith its two ends inmercury(see figure). . . . 3 2 q m 2gh 10 10 2 10 3 15 coulmb.MAGNETIC FIELD www.Make use ofthe fact that impulse of force equals .1Wb/m2. .physicsashok. m= 10 gm. Evaluate q for B = 0. . 2gh . l l But 02 = v0 2 2gh . . 0 v . assuming that the time of the current pulse is verysmall incomparisionwiththe time offlight. v . . thewirewill jump up.in 11 or dq Fdt B . .1 .idt . is sent through thewire. [g = 10 m/ s2] Sol. . I l B = F . × × × × × × × × × × × × × × × × × × × × × l m B i Calculate. that is. If a charge. Idt mv m 2gh B B . . . l or 0 q Fdt mv B B .which equalsmv. fromthe height h that thewire reaches. a current pulse q . Fdt = mv or I l Bdt = mv But 1 mv2 mgh 2 . m 2gh q B . the size of the Hg charge or current pulse.Thewire is ina homogeneous field ofmagnetic inductionB. The tension in newtons developed in the ring is : (A) 100 (B) 50 (C) 25 (D) 10 Sol. . T = IrB = 100 × 0. ..... . Tcos /2 /2 /2 Tcos /2 T T F + B or Ir. . .. .5 × 0.. We consider a small portion .2T carries a current I = 100A.l of the loop.Ametal ring of radius r = 0.. F 2T sin 2T T 2 2 . or I.Example 18.2 = 10 N Hence option (D) is correct .5mwith its plane normal to a uniformmagne tic fieldB ofinduction 0.. For equilibrium.lB = T.B=T. . e. Current or motion of positive charge Force F Field B v NOTE : To learn this rule.. Mother .1. central finger and thu mb of our left hand insuch away that these three are perpendicular to each other then. (iii) If . no force is exerted on a stationary charge. F ... When its velocityis along the z-axis.in 12 FORCE ON A MOVING CHARGE IN MAGNETIC FIELD : LORENTZ FORCE If a charged particlemoves in amagnetic field. if first forefinger is in the direction ofmagnetic field. Rules of Find the Direction of Force (i) Right hand palm rule : Ifwe stretch the right hand palmsuch that the fingers and the thumb are mutually perpendicular to each other and the fingers point the direction ofmagnetic field and the thumb points the direction ofmotionof positive charge.. (ii) If .N.v . Father . 3 j. when the charged particle is moving parpendicular to the field. remember the sequence of Father.physicsashok. then F = 0 i.e.e.Wh at is themagnetic field ? Sol. Field B Force F Current or motion of positive charge (ii) Fleming s left hand rule : Ifwe spread the forefinger. Magnetic field Central finger . . then thumbwill represent th e direction of force. .Whena protonhas a velocity v .2 i .. Mother. Current or direction of positive charge Example 19.106 m/ s .28 . second centralfinger is in the direction of current. × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × × F Cases : v (i) If v= 0.= sin 90º = 1 F = qvB × 1 = qvB i.. . . when the charge ismoving parallel to the field t hen no forcewillbe exerted bythe field. In vector form. then sin . Thumb .13 k . . then F = 0 i. it experiences a force F . q.10. the direction of forcewillbe along the out ward normalon the palm. it experiences a force along the x-axis.. child. B.= 0. then a force acting on it is given by F = qvB sin . the force exerted by the field will be maximum. Substituting proper values in.. . Child .= 90º.MAGNETIC FIELD www. in a m agnetic field.. Force Forefinger . . q. . ..10. . .6 . .28. We have. 3 j . 6 0 . .F .10.B j . . B.v . . 13 . . 1.10 . . . . k . 2 i . . 1. . 19 . 4 j. = 90º : Themagnetic force is F = Bqv sin 90º = Bqv. . MOTION OF CHARGED PARTICLE IN MAGNETIC FIELD The pathofchargedparticle inuniformmagnetic field dependsonangle betweenv . path is circle.MAGNETIC FIELD www. . . if r be the radius of the circle. . Here.physicsashok.Therefore. r mv qB . B . .T .Thismagnetic force is perpendicular to the velocity at every instant.6 × 2 × B0 or 0 B 1. . 1.in 13 .0. .. . . Therefore. . . The necessarycentripetal force is provided by the magnetic force hence. then + + q v v q or B B Fm= 0 mv2 qBv r . or T 2 m qB .When . Further. Hence.. time period of the circular pathwill be 2 mv 2 r qB 2 m T v v qB . This expression of r can bewritten in following different ways : mv p 2Km 2qVm r qB qB qB qB . .4 3.28 0. or the angular speed (.2 . . . . Case-II. following cases are possible : Case-I . P =momentumof particle K= KE of particle p2 or p 2Km 2m .. andB. . Hence. themagnetic field is. .When .28 = 1. is 0º or 180º : The magnetic force is F = Bqv sin 0º or sin 180º = 0. . ..) of the particle is 2 qB . path of the charged particle is a straight line (undeviated) when it ente rs parallel or antiparallel to magnetic field. . . . Frequencyof rotation is.T m . . qB m . . .1 f or f qB T 2 m . . . . Sol. ) gives a circular path and the compon ent parallelto field ( | | v ) gives a straight line path.67 × 10 27 kg]. Pitch is defined as the distance travelled alongmagnetic field in one complete cycle. . . 2 m qB . Substituting thevalues. . Fromthe relation R mv qB . Time period and frequency do not depend on velocity and so theyare given by T 2 m and f qB qB 2 m . and another perpendicular to B.4 × 106 m.MAGNETIC FIELD www. . p = v| |T or p .. There is onemore termassociatedwitha helicalpath. The radius ofthis helicalpath is. .. 27 7 8 19 6 1.e.physicsashok. . . r mv mvsin qB qB . . . p 2 mvcos qB .we have .. Example 20. . is other than 0º. q. 180º or 90º. m + B v v sin v cos The component perpendicular to field ( v. i. .in 14 IMPORTANT FEATURES If angle . . The resultant path is a helix as shown in figure.. Let the two components be | | v and v. .vcos . . mp = 1. . . Then v| | . that is pitch (p) of the helic alpath.. We have B mv qR . . . then velocity of charged particle can be res olved in two components one along B. What is the smallest value of B that can be set up at the equator to permit a proton of speed 107m/s to circulate around the earth ? [R = 6. . .67 10 10 . v cos. and v vsin . . uniform. horizontal and parallelto surface as shown. (B) m cosec . B m q (A) mcos qB .4 10 . qB (C) m cot . . .6 10 6. qB (D) none . .6 10 T 1. Example 21. . . . Find the time fromstart when block loses contact with the surface. Ablock ofmassm&charge q is released on a long smooth inclined planemagnetic field Bis constant.B 1. . . v mg cos qB .An electronhaving kinetic energyTismoving in a circular orbit ofradiu sRperpendicular to a uniform magnetic inductionB. But kinetic energy P2 T 2m .. the radiuswillbecome (A) 3R 2 (B) 3R 2 (C) 2R 9 (D) 4R 3 Sol. P . R mv P qB qB . Hence option (C) is correct. Example 23. 2mT .physicsashok. But v = u + at or v = 0 + gt sin . . . For losing the contects N = 0 .MAGNETIC FIELD www. qvB=mgcos. Example 22. option (C) is correct. . In space there is a . . 2m(2T) R´ q(3B) . 2mT R qB . . massm) has velocity v0 at origin in +x direction. . Hence. Acharged particle (charge q. t mcot qB . R´ 2 R 2 R 3 9 . . . . If kinetic energyis doubled andmagnetic induction triple d. gt sin mg cos qB . . .in 15 Sol. Themass of the particle is (A) qBd V (B) qB2d2 4V (C) qB2d2 8V (D) qBd 2V . 0 2mv y 2r qB . perpendicular to the B V S plane of the paper as shown in the figure.uniformmagnetic fieldB in z direction. O C + B .An ironwith electric charge q > 0 starts at rest froma source S and is accelerated through a potentialdifferenceV.. The particle is deflected by themagnetic field and emerges throughthe bottomhole at a distance d fromthe top hole. Sol. . Example 24. Find the y coordinate of particlewhenis cros ses y axis. It passes through a hole into a region of constant magnetic field B. The parth ofcharged particle is circularwhose radius is 0 mv r qB . Amass spectrometer is a devicewhich select particle of equal mass. 67 × 10 27 kg. If themaximumradius of the orbit of the particle is 0. (a)As in case ofmotion of a charged paricle in amagnetic field. or 2 2 d 2mV 4 qB .e. Acyclotron is operatingwith a flux density of 3Wb/m2. The ionwhich e nters the field is a proton havingmass 1. (b) the kinetic energy of the particle.. So. qB2d2 m 8V . . max max v qBr m . The speed of charged particle just before entering themagnetic field isV0. Hence. find : (a) themaximumvelocityof the proton. v qBr qB m . The radius of circular path inmagnetic field is 0 mv r qB .physicsashok.in 16 Sol. and (c) the period for a half cycle. 0 v 2qV m . 19 8 max 27 1.MAGNETIC FIELD www. Sol. option (C) is correct Example 25.6 10 3 0. 20 qV 1 mv 2 . So.5m . or 2 2 2 2 0 2 2 2 2 d m v m 2qV 4 q B q B m . . . or 0 d mv 2 qB . r mv i.5 v 1.43 10 m/ s . . T 1 (T) 1. . as 8 8 T 2 r 2 0. time for helf cycle. (b) .71 10 J 107 MeV 1.1. (c) In case of circularmotion. . . .67 10 27 1. .43 10 .5 2. . . . . . . . .43 108 2 2 2 . . . . .67 10 .71 10 K 1. . . . So. .09 10 8 s 2 . . . . . . .19 10 s v 1. . 11 11 19 1.6 10 . . i. .K 1 mv2 1 1.e. . . . . . Here 0 0 R mv qB .Aparticle of mass m . Sol. The region between x = 0 and x = Lis filledwith uniformsteadymagneti c field 0 B k . (A) . = 30º. Neglect the gravitythroughout the question (a) Find the value of L iftheparticle emerges fromthe regionofmagnetic fieldwithits finalvelocityat anangle 30º to its initialvelocity. × + + + + + + + + + + + + + + + + + + + + + + + P v0 X Y q A C x=0 x=L v0 L R B = B0 k or 0 0 1 qB L 2 mv .physicsashok.1 L. . if themagnetic field now expands upto 2.MAGNETIC FIELD www. 0 0 L mv 2qB . (b) Find the finalvelocity of the particle and the time spent by it in themagnet ic field. (b) In part (i) sin 30º L . 0 0 sin 30º L mv qB . positive charge q and velocity 0 v i travels along x-axis and enters the region of themagnetic field. .in 17 Example 26. sin L R . . . Themagnetic field at the centre P due to current inwireNMis 0 . deviation of the particle is . B i . Awire loop carrying a current I is placed in the x y plane as shown in fig. v . . Sol.v i . . ..1R > R Nowwhen L´ = 2. f 0 B v . find its instantaneous acceleration. L´ > R Therefore. or 1 L 2 R . 1 B µ I sin 60º sin 60º 4 r .1R 2 . . . and AB 0 t T m 2 qB . is applied find the force and the torque acting on the loop due to this field.1 L or 2 2.R . . V P +Q 120º M I N a y O x (b) If an external unfiormmagnetic induction field B . or L = R/2 × × × × × × × × × × × × B R v0 v0 L´ =2. = 180º is as shown. alongNP (see figure). (a) If a particlewith charge +Qandmass m is placed at the centre P and given a velocityv . Example 27. . .MAGNETIC FIELD www. . . MN . . 30º S r P a M N a . .. directed towards the reader perpendicular to the plane of paper . . . . . v P Q 120º M N a y 60º 30º 30º Q x V sin60º V 0 V cos60º 1 B µ 2I 3 4 a . . .in 18 0 1 µ B I 3 3 4 a / 2 2 2 . . r a 2 . directed awayfromthe reader perpendicular to the plane of paper. . . . . . . . . . 3 a Themagnetic field at the centre P due to current in arcMNis 0 0 0 2 µ 2 I µ 2 I 2 / 3 µ 2 I B 2 a 2 4 a 2 4 3a . cos30º MS a . .physicsashok. . . MS 3 a 2 . sin 30º r a . . . . . . . . . . . y x P B v M R Q Q N 120º The direction of acceleration is given by the vecotr product v .68) 4 a 3 4 a . . . . . (directed away fromthe reader perpendicular to the plane of paper) The force acting on the charged particleQwhen it has a velocity v and is instant aneously at the centre is F = QvB sin . . . . . . 0 0 µ 2I µ 2I B 3 (. . B .68) M m m 4 a .The netmagnetic field 0 0 1 2 B B B µ 2 3I µ 2 I 4 a 4 3a . . . . or byapplying Fleming s left hand rule . . . . . . . . . .. = QvB sin 90º = QvB The acceleration produced 0 F QvB Qv µ 2I A (0.. . 0 A 0.11µ IQv ma . . . . . .. . MAGNETIC FIELD www.in 19 . . . . . M . . . Find theminimumvalue ofBneeded tomake the electrons hit S. ... .e. Example 28. . . B . .. . .. .RPQ = 30º i.1m.. a2 . . . The force acting on the loop is zero. . . . . . . . .RPN = 90º and . .MPR = 120º 90º = 30º Since.. . .. Kinetic energyof electron. parallel toGS exists in the region outside 60º B S G X the electron gun. . .physicsashok. . . speed of electron. . . Sol. .Auniformmagnetic field B. . . . The electrons are requeired to hit the spot Swhere GS = 0. . .614 BIa2J 3 4 . A a2 3 k 3 4 . where M= IA where A= (area of PMQNP) (area of trangle PMN) A 1 . . . .. . 2 A a 1 3a a a2 3 3 2 2 3 4 .. v 2K m . Ia2 3 k iB 3 4 . . . The torque acting on the loop in themagnetic field is giving. . 60º B S .MPN = 120º . K 1 mv2 2keV 2 . . . . An electron gunGemits electrons of energy 2 keVtravelling in the positiveX-direction. 1 MN PS 3 2 .. BIa2 3 j 0. .MPQ = 60º . . . and the lineGS make an angle of 60º with the x-axis as shown in figure. .wthe acceleration vectormakes an angle of 30º with the negative x-axis. . . ..G v 16 31 v 2 2 1.65 × 107 m/s Since the velocity (v) . . v = 2. 2... . the pathwill be a helix. . . ... the particlewillhit S if GS = nP Here n = 1. of the electronmakes an angle of .1 10 .....6 10 m/ s 9.. . . So. 3. . .. = 60º with themagnetic field B. . .. p pitch of helix 2 m v cos qB . . 2 k . Substituting thevalues.2k.65 10 ) 1 B 2 tesla (1.e(2 k ) . it experiences a force 2 F . 1 . (. .e(v ) . 3 F . F1 .. 2e( i . .k ) . m/s at the same point. . If the electron ismoving with a velocity 2 V . .MAGNETIC FIELD www. j . . . GS p 2 m vcos qB .in 20 But for B to beminimum. 0 . . 2 i . .e(v3 . The force the electronwould experience if it weremovingwith a velocity 3 V . B) . . . .1 10 )(2. B ..e(2 j) . An electronmovingwith a velocity 1 V ... B .6 10 )(0. min B B 2 mv cos q(GS) . (.. . . eB( i) ... .we have 31 7 min 19 (2 )(9.N (C) . . j .physicsashok.2 j . . . B . .1) . n = 1 Hence. m/s at the same point is (A) zero (B) 2k. 2 i ..e(2 i ) . . B) . . . . . m/s at a point in a magnetic field experiences a force 1 F .Bk ) . .73 × 10 3 tesla Example 29. .2 j . or Bmin = 4. 2eBi eB = 1 F3 . .. eB = 1 2 F .Bk ) . . . . qv1 .2 jN . .2 i N . . . . .. (. 2 j .N (D) informationis insufficient Sol. q(v . qE . Combined forceF ... . then. . . is known as lorentz force (i) E || B||v . .... .. B) . F ..Net force experienced byit is given byfollowing equation. E B v In above situation particle passes underviated but its velocitywillchange due to electric field andmagnetic force on it is zero.. in an electric field E.Hence option (A) is correct Charged particle in uniform & . .. andmagnetic field B. E B When a charged particlemoveswith velocityv . . . ... . . m. points inthe x-direction. . 2 . B . . . . .physicsashok. Byz . Bzy . R c y v0 sin z Cycloid motion Suppose that B. and uniform.. . . . . . Their generalsolution is 1 2 3 2 1 4 y(t) C cos t C sin t (E / B)t C z(t) C cos t C sin t C ..yy . . . .. ma . z(t) E (1 cos t) B . . . B.E . y(t) E ( t sin t). . . . .. . . . . . qB m . . zz .in 21 (ii) E || B . . . . . particle is releasedwith velocity v0 at an angle .. T qB qB . z E y B . . y z.Ez . 0 r v cos t 1 qE t i Rsin t j R R cos t k 2 m . . . . . .. x . . q. . . . . v . . and E. v0 +q E.. ..... . . . . . B v0 x E. . . B v0 cos +q v0 sin y z 0 mv sin 2 m r . q. . . .. . .. .. ... inthe z-direction. . . .MAGNETIC FIELD www. E x B z a b c y 0 F . . . (y R. . . v. . R E B . . . Notice that the overallmotion is not in the direction ofE.. rolling down th e y axis at speed. but perpendicular to it. . ..t)2 + (z R)2 = R2 v R E B . .The curve generated in thisway is called a cycloid. The particlemoves as through it were a spot on the rimof awheel. . E i . (B) 0 0 16 B E . qE . . . .3)2 . y j) .. 0 0 0 x 25m 25 2qE 2E . E 0 0 0 0 0 W . . q m . . startsmoving fromthe or igin under the action of an electric field 0 E . Its velocity at (x0. .MAGNETIC FIELD www. (x i . . (C) 0 25 2.magnetic force is always per pendicular to instantaneous displacement. W T 1 mv2 1 mu2 2 2 .S .physicsashok. 0) is (4 i . Theworkdone bymagnetic force isWB = 0. y0. The value of x0 is : (A) 0 0 13 E 2 B . or 0 0 qE x 25m 2 . because. . . 42 . . . . 5m/ s According to work energytheorm. qE x . B k . The speed of particle at (x0. y0) is v . Sol. .E (D) 0 5 2B . . Aparticle of specific charge (charge/mass) . (. 3 j) . .in 22 Example 30. qE i . andmagnetic field 0 B .. . or 2 E B W W 1 m5 0 2 . Hence.1/2 [B/. . .Uniformelectricmagnetic fields exist inthe regionalong the+ydirection . . also. r mu qB . . y vt qE t2 2m . Aparticle of specific charge (q/m) is projected fromthe origin of co ordinates with initial velocity [u i .E] is an integer Sol. mu2 quB r . and y = 0 0 vt qE t2 2m . . option (C) is correct. y a qE m . x = 0. T 2 m qB . ofmagnitude EandB.E] in an integer (D) [uB/. v j] . .E] is an integer (B) (u2 + v2. . . Example 31.E] is an integer (C) [vB/. The particlewilldefinitelyreturn to theorigin once if (A) [vB/2.. For origin. in 23 . . Total force on length l of the short conductor is 0 1 2 F f µ 2i i 4 r . . Sol. r NOTE : The wires attract each other if currents in the wires are flowing in the same direction and they repel each other if the currents are in opposite directions.physicsashok. l l 7 10 2 2 10 2 5 F 8 10 N . . Aconductor of length 2 mcarrying current of 2Ais held parallel to an infinitely long conductor carrying current of 10Aat a distance of 100mm. . n VB E . . l . option (C) is correct FORCE BETWEEN PARALLEL CURRENT CARRYING WIRES Consider two longwires 1 and 2 kept parallel to each other at a distance r and carrying currents i1 and i2 respectively in the same direction. . Example 32. t 2mV qE . Find the force on small conductor . . For returning at the origin. F i1 l i2 1 2 The same force acts onwire 1 due to wire 2. We knowthat force per unit length of short conductor due to long conductor is given by 0 1 2 f µ 2i i 4 r .MAGNETIC FIELD www. Hence. n 2mVqB qE(2 m) . nT 2mV qE . 2n m 2mV qB qE . The force per unit length of thewire 2 due towire 1 is : 0 1 2 F µ i i 2 r . . 1 . each extending fromz = . Example 33. Astraight segment OC (of lengthL) of a circuit carrying a current i is placed along theX-axis as shown infigure. . . Force is attractive if the direction of current is same in both the parallel con ductors and is repulsive if the direction of current is opposite intwo parallel conductors. . . . . are fixed at y = a and y = + a respectively.. . If the × × A B Y X O i C Z wiresAandB each carry a current i into the plane of the paper. .0.What will be force ofOC if the current in the wire Bis reversed ? . to + . obtain the expression for the force acting on the segment OC. as shown in the figure.Two infinitely long straight wiresAand B. Total force onthewirewill be x L 2 L 0 2 2 x 0 0 F dF µ i xdx x a .in 24 Sol. 0 A B | B | | B | | B| µ i 2 AP . .. .. . . .x µ i x B .. (a) Let us assume a segment of length dx at a point P.. force on the element will be (F = ilB) 0 2 2 µ i x dF i dx a x . . . . . . a distance x fromthe centre shown in figure. .. . . . . . net magnetic field at Pwillbe along negative y= axis as shown and 0 net µ i x B 2 | B| cos 2 2 AP AP . . . . .. Therefore. .. . .MAGNETIC FIELD www. . (AP) (a x ) .. (in negative z-direction) . . × × A B y x O i y´ C I a a I dx x L Bnet BB BA is the positive Z-direction and × is the negative Z-direction and Magnetic field at P due to current in wiresAand B will be in the directions perp endicular to AP and BP respectivelyas shown. .physicsashok. .. . . .. . 0 0 net 2 2 2 µ i. .. Therefore. . . MBsin . . . . . whereM= NIA M B I Work done in rotating loop in uniformfield from.. . . . . A× B Y X i Bnet C BA BB P (b) If current inwire Bis reversed. (innegative z-axis) Hence 2 2 2 0 2 µ i L a F ln k 2 a . . . . . is not necessarily parallel to BP.e. . .. . | . .1 to .. . 2 2 2 0 2 µ i L a F ln 2 a . thenmagnetic fields due to AandBwillbe in the directions shown in figure. i. ..2 W=MB (cos . . net magnetic field net B . |. force onwireOCwill be zero. . B . willbe along positive x-axis and since current is also along positive x-axis. . CURRENT LOOP IN UNIFORM MAGNETIC FIELD . .2) . . Note : A B .. .1 cos . .. . M . .. . . . . . The cone is uniformlyrotated about its axis at angular velocity. . . Qcharge is uniformly distributed over the same surface of a right ci rcular cone of semi-vertical angle . . Figure shown a square current carrying loopABCDof side 10 cmand . . . . . . The convectioncurrent is I q qv 2 2 r . Angularmomentum L I 3 mR2 10 . P 3Q R2 20 .r (D) qv. I = Themoment of gnertia 3 MR2 10 .in 25 Example 34. Sol. . Themagneticmoment is M I r2 qv r2 2 r . . .MAGNETIC FIELD www. Example 36. . M qvr 2 . . But tan R h . . . and height h . . . (B) qvr 2 (C) qv. 3QR2 3Qh2 tan2 P 20 20 .. . . P Q 3 mR2 2m10 . . .r2 Sol. Hence option (B) is correct Example 35. Butmagneticmoment P QL 2m . Themagneticmoment of a circular orbit of radius r carrying a charge q and rotatingwith velocity v is given by (A) qvr 2. R = h tan. Calculated associatedmagnetic dipolemoment. . .physicsashok. Themagnitude ofmagneticmoment is M = Il2 = 10 × (10 × 10 2)2Am2 = 10 × 10 2 = 0.m2 y x z 30º D C B A i = 10 (C) (0. 3 k .A.m2 (B) (0.A.m2 Sol. 3 i .1 Am2 . Themagneticmoment M.A.05). k . k . j.. i . k .05).m2 (D) .05). of the loop is (A) (0.. i .A.current i= 10A. 1 i 3 j 2 ... Sol.Msin 60º j . . mgr . 2I (A) BnaI sin 60º (B) 8 BnaI cos 60º (C) 4 BnaI sin 60º (D) none Sol. . Here M= 2n(2I) (2a) M= 8 nIa .r2B . M . . . . . . 3 j. .. Example 39. . .in 26 The normal on the loop is in x z plane. (0. (refer figure).r B ... Example 37.. mgr .. B j. . B i .axis.5mis placed on a smooth horizontal plane.physicsashok. (b) Since. an area 2a and carries a current 2 I.. Mcos60º i . . x y B . Bz is parallel to dipolemoment .r B . = MB cos 60º .. . B i .Ahorizontalmagnetic field .At this particular place..= 8 nIa cos 60º Hence option (B) is correct Example 38.. x I mg rB . the earth smagnetic field is x y B . The torque onthe coil due tomagnetic force is 60º B coil 2n. M . The plane of the coil is at 60º to a horizontaluniformmagnetic fie ld of flux densityB.. M. ... .i .The ring carries a current i= 4A. =MB sin(90º 60º) .MAGNETIC FIELD www. Howlargemust I be before one edge of the loopwill lift fromtable ? (b) Repeat if. B k . I. 2a. . (a)Arigid circular loop of radius r &mass m lies in the xy plane on a flat t ble and has a current I flowing in it. . ..Am2 . B . .. 2 x I. . It makes 60º angle with x . M M i 3 M j 2 2 . M 0. B . Aconducting ring ofmass 2 kg and radius 0. . mgr or 2 2 2 x y I.05). (a)Torque due to magnetic force should be greater than torque due to weight .. . Arectangular coil PQhas 2n turns.. B = 10T is switched on at time t = 0 as shown in figure. . rad/s2 Sol. if a coilor a closed ofany shape is placed in uniform electric f ield. rad/s2 (B) 20. Aswe know. rad/s2 (D) 15. The initial angular B acceleration of the ringwillbe (A) 40. rad/s2 (C) 5.magnetic force on the coil or the loop is zero. The value of B if the sphere is in B equilibriumis (A) mg cos IR . . the centre ofmass of the coil remains in rest.physicsashok. Hence option (A) is correct Example 40. . Current in the coil is I . Asquare current carrying loopmade of thinwire and having a massm= 10 g can rotatewithout frictionwith respect to the vertical axisOO1. Sol. the torque on the coil is . . (D) mg sin IR . . . .R2B sin. . The torque due tomagnetic force is . The loop is placed i n a homogeneousmagnetic . .IR (C) mg tan IR . This torque balances the torque due to friction about centre ofmass. . . B mg IR .B = PB sin(180 . Hence option (B) is correct Example 41. . . = . net torque should be zero. . mg sin . passing throught he centre of the loop at right angles to two opposite sides of the loop.in 27 So. For equilibrium. .IRB sin . .MAGNETIC FIELD www. Themagneticmoment of the loop is perpendicualr to the plane. or mg sin .) 180º P B . . = I. FR = I. In the figure shown a coilof single turn iswound on a sphere of radi us R andmass m . 8B 8 10 40 rad / s2 m 2 . . (B) mg . = f Also.r2B or 2 mr 2 4 r B 2 . The plane of the coil is parallel to the plane and lies in the equatori al plane of the sphere. . .R2Bsin. = IAB = 4.B = PB sin. Sol. . . = PB sin.Acurrent I =2Ais flowing in the loop. . = IAB. or 0 IAB I .fieldwith an induction B = 10 1 T directed at right angles to the plane O+ B I O1 of the drawing. or I0. . . . . . Find the period of smallosc illations that the loop performs about its position of stable equilibrium. . . 2 0 IAB I . = IAB. . . . . . .. or . or 0 IAB I . . the frame is at rest in the position shown in the figurewith its sides parallel toXandYaxe s. Here 2 2 2 2 0 0 0 0 0 I m m m m 2 2 12 12 . PQRS is a rigid squarewire frame carrying a steady current I0. . .57 sec.At time t =0. T = 0. l l But 3 0 m m 2. Each side of the frame is of massMand length L. (a) What is the torque . is directed at an angle of 45º to the X-axis inX Yplane.MAGNETIC FIELD www.Given : the moment of inertia of the frame about an axis throughits centre perpendicular to its plane is 4ML2 3 .. T 2 I0 IAB .. .physicsashok. with its centre at the originO. . . I = 2 amp After putting the value. . l l l l 2 2 0 0 0 m m I 2 6 . 2 .t is so short that any variation in Y 0 X P Q S R I the torque during this intervalmay be neglected). . . . .t. l2 =A= area of loop.in 28 or 0 2 IAB T I . and the axis about which this rotation occurs (. .. Sol. Magneticmoment of the loop. . ... . . about Oacting on the frame due to the magnetic field ? (b) Find the angle bywhich the frame rotates under the action of this torque in a short intervalof time . . Example 42. .5 10 kg 4 .Auniformconstantmagnetic fieldB. . . i directiono r parallel toQS. . . . . (a)Torque acting on the loop. B (Bcos 45º ) i (Bsin 45º ) j B ( i j) 2 . . . . . . . I L B . |. . (b)Axis of rotationcoincideswith the torque and since torque is in j.. .. . 2 0 I L B( j i) 2 .... the loopwill rotate about an axis passing throughQand S as shown along-side : Angular acceleration. .. . . Magnetic Field. (iA)k . . 2 0 M B (I L k ) B ( i j) 2 . or 2 0 | . . . .0 M . | | I .Therefore. ... . (I L )k . . non-magneticmetal.t is 1 .physicsashok.The lower end of the coil is connected to a hair-spring s of phosphor bronze having onlya fewturns. N Core S P S Q R s T T1 T2 m Moving coil galvanometer In order to eliminate air-disturbance. MOVING COIL OR SUSPENDED COIL OR D´ ARSONVAL TYPE GALVANOMETER Principle : When a current-carrying coil is placed in magnetic field. ( t)2 2 . IQS + IPR = IZZ (Fromtheoremof perpendicular axis) Y X P Q S R But IQS = IPR .Asoft ironcylinder known as the core is placed symmetricallywithin the coil and detached fromit. Construction : It consists ofa narrowrectangular coilPQRS consisting of a large number of turns of fine insulated copperwirewound over a framemade of light.MAGNETIC FIELD www. . . . . . The coil is suspended b etween the two cylindrical pole-pieces (Nand S of a strong permanent horse-shoemagnet) by a thin flat phosp hor bronze strip. . ( t) 4 M . the upper end ofwhichis connected to amovable torsion head T. 2 QS ZZ 2I I 4ML 3 . thewhole arrangement is enclosed in a bra ss case having a glasswindow . .in 29 Where I =moment of inertia of loop about QS. . . 2 QS I 2ML 3 . . . . 2 0 0 2 | | I L B 3 I b I 2 ML 2 M 3 . it experie nces a torque. . Angle bywhich the frame rotates in time . or 0 2 3 I B . . the angle between Rad ial magnetic field the plane of the coil and themagnetic field is zero in all the orientations of t he coil. Themagnetic lines of forcewithin the air gap are along the radii. the plane of the coil remains N S always parallel to the direction ofthemagnetic field i. .The torsion head T is co nnected to a binding terminal T1.Aplanemirror or a concavemirror of larger radi us of curvature is rigidly attached to the phosphor bronze strip. This helps to measure the deflection of t he coil by lamp and scale arrangement. Levelling screws are provided at the base.on the front.e. The lower end of the spring s is connected to a binding terminalT2. So. the phosphor-bronze strip acts as one current lead to the coil. Radialmagnetic field : The magnetic field in the small air gap between the cylinderical pole-pieces is radial.. On account of this. F = NBIl Applying Fleming s left hand rule. Now. . or I k NBA . or I . . or . be the angular twist . I .. So.in 30 Theory : Let I = current flowing through the coil.e.we find that this force is normal to the plane o f the coiland directed inwards. I . Force on RS. they do not experience any force..] b F P S F When the coildeflects. So. .The sides PQandRS remain perpendicualr to the direction of themagnetic field. These sides experience forces perpendicular to the plane of the coil. . awayfromthe reader. On account ofelasticit y. . then Moment of restoring couple = k. the sides SP and QR remain parallel to the direction of the mag netic field. i. . The forces onthe sides PQandRS always remain perpendicular to the plane ofthe coil. ...MAGNETIC FIELD www.. is the galvanometer constant. . where K k NBA . So. NBIA= k. where k is the restoring couple per unit angular twist.. So. K.we find that this force is normal to the plane o f the coiland directed outwards. b = breadth of the coil N= number of turns in the coil. This couple tends to deflect the coil and is known as deflecting couple. It is also known as torsio nal constant.e. b F F P S Q R l Current-carrying loop in magnetic field Force on PQ. . Moment of deflecting couple =NBIl × b =NBIA [The field is radial. i. and (iii) act at different points. If . .This couple is proportional to the twist. the perpendicular distance between the forces is always equal to b as in fig. the two forces constitute a couple. For equilibriumof the coil. therefore. a restoring couple is set up in the fibre. B=magnetic field induction l = lengthof the coil. the suspension fibre gets twisted. the plane ofthe coilremains parallel to th emagnetic field inall the orientations of the coil.. towards the reader. A(= l × b) = area of the coil Since the field is radial. F = NBIl Applying Fleming s left hand rule.physicsashok. (ii) opposite in dir ection. The forces onthe sides PQandRS are : (i) equal inmagnitude. The scale is calibrated to give direct values of current. . the deflection of the coilis proportional to the current flowing through the coil.So. This explains as towhywe can use a linear scale in a galvanometer. Advantages : (i) The galvanometer can bemade extremely sensitive. I . . So. therefore. V RI . . the galvanometer will not respond toweak electric currents. li near scale canbe used. VOLTAGE SENSITIVITY OF A GALVANOMETER It is defined as the deflection of the meter per unit voltage.in 31 CURRENT SENSITIVITY OF A GALVANOMETER Agalvanometer is said to be sensitive if it gives a large deflection for a small current.This typeofgalvamometer is called aperiod ic or dead beat galvanometer. or NBA V kR .e. NBIA= k.. therefore. k can be decreased by using phosphor-b ronze for suspension. . So. The current sensitivity of a meter is the deflection of the meter per unit curre nt. T1 T2 N S 30 20 10 0 10 20 30 . But these require very careful handling. So. . for general use in the l aboratory and for those experimentswhose sensitivityis not required. therefore. i. . pointer type galvanometers are used .e. (v) The lamp and scale arrangement used to measure the deflection of the coilmak es the galvanometer very sensitive. the galvanometer coilcomes to rest quickly. The galvanometer can bemade ballistic bywinding the coil on a non-conducting fra me of ivory or ebonite. [. (ii) Since themagnetic field B is very high.] The sensitiveness can be increased by increasing N. Now. Theycanmeasure c urrents of the order of 10 9 ampere.physicsashok.. POINTER TYPE OR WESTON OR PIVOTED MOVING COIL GALVANOMETER The suspended typemoving coil galvanometers are very sensitive. It is given by NBA I k . the externalmagnetic fie lds cannot appreciably after the deflection of the coil.MAGNETIC FIELD www. V . (iii) Since the deflection of the coil is proportional to current. B can be increased by using a strongmagnet. the galvanometer can be used in any position. i. k can be further reduced by using quartz suspension fibre. . damping is produced by eddy currents. But N andA cannot be increasedmuch because thiswill increase the length and consequentlythe resistance of the coil. (iv) Since the coil is wound over metallic frame. In that case.Aand B and decreasing the val ue of k . NOTE : Numerical Examples based on Moving Coil Galvanometer Formulae used : 1.Alight aluminiumpointer is attached to themoving coil. Current sensitivity. there is a m agnetic S-pole near the geographicalnorth. EARTH S MAGNETISM Afreelysuspendedmagnet always points inthe north-south direction evenin the abse nce ofanyothermagnet. = NBIA 2.MAGNETIC FIELD www. they are spread over an area. the coilis pivoted between two ball-bearings. = NBIA 3.The positions of the earth smagnetic poles are not well defined on the globe. . k. k in N m rad 1. Voltage sensitivity. Magnetic equator : The great circlewhose place is perpendicular to the earth smagn etic axis is called earth s magnetic equator. the following three ele mentsmust be known : . Magnetic Elements To have a complet knowledge of earth smagnetismat a place. Geographicalmeridian : The line joining the geographical north and south poles i s called the geographic axis and a vertical plane passing through it is called the geographicalmeridian.physicsashok.The shape ofe arth smagnetic field resembles that of a barmagnet of length one-fifth of earth s diameter buried at its centre. Thus. a = NBA I k 4.in 32 Inthis typeof galvanometer. w hile the north pole of earth smagnet is towards earth s south pole (geographical south). S N Magnetic axis Magnetic N-pole Geographic S-pole Magnetic equator Equator Magnetic S-pole Geographic N-pole Geographic axis The south pole of earth smagnet is towards earth s north pole (eographical north). R in ohm. This suggests that the earth itself behaves as amagnet which causes a freely sus pendedmagnet (ormagnetic needle) to point always in a particular direction : northand south. Magnetic meridian : The line joining the earth smagnetic poles is called themagnet ic axis and a vertical plane passing throughit is called themagneticmeridian. The controlling couple is providedwith the help of a spring. A in m2. a = NBA V kR Units used : B in tesla. Geographical equator : The great circlewhose plane is perpendicular to geographi cal axis is called geographical equator. and amagneticN-pole near the geographical south. (i) Angle of declination : The angle between the magnetic meridian and geographi calmeridian at a place is called the angle of declination (or simplythe declination) at that place. .(i)Angle of declination (ii)Angle ofdip or inclination (iii)Horizontal component of earth s field. H . (iii) Horizontal component of earth s field : The direction of earth s field at them agnetic poles is normal to the earth s surface (i. N S C i V D A B H In fig. in horizontaldirection).).ABCDis themagneticmeridian andAB´C´Dis the geographicalmeridian.e. AgainEqs.(ii) . is the angle of declination.. The angleB´AB = . e e V B sin tan H B cos .(i) and vertical component V = Be sin . Fromfig. (i) and (ii) give 2 2 2 . is called angle of dip (.. or V =Htan .ACshows the direction of resultantmagnetic field of earth and the angleBA C(=. or 2 2 e B . .e.in 33 Magnetic meridian B H C D C´ B´ Geographical meridian Geographic north Magnetic north In fig... V . . the resultant earth s field can be resolved in two comp onents as shown in fig.physicsashok. V . in vertical direction) and at magnetic equator it is parall el to the earth s surface.) between it and the horizontalABis the angle of dip.. (ii) Angle of dip or inclination : The anglewhich the axis of needlemakeswith th e horizontal.MAGNETIC FIELD www. In otherwords. e H . . (a) the horizontal component HalongABand (b) the vertical componentV..Thus. . B cos . Horizontalcomponent H = Be cos . .. sin . . (i. . the angle ofdip at a place is the anglewhich the resulta ntmagnetic field ofearth at that placemakeswith the horizontal. 2 2 . alongAD. Does it experience anyforce? Ifit does.Explainwhy? 20. Avoltmeter is aWeston galvanometer ofveryhigh resistance. . the curr ent in the electrolyte is constituted bythemovement of positive and negative ions in opposite directions. Two parallel wires currying current in the same direction attract each other while two beams of electrons travelling in the same direction repel each other. Can a charged particle entering a uniformmagnetic field normally fromoutside complete a circle? 21.Willth ere be anychange in its length . a potentiometer or a voltmeter? 13. . Of the three vectors in the equation F . Why is an ammeter connected in series? 14. Is this true or false?Give reasons in support of your answer. Howdo you knowthat the current inside a conductor is constituted by electron s and not by protons? 18. Whyis themagnetic needle short in a tangent galvanometer ? 8. 16..in 34 THINKING PROBLEMS 1.what is this force if its radius is a and the current passing through it equals? 17. Acurrent-carrying circular conductor is placed in a uniformmagnetic fieldwit h its plane perpendicular to the field.MAGNETIC FIELD www. Acopper pipe is filledwithan electrolyte. Arectangular current loop is in an arbitary orientation in an externalmagneti c field. An ammeter is aWeston galvanometer whose resistance ismade negligible byshun ting the coil. Averystrong current ismade to flowfor a short time through a solenoid. Is anywork required to rotate the loop about an axis perpendicular to its plane? 5. If an electron is not deflected in passing through a certainspace.B. Is this true of Fa lse? 12.Willsuch a pi pe experience a forcewhen placed in amagnetic field perpendicular to the current? 19. Why is the field in amoving coilgalvanometer radialin nature? 9.physicsashok. Is this true or false? 11.What happens to the electrolyte in these circumstances? 22. At that deflections is a tangent galvanometermost sensitive? 7. canwe be s ure that there is no magnetic field in that region? 3. What is the greatest disadvantagewith a suspended-typemoving coilgalvanometer ? 10. Could this deflection be caused (a) by an electric field? (b) by a magnetic field ? (c) If either is possible. Abreamof protons is deflected side ways. . howcan you tellwhich one is present? 4. Will a tangent galvanometer work in the polar region ? 6. What is a faraday? 15. Which gives amore accurate value of a potential difference. There is no charge in the energyof a charged particlemoving in amagnetic fie ld althrough amagnetic force is acting on it.One electrode is at the axis ( ) and the other electrode is at the edge of the bath (+). Acylindrical electrolytic bath containing a solution of copper sulphate betw een two electrodes ismounted above the northpole of a strong electromagnet. which pairs are always at right angles?Whichmayhave any angle between them? 2.When a voltage is applied. qv. Why is this so ? . It is found thatmore low-energycosmic rays eachthe earth at the northand southmagnetic poles thanat t hemagnetic equator. Cosmic rays are charged particles that strike the atmosphere fromsome extern al source. 23.and diameter?Explain. . If the force vanishes of stopping themotion. No.in 35 SOLUTION OF THINKING PROBLEMS 1. Afaradayis the amount of charge required to liberate 1 gramequivalent of any sustance. and v . B. 14. No. they do not cancel out each other. 11.When a current-carrying flet conductor is placed inamagneti c field perpendicular to the flat face. (b)Ys. v . it is due to an electric field. The field ismade radial in order to have a linear relation between the curren t and the deflection. (a)Yes. The field due to the circular coil is uniformover a very small region about t he centre of the coil. Apotentiometer gives amore accurate value of a potential difference than a v oltmeter as it draws no current fromthe cell. The current constituted bynegtive ions in the positive directionis the same as the current formed bythe positive ions and so both form current in the same direction. and F . Ifit sill experience s a force in the same direction. are always at right angle. The greatest disadvntage is that this type of galvanometer is not portable. 12. a transversevoltage is developed. it canbe determined whether the current is due to negative or positive charge carriers. So the pipe will experience a force.). 16. 3. there is amagnetic field.. No. we cannot be sure that there is no magnetic field because the force will be zero when the direction of motion is along the direction ofthe field. and so nowork is done. 7.physicsashok. So there is no charge in the energyfo the particle though amagne tic force acts on it. It is true. 13. It is true. B. because the earth smagnetic field is vertical there.MAGNETIC FIELD www. . 10. 18. Here t here is no charge in . 8. may have any angle between them. So the needle must be short so that it maybe assumed tomove in uniformmegnetic fields. 5. It does not experiencemayforce because the forces on the elements are direct ed radially away and sumup to zero. BytheHall effect. it must be connected in series because it is a characteristic property ofseries connection that the same curren t passes through allparts of the circuit. True.Acharged particle experiences a force at right angles to the velocity a nd so it moves in a circle with a constant speed. it could be due to a magnetic field. no work is done in rotating the coil becausework doneW=mB (1 cos. 19. (c) Stop the proton and keep it stationary. Acurrent inonwire produces onlyamagnetic field and on electric field (becaus e a current-carrying conductor is electricallyneutral) over the other current-carrying conductor and so onlyama . 17. it could be due to an electric field directed perpendicular to the mo tion. The pairsF . 15. 2. Fromthe directionofthis voltage calledHa llvoltage. 6. Since the current following through the circuit has to pass through the amme ter. 4. 9. Atengent galvanometer ismost sensitivewhen the deflections are near about 0º. . The f orce experienced is given by F = qvb sin(v . . Thus there is contraction along the length ofthe solenoid. The current in the adjacent turns flow in the same direction and so they att ract each other. the boundaryline of themagneitc field has to be the diameter of the circ le. themagnetic field is parallel to the direction ofmotio n of the cosmic rayparticles. On account of this repulsive force t he diameter of the solenoid increases. So there arise bothmagnetic force (attractive) and electric force (repulsive) betwe en them. it will complete a semi-circle. they repel each other. 23. . 20. Hence the cosmic ray particles do not experience anyforce. both being vertical.At the equator these particles experiencemaximumdeflecting force and henc e low-energy particles cannot reach the earth. The repulsive force being in excess of the attractive force. No.whatever be the velocity of the charged particle. But an electron beamis a source of both. theyrepel each other.As the same currents in the diametri callyopposite elements flowin opposite directions. 21.Hence. Because at the poles.B. It moves counterclockwise (viewed fromabove) 22. an elect ric and amagnetic field.. ).gnetic forcewhich is attractive is nature arises between them. 9. Statement-2 : A moving charge produces a magnetic field. Statement-2 : The mass of proton is 1837 times more than the mass of electron. then path of electron is more curved than that of proton. the combination of the wires forms a current loop. 4. the time period of revolution of . Statement-2 : Electron has a tendency to form large curve due to small mass. 1. 7. Statement-1 : In electric circuits. wires carrying currents in opposite direc tions are often twisted together. with the same speed. The magnetic field generated by the loop might affect adjacent circuits or component s. Statement-1 : Free electrons always keep on moving in a conductor even then n o magnetic force act on them in magnetic field unless a current is passed through it. 3. Statement-2 : The average velocity of free electron is zero. then only possibility is that there is no magnetic field in this region.physicsashok. 11. Statement-1 : Electron cannot be accelerated by the cyclotron.mark the correct answer as (A) If both Statement-1 and Statement-2 are true and Statement-2 is the correct explanation ofStatement-1. the force experienced by proton will be more than electron. 5. Statement-1 : An electron and proton enters a magnetic field with equal velo cities. then.in 36 ASSERTION&REASON Astatement of Statement-1 is given and a Corresponding statement of Statement-2 is given just belowit of the statements. (E) IfStatement-1 is false but Statement-2 is true. Statement-2 : Cyclotron is suitable only for accelerating heavy particles. it is deflected towards west. 2. Statement-1 : If an electron and proton enter in an electric field with equal energy. Statement-2 : Force is directly proportional to magnetic field applied. (C) If Statement-1 is true but Statement-2 is false. Statement-1 : The magnetic field produced by a current carrying solenoid is independent of its length and cross sectional area. 10. Statement-2 : Electron has negative charge. Statement-1 : Magnetic field interacts with a moving charge and not with a st ationary charge. -particle is double that of proton.MAGNETIC FIELD www. (B) If both Statement-1 and Statement-2 are true and Statement-2 isNOT correct e xplanation ofStatement-1. (D) If both Statement-1 and Statement-2 are false. . Statement-2 : The metallic frame help in making steady deflection without any os cillation. -particle enter a uniform magnetic field p erpendicularly. Statement-2 : In a magnetic field. the time period of revolution of a charged pa rticle is directly proportional to the mass of the particle and is inversely proportional to charge of particle. Statement-1 : If an electron is not deflected while passing through a certain region of space. Statement-1 : The coil is wound over the metallic frame in moving coil galvan ometer. 8. Statement-2 : If the wire are not twisted together. Statement-1 : If an electron while coming vertically from outer space enter t he earth s magnetic field. 6. Statement-1 : If a proton and an . Statement-2 : The magnetic field inside the solenoid is uniform. 12. [JEE 2008] . Statement-1 : The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Statement-2 : Soft iron has a high magnetic permeability and cannot be easily ma gnetized or demagnetized. D and G carrying currents are arranged as shown in Figure. The particle then passes through a uniform transverse magnetic field in which it experiences a force F. In a hydrogen atom. . . (D) e 2 rB 0 . The magnetic field due to orbital motion of the electron at the site of the nucleus if B. .ner2M (C) (M m) ner2 m . The angular velo city . An electron of charge e moves in a circular orbit of radius r around a nucleu s. straight and parallel wires C. . an electron of mass m and charge e is in an orbit of radi us r making n revolutions per second. . 4. (B) r 0 eB . . of the electron is (A) 4 r 2 0 eB .4 N (B) 0. If the pote ntial difference between the anode and the cathode is increased to 2 V.ner2 m (B) m . An electron is accelerated to a high speed down the axis of a cathode ray tub e by the application of a potential difference of V volts between the cathode and the anode. . The force experienced by a 25 cm length D C G 30 A 10 A 20 A 3 cm 10 cm of wire C is (A) 0. A charged particle of specific charge s passes through a region of space show n. (D) . .physicsashok. .04 N (C) 4 x 10 3 N (D) 4 x 10 4 N 5.MAGNETIC FIELD www. . the electron will now experience a fo rce (A) F 2 (B) F 2 (C) 2 F (D) 2 F 2. . Three long. .in 37 Level # 1 1. . the magnetic moment associated with the orbital motion of the electron is (A) M . (C) e 4 rB 0 .ner2 3. If the mass of the hydrogen nucleus is M. . E C D B B = 0 below this line . = ..0 .. (C) ...... (C) The radius of the trajectory of the charged particle in REGION 2 is s. 2EBR (C) kR x . 6. x = 4x0. x = 6x0. . . x = 5x0. . EBr (D) 2kR x . .1 .. (B) Work done to move the charged particle in REGION 1 is ZERO. . v . X If r is the radius of the semicircle and k is the force constant for each spring . x = 3x0.. (D) 0 . then the extension x in each spring is (A) kR x . A wire of resistance R in the form of a semicircle lies on the top of a smoot h table. inwards perpendicular to p aper are placed at x = 2x0. . EBr 7. ad infinitum Here x0 is a positive constant. . (D) The particle emerges from REGION 2 with a velocity ' . outwards perpendicular to paper are placed at x = x0. The magne tic field at the origin due to the above collection of current carrying conductors is (A) ZERO (B) 4 x log 2 0 e 0 .. . A uniform magnetic field B is confined to the region as shown.. ad infinitum on the x axi s. Another infinite collection of current carrying conductors each carrying a current .(A) Velocity of the particle in REGION 1 is B . An infinite collection of current carrying conductors each carrying a current .. 2 B S E B0 xxxxxxxx xxxxxxxx xxxxxxxx xxxxxxxx x x x x x x x x REGION 1 REGION 2 where ' . ... The ends of the semicircle are attached to springs C and D whose other ends are fixed. .. 2EBr (B) kr x . .. 0 e 2 x log 2 . . . . 5 x 10 4 T out of the page. . Th e A B S 1. in a uniform magnetic induction (field) B. with each side of length L. An electron is fired from the point A with a velocity V0 = 2 x 108 m/s. Wire A carries a current of 10 A directed into the plane of the paper. ( . angle will the loop rotate by itself before the torque on it becomes zero ? . Wi re B carries a current such that the magnetic field at P at a distance of 0. is bent as shown. carrying a current . < 90°) with the direction of B. are perpendicular to the plane o f the paper. (C) 3 BIL (D) zero 9. Through what A B B X X Y Z C D . as shown. (D) 180° . parallel to the positive y-direction.5 x 10 2 T out of the page 11.5 x 10 2 T into the page (D) 1. The force experienced by the wire is B A C D E L L Z B Y X O F (A) BIL in the positive y-direction. (B) 1.The magnitude and direction of the current in wire is zero.in 38 8. The loop can rotate about the axis XX . A conductor ABCDEF.physicsashok. (B) 90° (C) 90° + . (B) BIL in the negative z-direction. is placed in a uniform magnetic field B. The plate of the loop makes an angle .5 x 10 3 T into the page (C) 1. 10. Two long straight parallel wires 2 m apart.MAGNETIC FIELD www. The magnitude and direction of magnetic field that will cause the A B 15 cm x V0 y electron to follow a semicircular path from A to B is (A) 1.2 m . The square loop ABCD.8 m fro m this wire is zero. (A) . It is carrying a current . 12. The magnetic moment of A +++ B ..8 x 10 6 A into the page. r 0 14. .. A non-conducting rod AB of length . (B) 2 3 . . has a linear charge density . the rod is (A) 2 2 . as shown in the figure. r 2 (D) T . The ratio of the magnitudes of the magnetic moment of the system and its angular mom entum about the centre of the rod is (A) 2m q (B) m q (C) m 2q (D) m q .08 A out of the page. (D) 3 6 .. Two particles each of mass m and charge q. are attached to the two ends of a light rigid rod of length 2 .1.6 m 2 m P magnitude and direction of the current in wire B is (A) 2.. (C) 3 3 2 . . The rod is rotated about an axis passing through point A with constant angular velocity . . r 3 (B) T 2 . r (C) T .. (A) T 2 . 13. (D) 2. The rod is rotated at a constant angular speed about a perpendicular axis passin g through its centre. If a charged particle is describing a circle of radius r in a magnetic field with a time period T then.8 A out of the page (C) 2...8 A into the page (B) 2. . . Two particles Y and Z emitted by a radioactive source at P made tracks in a could chamber as illustrated in the figure. 0) along an electric field 0 E j . 17. flows through a long conducting wire bent at right angle as shown in figure. perpendicular to the wall. Careful measurements showed that both tracks P Y were circular.in 39 15. What minimum magnetic field must exist in this reg ion for the particle not to hit the wall? (A) v/sd (B) 2v/sd (C) v/2sd (D) v/4sd 19. The magnetic field at a point P on the right bisector of the angle XOY at distance r from O is (A) 0 r . The resistances of three parts of a circular loop are as shown in the figure. . (D) Zero A B R C R a 2R O I 120° 120° 18. Current . A magnetic field acted downward into the paper. (B) 0 3 I a . Z Which one of the following statements is certainly true? (A) Both Y and Z particles carried a positive charge. (D) The charge of the Z particle was twice that of the Y particle. The magnetic field at the centre O is (A) 0 6 I a . the radius of Y track being half that of the Z track. A particle with a specific charge s is fired with a speed v towards a wall a t a distance d. The angular momentum of the particle about origin (A) is zero (B) is constant (C) increases with time (D) decreases with time 16. (C) 0 2 3 I a .physicsashok. (C) The mass of the Z particle was twice that of the Y particle. (B) The mass of Z particle was one half that of the Y particle. A charged particle of mass m and charge q is released from rest from (x0.MAGNETIC FIELD www. (B) 0 2 r . Two mutually perpendicular conductors carrying 1 . . and 2 . Find locus of points at which magnetic induction is zero? (A) 22 2 . then if we represent the magnetic induction due to bigger coil Y at O as By and that due to smaller coil X at O as Bx. . . (D) 1 4 y x B B . 2 1. Two circular coils X and Y having equal number of turns and carry equal currents in the same sense and subtend same solid angle at point O. (D) 0 . then (A) 1 y x B B . 20. . 21. 2 2 1 4 r . 2 r . . . . . (C) 1 2 y x B B . (B) y 2 x B B . lie in the x-y pl ane. If the smaller coil X is midway d X Y d O be between O and Y. . X Y O I P x 45° r (C) 0 . . . .y . 2 2 x . . 1 2 x . . x (B) y (C) x (D) 2 1 y . . . 1 2 y . . . A charge q is moving with a velocity 1 v. 3 j. .7N . .in 40 22. perpendicular to the magnetic field.1i m s at a point in a magnetic fi eld and experiences a force . Two charged particles A and B enter a uniform magnetic field with velocities normal to the field.02 m is suspended from a vertex such that it is hanging in a vertical plane between the pole pieces of a permanent magnet producing a horizontal magnetic field of 5 x 10 2 T. . (C) the ring may tend to contact (D) none of these. . 0) is . It s velocity at (x. (C) The specific charge of A is greater than that of B (D) the speed of A is les s than that of B.m (B) 5 2N .7N . (B) the ring may tend to expand.physicsashok. A particle of charge q and mass m starts moving from the origin under the ac tion of an electric field 0 E . The possible reasons are: (A) The momentum of A is greater than that of B (B) the charge of A is greater t han that of B. A coil in the shape of an equilateral triangle of side 0. 26. (B) the momentum of M is greater than that of N. N. (A) 5 2N . 1 F . Their paths are shown in the Figure. (D) the speed of M is less t han that of N.MAGNETIC FIELD www.4N . m 23. E i . then: (A) there is no net force on the ring. B k .m (C) 5 3 . The value of x is: (A) 0 0 36E B qm (B) 0 25 2 m q E (C) 0 10m q E (D) 0 0 25E B m 24. it e xperiences . The direction of the field is perpendicular to the plane of the ring. with v elocities. 25. The possible reason are: (A) the charge of M is greater than that of N. the paths are as shown in the figure. q . Find the couple acting on the coil when a current of 0.1 j . and magnetic field 0 B .1 k . .1 N S ampere is passed through it and the magnetic field is parallel to its plane. 3. m (D) 10. 27. If the charge is moving with a velocity 2 v.4i .10.10.1 j ma/s at the same point. (C) specific charge of M is greater than that of N. Two charged particle M and N enter a space of uniform magnetic field. A current-carrying ring is placed in a magnetic field. MAGNETIC FIELD www. initiall y parallel to and against the field with a small initial speed.1i . (out of the page) and is deflected a distance d after travelling a horizontal distance a.physicsashok. . 28. . (C) move parallel to the x-axis at least once during every rotation if . 31. When a steady current is set up in each of them: (A) The two rings rotate towards a common plane. (D) never move perpendicular to the x-direction. one of slightly smaller diameter then the other. (B) move parallel to the x-axis once during every rotation for all values of . initially the planes of the rings are mutually perpendicular. k . The ratio of the energy required to set up in a cube of side 10 cm a uniform magnetic field of 4 wb/m2 and a uniform electric field of 106 V/m is: (A) 1. Let T 2 m j . k . .1k .4 x 106 (D) 1. A particle with charge +Q and mass m enters a magnetic field of magnitude B.4 x 103 30.in 41 a force 2 F . j . A charged particle q enters a region of uniform B . (D) The outer ring stays stationary while the inner one moves into the plane of the outer ring. A charged particle is fired at an angle .4 x 105 (C) 1.i . . The direction of the motion of th e particle is perpendicular to the direction of B. 29. q . (B) The inner ring will oscillate about its initial position. k .Wb m2 (B) .i .Wb m2 (C) . N . (C) The inner ring stays stationary while the outer one moves into the plane of the inner ring. j . . are su spended along their common diameter as shown in the figure. . (B) 2 qBa (C) Zero (D) not possible to be determined as it keeps changing.4 x 107 (B) 1. (D) the total energy of the beam is conserved. The magnitude of the momentum of the particle is: (A) 2 2 qB a d d . 32. Then: (A) the beam will pass through the field accelerating down the field without cha nging its width. .Wb m2 (D) .i . . j .i . .. 33. in a uniform magnetic field direct ed along the x-axis. A parallel beam of electrons is shot into a uniform electric field. During its motion along a helical path. the particle will: (A) never parallel to the x-axis. The magnetic induction B . Two insulated rings. . existing only on the right of the boundary YZ. 45°. . at that point is: (A) . k . (C) the beam tends to narrow down at the beginning and to spread out later. (B) the beam tends to spread out at the beginning and to narrow down later. . (B) 0 2 2 a . . . The time spent by the particle in the field will be: (A) T. . . (D) 0 2 a . . . .QB . . . . (C) 0 2 a . . (C) 2 2 T . . The magnetic induction at the centre of the loop is: (A) 0 2 a . . . . . . . . . . (B) 2T. The current I flows through a square loop of a wire of side a. . . . . 34. (D) 2 2 T . . . If rp. and r.in 42 35. qfR2B. (D) acceleration of the particle will never become equal to zero. denote respectively the radii of the trajectories of these particles. A semi-circular wire of radius R is connected to a wire bent in the form of a sine curve to form a closed loop as shown n the figure. If the loop carries a current . (D) p d r r r. . at a point in the space: (A) particle can not remain in static equilibrium. A circular loop of radius R carries a charge q uniformly distributed on it.m s . A charged particle of unit mass and unit charge moves with velocity of v . . (C) The path of the particle may be y2 + z2 = 25.-particle having the same kinetic energy are mo ving in circular trajectories in a constant magnetic field.physicsashok. . A proton. . qfR2B (D) None of these 37. . (B) The path of the particle may be x2 + y2 = 25.14 s. and is placed in a uniform magnetic field B. Choose the correct alternative(s): (A) The path of the particle be x2 + y2 4x 21 = 0. (B) d p r r r . 0 (C) 2. qfR2B . 40. (D) The time period of the par ticle will be 3. then the total force acting on the sine curve is: B (A) 2B. They carry steady equal currents flowing out of the plane of the paper.R (downward) (B) 2B. The maximum and minimum torques acting on the loop due to the magnetic field are. It is rotated at a frequency f about one of the diameters. .R (upward) (C) B. respectively: (A) . 36. 6 j. The particle will move in a (JEE 1999) (A) straight line (B) circle (C) helix (D) cycloid 41.MAGNETIC FIELD www. . (C) d p r r r . . . If a charged particle is released from rest. (B) first particle will move along a curved path but after some time its velocit y will become constant. A charged particle is released from rest in a region of steady and uniform e lectric and magnetic fields which are parallel to each other. 8i . (C) particle will come to rest at regular interval of time. Two long parallel wires are at a distance 2d apart.R (upward) (D) Zero 38. Suppose a uniform electric field and a uniform magnetic field exist along mu tually perpendicular directions in a gravity free space. in a magnetic field of B . rd. . 0 (B) 0. 2 k T . . 39. as shown. a deuteron and an . The variation of the magnetic field B along the line XX is g iven by (JEE 2000) (A) d d X X B (B) d d X X B (C) d d X X . (JEE 1997) (A) p d r r r . then. A uniform magnetic field B exists along its diameter . An infinitely long conductor PQR is bent to form a right angle as shown in f igure. The ratio 1 2 H H is given by (JEE 2000) (A) ½ (B) 1 (C) 2 3 (D) 2 . The magnetic field due tot his current at the point M is H1. The magnetic field at M is now H2. A current .B (D) d d X X B 42. another i nfinitely long straight conductor QS is connected at Q so that current is . flows through PQR. the current in PQ remaining unchanged. 2 in QR as well as in QS. Now. . If it is subjected simultaneously to an electric field along the +x direction and a magnetic field along the +z direction. 44. . j . . 0. A uniform magnetic field exists perpendicular to A this plane. The magnetic field due to this loop at the point x y z I 2a P(a. k (D) 1 . When a current . (D) . The speeds of the particles are V B A and VB respectively and the trajectories are as shown in the figure. . passes through the coil. j . . . A coil having N turns is wound tightly in the form of a spiral with inner an d outer radii a and b respectively. (C) all ions deflect towards y direction (D) positive ions deflect towards y direction and negative ions towards +y direct ion. An ionized gas contains both positive and negative ions. 3 . j . 2 . . a) points in the direction (JEE 2001) (A) 1 . k (B) 1 . 2 i . k . (C) . i (C) . is placed as shown in the figure. the magnetic field at the center is (J EE 2001) (A) 0 N b . 0 ln 2 N b b a a . .in 43 43. . 1 3 i . Then (JEE 2001) (A) mA vA < mB vB (B) mA vA > mB vB (C) mA < mB and vA < vB (D) mA = mB and vA = vB 46. (B) 0 2 N b . . .physicsashok.MAGNETIC FIELD www. A non-planar loop of conducting wire carrying a current . then (JEE 20 00) (A) positive ions deflect towards +y direction and negative ions towards y direct ion (B) all ions deflect towards +y direction. Each of the straight sections of the loop is of length 2a. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. . 0 ln 2 N b b a a . k 45. 2 q b a B m . . . . . in the negative z direction. A long straight wire along the z-axis carries a current . . B m . 47. . . 48. . The minimum value of v required so that the particle can just enter the r egion x > b is (JEE 2002) (A) qbB m (B) q . extending from x = a to x = b. . (B) . . 0 2 2 x x . (C) .b a. . . 0 2 i x j y . It enters a region containing a uniform magnetic field B directed along the negative z direc tion. 0 2 2 2 y x . y) in the z = 0 plane is (JEE 2002) (A) . 2 i y j y .. . . The magnetic vector field B . . (C) qaB m (D) . A particle of mass m and charge q moves with a constant velocity v along the positive x direction. 2 2 x j y i x y . . . . at a point having coordinates (x. . Which one of the P following combinations is possible? (JEE 2003) (A) E . a i . B . a i . (B) E . . B . 0. . 0 2 . B . . c k . . c j . For a positive charged particle moving in a x-y plane initially along the x-axis. 0.. (C) E . . 2 2 x i y j x y . b k . . b j . The curved path is shown x y in the x-y plane and is found to be non-circular. c k . B . . . (D) . a i . . . (D) E . there is a sudden change in its path due tot he presence of electric and/or magnetic fields beyond P. b i . . 49. c k . . A magnetic needle is kept in a nonuniform magnetic field.physicsashok. If steady current . Two particle X and Y having equal charges. 0 (C) E .MAGNETIC FIELD www. is placed in a uniform magnetic field pointing into the plane of the paper as shown. 0. 0 55. B . 0 B . 51. 0.in 44 50. arrange them in the decreasing order of Potential Energy (JEE 2003) B n I B n II B n III B n IV (A) I > III > II > IV (B) I > II > III > IV (C) I > IV > II > III (D) III > IV > I > II 52. III & IV. enter a region of uniform magnetic field and describe circular paths of radii R1 and R2 respectively. A current carrying loop is placed in a uniform magnetic field in four differ ent orientations. It comes out B x y e u of the region with speed v then (JEE 2004) (A) v = u at y > 0 (B) v = u at y < 0 (C) v > u at y > 0 (D) v > u at y < 0 53. B . is established in the wire as shown i I in the figure. B = 0 (B) E = 0. It experiences (JE E 1982) (A) a force and a torque (B) a force but not a torque (C) a torque but not a force (D) neither a force nor a torque 54. An electron traveling with a speed u along the positive x-axis enters into a region of magnetic field where . x Y I B The loop will have a tendency to (JEE 2003) (A) contract (B) expand (C) move towards +ve x-axis (D) move towards ve x-axis. If E and B represent the electric and magnetic fields respectively. after being accelerated through t he same potential difference. the loop will : (JEE 1985) (A) rotate about an axis parallel to the wire (B) move away from the wire (C) move towards the wire (D) remain stationary 56. II. B = 0 (D) E . I. . .B k x . This region of space may have:(JEE 1985) (A) E = 0. A conducting loop carrying a current . 0 . A proton moving with a constant velocity passes through a region of space wi thout any change in its velocity. A rectangular loop carrying a current i is situated near a long straight wir e such that the wire is parallel to one of the sides of the loop and is in the plane of the loop. The ratio of the mass of X to that of Y is (JEE 1988) . . A particle of charge +q and mass m moving under the influence of a uniform electric field E i and uniform magnetic field B k follows a trajectory from P to Q as shown in Figure.2 1 2 R R (D) 1 2 R R 57.1 2 1 2 R R (B) 2 1 R R (C) . (C) Rate of work done by the electric field at P is zero (D) Rate of work done by the electric field at Q is zero .2 j 2v P a Q x 2a B V E which of the following statement(s) is/are correct? (JEE 1991) (A) 3 2 4 E mv qa . . . . . The velocities at P and Q are v i and . . (B) Rate of work done by the electric field at P is 3 3 4 mv a . . . . . .(A) . . . but not zero (B) The magnetic field at any point inside the pipe is zero. Then: (JEE 1993) (A) The magnetic field at all points inside the pipe is the same. Two very long. A current 1 flows along the length of an infinitely long. Its instantaneous velocity v is perpendicular to this plane. . (C) As seen from O. respec tively. straight. .in 45 58. straight. The: (JEE 1994) (A) H+ will be deflected most. and .. . . An infinite current carrying wire passes through point O and in perpendicula r to the plane containing a current carrying loop ABCD as shown in the figure. He+ and O2+ are 1 amu. the loop rotates anticlockwise. The masses of H+. 60. (C) The magnetic field is zero only on the axis of the pipe (D) The magnetic field is different at different points inside the pipe. a point charge q is at a point equidis tant from the two wires. He+ and O++ all having the same kinetic energy pass through a region in which there is a uniform magnetic field perpendicular to their velocity.B j . At a certain instant of time. the loop rotates clockwise. thin-wal led pipe. (D) 0 61. in the plane of the wires. Choose the correct option (s) (JEE 2006) (A) Net force on the loop is zero C B O O´ A D (B) Net torque on the loop is zero. (B) 0 qv d .MAGNETIC FIELD www. . 59. . H+ . (C) 0 2 qv d . (B) O2+ will be deflected most (C) He+ and O2+ will be deflected equally (D) All will be deflected equally. 4 and 16 amu respectively. exists in the region a < x < 2a and 0 B .. The distance between the wires is d. B j . 62. .physicsashok. The magnitude of the force due to the magnetic field acting on the charge at this instant is (JEE 1998) (A) 0 2 qv d . (D) As seen from O. parallel wires carry steady currents . A magnetic field 0 B . . . in the region 2a < x < 3a. The trajectory of the charge in this region can be like . l (D) Time spent in Region Ii is same for any velocity V as long as the particle r eturns to Region I . enters the 0 a 2a 3a x B0 B0 magnetic field at x = 0. l (C) Path length of the particle in Region II is maximum when velocity v q B m . A particle of mass m and charge q. The length of the Region II is l. moving with velocity v enters Region II normal to the boundary as shown in the figure. where B0 is a positive constant. l (JEE 2008) (B) The particle enters Region III only if its velocity v q B m . Region II has a uniform magnetic field B perpendicular to the plane of the paper. Choose the correct choice(s) × × × × × × × × × × × × × × × × × × × × × × × × Region I Region II Region III 0 v (A) The particle enters Region III only if its velocity l v q B m . (JEE 2007) (A) z a 2a 3a x (B) z x a 2a 3a (C) z a 2a 3a x (D) z x a 2a 3a 63. where v0 is a positive constant.. A positive point charge moving with a velocity 0 v . v j . The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current 1 as shown. it is found that the filter transmits . the electron moves in an orbit of radius 0. The magnetic moment associated with the orbital motion of the electron is ________ .5Å making 1016 r evolutions per second. as shown in Figure.3 MeV. A metallic block carrying current . A uniform magnetic field with a slit system as shown in figure is to be used as momentum filter for high-energy charged particles.3 B Tesla and deuterons are passed into the filter. (JEE 1988) 5. In a hydrogen atom. A neutron a proton.MAGNETIC FIELD www. A wire of length L meters carrying a current i amperes is bent in the form of a circle. A wire ABCDEF (with each side of length L) bent as shown in figure and carrying a current . The magnitude of its magnetic moment is ________ in MKS units. -particles Source Detector B each of energy 5. (JEE 1984) 2. (JEE 1988) 4. is subjected to a uniform magnetic induct ion as B . (JEE 1996) 7. (JEE 1997) . the energy of each deuteron transmitted by the filter is _____ MeV. The magnitude of I I R2 S R C Q P R1 the magnetic induction at the centre C is ________ . The electron follows track ________ D and the alpha particle follows track ________. and an electron and alpha particle enter a region of constant magnetic field with equal velocities. (JEE 1990) 6. The moving charges experience a force F .in 46 FILL IN THE BLANKS 1. (JEE 1987) 3. The force experienced by the wire is ________ in the ________ direction. The tracks of the X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X X B C particles are labelled in Figure. The magnetic field is increased to 2.physicsashok. given by _____ which results in the lowering of the potential of the face _____ Assume the sped of the carriers to be v. With a field B Tesla. is placed in a uniform magnetic induction B B A C D E F Z X Y I parallel to the positively-direction. The magnetic field is along the inward normal to the plane of the paper. A current flows along length of a long thin cylindrical shell: Column I Column II (A) Magnetic field at all points lying inside the shell (P) Inversely proportion al with distance from axis of shell (B) Magnetic field at any point outside the shell (Q) Zero (C) Magnetic field is maximum (R) Just outside the shell (D) Magnetic field on the axis of the shell 15. Column I Column II (A) Unit of magnetic field (P) Am2 (B) Unit of magnetic permeability (µ0) (Q) N Am (C) Unit of magnetic flux (. No net force acts on a rectangular coil carrying a steady current when suspen ded freely in a uniform magnetic field. When they pass through a uniform magnetic field perpendicular to the direction of their motion. they describe circular paths of the same radius. A charged particle enters a region of uniform magnetic field at an angle of 85° to the magnetic line of force. (JEE 1983) 10. Two long parallel wires carrying equal currents in opposite directions are p laced at x = . Then: Column I Column II (A) Magnetic field B1 at origin O (P) 0 3 i a . charge being at rest (T) Parabolic path 14. match the following: Column I Column II (A) In electric field (P) Straight line path (B) In magnetic field (Q) Circular path (C) In crossed field (R) Helical path (D) In mutually perpendicular electric and (S) Parabolic path magnetic field. An electron and proton are moving with the same kinetic energy along the sam e direction. 0.physicsashok.MAGNETIC FIELD www. (B) Magnetic field B2 at P (2a. There is no change in the energy of a charged particle moving in magnetic fie ld although a magnetic force is acting on it. The path of the particle is a circle. Column I Column II (A) Electric field (P) Stationary charge (B) Magnetic field (Q) Moving charge (C) Electric force (R) Changes the kinetic energy (D) Magnetic force (S) Does not change kinetic energy 13. 0) (Q) 0 4 i a .in 47 TRUE / FALSE 8. (JEE 1981) 9. a parallel to Y-axis with z = 0. (JEE 1985) TABLEMATCHING 12.) (R) N A2 (D) Unit of magnetic dipole moment (S) Nm A 16. . . Regarding the trajectory of a charged particle. . (JEE 1983) 11. 0) (S) Zero . 0) (R) 0i a . 0. 0.. . (D) Magnetic field at N ( a. (C) Magnetic field at M (a. . which have same radii.MAGNETIC FIELD www. each of the same magnitude q. Let B be the magnetic field at M and µ be the magnetic moment of the system in this condition. are arranged in different m anners as shon in Column-II. Assume each rotating charge to be equivalent to a steady current. Now. M is the mid-point between the two innermost charges. Let E be the ele ctric field and V be the electric potential at M (potential at infinity is zero) due to the given charge distribution when it is at rest. M is the common centre + Q P M + + of the rings.physicsashok. 0 (Q) + + + Charges are on a line perpendicular to PQ at Q P M equal intervals. (C) Point P is situated at the (R) There is no magnetic field at P mid-point of the line joining the centres of the circular P wires. Six point charges. Match the statements is Column-I with the st atements in Column-II. PQ is + + M + P Q perpendicular to the plane of the hexagon. common center of the wires. the whole system is set into rotation with a constant angular velocity about the line PQ. Two wires each carrying a steady current I are shown in four configurations in Column-I. (B) V . (B) Point P is situated at the (Q) The magnetic fields (B) at P due to the curre nts in mid-point of the joining the the wires are in opposite directions. [JEE 2009] Column I Column II (A) E = 0 (P) Charges are at the corners of a regular hexagon. (C) B = 0 (R) Charges are placed on two coplanar insulating rings at equal intervals. centers of the circular P wires.in 48 17. PQ is rependicular to the plane of . In each case. P 18. (D) Point P is situated at the (S) The wires repel each other. M is at the centre of the hexagon. which have same radii. Some of the resulting effects are deseribed in Column-II. a point M and a line PQ passing through M are shown. Column I Column II [JEE 2007] (A) Point P is situated P (P) The magnetic fields (B) at P due to the currents midway between the wires. in the wires are in the same direction. PQ is perpendicular to the line joining the centres and coplanar to the rings. PQ is parallel to the longer sides. . M is the mid-point betP Q M + + + ween the centres of the rings. M is at the centre of the rectangle. (T) Charges are placed on two coplanar. (D) µ . 0 (S) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid points of the P M Q + + longer sides.the rings. insulating rings at equal intervals. extra electrons) built up on a stationary door kno b. in which direction would the particle travel (until crashing)? (A) North (B) East (C) West (D) Now here. it would stop moving 3. and has strength Fmag = qvB where q is the particle s charge. If the magnetic field in figure were turned off. In order for a cyclotron to work properly. North West South East particle where m denotes the particle s mass. and B is the magnetic field stre ngth. An alpha particle has charge 2e and mass 4 amu. the magnetic field must make the p article move in a circle. . In figure 1. In figure 1.e. Small charged particles are deposited at high speed into a circular pipe. a magnetic field keeps it moving in a circle at constant speed. (D) Static electricity (i. A cyclotron is a device used by physicists to study the properties of subatomic particles. If both pa rticles experience he same magnetic field in the same cyclotron. 1. when the particle reaches the desired speed. Which of the following cannot create a magnetic field? (A) Electrical current flowing through a well-insulated straight metal wire. As a result of this force. (C) Electric current flowing around a superconducting ring.MAGNETIC FIELD www. (B) A beam of electrons moving across a cathode ray tube. Let falpha and f proton denote the frequencies with which those particles circle a cyclotron. v is its speed. The magnetic force on the particl e points towards the center of the circle. electric and magneti c field then accelerate the particle to an even higher speed. what is alpha proton f f ? (A) 2 (B) 1 (C) 1 2 (D) 1 4 4.in 49 PASSAGE TYPE PASSAGE = 1. m . Finally.physicsashok. amagnetic field pointing into the page keeps a charged particle trave lling in a counterclockwise circle at constant speed inside the cyclotron. what is the direction (if any) of the particle s acceleration? (A) North (B) East (C) West (D) It has no acceleration 2. Which of the following particles would not work in a cyclotron? (A) lithium atom (Li) (B) positive lithium ion (Li+) (C) negative lithium ion (Li+) (D) all of the above particle would work . In the following questions. neglect gravity. A proton has charge e and mas s 1 amu. 5. The frequency (in hertz) is the number of rev olutions completed by the particle per second.. the particle moves in circles in the cyclotron at frequency 2 f qB . bullet trains in Japan ha ve a maximum speed of about 250-300 km/h while a MAGLEV train under development has reached a speed of 411 km/hr. . For comparison. The idea of MAGLEV transportation has been in existence since the early 1900s. Magnetically levitated (MAGLEV) trains are considered to be important future tra velling machines. The MAGLEV train uses powerful on board superconducting electromagnets with zero ele ctrical resistance to support the train above the rails.PASSAGE = 2. The basic idea of a MAGLEV train is to levitate it with magnetic fields so that there is no physical contact between the train and the rails. The b enefits of eliminating friction between the wheel and the rail to obtain higher speeds and lower mainte nance costs has great appeal. Find the time after which collision occurs between the particles if projection speed equals 2vm . In addition. 1. 1. (A) MAGLE trains are operating in which country (B) USA Super conducting electromagnets are installed in (B) on the railroads Coils installed on the wall will carry currents only if train is moving very fast (B) all times whether train is standing or run ning only if train is moving irrespective of whether fast or slow. PASSAGE = 3. (A) Japan 4. The electromagnets on the train and outside produce forces that levitate the train and keep it centered above the rails. the superconducting magnet on the train induces a current in t hese coils and makes them electromagnets.(B) small size (D) zero electrical resistance. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. MAGNETIC FIELD The walls along the track contain a continuous series of vertical coils of ordin ary wire. Find the maximum value of projection speed vm so that the two particles do not collide. What is the major advantage of using super conductors in the electromagnets (A) very low resistance for electric current (C) low hysteresis loss 2. as shown in Figure. (C) China (D) None of these (C) on the walls (D) on the station another charged particle 2q of mass 2 m in a uniform magnetic field B as shown in figure. A charged particle +q of mass m is placed at a distance d from xinside the train 5. They have low maintenance cost but high running cost they have low maintenance cost but high energy efficiency they have low maintenance and running cost as well as low installation cost. Choose the correct statement (A) MAGLEV trains have high maintenance cost (B) (C) (D) 3. . electric current flowing through coils outside the train propels the train forward. As the train passes each coil. If the particles are projected towards each other with equal speeds v. (C) China (D) None of these (C) on the walls (D) on the station (B) small size (D) zero electrical resistance. 2. (D) None of these. 3. Assuming the collision to be perfectly inelastic find the radius of particle in subsequent motion. in 50 .(A) (C) xxx q.m -2q.2m VV xxxx xxxx d xxxx (Neglect the electric force between the charges) www.physicsashok. as shown in figure. Determine the maximum distance from the wire along x-axis up to X Z e O V0 i which electron can move. T he surface charge density is . . which is free to move in a vertical plane and carr ies a steady current of 20 A. A long wire of radius a is placed along z-axis and carries current i as indic ated in the figure..j i 2 V V . 2.physicsashok. A particle of charge q and mass m is projected from the origin with velocity v . Here . A flat dielectric disc of radius R carries an excess charge on its surface. 5. Find the (a) instantaneous acceleration. 3. is in equilibrium at a height of 0. A wire loop ABCDE carrying a current . = 0 . The disc rotates about an axis perpendicular to its plane passing through the ce ntre with angular velocity . B0 is applied.01 m over another parallel long wire CD which is fixed in a horizontal plane and carries a steady A B current of 30 A. Show that when AB is slightly depressed.in 51 Level # 2 1. C .MAGNETIC FIELD www. r/2 4. is placed in the x-y plane as shown in figure. .B0x . i in a nonuniform magnetic field k B .0 directe d along x-axis. m/s. (B) If an external magnetic field i B . find the x y O A B C D E 90° force and torque acting on the loop due to this field. . An electron escapes from the surface of the wire with velocity . Find the torque on the disc if it is placed in a uniform magnetic field B direct ed perpendicular to the rotation axis. y-axis is taken perpendicular to the plane of paper directed into the paper. A particle of mass m and charge q is projected from origin with velocity . A long horizontal wire AB.0 and B0 are positive constants of proper dimensions. . 0 . Find the maximum positive x coordinate of the particle during its motion. 0 cm.0 A.0 cm is tightly wound in the shape of a very long coil with crosssection radius R = 2. Find the period of oscillations. .D it executes simple harmonic motion. 10. 7. Find the magnetic induction inside and outside the solenoid as a func tion of the distance r from its axis.T. 8. = 0. the number of turns is N = 500. wound tightly on half a tore (Figure). A current . Find the magnetic induction at the centre of the polygon. = 5. .Ana lyse the obtained expression at n. = 30°. = 5. Find the magnetic moment of a thin round loop with current if the radius of t he loop is equal to R = 100 mm and the magnetic induction at its centre is equal to B = 6. flows along a thin wire shaped as a regular polygon with n sides which can be inscribed into a circle of radius R.5 cm to make a single-layer straight solenoid. Find the magnetic induction at the centre of rectangular wire frame whose dia gonal is equal to d = 16 cm and the angle between the diagonals is equal to ..0 A flows through the strip. The diameter of the cross-section of the tore is equal to d = 5. the current flowing in th e frame equals . Calculate the magnetic moment of a thin wire with a current .0 . A thin conducting strip of width h = 2.8 A. a direct current . 6. 9. 12. Find : (a) the magnetic induction at the centre of the disc . rotates about its axis with an angular velocity .in 52 11. a B I Find the gauge pressure produced by the pump if B = 0.MAGNETIC FIELD www.0 = 5. The field occupies a region of space d = 10 cm in thickness (Figure).0 cm. Find : (a) Ampere force acting on the frame . Find the magnetic induction at the spot wher e the coil is located. A slightly divergent beam of non-relativistic charged particles accelerated by a potential difference V propagates from a point A along the axis of a straigth solenoid. . Find the force with which the beam acts onthe ta rget if the beam current is equal to . = 0. = 100 A.90 A is located in the same plane as a long straight wire carrying a current . The cross-section area of the coil is S = 1. Find the d . = 1. 14. (B) the mechanical work to be performed in order to turn the frame through 180° ab out its axis. When there is no current in the coil the balance is in equilibrium. A beam of non-relativistic charged particles moves without deviation through . with the currents maintained constant. 16. is made to flow across this pipe section in the direction perpendicular both to the vector B and to the axis of the pipe. . 17. The axis of the fram e passing through the midpoints of opposite sides is parallel to the wire and is separated from it by the distance which is . A non-relativistic proton beam passes without deviation through the region o f space where there are uniform transverse mutually perpendicular electric and magnetic fields with E = 120 kV/m and B = 50 mT.0 cm. A current . A square frame carrying a current . A small coil C with N = 200 turns is mounted on one end of a balance beam and introduced between the poles of an electromagnet as shown in Figure. and a = 2. 13. A proton accelerated by a potential difference V = 500 kV flies through a uniform transverse magnetic field with induction B = 0. = 0. 18. The frame side has a length a = 8. = 30 cm. the length of the arm OA of the balance beam is .51 T.5 times greater than the side of the frame. through which the proton deviates from the initial direction of its motion. (B) the magnetic moment o f the disc. In an electromagnetic pump designed for transferring molten metals a pipe section with metal is located in a uniform magnetic field of induction B (Figure).10 T. from the point A at two successive values of magnetic induction B1 and B2.0 A. The beam is brought into f ocus at a distance . = 22 mA through the coil the equilibrium is restored by putting the additional counterweight of mass . 15.80 mA. + B angle .physicsashok.m = 60 mg on the balance pan.0 cm2. A non-conducting thin disc of radius R charged uniformly over one side with surface density . On passing a current . Then the beam strikes a grounded target. Find the speci fic charge q/m of teh particles. the trace of the beam on the screen S A S .the region of space A (Figure). shifts by . When the magnetic field is switched off.x . . Knowing the distances a and b. where there are transverse mutually perpendicula r electric and magnetic fields with strength E and induction B. find the specific charge q/m a b of the particles.. The segments LR and MS are along the x. A pair of stationary infinitely long bent wires are placed in the x-y plane a s shown in Figure. with which the wheel can be rotated. The wires carry currents of . (JEE 2003) 4. uniformly distributed on the rim of the wheel is free to rotate about a light horizontal rod. (JEE 1996) (i) Obtain an expression for the orbital magnetic dipole moment of the electron. An electron in the ground state of hydrogen atom is revolving in anticlock-wi se direction in a circular orbit of radius R. L R O P S M x y .12 m. A current of 10 A flows around a closed path in a circuit which is in the hor izontal plane as shown in the figure. = 10 A each as shown. n 30° B (ii) The atom is placed in a uniform magnetic induction B .0 B T0 T0 d angular velocity 0 .08 m and r1 = 0. . .MAGNETIC FIELD www. (A) Find the magnetic field produced by this circuit at the center. the segments RP and SQ are parallel to the y-axis such that OS = OR = 0. The rod is suspended by light inext ensible strings and a magnetic field B is applied as shown in the figure. find the m aximum . find the torque experienced by the orbiting electron. (JEE 2001) A D C r1 r2 (B) An infinitely long straight wire carrying a current of 10 A is passing the center of the above circuit vertically with the direction of the current being into the plane of the circuit. Find the magnitude and . 2. A wheel of radius R having charge Q.02 m. If the breaking tension of the strings are 3T0/2. The circuit consists of eight alternating arcs of radii r1 = 0. The initial te nsions in the strings are T0. such that the plane-normal of the electron-orbit makes an angle of 30° with the magnetic induction. What is three force acting on the wire at the center due to the current in the circuit? What is the force acting on the arc AC and the straight segment CD due to the current at the center? 3. .physicsashok.axis. Each arc subtends the same angle at the center.in 53 Level # 3 1. I. Find (a) k in terms of given parameters N. area A and moment of inertia I is placed in magnetic field B. A thin coil having 10 turns of wire and of radius 0. A solenoid of length 0. find th e frequency of oscillations. (c) the maximum angle through which coil is deflected. (JEE 2004) 8. A and B. if a current i0 produces a deflection of . each carrying current in the same direction are in the x-y plane of a gravity free space. Three infinitely long thin wires. enter a uniform magnetic field the direction of which is perpendicular to their velocities./2 in the coil in reaching equilibrium position. (JEE 1990) 6. (JEE 1989) 5.01 m carries a current of 0. (JEE 1997) (a) Find the locus of points for which the magnetic field B is zero. In a moving coil galvanometer.4 A.4 m and having 500 turns of wire carries a current of 3 A.Q direction of the magnetic induction at the origin O. . Find the ratio of radii of the circular paths of the two particles. whre i is current through the wire and k is constant. 7. If the linear density of the wire is . A proton and an alpha particle. (B) If the central wire is displaced along the z-direction by a small amount and released. after being accelerated through same potentia l difference. (Ignore the damping in mechanical oscillations) (JEE 2005) . show that it will execute simple harmonic motion. torque on the coil can be expressed as b = ki. (b) the torsional constant of the spring. Calculate the torque requi red to hold the coil in the middle of the solenoid with its axis perpendicular to the axis of the solenoid. d. id charge Q is passed thr ough the coil almost instantaneously. The rectangular coil of the galvanometer having numbers of turns N. The central wire is along the y-axis while the other two are along x = . (C S).0185 eV 8. (C P). 21 22 23 24 25 26 27 28 29 30 Ans. (C Q). D. (D Q)] 17. 51 52 53 54 55 56 57 58 59 60 Ans. . . 2 4 iL . (C QR). (B Q). (B P). evB. [(A Q). F alse 12. [(A Q). (D RS)] PASSAGE TYPES Passage = 1. (D P)] 16. . (C PQT). Bq 6 t m . . AC A ACD FILL IN THE BLANKS / TRUE-FALSE / MATCH TABLE 1. (B P). True 9. (D S)] 14. C 4. 10 11 12 Ans. 3. 1. A B A ABD C C ABD B AC D Q. (C R). 1. 2. A 5. 3d. 14. True 10. B A D D C A D A D C Q. 41 42 43 44 45 46 47 48 49 50 Ans. D 4.in 54 Answer Key ASSERTION&REASON Q. 2m V Bqd m . [(A QR). B C B D B C B A B B Q. 1 2 3 4 5 6 7 8 9 10 Ans. A 3. (B RS). 3. D 2. 1 2 3 4 5 6 7 8 9 Ans. +z direction 6. 11 12 13 14 15 16 17 18 19 20 Ans. 61 62 63 Ans. (B PQR). . [(A R). C Passage = 3. [(A PQ). . 31 32 33 34 35 36 37 38 39 40 Ans. [(A PRS). ABCD 7. False 11.lB . (C PQR). 5. . E B C Level # 1 Q. 1. [(A PT). . (D QS)] 18. (D Q)] 15. 1. B 2.25 x 10 23 Am2 4.physicsashok. A 5. D Passage = 2. (B R). R . (B P). A E A B A A D A B Q. C C B BCD ACD ABC A BD C AD Q. (D QS)] 13. .MAGNETIC FIELD www. C D C D ABCD C D B C C Q. . A A C B ACD B B ABD A A Q. C 2. 0 1 2 1 1 4 I R R . C 3. .. .4 2 1. . 0 0 q a 4 2 B V j i 8 2 rm . 0. . . . 30mA. . (b) A ( 0a 0 / ) ln [2 1) /(2 1)] 0. 8.40 N 0 0 .BR4 3. .p . 4 .3mT. 14. 13. . . B 4 0 / d sin 0. . 1/ 2N.B/ a . /N. . 2 0 3 pm .4 T. . . 0.MAGNETIC FIELD www.. . 6. . . . . 30 . 0. m .. .S . . 0 nµ I tan 2R n . . . . .in 55 Level # 2 1. . (b) p 1/ 4 R4. .. . . B q 2mv 0 0 2. 12. .. 2. . (a) F 2 / .m 11. (a) B 1/ 2 0 R . 4. . B .mg. . . . . B /h (1 (h / 2 R) 0. 15. 7.m 10. .. . 2 2 pm .5 A. . .. . 0.5 kPa. . . . 0 2 0 9. . 5. . (B) Zero. . .r R. .physicsashok. .. . . . 0 2 B 2 1 4 r . . r R.10 J.10mT.R B/.. .. . . . . / 4 2 / r. . . 0. (a) 0 . . ei (2 mv ) ae 0 0 m .2 Sec. . .d . . .. . . Force on segment CD = 8. 18. . (i) M eh 4 m . . . . /qB = 20. . . . 0 man 2 W DT BQr . (a) 6. . . (ii) ehB 8 m . . . F = mE . 4. 2..N. .1 × 10 6 N (inwards) 3. r m q 2 . 5.. . . . .. 6. . . .2mV arcsin dB q . . Level # 3 1. . 7. . .54 x 10 5 Tesla (b) 0. . 3 x . 2E x q/m a(a 2b)B 2 . . p p p r m q 1 . .. z 0. . 2 2 1 2 2 (B B ) q/m 8 V . . . . .. (a) . . . 1 x 10 4 T or Wb m 2. and B. . = 5. (b) 1 2 0 d 2 .. Force on arc AC = 0.9 x 10 6 Nm. 17. 16. d . directed perpendicular to the both M . . X X X X . . 2i NAB . (a) k = NAB. (c) 0 Q NAB 21i . (b) 0 C .8. ATOMIC PHYSICS DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. TIRUPATI. PH NO. 9440025125 . Intensity of light source at a point The amount ofenergyincident ona point per unit area at that point ina unit time is called the intensityoflight . m = 2 E c = 2 h c . p . n = numbers of photons emitted per sec by the source. m (c)de-brogliewavelength of photon . Energy emitted per sec. = frequency of emitted photons.c m.in 1 MASS OF A PHOTON Rest mass of photon is zero. straight line tan.physicsashok. p =photon momentum . p O m . = h p (d) Effectivemass of photon . (d) Effectivemass versuswavelength for a photon.. = c (b) E = mc2 . = h c . then Thus. Rectangular hyperbola Power of a light source Suppose P = Power of the light source. (c)wavelengthversusmomentum. Energyof single photon= h.E =mc2 m= 2 E c but E = h. straight line tan. = c2 . E O m (a) p = mc . Energy of photon. Effectivemass : Ifwe assume thatmis the effectivemass of photons inmoving condit ions.p = h . then according to Einstein s theory. by the source =W= h. = pc. = p c C1: Drawthe graph of (a)momentumversus effectivemass of photon (b) Energyversus effectivemass. .ATOMIC PHYSICS www. Rectangular hyperbola m = 2 E c = 2 h c . Let. = 2 hc . Sol: . at that point. . physicsashok. intensity of sunlight received by earth = I = 1400W/m2..49 × 1011m. . Suppose. Intensity of light at a distance r : Let. =meanwavelength = 6000Å. I = 2 W 4. of photon incident / sec /Area .in 2 Thus. . n = 4 r2 I hc .. I = 2 W 4. Power. [. calculate the number of photons emitted fromthe sun per sec ond assuming the average radius of earth s orbit around sun is 1.. = .r Example 1: The intensityof sunlight on the surface of earth is 1400W/m2.where n is number of photons emitted bysun per sec.(i) Again. intensityat a point is energyincident per unit area unit time. intensityat a distance r froma point source of powerW.49 × 1011m.. . I = Intensityof light at a point . Sol: Average radius.W= nh. = hc .Assuming themeanwavelength of sunlight to be 6000 Å. = I 4 r2 hc .r Thus.r Sun r Earth Power = P I = 2 nh 4 r . [Fromequation (i)] n = I 4 r2 h . SI unit ofintensityis J/secondm2. I = n0h. Intensityat distance r = W Area = 2 W 4. h. . ] . .. = Frequency of photon n0 = no. . Energy emitted per second by the sum= Power of the sun =W but. . W= Power of a point source r Amount of energy received byspherical surface per second P = Amount of energy emitted by light source per second =W Hence.ATOMIC PHYSICS www. . r = 1. e. . . .. is the frequency of radiation.. .63 10 3 10 . . = IA = 2 W 4 r . . A If . Photon Flux (i. . then the energy is photon is given by h.. n = 1. IfAis the area of themetal surface onwhichradiations are incident. . .. . . photon/sec) Suppose. . thenthe power received bythe plate is W.49) 10 6 10 1400 6. .18 × 1045 photons per sec. .2 22 7 34 8 4 (1.W= Power of a source. = n. .ATOMIC PHYSICS www. .Assuming themeanwave lengthof sunlight to be 6000 Å. is the number of photon incident per sec or photon-flux .concentration = n c . .physicsashok. Photon flowdensity. where c is the speed of light.in 3 If n. n0. n.e.W= 3. Photon-concentration : The number ofphotons per unit volume of the space. h. .. . = W' h. = n ' A = 2 W 4. .9× 1026W. . then Photon . where n. A=N.. .r Power received by areaAon earth.W. .r h. .. . h. W.r × A but. n. i. Example 2: The power of light emitted bythe sun is 3.W. h.49 × 1011m. . calculate the photon flux arriving at 1m2 area on earth perpendicular t o light radiations.9 × 1026W r = earth sunmean distance = 1. . = 2 W A 4 r h . I = 2 W 4. = 2 W 4. = n. is the number of photons incident on the plate per second. . . . Photon flow density (n0) : The number of photons incident on unit area of the plate in one second is called the photonflowdensity. . we have W. . Sol: Power of light emitted by the sun. 2 W 4 r .49 × 1011m . = meanwavelength of sun light = 6000 Å= 6 × 10 7m P Sun r Earth Intensityoflight an earth. then. If n is the number of photons incident per unit area per second. = Plate 2 W A 4 r h . n´ is called photon flux. . r P Area(A) . = I ×A .. . The average radius of earth s orbit around sun is 1. 9 10 ) (6 10 ) 1 4 (1. . . .h.22 × 1021 photon/sec . . = 2 W A 4 r hc . . . . . . ..63 10 3 10 . . . .] n´ = 26 7 2 22 34 8 (3.49) 10 6. . . photon /sec = 4. . [. . . . .. . . n. = hc/. i. . × (change inmomentumof each photon) but. . ] = Radiation pressure .. P A t . P A t . .e. . .t = I h.we get P A t ... . ... . . .. (b) change in momentumof light takes place. change ofmomentumsuffered by each photon= h . .. i. force exerted = P t .(ii) Using equation (i)&(ii).in 4 PRESSURE EXERTED BY PHOTONS OR RADIATIONS PRESSURE (a) Each photon carries energyandmomentum..ATOMIC PHYSICS www. = I h .(i) Momentumdelivered to unit area inunit time. . . [. × (change inmomentumof each photon) ... . × (change inmomentumof each photon) . . = N A t . Energyincident onunit area in unit time = I(by the definition of intensity).e..(iii) Radiation pressure for perfectly absorbing surface : In this case.. which causes impulse or force on the surface resulting in a pressure called radiation pressure. Each photon carries energy h. Number of photons incident on unit area in unit time. = Force exerted A . andmomentum= h c . . N A. . the light is either absorbed or reflected or both is done by the surface.. . .physicsashok.Hencewhen photons of light is incident on a surface. Let light of intensityI is incident on a surface. Pr= Radiation pressure = I h. × 2h c .we get hv c Perfectly absorbing surface Radiation pressure = I h. . change inmomentumof each photon = h c . . hv c Perfectly reflecting surface hv Hence. . . . . . h c . = I c . . . . . . . = 2I c . . . using the equation (iii).c . Radiation pressure = I c for a perfectlyabsorbing surface. . . we get c Radiation pressure = I h . = 2h c . . . . Radiation pressure for a perfectly reflecting surface : In this case. . Using the equation(iii). × h c . Radiation pressure = 2I c for a perfectlyreflecting surface. . Radiation pressure = I c (ii) For a perfectly reflecting surface . c In this case. . Change inmomentum= h c .. = 1 . by the definition of radiation pressure byphotons of light . Thus.13 . Radiation pressure = (1 + .13ms and energyE = 10J. . . = (1 + .in 5 Radiation pressure for a surface of reflection coefficient (. . d2 Pulse duration. .physicsashok. Radiation pressure = (1 + 1) I c = 2I c C2: Alaser emits a light pulse of duration . E = 10J Plate area = 4 Laser d d = 10 . . = reflection coefficient .we get. .) I c .50 I = Intensityoflight c = speed of light = 3 × 108m/s . = 0. Radiation pressure = (1 + .) : hv c surface hv .) h c . = reflection coefficient = 0. = 0 . .50.ATOMIC PHYSICS www.m . h c .. .mon a surf ace perpendicular to the beamand possessing a reflection -coefficient . using equation(iii). h c . = 0.) I c for photons falling normallyon a surface.(i) Here. . Sol: Laser energy. Momentumof reflected photon= .e.s Pressure exerted bylight pulse i. .= 0. Find the mean pressure exerted by such a light pulsewhen it is focussed into a spot of diameter d = 10 . NOTE : (i) For a perfectly absorbing surface ....Momentumof incident photon= h c . . . .. . . . . . . .Bythe definitionof intensityof light. I = 2 E d 4 .) 2 4E c.9 × 107 Nm 2 = 490 atm .50) 8 3 12 4 10 3 10 (0. .. .(ii) . . I = Laser energy (pulse duration). ) 100 10.13 10. = 4.e. .d = (1 + 0. .. . .(Area of plate) i. From(i)&(ii) Radiation pressure = (1 + . . = 3.6 10 3 10 .1 10 10 1.physicsashok.ATOMIC PHYSICS www. Pressure = force / area = i Ae . Pr. Number of electrons incident per second = i/e = no. . . . . . . . . . . . . .6 10 . . . . . . . (p1 + p2) . . .7 × 10 6 Nm 2 . . Find the pressure on the target. 2mE E c . 4 19 31 4 19 1/ 2 8 (2 9. of photons emitted per sec ond. . .Consider the idealsituationwhere eachincident electron gives rise to a photon ofthe same energy. assuming that the electrons strike the target normallyan d the photons leave the target normally. . . . . . Sol: Energy of each electron = energy of each photon= E = 104 eV Momentumdelivered by each electron = p1 = 2mE (photon) p e 1 p2 Plate Momentumtaken away byeach photon = p2 = E c Change ofmomentumdue to each electron-photon pair = p1 ( p2) = p1 + p2 Current incident on target = i= 10 6A .Acurrent flows in an x-ray tube operated at 10. . .6 10 ) 10 1. . . . Pr = 6 4 19 10 10 1. .in 6 Example 3: 1 . Totalmomentumchange per second = force = i e . . . . The target a rea is 10 4 m2. . . (p1 + p2) = i Ae . . .000 V. .FORCE EXERTED BY PHOTONS ON A SURFACE We know that.. and. . . . .when photons of light is incident on a surface.(ii) Momentumdelivered inunit time. × (change inmomentumof each photon) = P h. the change inmoment umof photons takes place resulting in a force exertion by the photons on the surface. P t . of photons incident in time . (change inmomentumof each photon) but.... = N t .t. = rate ofchange ofmomentum . Let light ofpowerWis incident on a surface. . power is given by W = N .t (h.. whereN= total no. . .(i) Energyincident in unit time =W(bythe definition of power) Power =W Surface But. .t = W h. . P t . . Each photoncarries energy h.momentum= h c . N .)... . force = P h . hv c surface hv . × (change inmomentumof each photon) . . . Force exerted by photons on a perfectly reflecting surface: In this case. ... .. . . . . Thus. . . .(iii) Force exerted by photons on a perfectly absorbing surface : In this case.. . ...we get hv c Perfectly absorbing surface force = W h . × 2h . . h c . . . = F = Force exerted by the photons. . using equation(iii).. . . F = W c for a perfectly absorbing surface..ATOMIC PHYSICS www. P t . . .physicsashok. .we get Force = W h . change ofmomentumsuffered by each photon= h c . . . = W c thus. = 2h c . × h c . c using equation ( iii).in 7 Hence. . change ofmomentumsuffered by each photon = h c . . = (1 + .c . . . . . . . F = 2P c .. . . . . . h c . . . = (1 + . .: In this casemomentumof incident photon= h c .e.) h c . Change inmomentum = h c . . c . = 2W c thus. . for a perfectlyreflecting surface. h c . . using equation(iii). . Thus. . . hv c surface hv Momentumof reflected photon= . F = (1 + .we get Force = h . . Force exerted by photons on a surface of reflection coefficient .) h c . .) W c i. × (1 + .) W c for photons falling normally on a surface. change ofmomentum= (E/c) Force exerted = [(1 + . . ./c . . rate of change ofmomentum= force = (1 + . Find th emean pressure exerted by such a pulsewhen it is focused on a spot of diameter d =10 µmon a surface perpendi cular to the beamand with a reflection coefficient . If S is the area of the mirror onwhich light falls. Momentumof the incident photons =(I/c)(S cos .physicsashok. .5.50Wcm 2 falls on a planemirror of reflecti on coefficient .ATOMIC PHYSICS www.in 8 C3: If a point source of light of powerWis placed at the centre of curvature of a hemispherical surface..) E/c]/T./c) in the opposite direction.) IS cos2.E/c) = (1 + . .(IS cos2. . = 6. Sol. = 0.whose inner surface is completelyreflecting. the transverse section of the incident beamis S cos .5 × 104 cos2 45º)/3 × 108 = 1.d2 = 2 8 4 10 4(1 )E 4(1 0.) I cos2. ./c) cos .37 × 106 Nm 2 C7: Aplane light wave of intensityI = 0. Sol. required pressure = (1 + 0. = 0. .25 × 10 5 Nm 2 Example 4:An isotropic point source of radiationpower Pis placed on the axis of an idealmirror./c.= IS cos 2. p.The angle of incidence . .The distance between the source and themirror is n times the radius of themirro. .) E/c]/T Pressure = 4 [(1 + . Momentumof reflected photons = . Find the for . = 45º. normal pressure = force/area = (1 + . then net force exerted on the plate is given by Fnet = 2W c cos.5) (0. . W (Source) C4: If a perfectly reflecting solid sphere of radius r is kept in the path of a parallel beamof light of large aperture and having an intensityI..10ms and energyE = 10 J.).)E/c .momentumof a photon = hv/c = E/c Momentumof reflected photon = .5) 10 cTd 3 10 10./c .8. . . C5: If photon of light of powerWfalls at an angle . I r Note that force is equal to the product of (I/c) and the projected area of the sphere. C6: Alaser emits a light pluse of durationT = 0.(E/c) ( .. to a perfectly reflecting surface. 10. then the force exerted by the beamon the sphere is given F = pr2I c .Normal component ofmomentum flux = (IS cos . . Find the normal presure exerted by light on that surfa ce. then the force onthe hemisphere due to the light falling onit is given byF = W 2c . .ce that light exerts on the mirror. in verticaldirection. plane strip ofmassmsuspended froma fixed suppor t through a string.ATOMIC PHYSICS www. momentumof flux = P sin . where . + d./2c d S nR B B R A . .) (2. . d.A continuous beamofmonochromatic light is incident horizontally on the strip and i s completely absorbed. = (P/4. .) . and . T F m mg T = tension in the string and Tcos. find the deflection of the str ing fromthe vertical.physicsashok.in 9 Sol. hence force exerted by the photons of light. force onmirror = 0 p sin cos d c . Tsin. . F = N h c . Dividing the above equations. = P sin ./c . = (Nh ) c . in horizontaldirection. The energyfalling on the strip per unit time isW. Energyfluxthrough the annular space between two cones ofhalf-angles .we get . by the definition of power . but. rate of change ofmomentumalong the normal. = F. cos . ./2c) cos . ifthe strip stays inequilibrium. =mg. t henfor the equilibriumofthe strip. Sol: Force exerted by the photons of light = (number of incident photons per sec) × (change inmomentumof each photon) Let. F = W c If the stringmakes an angle . d. F = P sin2 . W = Nhv. sin .withthe verticalwhen the strip is at equilibrium. = half-angle of the cone subtened by themirror = tan 1(1/n) . 2(P sin . d. d. Number of incident photons per sec =N Light change inmomentumof each photon= h c . / 2c = P/2(n2 + 1)c Example 5: Figure shows a small. . = F mg = W/c mg = W cmg . . . Find the force that the light exerts on the sphere. . cos. p.. = W cmg .momentumof the incident pulse = E/c. . Example 6.. i p . with respect to a frame of reference with the outward normal as the y-axis and a line perpendicular to it and lying in the plane of themirror as the x-axis . i . . tan.0 cm..tan. j) .80Wcm 2 illuminates a sphere of ra diusR = 5.. Sol.. (E / c) (sin . = tan 1 W cmg . Aplane light wave of intensity I = 0. . (. = 2(N'h ) c . . 0. 2. 1..h.. E / c (. E = total energy=N. = 2E c [.we get Impulse = Total change inmomentumof photons = (Total number of photons) × (change inmomentumof each photon) . .82 . change inmomentumof each photon = h c .. .. .physicsashok. 2 . × (change inmomentumof each photon) .. . E / c 1. Impulse =N. × (change inmomentumofeach photon) = N. p .108 . f i . This change inmomentumofthe photons causes impulse. Impulse = E c for a perfectlyabsorbing surface. . | . using equation(i).E / c j . (.(i) Impulse on a totally absorbing surface: In this case.1) cos. (.10 / 3 . cos.. p | . 0.1)2 sin2 . j) . .E/ c) (sin. p . .8cos60º . i .cos 2. Impulse = .ATOMIC PHYSICS www. Using impulse momentumtheorem. .in 10 pf = . | .we get Impulse = N'h c . .. = 5.1)E / c sin .. × h h c c ..1)2 cos2 .] . .2 × 10 8 Nm 2 Impulse applied by photon on a surface Let hn be the energy of photons of a light incident normallyon a surface. Momentumof anincident photon= h c . . p . . . where E is the total energyof the light . = E c . . .E/c and f p .2 . . i . (. . ... . . Impulse on a totally reflecting surface: Impulse = N. p | . Change inmomentumof the photon takes place due to impact ofthe photonswith the s urface.. Impulse on a surface of reflection-coefficient . = (1 + .we get . . . change inmomentumof each photon = h c . Using equation(i).2E c for a perfectlyreflecting surface..: In this case. . h c . .) h c . . . . . ) = H .. finalmomentumof themirror =mv change inmomentumof themirror due to impact =0 ( mv) =mv. Sol: Impulse applies bylaser on themirror. Loss in kinetic energy=Gainin potential energy 1 2 mv2 = mgh . Impulse =change inmomentum 2E c = mv or v = 2E mc . (g = 10m/s2). Find the angle throughwhichthe threadwillbe deflectedwhen a short laserwith energyE =13J is shot in horizontaldirection at right angles to themirror.ATOMIC PHYSICS www.(1 + . m m Laser . = (1 + .(1 Let. .. asmirror is perfectly reflecting. = (1 + . v2 = 2gH 2 2 2 4E m c = 2gH. [E =total energy] ... cos.physicsashok.. Impulse =(1 + . Thus.in 11 Impulse = N. .. ..) N'h c .where v is the speed ofmirror just after impact.) E c for photons falling normally on the surface. .. . .) h c . Impulsemomentumtheoremgives. But. = 10 cmas shown in the figure. . Initialmomentumof themirror = 0 v = 0 v .) E c . Impulse= 2E c .(i) Totalmechanical energyof themirror willbe conserved after impact. Example 7: A small perfectly reflecting mirror of mass m = 10 mg is suspended byaweightless thread oflength . [as v = 2E mc fromequation (i)] 2 2 2 2E m c . = 5 8 13 10. . 1 cos.= gH .3. (1 cos.1 m .10 10. H = .0. . . = E2 mc g. .).(1 cos. ] sin 2 . . . sin 2 .1 = 0. = 2 2 2 2E m c g. . = 0. . m= 10 × 10 6 kg. = 2 2 2 2E m c g. . 2sin2 2 .0043 or . . [. . .)] 1 cos.5º PHOTONS UNDER GRAVITY . . = 0. . 2 2 2 2E m c = g. = 2sin2 2 . Here E = 13J. [. but wewill consider themass to be co nstant as difference in v´ and v is very small. Sol: Change in frequencyof the photontakes place. E = hv Ground H Let v. Sol: Let v. . m = 2 E c = 2 h c .be the frequencyphoton at a infinite distance. E = hv + mgH v. .in 12 Photons can be considered as a particle ofmassm. v. . Example 8:Aplanet ofmassMand radius R emits a photon of frequency v. thenwh at will be its frequency on the surface of earth. . .ATOMIC PHYSICS www.. a photonof frequency v acts gravitationally like a particle ofmass 2 h c . be the frequency of photon on the surface of earth.physicsashok. hv + 2 h c . = v 2 1 gH c . C8: If a photon frequency v falls on the surface of earth froma height h. v. hv + mgH = hv.What will be the frequency of photonwhen it covers an infinite distance. m= 2 h c . Energy of photon on the surface = hv G Mm . . then mc2 = E = hv . . R at . v. M G=Gravitational constant v Mass of the photon.. Mass of photon = 2 h c . If m = mass of photon v = frequencyof photon E = energy ofphoton. . gH = hv. . Mass of the photon depends onits frequency. Energyconservation gives Initial energy= final energy . Thus. hv 2 GM h R c . NOTE: . .. PHOTOELECTRIC EFFECT .. . = 2 GM Rc is called frequency shift. . . Energyof photonat infinity= hv. . Energymust be conserved. = hv´ .. . . .. = v 2 1 GM Rc .. . v.... ..R = hv 2 GM h R c . . . even then photoelectrons are produced but theyare not allow ed to leave plate C (i) firstly. Hence..According to this explanationwhena single photon is inci dent on ametalsurface.However. the incident photons eject electrons by collisionwith its atoms. current starts flowing. no current is found to flowwhen light falls on plate C. This phenomenonwas first discovered byHeinrichHertz in 1887. it loses its negative charge. Experimental Study of Photoelectric Effect Quartz C A + E V Photoelectric phenomenoncanbe studiedwith the help ofa simple apparatus showninfigure.Ten years later. Einstein proposed an expla nation of photoelectric effect as early as 1905. no current is found to flowin the circuit. The explanation of the above behaviour lies in the fact that whenE is irradiated with light. It consists oftwo photosensitive surfaces E andC enclosed in an evacuated quartz bulb.(i) It is knownasEinstein s photoelectric equation. hf =W0 + 1/2mv2 .Lenard showed that the action of light was to cause the emission of free electrons from metal surface. Einstein s Photoelectric Equation: Following Planck s idea that light consists of photons. This energyis known as the photoelectricwork function of themetal and is represented byW0. potassium. In the absence of light. In case. WhenC is irradiated...physicsashok. These photoelectrons are imm ediatelyattracted bythe collector plateC thereby starting current flowas indicated bythe ammeter. Thomson and P. (ii) the balance of the photon energyis used up in giving the electron a kinetic energyof 1/2mv2. there is no flowof current in the circuit and the ammeterA reads zero. itwas discoveredthat alkalimetals like lithium.Although these electrons are no different fromallother electrons. because of the pulling effect of positive potentialofCand (ii) secondly.Afterwards. sodium.Hallw achsmade the important observation thatwhen a negatively-charged zinc plate is irradiatedwith ultra-violet light.J. the photon energy is just sufficient to liberate the electrononly then no energywould be available for imparting kinetic energy to the electron. When plate E is exposed to some source of monochromatic light.Hence.. rubidium and caesiumeject electronswhen visible light falls on them. The photon energy is utilised for two purposes: (i) Partlyfor getting the electronfree fromthe atomand awayfromthemetal surface.ATOMIC PHYSICS www.in 13 Ejectionofelectrons fromametalplatewhenilluminated bylight or anyother radiation ofsuitablewavelength (or frequency) is called photoelectric effect. J.One year later. it is customaryto refer to themas photoelectrons. the above equationwould redu . it is completely absorbed an imparts its energyhf to a single electron. due to repulsion fromthe negative plate E. . . the Einstein s photoelectric e quation becomes hf = hf0 + 1 2 mv2 or hf = hf0 + K. therewould be no emission of electronsw hereas for frequencies greater than f0.0) ..(ii) where f0 is calledthe thresholdfrequency.. It is definedas theminimumfrequencywhi chcancause photoelectric emission. electronswould be ejectedwith a certain definite velocity(and h ence kinetic energy). Long Wavelength Limit (.ce to hf0 = W0 .E. For frequencies lower than f0.. Substituting this value ofW0 in equation (i) above. 4 10 W .0will. By analogy.physicsashok. = 26 0 19. .0 . . . . < .in 14 It is thewavelength corresponding to the threshold frequency f0. 0 1 f = 0 h W . it represents the upper limit ofwavelengthfor p hotoelectric phenomenon. Its physical si gnificance is that radiations havingwavelength longer than .0would not be able to eject electrons froma givenm aterialwhereas those having .602 10 W .625 10 W . Now. metre = 0 12. .625 10 1..ATOMIC PHYSICS www. .0 = 0 c f Also. it is also referred to as thresholdwavelength. = 7 0 12. .0 = 8 34 19 0 3 10 6. Inotherwords. . metre (ii)WhenW0 is electron volts (eV) . .0 = 0 ch W (i) WhenW0 is in joules .875 10 W . c = f0 .. W0 = hf0 . 400 . . .0 = 8 34 0 3 10 6. W Å Kinetic energy of Photoelectrons Einstein s photoelectric equationcan be used to find the velocityand hence the kin etic energyof an ejected photo-electron. . . .0 in Å . and f0 = 0 c . . f = c .E. and . . 1 1 . .875 × 10 26 0 . and .E. . = 3 × 108 × 6. . 1 1 . 1 1 . . . joules .0 inmetres = 19. . K. 1 1 . = ch 0 . . . joules . . . = hf hf0 = h(f f0) Now. .625 × 10 34 0 . . . . . .E. . hf = W0 + 1 2 mv2 = hf0 + K. . = 19. . . . . K. .875 × 10 26 0 . . . eV . . . . . .in 15 = 16 19 19. .0 inmetres = 16 0 19. . . . joules .10 .875 10 1.602 10 . and . 1 1 .400 0 . . . and .0 in Å Incidentally. K..ATOMIC PHYSICS www.875. . .625. . 0 . itmaybe noted that this also represents themaximumvalue ofthe kine tic energya photoelectron can have.6. . Emass = h(f = 12. . . .0 in Å . and . .400 0 . . .f joules . .602 10 .10. = 26 0 19.875 10 1. 1 1 . eV .875. f0) = h. . Its value is given by W0 = hf0 = 0 ch . .E. . = 8 34 0 3. 1 1 .10. eV .10. joules .physicsashok. .0 in Å Photoelectric Work Function As explained above it is defined as the energywhich is just sufficient to libera te electrons froma bodywith zero velocity. . = 12. . . .0 in Å = 16 19 0 19. W0 = 0 12. eV .0 in Å Laws of Photoelectric Emission R B V Quartz C A The apparatus shown in figurewas used byMillikan to studythe photoelectric effec t and the various laws governing it.. eV . Bo th these electrodes are enclosed in an evacuated glass envelopewith a quartzwindowthat permits the passa ge of ultraviolet and . E is the emitting electrodemade of the photosensitivematerialand C is the collecting electrode. 400 . S is a source of radiations of variable but known frequency f and intensity I.0 in Å . . number of electrons emitted per second) is dire ctly proportional to the intensityofthe incident light (or radiation).Areversing switch RS enables the polarities of the two electrodes to be reversed. However. if ultraviole t or visible light is allowed to fall on the emitting electrode.No photo-electrons are emitted for wavelength greater than . electrons are liberated and circuit c urrent is set up. The graph is similar to one shown in figure. If the tu be is in the dark. (iii) Themaximumvelocityofelectron emission (and hence kinetic energy) varies . then no photoelectrons are emitted and themicroammeterAread zero.As seen fromf0 =W0/h. Increase in intensity means more photons and hence greater ejection of electrons. Light Intensity I Photoelectric current Frequency constant This can be verified by increasing the intensity of light and measuring the corresponding photoelectric current while holding the frequency of the incident light frequency of the incident light and the accelerating potential V of the collecting electrode C constant. photoelectr ic emissionwas found to be governed bythe following laws: (i) Photoelectric current (i. f0 Frequency Photoelectric current Intensity constant Light oflower frequency(or longerwavelength). Thewavelengthcorrespondingto be thresholdfrequencyf0 is called thresholdwaveleng thor longwavelength limit. It is called threshold frequency and its value depends on the nature of thematerial irradiated because for eachma terial there is a certain minimumenergynessecaryto liberate anelectron.As shown. however strong. This fact can be verified by keeping the light intensity constant while varying the frequency. (ii) For each photosensitive surface. This energy is known as photoelect ricwork function or threshold energyW0. willnot be able to produce any electron emission.e. it is seen that there is a limiting or critical frequency below which no photoelectons are emitted.physicsashok.0. Moreover. Fromthe experimentaldata collected byRichardson and Compton in 1912.The graph so obtained is shownin figure. there is aminimumfrequency of radiation (called threshold frequency) at whichemission begins. anypotentialdifference can be established between the tw o electrodes.The photoelectric or quantumyield (wh ich is definedas the photoelectric current in amperes per watt of incident light) depends on the frequency (and not intensity) of the incident light.ATOMIC PHYSICS www.in 16 visible light. nomatter ho wlong the light falls on the surface or howgreater is its intensity..The current is found to increase (though non-linearly) with the frequency of the incident light. The same fact is illustrated byfigure. If the intensityof the incident light is increased (keeping frequencyconstant). f f0 Frequency Emax . As seen fromEinstein s photoelectric equation of 2 max 1 mv 2 = h(f f0) or Emax . Incidentally.m ore photonswillbe incident . itmay be noted that the slope of the curve gives the value ofPlanck s constant h. increase in the frequency ofthe incident light increase the velocitywithw hich photoelectrons are ejected.linearlywith the frequencyo the incident light but is independent of its intensi ty. Intensity constant Hence. doubling the intensity of the incident light mer ely doubles the current but does not affect the value ofV0.However. then 2 max 1 mv 2 = eV0 or vmax = 0 2.eV m . This velocity is such that it gives enough kinetic energy to the electrons so as to surmount the retarding electric field between the two electrodes. Up to some stage as this p.ATOMIC PHYSICS www. IfVis reversedwith the help of the reversing switchRS i. The proportion of the electrons having a particular velocityis independent of the intensityof r adiation.Obviously. E ismade positive and Cnegative.e. (v) Electrons are emitted almost instantaneously evenwhen the intensityof incide nt radiations is very low. Since an electron can absorb only one photon. The time lag between the incidence of radiation and emission of first electrons is less than 10 8 second.more photo electro nswill be ejected.d. the frequencyand intensityofincident light are held constant but the potential difference V between the two electrodes E and C is increased.physicsashok. is increased photoelectric current is also increased. soon some value ofVis reachedwhen the current reaches a limiting or saturation value when al the photoelectrons emitted by E are immediately collected byC. each photon having the same energy. Further increase inVhardlyproduces anyappreciable increase in current as shown by the flat portion of curve I in figure. some electrons do succeed in re achingCdespite the fact that the electric field opposes theirmotion. the current i does not immediatelydrop to zero proving that electrons are emitted fromE with some definite velocity.Hence. v i v0 O a b I H Suppose in figure. Now. stopping potentialV0 isdirectlyproportional to fre quencybut is independent of the intensityof the incident light. each photoelectronwillhave the sam e energyandwill be ejectedwiththe same velocity. increase inintensityonlyincreases the nu mber of photo-electrons ejected but not their kinetic energy. When reversedVismade large enough. As seen fromcurve II of figure.in 17 on themetal surface. (iv) The velocities of emitted electrons have values betweenzero and a definitem aximum. if vmax is themaximumvelocityofemission of a photoelectron andV0 the stoppi ng potential. (vi) For a givenmetalsurface. Stopping potentialis that value ofthe retarding potentialdifference betweenthe t wo electrodeswhich is just sufficient to half themost energetic photoelectrons emitted. a valueV0 (called stopping or inhibiting pote ntial) is reachedwhen current is reduced to zero.Hence. But as frequencyis inc reased above f0.1. it is found that the stopping potential varies linearlywith frequency as shown in figure.76. the stopping potentialvaries linearlywith the frequencyofincident light. hence stoppingpotential is zero for that reason. the experiment is repeated by varying the frequency of the light. Below threshold freq uency.V0 = 5.10 .93 × 105 0 V m/s Obviously Emax = eV0 joules = V0 electron-volt If however. .or vmax = 11 2. no electrons are emitted. and . V0 = 12.0 in Å C9: Photoelectrons with a maximumspeed of 7 × 105m/sec are emitted froma metal sur facewhen light of frequency 8 × 1014Hz falls on it. . . . Sol: (i) W0 = 0 . . W0 = hf0 and 2 max 1 mv 2 = eV0 .1 × 10 31 × (7 × 105)2 = 6. V0 = ch e 0 . 2 max 1 mv 2 = h(f f0) or 1 2 × 9. . . = 12. . . Calculate (i) the maximumenergy of the photoelectrons emitted and (ii) thew ork function for tungsten. . .0 inmetres . Calyculate the threshold frequency of the surfac e. . Now. . . .625 × 10 34 (8 × 1014 f0) . 1 1 . . 1 1 . V0 Now.in 18 Einstein s photoelectric equationmay be expressed interms of stopping potentialas given below: hf = W0 + 2 max 1 mv 2 f0 Frequency Stopping potential.4 × 10 7 0 .ATOMIC PHYSICS www. and f0 = 0 c . volt . . Sol: Emax = h(f f0) .635 × 1014 Hz C10:Atungsten cathodewhose thresholdwavelength 2300Åis irradiated by ultraviolet l ight ofwavelength 1800Å.400 0 . .physicsashok. . hf = hf0 + eV0 or V0 = 0 h(f f ) e . f0 = 4. 1 1 . . f = c . . volt . . and . bothin electron-volts. . .12. . . . 400 . = 12. . . . 400 2300 = 5.400 0 .400 1 1 1800 2300 . eV = 12. 1 1 . . . . . .what is photoelectric thresholdwavelength ? .= 1. = 6000Åfalls on ametalsurface and emits photoelectronswith a ve locityof 4 × 105m/s.4 eV (ii) Emax = 12.5 eV C11: If light of . Out of this. . 4.0 = 12. 000 6000 = 2.625 10 . Sol: (i) f = c .625 × 10 34 × 7.0 = 12.7 1. 1 1 .1 0. .1 × 10 31 × 5 2 19 (4 10 ) 1.5 × 1014 Hz (ii) E = hf = 6. . 000 1. = 6000Å= 12. Hence.8 eV would be used for dislodging the electron and the balancewould represent its kinetic energy.E. = 4000 Au strikes the cathode of a photocell a retard ing potential of 0. = 7. eV.95 × 10 19 J = 3.5 × 1014 = 4.445 eV Energy content of photon of .625 eV . wemay use . . W0 = 2.what will bemaximumkinetic energyof the photoelectro ns coming out? Sol: Energy of the light photon = 12.1 eV (iii) W0= hf K. .E. . = 8 10 3 10 4000 10.625 = 7631 Å C12: Calculate the threshold frequency for gold having photoelectricwork functio nequal to 4.. Emax = 5.physicsashok.Calculate (i) light frequency(ii) photon energy(iii)work function (iv) threshold frequencyand (v) net energyafter the electron leaves the surface.8 = 0.07 0.4 = 2.in 19 Sol: K. = 0. 400 4. . C13:When violet light of .4 V is required to stop emission of electrons. Emax = 12. . If light of wavelength 2220Åfalls on gold. of photoelectrons = 1 2 × 9.6 10.07 eV . .7 eV (iv) f0 = 0 W h = 19 34 2.445 = 1.8 eV.58 4.6 10 6. .400 0 .58 eV.78 eV Alternatively.ATOMIC PHYSICS www. = hf V0 = 3.8 = 2583 Å. . 000 2220 = 5. . 5 × 1014 Hz (v) Net energy hf W0 = 3. or hc .972 × 10 19 J.E.. .1 2. Calculate the value of Planck s constant and the thresholdwavelength for the photo cathode. Sol: hf = W0 + K.7 = 0.4 × 10 20 J Example 9:Aphoton ofwavelength 3310Åfalling on a photo cathode ejects an electron of energy 3 × 10 19 J and one of wavelength 5000 Å ejects an electron of energy 0. . . = W0 + K. . = 6.4 eV = 6. .E. physicsashok. = 1. . Find the stopping pote ntialwhen themetal is irradiatedwith (a) monochromatic light having awavelengthof 4000Å. 2000Å V0 = 12. . .3 V (c) Ifwork function is double.W0 = 3 × 10 19J . vmax = 7. .ATOMIC PHYSICS www.2 volt (b) Stopping potential is independent ofthe intensity of the incident light but varies directlyas frequencyf provided it ismore than f0. . .62 × 10 34 J-s Substituting this value of h. . . then .400 1 1 2000 6525 . 8 10 h 3 10 5000 10. = 4.0 is reduced to half i. what would be the a nswer to (a) and (b) above? Sol: (a) V0 = 12. = W0 + 3 × 10 19 In the second case. 8 10 h 3 10 3310 10.62 10 3 10 . . (b) light having twice the frequencyand three times the intensity of that in (a) above.12 × 105 m/s Example 10:Acertainmetalhas a thresholdwavelength of 6525Å. . (c) If a material having double the work function were used. . = W0 + 0. . = 6620 × 10 10 m.972 × 10 19 Subtracting one fromthe other. . .e. .0 = 0 ch W = 8 34 19 3 10 6.e. . .0 = 6525 2 =3262.400 1 1 4000 6525 . . . h = 6. . Since the incident light . thewavelength of the light is half i. . Since frequencyis twice. . .in 20 In the first case. . .5Å. . . a stopping poten tial of 3. .400 1 1 5000 .1 V. . = 2000 Å. . V0 = 12.1 = 12. Find the unknownwavelength. V0 = 12. = ? Now. . . . 1 1 . . . 3. = 2. . .0 = 5000 Å. . .With an unknownwavelength.has . . V0 = 3. In the second case. . Sol: Here.5 . . .400 1 1 2000 3262. . . . .4 volt C14:Acertainmetallic surface is illuminated bymonochromatic light of variablewav elength. .1Vis necessary to stop photoelectric current.400 0 . . it would not be able to produce photoemission. = 2.222 Å . = 4000 Å. . .No photoelectrons are emitted above awavelength of 5000Å. . 3 1.04% (C) 20% (D) 10% Sol. In aluminium.(1) . I ne t .9V.. .in 21 C15: Light ofwavelength 2000Åfalls on analuminiumsurface. the percentage of incident photonswhich produce photoelectrons. .6e = h (2 × 1015 f0) 12. The percentage of incident photonswhichproduce photoelectrons is = n / t 100 N/ t .6 10 2 10 . If the current i n the cell is 0. h = 19 15 8.2 × 10 19 J (ii) Emin = 0 (iii) Ve = Emax = 2 V (iv) .9e = h(4 × 1015 f0) Subtracting one fromthe other. .2 eVare required to remove an electron.0 = 0 12400 W = 12400 4. 10 3Wof 5000Ålight is directed on a photoelectric cell. . . = 6.When light of frequency 4 × 1015Hz is used.4% (B) . . is (A) 0.6 10 .we get 4.4 Å C16: The stopping potential is 4.44 × 10 34 Js Example 11. .ATOMIC PHYSICS www.2 = 2952.3 × 1. .16 10 t e 1. 16 19 n I 0.we have 8.2 eV (i) Emax = (6. Sol: eV0 = h(f f0) Substituting the two given values.3e = 2h × 1015 8. Calculate the value ofPlanck s constant.2) eV = 2 eV = 2 × 1. the stopping potential is 12.2 4.6 × 10 19 = 2h × 1015 .16 µA.. 4.physicsashok.6Vfor light offrequency 2 × 1015Hz.6 × 10 19 = 3. Determine (i) KE of the fastest emitted photo-electron (ii) KE of the slowest emitted photoelectron (iii) stopping potentialand (iv) cut-offwavelength for aluminium. Sol: Photon energyofthe incident light is = 12400 2000 = 6. . n 1012 t .(2) and W N hc t . . .. . (2) and (3)] Persentage .. 0.. . . . . .. . N W 1016 t hc 4 .. . .(3) Persentage = 12 16 10 4 100 10 . . . . . (1). . [fromeq.04% Hence (B) is correct. (2) Substracte eqn.E. . (2) 4hc hc K´ K 3 . K . . K K´ K 3 . 2mK P .2.. .physicsashok.(1) hc K´ 3 / 4 . hc . then : (A) 2K1 = K2 (B) K1 = 2K2 (C) K1 < 2 K 2 (D) K1 > 2K2 Sol. .... = P2 2m P . with excitingwavelength . 4K K´ 3 .. . K h . Hence (D) is correct.ATOMIC PHYSICS www. K´ K K 3 . (1) fromeq. . 1 K . . If the excitingwavelengthis changed to 3 4 .in 22 Example 12. . . . . . Example 13. . . K. .K . .. themaximumkinet ic energyofelectron isK.1= 2. .2. hc K´ K 3 . In a photo-emissive cell. . If . 4hc 3hc K´ K 3 . 1 2 2 1 K . the the kinetic energyof the fastest emitted electronwillbe : (A) 3K/4 (B) 4K/3 (C) less than 4K/3 (D) greater than 4K/3 Sol.1and K2 corresponding to . Let K1 be themaximumkinetic energy of photoelectrons emitted by a li ght ofwavelength . . . .K . . . Hence (C) is correct. + K2 5 . and E2 = . + K1 K1 = . = . . . 2 max1 2 max 2 v 1 v 4 .2 .ATOMIC PHYSICS www. .1 = 2. Example 14. The ratio of themaximumvelocity of photoelectro ns emitted is the two caseswill be (A) 1 : 2 (B) 2 : 1 (C) 1 : 4 (D) 4 : 1 Sol.physicsashok. 2 1 K 4 K . + K1 2 . E1 = 2 . + K2 K2 = 4 . . E1 = .in 23 . 2 max 1 2 max 2 1 mv 1 2 1 mv 4 2 . 2 1 K K 4 . 1 2 K 1 K 4 . 2 1 K K 2 . = . 2 2 2 1 2 K K . E2 = 5 . Radiation of two photon energies twice and five times thework functi on ofmetal are incident sucessively on themetal surface. b= 2. the eject ed photoelectrons have maximumkinetic energyTa eVand de-Brogliewavelength.. themaximumkinetic energyof photoelectrons liberated from another metal B by photones of energy 4. then find (a) Thework function of a (b) Thework function of b is (c) Ta and Tb . max 1 max 2 v 1 v 2 . Hence (A) is correct.a.25 eVstrike the surface ofametalA.5) eV.a.7 eV is Tb = (Ta 1. Example 15. If the De-Broglie wavelength of these photoelectrons is .When photons of energy 4. . = .7 = .in 24 Sol. + eV or h . A a P .physicsashok.0 eV. 2 2 A A max 1 P mv 2 2m . .ATOMIC PHYSICS www.b = 2. . h . .(2) But .25 = 2 A 2a h 2m . . 2 A Amax h 1 mv 2 . wheremismass of electron. 4. . . and a A h P . . . .2 eV (c) 2 eV and 0.Thework function ofth emetal is 3. the frequency of the incident light is __________. .B + Tb 2 B 2b h 4. . = 3 + 1 = 4 eV ... . h .5 eV C17: An isolatedmetal body is illuminatedwithmonochromatic light and is observed to become charged to a steadypositive potential1.25 eV (b) 4. or 4. Sol.a 1.0Vwithrespect to the surrounding. . .5 and Tb = Ta After solving (a) 2. . . . .(1) For B. 2A A P h 2m .7 2m .. . .96 × 1015 Hz Example 16. . . . . If only 1 of each 5 × 109 incident photons is absorbed and causes an electronto be ejected fromthe sur face. . .t = no. Sol. . of photon incident per second. . .63 10 . . 663mWof light froma 540 nmsource is incident on the surface of ameta l. 663 10 3 N hc t . .6 10 h 6. N . the totalphotocurrent in the circuit is _______. . = 0.19 34 4 eV 4 1. ejects electronswhichrequire twich the voltageV0 stop theminreaching a colle ctor. 0 V 15 volt 8 .2 = 220 nm After solving. I ne 5.ATOMIC PHYSICS www. . . . .1 mfroma metal surface. Light ofwavelength 330 nmfallingon a piece ofmetal ejects electronsw ithsufficient energywhich requires voltageV0 to prevent a electron fromreading collector.6 × 10 34 Js and 1 eV= 1. . . eV . . . .6 × 10 19 J) Sol. . . 0 1 hc . . and 0 2 hc . . . (take plank s constant. 3 9 9 n 1 N 1 663 10 540 t 5 10 t 5 10 1242 nmeV .1 = 330 nm and . . light ofwavelength 220 nm. . 2eV .physicsashok. Asmall 10Wsource of ultraviolet light ofwavelength 99 nmis held at a distance 0.76 1011 A t .in 25 . . . 9 n / t 1 N/ t 5 10 . . . . Example 17. .05 nm. . Find (i) the average number of photons striking an atomper second. . N 663 10 3 663 10 3 t hc 1242 nmeV 540 . Sol. Find the numerical value of voltageV0. Here . . . In the same setu p. Example 18. . h = 6. The radius of an atomof themetal is approximately0. . (ii) thenumber ofphotoelectrons emitted per unit area per secondifthe efficiency ofliberationofphotoelectrons is 1%. . . (i) I = intensity = 2 10 . . . . . .w= power incident on atom = . .1) .05 10 4 10 . . . . w n hc t . . . . n w 5 t hc / 16 . .(0. . . . .4.2 9 2 2 I r 10 0. . . no. . . 1 2 0 1 2 0. . = . Fromgraph 15 h 2 e 0.ATOMIC PHYSICS www.49 0. 19 15 h 2 1. .0 eV and of radius 8. = 2 eV Example 20. . .49 10 . estimate the value ofwork functionof the cesiumand Planck s constant. The surface ofcesiumis illuminatedwithmonochromatic light of variouswavelengths and the stopping potentials for the wavelengths are measured. h .physicsashok.6 10 0. . (b) Find the ratio of thewavelengthof incident light to the de-Brogliewavelength . .Assume that the sphere is isolate d and initially neutral and that photoelectrons are instantlyswept awayafter emission. 2 × 10 3Wemitsmonoenergetic photons of energy 5. The efficiency o f photoelectron emission is one for every 106 incident photons. . of ejected electrons per unit area per second I 1 1020 hc / 100 80 . .5 1. h = 6. + eVs or s V h e e . of photons incident per unit area per second I hc / . (a) Calculate the number of photoelectrons emitted per second. .0 eV. .8 mfrom the c entre of a stationary metallic sphere of work function 3. . . .0 × 10 3 m.0 1. Example 19.53 × 10 34 Js But 2 e . .49 10 . .in 26 (ii) no. In a photoelectric effect set-up a point source of light of power 3. .5 v × 1015Hz supporting potential (volt) Sol. The results of this experiment is plotted as shownin the figure. The source is located at a distance of 0. . . 0×10 3m Hence number of photons emitted per second . energyemitted per second. (c) It is observed that the photoelectronemission stops at a certain time t afte r the light source is switched on why? (d) Evaluate the time t. (a) Energyof emitted photons E1 = 5. E2 = 3.2 × 10 3 watt or 3. Sol.8 m r = 8.0 × 1.of the fastest photoelectrons emitted.2 × 10 3 J/s Therefore.6 × 10 19 J E1 = 8.2 × 10 3 J. s 0.0 eV = 5.0 × 10 19 J Power of the point source is 3. 0 10 .01 × 10 4) . .0 eV = 2. . (Here h = Planck s constant andm=mass of electron) .0mfromthe source S wi ll be n2 = 1 2 n 4.01 × 10 4 m2 Therefore.0 × 1015 photons/sec.6 × 10 19 J Kmax = 3.r2 = . or 3 1 19 3.0 1015 4 (0. number of photons incident on the sphere per second are n3 = n2 A= (5.ATOMIC PHYSICS www. . 5.0 × 1014 photon/sec The area ofmetallic sphere overwhich photonswill fallis : A = .2 10 n 8.0 × 1014 × 2.physicsashok. 2. . m2. . . .2 × 10 19 J The de-Brogliewavelength of these photoelectronswillbe 1 max h h p 2 K m .0 3.8) = 4.64) . n = 3 6 n 10 = 11 6 10 10 = 105 per second or n = 105 per second (b)Maximumkinetic energyof photoelectrons Kmax = Energy of incident photones work function Kmax = (5.in 27 2 1 1 E n E .0) eV = 2.(0.0 × 1. . . 1011 per second But since one photoelectron is emitted for every 106 photons hence number of pho toelectrons emitted per second. Number of photons incident on unit area at a distance of 8. n1 = 4.(8 × 10 3)2 m2 . .0 eV . . (b)As discussed in part (c). emission of photoelectrons is stoppedwhen potential on themetal sphere is equal to the stoppeing potentialof fastestmoving electrons. . . Since Kmax = 2.34 1 19 31 6. . . . . . the desired ratio is 2 1 2475 285.68×10 10 m = 8.2 10 9. .2 (inÅ) = 1 12375 E (in eV) or .68 . .1 10 .2 = 12375 5 Å = 2476 Å Therefore. (c)As soonas electrons are emitted fromthemetal sphere.1 = 8.1 8. Emissionof photoelectrons is stoppedwhen its potential is equal to the stopping potential required for fastestmoving electrons.68 Å Wavelength of incident light . . it gets positively charg ed and acquires positive potential.63 10 2 3. The positive potential graduallyincreases asmore andmore photoelectro ns are emitted fromits surface. . [Given. . W0.Assuming efficiency of photo-electron ge neration per incident photon to be same for both the cases.where .When anoth ermonochromatic radiation ofwavelength .1 is given by 1 2 mv1 2 = 1 hc . Maximumkinetic energy of photoelectrons is given byEk = hc .1. Maximumkinetic energyof photoelectrons emitted byradiation ofwavelength . .0 10 9 4 r 8.6 × 10 34 Js. q 9. .2 =1650Åand power P = 5mWis incident. . Similarly. .(1) wheremismass of an electron and v1 ismaximumvelocityof photoelectrons. Then .in 28 Therefore. 9 .1= 3000Åfalls on a photocellopera ting in saturation mode.6 × 10 19 coul.. . . time required to acquire to charge qwill be 2 14 q 1. or t . W0 or 2 1 0 1 mv 2 hc W . iswavelength of incident radiation andW0 iswork function of the surface onwhich radiation is incident.] Sol. 22 0 .728 10 t sec sec 1. 111 second Example 21.Monochromatic radiation ofwavelength . . . . for radiation ofwavelength .The corresponding spectral sensitivityof photocellis J = 4. stopping potentialV0 =2 volt. c = 3 × 108 ms 1 and e = 1. calculate (i) thresholdwavelength for the cell and (ii) saturation current in second case.8mA/W. .Let q be the charge required for the pot entialon the sphere to be equal to stopping potentialor 2 volt. . It is found thatmaxi mumvelocityof photoelectrons increases to n = 2 times. 1.physicsashok. 3 0 2 1 .6 10. .0 10..ATOMIC PHYSICS www. . h = 6.78 × 10 12 C Photoelectrons emitted per second = 105 [part a] or charge emitted per second = (1.. q = 1.6 × 10 14) C Therefore. . .6 . . .6 × 10 19) × 105 C = (1. . . . . . .(2) But v2 = 2v1. . 0 0 1 2 4 hc W hc W .. .2 1 mv hc W 2 . .. hc . .. or W0 = 3 eV . .W . . .. .(3) Fromequations (1) and (3). therefore fromequation (2). . 2 1 0 2 2mv . . . . E2 = 2 hc .ATOMIC PHYSICS www. Energy of each photon ofwavelength .1 =3000Åis J = 4.2 in a radiation of power P 2 2 P P E hc .0198 ( / hc) .6 10 . . (i) In saturationmode.1= 3000Åis incident. Number of photons in 1 joule radiation ofwavelength .2.1 1 1 1 E hc . spectalsensitivitywithwavelength . . . 16 1 3 10 0. .physicsashok. = 3 × 1016 . a charge of 4.8mCflows i n saturationmode or 4. . . of electrons ejected by these photons = J e = 3 19 4. .1 is E1 = 1 hc . Rate of incidence of photons ofwavelength . Energy of eachphoton ofwavelength . .0 is thresholdwavelength. No.in 29 But work-functionW0 = 0 hc . . Itmeans when 1 joule radiation ofwavelength .8 mC e electrons are ejected.8mA/Wor 4. . .8 10 1. where .0 = 0 hc W = 4125 Å Ans. .8m C/J. Efficiencyof photo-electrongeneration per incident photon. e Cs 1= 13. P hc . (iii) Ifwork function of photo-emissive surface isW0 = 1 eV. per second Since.Aphotoemissive detector D of surface area S = 0. per second . D A L S 60cm 6 m (i) calculate photonflux at centre of screen and photo current in the detector.9. Ans. calculate value ofs topping potential intwo cases (without andwiththe lens in aperture). efficiency .2 µA. ther efore.. Hence. c = 3 × 108 ms 1. . of photo-electron generation is same for both the case. . (ii) Example 22. . . rate of ejection of electrons in later case 2 . = 0. . A monochromatic point source S radiating wavelength . h = 6. = 6000Åwith power P = 2watt.625 × 10 34 J-S. Given. Rate of flow of charge in saturationmode = 2 P hc . Efficiency of detector for photoelectric emissionper incident photon is .2 µ Cs 1 But rate of flowof charge is current. saturation current is second case = 13. calculate new value of photon flux and photo current assuming a uniformaverage transmission of 80%fromt he lens.5 cm2 is placed at centre of the screen. an apertureAof radius R = 1 cm and a large screen are placed as shownin fig. (ii) If a convex lens L of focal length f = 30 cmis inserted in the aperture as shown. Energy of each photon is E . average transmissionfromlens is 80%. rate of transmissionofpho tons fromareaAof lens = 0.33 × 1018 photons/m2s Considering a very small areaAof the lens. Distance of lens fromsource is r2 = 0.1 . Photon flux at lens is . Rate of emission of energy fromthe source is P = 2 watt = 2 Js 1 . f = + 30 cm v = ? Using lens formule. Photon flux at detector.´ .6 × 10 8 amp (ii) When a concave lens is inserted in the aperture. or n = 6. = (0.6) . 16 1 2 1 n 1. therefore.33 10 4 r . .r = 1.in 30 Sol. .60m . Rate of incidence of photons on this area of lens =A.8 A. 2 2 A A v (0. Therefo re.´ Nowconsidering refractionthrough the lens. hc . Therefore.8 A ´) . to calculate photon flux. it refracts incident rays. Rate of emission of photons fromthe source is n P P E hc .´) (0. .physicsashok. u = 60 cm. photon flux and hence photo-current changes.ATOMIC PHYSICS www.288 . first rate of emission of electrons fromthe source should be calculated. v = + 60 cm Since. .´= 2 2 n 4..1S per second Since. Rate ofphotons transmitted per unit solid angle is T (0. S Rate of emission of electrons fromdetector = . .8 . This solid angle. 1 1 1 v u f .1S)e = 9. . Photon flux is rate of incidence ofphotons per unit are ofdetector.6)2 = 0. . . current is charge flowing per second. ..04 × 1018 photons per second (i) Distance of detector fromsource is r1 = 6m . . .´ But these photons are transmitted in a solid angle subtended by the areaAat P as shown in fig. therefore photo current = (. . . photons/m2s Rate of incidence of photons on detector = . 80m .D S P v 4. It takes time for a particular plant to receive a certain amount ofwater. . . loosely-tiedwater bags and a particle physicist throws the bags ra ndomly at the plants. .There should not be anythreshold wavelength according to thewave theory. In part (a).. In part ( b) of the figure. when light falls on ametal surface. it sprays all itswater on that plant ina very short ti me. photoelectrons are ejected and that too withou t any appreciable time delay. No matter howsmall is the intensity.in 31 Solid angle subtended by unit area of detector at P.W .An electronmaybe ejected onlywhen it acquired energymore than thework fun ction. In the photon theory.water is filled in identical. ge ts the fullenergyof the photon andmaycome out immediately. .2S .The dependence ofmaximumkinetic energyonwavelength is also against thewave theory. This fact is also not understood bythewave theory. . Photon flux at D.. 0 0 V 1 hc W 1. stopping potentialV0 is given bye.80) . water is sprayed froma distance on an area containing several plants. lowintensitymeans less number ofphotons and hence l ess number of electrons get a change to absorb energy. Each plant receiveswater at nearly the same rate. .2 = T.According to this theory. hc . therefore.But anyfortunate electrononwhich a photonfalls. energy is continuou slydistributed over the surface..´ = 1.physicsashok. stopping potential is same for boththe cases.more inten sitymeansmore energy and themaximumkinetic energymust increasewiththe increase inintensitywhichis not true.07 volt e . In the sameway.ATOMIC PHYSICS www. Ifwe use a low-intensity source. .67 × 1016 photons/m2s Rate of incidence of photons on detector = . energyis continuouslydistributed over the surface. byusing suffic ientlyintense light of any .´= 2 1 (4.2S Rate of emission of electrons fromdetector = . .we illustrate an analogyto thewave the particle behaviour of light. Thiswill result in no photoelectron.20 × 10 7 amp (iii) Since.All the free electron s at the surface receive light energy. PHOTOELECTRIC EFFECT AND WAVE THEORY OF LIGHT According towave theory.V0 = 0 .2S)e = 1. . .When a bag collideswith a plant. there will be many collisions and any extra energy accumulated so far wi ll be shared with the remainingmetal.. Photo current = (. In figure...whole of the energyassociatedwith a photon is absorbed by a free electronwhen the photon hits it. andW0 bothremainunchanged. itmaytake hours before an electron acquires thismuch energ y fromthe light. . . According to this theory. . and . This is contrary to experim ental observations. In this period. (a) (b) Themaximumkinetic energyof a photoelectrondoes not depend on the intensityof the incident light. an electronmaybe given the required amount of energyto come out. showthe existence of thresholdwavelength. however.wavelength. . Exp eriments. ATOMIC PHYSICS www.physicsashok.in 32 DUAL NATURE OF LIGHT (a) Wave nature: Wave nature of light can be explained on the basis of reflection, refraction, in terference, diffraction and polarization. (b) Particle Nature: Energy is transported byenergy particles, photons. It could be explained byphoto electric effect, Zeeman effect, Comptoneffect etc. MATTER WAVE THEORY OR DE-BROGLIE S THEORY (a) This theorywas given on the basis of duelnature of light. (b) According to de-Broglie theoryeach and everymoving particle has somewave nat ure associatedwithitself which is calledmatter waves. (c) Thus,moving particles like e , proton, neutron, .-particle etc, also behave li kewaves. (d) Thesewaves arewaves of probability. (e) The wavelength associatedwith a moving particles is given by . = h/p, where p is themomentumof the particle. (f) Thiswavelength is known as the de-Brogliewavelength of the particle. DE BROGLIE WAVELENGTHS For Photon For moving particle Rest mass Zero m Effectivemass m = 2 E c = 2 h c . m= 0 2 2 m 1. v / c Energy(kinetic) E = hv = hc . E = 1 2 mv2, E = p2 2m Momentum p = E c = h c . = h . p = mv = 2mE Wavelength . = h p = hc E . = h p = h 2mE Wavelength for charged particle . = h 2mE accelerated byVvolts K.E. = qV Speed c = 3 × 108 m/s v = 2E /m = 2qV/m De-Broglie s explanation for stable Bohr s orbits : (a) De-Broglie suggested that non-radiation ofenergybythe electrons circling in aBohr s orbit can be explained on the basis on the basis ofthe formation of stationarywaves bythe electrons in circularmotion inBohr s orbits. . r . O Fourth Bohr-orbit (b) Comparing this to the vibrations ofawire loop suchstationarywaveswould be formed ifeachwave joins smoothlywith the next. ATOMIC PHYSICS www.physicsashok.in 33 (c) In otherwords the number ofwavelengthsmust be an integer. (d) Condition for stable orbits : An election cancircle around an atomic nucleuswithout radiation energyif the cir cumference ofits orbit is an integralmultiple ofthe electronswavelength. i.e. 2.r = n... condition for stable orbits. 2.r = n h p . pr = nh 2. . mvr = nh 2. [. p = mv] mvr = nh 2. Asmvr is the angularmomentumof the circling electron. Bohr s postulate is justifie d. C18: Calculate the de-Broglie wavelength associated with themotion of earth (mas s = 6 × 1024kg) orbiting around the sun at a speed of 3 × 106ms 1. Sol: . = h mv = 34 24 6 1 6.63 10 (Js) (6 10 ) (3 10 )(kg ms ) . . . . . . . . . = 3.68 × 10 65 m NOTE: The wavelengths associated with the motion of macroscopic objects like ear th, train etc, are negligible compared to their sizes. This is why the wave-like character of these objects is not observable in our daily life. C19: Calculate the de-Broglie wavelength of an .-particle ofmass 6.576 × 10 27 kg an d charge 3.2 × 10 19 coulomb, accelerated though 2000V. Sol: E = kinetic energy of a-particle = qV . E = 3.2 × 10 19 × 2000 J = 6.4 × 10 16 J deBrogliewavelength, . = h 2mE , [. 2mE =momentumof photon] = 34 27 16 6.63 10 (J s) 2 6.576 10 6.4 10 (J kg) . . . . . . . . . . = 2.28 × 10 13 m NOTE: The wavelength associated with .-particles is of the order of size of the .-particle. That is why the wave like character of a-particle is observable. C20:Aparticle ofmassmand charge q is accelerated through a potentialdifferenceV. Find (a) its kinetic energy (b)momentum, and (c) de-Brogliewavelength associatedwith itsmotion. Sol:When the particle is accelerated througha potentialdifferenceV, gain in kine tic energyis given byK= qV. (a) Thus, kinetic energy,K= qV. (b) Momentumof the particle (p) is given byK= p2 2m p = 2mK = 2mqv . Momentum= 2mqv ATOMIC PHYSICS www.physicsashok.in 34 (c) De- Brogliewavelengthis given by, . = h p = h 2mqV Thus, wavelength= h 2mqV Example 23. Assume that the de-Broglie was associatedwith an electron can forma standing wave between the atoms arranged in a one dimensional arraywith nodes at each of the atomic si tes. It is found that one such standingwave is formed if the distance d between the atoms of the array is 2 Å.Asimilar standing wave is again formed if d is increased to 2.5 Åbut not for any intermediate value of d. Find the energy of the electron in eV and the least value of d for which the standing wave of the t ype described above can form. Sol. Fromthe figure it is clear that p . (./2) = 2 Å (p + 1) . ./2 = 2.5 Å 2.0) Å = 0.5 Å . ./2 = (2.5 or ..= 1 Å = 10 10 m (i) deBrogliewavelengthis given by 2 Å 2.5 Å N N p-loops (p + 1) loops /2 h h p 2 km . . . K= kinetic energy of electron 2 34 2 17 2 31 10 2 K h (6.63 10 ) 2.415 10 J 2m 2(9.1 10 )(10 ) . . . . . . . . . . . 17 19 K 2.415 10 eV 1.6 10 . . . . . . . . . . . . K = 150.8 eV (ii)N N The least value of d will bewhen only one loop is formed . dmin = ./2 or dmin = 0.5 Å C21: Find the de-brogliewavelength associatedwith an electron accelerated throug h a potential difference of 30 kV. Sol: Kinetic energyof the electron K = qV = e(30 kV) = 30 keV = 30 × 103 × 1.6 × 10 19 J = 4.8 × 10 15 J Now,momentumof the electron = 2mK = 2.9.1.10.34 . 4.8.10.15 kg.J = 2.96 × 10 24 kg.m /s . De-brogliewavelength, . = h Momentum = 34 24 6.63 10 J s 2.96 10 Kgm/ s . . . . . = 2.24 × 10 10 m = 2.24 A ATOMIC PHYSICS www.physicsashok.in 35 ATOMIC STRUCTURE ATOMIC MODEL By now, it iswell-known thatmatter, electricity and radiation etc. are all atomi c incharacter.Although, no one has so far seen individualatoms, there is no doubt that they reallyexist. In 1895, it was discovered by J.Perrin in Paris that the cathode rays consist of negatively-charged practices called electrons. In 1897, J.J. Thomson measured the e/mratio for an electronwhereas its chargewasmeasured byMillikan in 1906 byhis famous oil-drop experiment.Mass of the electronwas found by dividing charge e bythe ratio e/m.Discoveryofpositive rays during the latter part of 19th centuryindicated tha t a normalatomconsisted of both negative and positive charges. But howthese charges are distributed in a n atomwas not known at that time. Consinuous efforts have beenmade since then to studythe physical stru cture of anatomsuch as its extra-nuclear electronic structure chieflywiththe help of spectralproperties of atoms.To account for the spectroscopic dataobtained experimentallyover the years, several theories regard ing atomic structure have been proposed fromtime to timewhich are called the atomicmodels.Various atomicmo dels proposed by scientists over the last fewdecades are: (i)Thomson sPlumpuddingmodel, (ii) Rutherford sNuclearmodel, (iii) Bohr smodel(iv) Som merfeld s Relativisticmodel (v)Vectormodel and finally(vi)Wave-mechanicalmodel. These differentmodels have been suggested one after the other inaneffort to get a satisfactoryinterpretation of the experimentaldatawhich, it is hoped, willultimately lead to a perfect and complete understanding of the physical structure of an atom. Thomson s Plum PuddingModel According to thismodel, the atomis regarded as a heavy sphere of positive charge seasonedwith enough electron plums to make it electricallyneutral. Thomson visualized the positive charge ofanatombeing spread out uniformly throughout a sphere of about 10 10 metre radius with electrons as smaller particles distributed in circular shells as shown infigure.Whereas the net force + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + + ++ + + + + + + + + + + + + + + + + + + + exerted bythe positively-charged sphere on each electron is towards the centre o f the sphere, the different electrons experiencemutual repulsion and get arranged in the formof circular she lls. This atomicmodelwas given up after some time because it could not provide any sa tisfactorymechanism for explaining the large deflection suffered by.-particles inRutherford s experime nt. Rutherford s Experiment on .-particle Scattering As showninfigure high-speed a-particle (i.e. heliumnucleieachwitha charge of+2e) fromsome radioactive material like radiumor radon, confined to a narrowbeamby a hole in a lead block, weremade to strike a verythin goldfoilG.Whilemost ofthe .-particleswent straight throughthe foilas if therewerenothing there (and produced scintillations on a fluorescent screen), some of them collided with th e atoms ofthe foil and were scattered around at various angles -a fewbeing turned back towards the sour ce itself. 150º 45º 60º . . . . . . Radium Lead block + Beam of .-particles ATOMIC PHYSICS www.physicsashok.in 36 Althoughsome smallangle scattering could be expected fromThomson smodel, large-ang le scatteringwas absolutelynot expectedat all. Further detailedexperiments inRutherford s laborator ybyGeiger andMarsden showed conslusivelythat large-angle scattering canbe expected onlyifone assumes that amassive positive point charge exists at the centre of each gold atomas shown in figure. According to thismodelproposed byRutherford in 1911, the positivemassive part of anatomis assumed to be concentrated in a very smallvolume at its centre. This central core, nowca lled nucleus, is surrounded by a cloud of electronswhichmakes the entire atomelectrically neutral. . . . P +Ze Nucleus Asymptote of trajectory Trajectory of .-particle The large-angle scattering of positively-charged .-particles could be easily explained on this atomic model as shown in figure. This scattering is due to themutual repulsion (as per Coulomb s law) between the .-particles and the concentrated positive charge on the nucleus. The .-particle approaches the positively-charged nucleus alongAO. If there were no repulsion fromthe nucleus, it would have passed at a distance of p fromit.However, due to coulombic force ofrepulsion, the .-particle follows a hyperbolawith nucleus as i ts focus.The linesAOand A.Oare the asymptotes of the hyperbola and present approximatelythe initial and finaldirections of the .particlewhen it has passed out of the effective range of the nuclear electric fi eld. As seen, the .-particle is deflected through an angle .. The perpendicular dista nce fromthe nucleus to the lineAOis called the impact parameter and is denoted by p.The Rutherford s scatteri ng formula is tan 2 . = 1 2 Q Q 2pE where Q1 = Charge of the incoming .-particle Q2 = Charge ofthe scattering nucleus E = Kinetic energy of the incident .-particle p = Impact parameter Distance of Closest Approach Suppose that an .-particle approaches a positively-charged nucleus for a head on collisionwith a kinetic energy ofK.As shown in figure at pointA, the repulsive force ofthe nucleus is so strong as to stop the .particlemomentarily.At this point, all the kinetic energyof the .-particle is co nverted into potential energy. Let Dbe the distance of closest approach of the .-particle. The potential at poi ntAdue to nuclear charge Ze is = 0 Ze 4.. D + D A . Potential energy of the .-particlewhen at pointAis = 0 Ze.2e 4.. D = 2 0 2Ze 4.. D . K = 2 0 2Ze 4.. D or D = 2 0 2Ze 4.. K Since .-particles are generally obtained fromnatural radioactive substances, the ir kinetic energyK is known. Hence, value ofDcan be found easily fromthe above equation. ATOMIC PHYSICS www.physicsashok.in 37 C22: InGeiger-Marsden experiment on .-particle scattering fromgold foil, the kin etic energy of .-particles usedwas 7.68MeV. Calculate the distance of closest approach of .-particle if ato mic number of gold in 79. Sol: D = 2 0 2Ze 4.. K Here, Z = 79, e = 1.6 × 10 19 C; .0 = 8.854 × 10 12 F/m K = 7.68 MeV = 7.68 × 1.6 × 10 13 J D = 19 2 12 13 2 79 (1.6 10 ) 4 8.854 10. 7.68 1.6 10. . . . .. . . . . = 2.96 × 10 14 m Major Deficiencies in Rutherford s Nuclear Model It was found later on that Rutherford smodel had two serious drawbacks concerning (i) distribution of electrons outside the nucleus and (ii) the stabilityof the atomas awhole. It canbe shown that electrostatic forces between the positive nucleus and the st atic negative extra-nuclear electrons are not enough to produce equilibriumin such a nuclear atom. For examp le, consider the case of an atom* having two electrons and a nucleuswith a charge of+2e. Suppose the elec trons are symmetrically placed at a distance of r fromthe nucleus and are stationary.The force of attrac tion between the nucleus and each of the electrons is F = e. 2 2e 4. .r = 2 2 2e 4..r while the force of repulsion between the two electrons is 2 2 e 4...4r = 2 2 e 16..r . Since the force of attractionis eight times the force of repulsion, the electronswill fall into the nucleus therebydestroying the stable structure of the atom. + + + + 2e + + + + 2e v v r m e + Photon (a) (b) (c) 2e (hv) To overcome this difficulty, Rutherford suggested that stability can be achieved (as in a solar system) by assuming that electrons, instead ofbeing static, revolve round the nucleuswithsu cha speedthat the centrifugal force balances the attractive force exerted bythe nucleus onthe electrons.As see n fromfigure (b), condition for stabilityis achievedwhen mv2 r = 2 0 2e.e 4.. r or mv2r = 2 0 2e 4. . In general, ifZ is the atomic number, then nuclear charge isZe, so that the abov e relation becomes mv2 r = 2 0 Ze.e 4.. r or mv2r = 2 0 Ze 4. . ATOMIC PHYSICS www.physicsashok.in 38 Incidentally, itmaybe noted that according to the above relation, it is possible to have an infinite number of orbits inwhich electrons can rotate. But this assumptionofrevolving electrons ledto serious difficultyfromthe point o fviewo the electromagnetic theoryaccording towhich an accelerated chargemust continuouslyemit electromagnet ic radiation or energy. Since an electronrevolving in a circular orbit has centripetal acceleration (= v 2/r), itmust radiate energy as per the laws of classical electrodynamics. Due to this continuous loss of energy, the electronswillgraduallyapproachthe nuc leus bya spiral path and finally fall into it as shown infigure (c). Hence, it is seen that the orbitalmo tion of the electron destroys the verypurpose forwhich itwas postulatedi.e. the stabilityofthe atom.Obviously, eit herRutherford s nuclear atomicmodelwithrevolving electrons isdefective or the classicalelectromagnetic t heoryfails inthis particular case. This dilemmawas solved in 1913 byNeils Bohrwho proposed an improved versio n ofRutherford s atomicmodel. Bohr s Atomic model Thismodel (first proposed for hydrogen atombut later applied to other atoms as w ell) retains the two essential features ofRutherford s planetarymodel i.e. (i) the atomhas amassive positively-charged nucleus and (ii) the electrons revolve round their nucleus in circular orbits the centrifuga l force being balanced, as before, by the electrostatic pull between the nucleus and electrons. However, he extended thismodel further byutilizing Planck sQuantumTheory. Hemade t he following three assumptions: (iii) an electron cannot revolve round the nucleus in any arbitrary orbit but in just certain definite and discrete orbits.Onlythose orbits are possible (or permitted) forwhichthe orbital angularm omentum(i.e.moment of momentum) ofthe electron is equal to anintegralmultiple of h 2. i.e. orbitalangularmomentum= nh 2. where n is an integer and h is Planck s constant. Such orbits are also known as stationa ry orbits. (iv) while revolving in these permitted stationary (or stable) orbits, the elect ron does not radiate out any electromagnetic energy. Inother words, the permissible orbits are non-radiating paths of the electron. (v) theatom radiatesout energy only when an electron jumps from oneorbit to anot her. If E2 and E1 are the energies corresponding to two orbits before and after the jump, the frequencyof the emitted photon is given bythe relation E2 E1 = hv or .E = hv where v is the frequency of the emitted radiations. C 23: If I is the moment of inertia of an electron and . its angular velocity, t hen as per assumption (iii) given above .I = nh 2. or (mr2).= nh 2. or (mr2 )v r = n.h 2. or mvr = n.h 2. Alternatively, sincethemomentumoftherevolvingelectronismv, itsmoment about the nucleus is =mvr Hence mvr = n.h 2. .....(i) e m v +ze r ATOMIC PHYSICS www.physicsashok.in 39 where, n= 1, 2, 3 for the first second and third orbits respectively. It is call ed the principalquantumnumber and because it can takewhole number values only, it fixes the sizes ofthe allowe d orbits (also calledBohr s circular orbits). r1 r2 r3 n = 1 n = 2 n = 3 +Ze E1 E2 E3 Permitted orbits (a) K LM +Ze E1 Electron Jump (b) E2 .E E3 Let the different permitted orbit have energies ofE1, E2, E3 etc. as shown in fi gure (a). The electron can be raised from n = 1 orbit to any other higher orbit if it is given proper amount o f energy.When it drops back to n =1 orbit after a short intervalof time, it gives out the energydifference . E in the formof a radiation as shown in figure (b). The relationbetween the energy released and frequency of th e emitted radiation is E2 E1 = hv or .E = hv Expressions for velocity, radius, energy of electron and orbital frequency in Bo hr s orbit Here it should be kept inmind that Bohr smodelis valid onlyfor hydrogen atomand hy drogen-like ions. In otherwords,we can saythat is applicable to hydrogen atomand ions having just one electron. Examples of such ions are He+, Li++, Be+++ etc. According to Bohr s + z e r e, m First postulate, v 2 2 0 Ze 4.. r = mv2 r where 0 1 4. . = 9 × 109 Nm2 C 2 v = velocityof electron and .0 = 8.854 × 10 12 C2/Nm2 m = mass of an electron = Absolute permittivity Z = atomic number .of vacuumor free space r = radius ofthe orbit e = magnitude of charge on an electron n = principalquantumnumber i.. 2... v = 2 0 ze 2. we have mvr = nh 2.. mv = nh 2... 3..(ii) where. r = 2 2 0 Ze 4.we get mv × 2 2 0 Ze 4.. let us substitute the value of v in equation (ii)..e. orbit number . n = 1. .. 4... .. mv .e. i. .(i) Fromfourth postulate ofBohr.... . n is a positive integer... nh Now. Fromequations (i) and (ii). 53Å. then rn = a0 n2 Z K. r . = nh 2. h n = K = kZe2 2r .. v . Vn = V1 Z n .0] U = 2 4 2 2 2 0 mZ e .E. 2 2 0 2 r n h mZe . If a0 = first Bohr radius = 0. . 1 Z For fixed z. r = U = kZe2 r [k = 1/4. 1 n . of the electron = 1 2 mv2 = 2 4 2 2 2 0 mZ e 8.ATOMIC PHYSICS www. .. Note that for fixed n. . v .physicsashok.. Z. n2 IfV1 is the speed of the electron in the 1st orbit Then.in 40 m × 2 0 ze r 2 nh . Potential energy of the atom= 2 0 Ze 4. where V1 = c 137 . r . where c is speed of light. . v = V 2.18 × 106 ms 1 . Orbital frequency for the electron. a0 = 0. Bohr radius. h n . E = K + U E = 2 4 2 2 2 0 mZ e 8. .r = 2 2 2 2 0 0 Ze mZe 2 2 hn h n . v = 2 4 2 3 3 0 mZ e 4. NOTE : It is assumed that the acceleration of the nucleus is negligible on accou nt of its large mass. .6 eV = 1 U 2 . En = 13. Some important results for H-atoms when n = 1 1. c/137 3. E1 = 13.4.52 × 10 16 s = 0 1 2 a v . .53 Å rn = a0n2/Z 2. .6 2 2 Z n in eV. h n In general. n h Time period of revolution (T) is given by T = T1 3 2 n Z where T1 = time period of revolution in the 1st orbit = 1. Total energy of the atom. v1 = 2... The stateswith higher energies are called excited states. where B1 is themagnetic field at the centre ofBohr atomdue t o the current generated by themotion ofelectron in 1st orbit. i = ev . For the first excited state. for the 2nd excited state. v1 = 6. This current will lead to self-generatedmagnetic field in the atomand alsomagnetic cu rrent (m) (a)Magnetic field(B) IfBis themagnetic field generated at the centre of atom. The potentialdifference throughwhich an electron should be accelerated to acquir e the value of excitation energyis called excitation potential.2 eV 6.6 eV.in 41 4.physicsashok. Ground state and excited states The state ofan atomwith the lowest energyis called its ground state or normal st ate. n= 1. n = 3 and so on. Formth excited state. B1 = 12. Excitation energy and excitation potential The energyneeded to take the atomfromits ground state to an excited state is cal led the excitation energy of that excited state. where i is the current due to motion of the electron and a0 is the 1st Bohr-radius. 1 rydberg = 13.ATOMIC PHYSICS www. = E. (B) Movement of electron in circular orbits in aBohr atomcauses electric current inthe orbit.5 tesla. identical to its ionization energy . . E1 = E1 = Binding energy of theH-atom.6 × 1015 Hz 7.6 eV.6 eV = 1 U 2 5. The value of ionization energyofH-atomin ground state is 13. Binding energy Binding energyof a systemis the energyneeded to separate its constituents to lar ge distances or itmay be defined as the energy releasedwhenits constituents are brought frominfinityto fo rmthe system.E. n =m+ 1 Ionization energy and ionization potential Theminimumenergyneeded to ionize an atomis called ionization energy. n= 2.6 eVthat of ionizati on potential is 13. then H-atom +e a0 e i B = 0 0 i 2a . U1= 27. K1= 13. Hence.6 eV 8. NOTE : (A) I. The potentialdifference throughwhich anelectron should be accelerated to acquire the value of ionization energyis called ionization potential. The value ofbinding energyofH-atomis 13. For ground state. .hence B = 0 0 e 2a . . . . .ev 20 a . L . (angularmomentumvector) µ = iA = . L .physicsashok. . . n = 435 Example 25:The quantumnumber n in theBohr smodel ofH-atomspecifies: (a) radius of the electron (b) energyof the electron . L . = e 2m . . = iA = . ) .ATOMIC PHYSICS www. 20 a ] L = mV1a0 = m2..va0] . Example 24: The quantumnumber ofBohr orbit inH-atomwhose radius is 0..01 10 0. = 20 2 0 e a m2 a . [i = ev. = iA . A= . we get B = 12.in 42 Putting the values of v and a0. = e 2m Vectorially. ...v 20 a [V1 = 2. = e 2m .01mmis (a) 223 (b) 435 (c) 891 (d) none of these Sol: (b)We know that rn = a0n2 n2 = n 0 r a n = 3 10 0. . . .5 Tesla (b) Magneticmovement vector ( .529 10 .ev 20 a (c) Relation between . & L . . . .28 × 10 10 cm Sol: (a) . .86 10 ) 3.6 10 ) (8. . . .0528 Å (c) 5. = 0. . Example 26: The radius Bohr s orbit in ground state for H-atomis (Take . ..528 Å (b) 0. . Z = 1. . h = 6.0 = 8.1 10 ) (1. . a0 = 2 0 2 h me . .86 × 10 12 C2/Nm2. .28 × 10 10 m (d) 5.(c) angularmomentumof the electron (d) allof these Sol: (d) rn = 2 0 2 h me . . (in this case. . . . h n .14 (9.528 Å . . × n2 En = 4 2 2 2 0 me 8.6 × 10 34 J-s) (a) 0. . .6 10 ) . Ln= n h 2 . n = 1) a0 = 34 2 12 31 19 2 (6. . . in 43 Example 27: The speed of the electron inthe first Bohr orbit ofH-atomis (Take c = speed of light in vacuum) (a) c (b) c 13. Z = 1 and n = 1] = 2 0 e c 2. h n = 1 3 2E hn where E1 = 2 4 2 2 0 . hc But 2 0 e 2.ATOMIC PHYSICS www. binding energy of the electron in the ground state is E 1. v = c 137 Example 28: In a H-atom. hc = 1 137 = fine structure constant .we knowthat v = 2 0 e 2.physicsashok.6 (c) c 137 (d) 137 c Sol: (c) For H-atom. h [since n this case. Then the frequency of revolution of the electron in the nth orbit is (a) 1 3 2E n h (b) 3 1 2E n h (c) 1 3 2mE n h (d) none of these Sol: (b) The frequencyof revolution of the electron in the nth orbit is given by v = 2 4 2 3 3 0 mZ e 4. An electron in a hydrogen like atomis in an excited state. . Z = 2.4 1. . n = 2 .6 10 J)(9. . . h C24: Calculate the energy of aHe+ ion in its first excited state. (i)Kinetic energyof electron in the orbits of hydrogen and hydrogen like at oms = | Total energy | . It has a total e nergy of 3. .6 10 J s) 2(3.we have 34 19 31 (6.63 × 10 10m or . Calculate : (i) the kinetic energy (ii) the de-Brogliewavelength of the electron Sol. .6 2 n . .4 eV (ii) The deBrogliewavelength is given by h h P 2 Km . Sol: . .1 10 kg) . . En = 2 2 13.mZ e 8. Kinetic energy = 3. . . .4 eV. = 6. = 6.6eV)Z n Here. (K= kinetic energyof electron) Substituting thevalues. eV = 13.63 Å .6 eV C25. En = 2 2 (13. . M 3 a a K 81 . . . n = 3 In the electronic third orbit. . L = 2} M L a 16 a 81 .53 eV (B) 1. rn × n2 and vn × 1 n 2n n n v a r . The electron in a hydrogen atommakes transition fromMshell to L. The ratio ofmagnitudes of initial to final centripetal acceleration ofthe electron is (A) 9 : 4 (B) 81 : 16 (C) 4 : 9 (D) 16 : 81 Sol.8 eV Sol.physicsashok. the energy of electron 2 2 Z E 13. Example 31.Here h is Planck s constant. The kinetic energyof this electronis : (A) 4. {. . n 2 2 a 1 n n . M = 3} L 2 a a K 16 .ATOMIC PHYSICS www.in 44 Example 30. The angularmomentumopf an electron in the hydrogenatomis 3h 2. .51 eV (C) 3.6 . {. Hence (D) is correct. . L nh 3h 2 2 . . n 4 a 1 n . . n 4 a K n .4 eV (D) 6. Using theBohr s quantization condition. E 13. .51 eV 9 . .n . Example 32. E 13. find the permissible orbital radii and en ergy levels of that particle. .6 1 9 .6 1. Hence (B) is correct. A particle of mass m moves along a circular orbit in a centrosymmetr ical potential field U(r) = kr2 2 . . = En Example 33.we get E = n h ..in 45 Sol. Hence mv2 r = kr . the orbita l velocity ofmassm when it is nearest to heavyparticle is (A) 2 0 3q 2. .. h Sol. h (B) 2 0 3q 4.AssumingBohr smodel to be true to this system.. .(2) solving equations (1) and (2) we get r = rn = n h m.(1) and mvr = n h where h h 2 . . The necessary centripe tal force to the particle is being provided bythis force F. h (D) 0 3q 4.. . = k m and total energy E = U + K= kr2 2 + 1 2 mv2 Substituting the values.. 2 2 0 . 2 e mv F r . In a hypothetical systema particle ofmass mand charge 3q is moving ar ound a very heavy particle having charge q.physicsashok. where . . h (C) 0 3q 2. F = dU dr = kr Negative signimplies that force is acting towards centre.ATOMIC PHYSICS www. . . . . . . ..q 3q mv 4 r r . Hence (A) is correct. . . . . 2 0 3 q h v 4 2 . 2 0 v 3 q 2 h .. .. . . 2 0 3 q mvr v 4 . . mvr h 2 . a brig ht spectrumcontinuously distributed ona dark background is obtained.Th e spectrumlooks like separate bands of varying colours. is called continuous emissionspectrum. Such type of spectrumis calle d absorption spectrum.in 46 SPECTRUM Dispersed light arranging itself ina patternofdifferent wavelength is referred t o as a spectrum. Molecular energy levels Energy Band spectrum . one obtains two bright yellowlines on a dark background. it is called an emission spectrum. Whenwhite light falls on a prismand the transmitted light is collected on awhite wallorwhite paper then a spectrumis obtainedwhich consists ofdifferent colours fromred to violet. Light emitted in such a process has certain fixedwavelengths. t hematerialmayabsorb certainwavelengths selectively. can emit light to lower its energy.Whenthis light is dispersed bya high resolution grating . Such a spectrumis called line emission spectrum. An emission spectrumcan be three types : (a) Continuous spectrum: That emission spectrumwhich is obtained by continuously varyingwavelength. when electric discharge is passed through sodiumvapour. in an excited state. (c) Band spectrum: Thewavelengths emitted bythemolecular energylevelswhich are g enerallygrouped into severalbunches. Light coming froma sourcemay be dispersed bya prismor by any other dispersingmedium. a candle or a red hot iron piece comes under this category. For example.0 nmand 589. Kinds of spectra : (A) Emission spectra:When a light beamemitted by certain source is dispersed to get the spectrum. Atomic energy levels Energy Line spectrum .An atomormolecule. certain sharp bright lines on a dark background is obtained. (B) Absorption spectrum:Whenwhite light is passed throughan absorbingmaterial. dark lin es or bands at the positions of themissing (absorbed) wavelengths are obtained.physicsashok. eachgroup beingwell separated fromthe other.when light is dispersed.When the transmitted light is dispersed.6 nm.When such a light is dispersed. White light Absorbing material Absorption spectrum . are also grouped. Such a spectrumis called band emission spectr um.ATOMIC PHYSICS www. Light emitted froman electric bulb. (b) Line spectrum: The atoms and molecules can have certain fixed energies. In this case. they vapour emits light of the wavelength 589. physicsashok. (d) According toBohr. Hydrogen Spectra : If hydrogen gas enclosed in a sealed tube is heated to high temperature. There missing lines are called Fraunhofer lines. instead of dark lines we get few characteristic dark bands (against coloured bac kground) called band absorption spectra. CO2 o r KMnO4 solution.When light coming fromthe sun is dis persed. (b) Band Absorption spectra : If absorbing media is polyatomic such as H2.As a result. (b) The electronremainsonlyfor a short intervaloftime (generallyin theorder of10 8 s) in the excited state and comes back to the ground state finally. it emit s radiation.ATOMIC PHYSICS www. This radiation consists of components of different wavelengths which deviate by different amoun ts. Series limit Energy Lyman series Balmer series Paschen series Brackett series n =1 n = 2 n = 3 n = 4 n = 5 n = . (c) The electroncan reach the ground state fromanyone of the excited states inma nyways. n = 1 state).in 47 An absorption spectrum may be of two types : (a) LineAbsorption spectrum : Light may be absorbed by atoms to take themfromlow er energy states to higher energystates. This shows that certainwavelengths are absent. it shows certain sharply defined dark lines. t he gas is found to absorb light of certainwavelength. The absorption spectrumconsists of dark lines on bri ght background. Explanation of hydrogen spectra by Bohr (a) The electron in aH-atomif not disturbed remains in the ground state (i. The radiationwith different amounts ofdeviation formsH-spectrum.many electron transitions take place. E = 0 (e) Lyman: nf = 1: 1 . In the similarwaywhenwhite light is passed through a gas. all electron transitions terminating at a particular state give rise to a particular spectral series.When the electron receives energyfromoutside.e. Such a spectrumis calleda line absorption spectrum. it is elevated to anyone of the higher permitted sta tes (excited state). = 1 E ch 2 2 1 1 1 n . .... . 3. .. . = 1 E ch 2 2 1 1 5 n .. 7.. = 1 E ch 2 2 1 1 4 n . n = 4... = 1 E ch 2 2 1 1 2 n . . 4.. . Brackett: nf = 4: 1 .. .... . . ..... 8. . Pfund: nf = 5: 1 ... . . 4. n = 6. . . ... .. . . = 1 E ch 2 2 1 1 3 n . . Balmer: nf = 2: 1 . ... .. ... .. 7 . . n = 3.... Paschen: nf = 3: 1 . 6... . .. n = 5. n = 2.. 5... . . .. 5. 6. .. . .... . . . According to Bohr. is calledwave number ( . . . . ) of the line and 2. . (g) The value of R is 1. 1 .6 eV (h) 1 . .physicsashok. . = i f E E hc . .. p = p E c Important points regarding H-spectra . . The correspondingwavelength is given by 1 . . the extra energyEi Ef is emitted as a photon of electromagnetic radiation. 1 rydberg = 13. . .in 48 (f) If an electronmakes a jump fromthe nith to nf th orbit (ni> nf). where c = speed of light in vacuum. (i) Photon energy= Ep = hv (j) Momentumof a photon = p = Photon Energy speed of light .ATOMIC PHYSICS www.we canwrite 1 . where R = 4 2 3 0 me 8. ch is called theRydberg constant. is called angular wave number of the line. = RZ2 2 2 f i 1 1 n n . = 2 4 2 3 0 mZ e 8. ch 2 2 f i 1 1 n n . . . Rhc = 13.6 eV.0973 × 107m 1 NOTE: En = 2 2 RhcZ n . e. 2) lines are violet. 2) is red. and the other lines are in the near ultraviolet (uv) .(6 . (f) The series corresponding to uv region. ( 5 . (b) Ahydrogen sample emits radiationwithwavelengths less than those inthe visibl e range (i.(a) The sharplydefined. theH. (c) the linesmay be grouped in separate series. line (4 . discretewavelengths exist in the emitted radiation in th eH-spectrum. 2) is blue. theH. the separation between the consecutive wavelengths decreases as wemove fromhigher wavelength to lowerwavelength. of absorption spectral lines = (n . visible region and infrared region ar e known asLyman. (e) Aparticularminimumwavelength in each series approach a limiting value knowna s series limit. theH. infrared).e. 2) and H. (g) In the Balmer series of hydrogen. uv light) and alsowithwavelengthsmore than those in the visible range (i. Balmer and Paschen series respectively. (d) In each series. (h) Theoretically possible no. 1) (i) Theoretically possible no. of emission spectral lines = n(n 1) 2 . . line (3 . kinetic energy(K). . some quantities are decreased and some are increased.g. The table given belowshowswhich quantities are increased and which are decreased.ATOMIC PHYSICS www. C27. . (c) Intensity of spectral lines could not be explained. He+.. : H. (b) Orbitswere taken as circular but according to SOMMER field these are ellipti cal. For example total number of lines fromn1 = n to n2 = 1 are n(n 1) 2 . C26. (e) It could not be explained theminute structure in spectrumline. (d) Nucleuswas taken as stationary but it also rotates onits own axis. Whenever the force obeys inverse square law 2 F 1 r . Total number of emission lines fromsome excited state n1 to another energy state n2(< n1) is given by 1 2 1 2 (n n )(n n 1) 2 . . .in 49 (j) Approximate range ofwavelength for different colours of visible light Colour Wavelength Range Violet + Indigo 3800 Å to 4500 Å Blue 4500 Å to 5000 Å Green 5000 Å to 5500 Å Yellow 5500 Å to 6000 Å Orange 6000 Å to 6500 Å Red 6500 Å to 7200 Å Infrared rays: 720 nmto 50 mm Ultraviolet light: 10 Å to 3800 Å Limitations of Bohr s Model (a) It is valid onlyfor one electron atomand hydrogen-like ions e. If force is not proportional to 2 .. . . (g) This does not explain the doublets in the spectrumof some of the atoms like sodium(5890Åto 5896Å). Table Increased Decreased Radius Speed Potentialenergy Kinetic energy Totalenergy Angular speed Time period Angularmomentum C28. .physicsashok. . and potential energy is inversely proportional to r. As the principalquantumnumber n is increased in hydrogenand hydrogen like a toms. K = | U | 2 and E = K = U 2 . (f) This does not explain the ZEEMANeffect (splitting up of spectral lines inmag netic field)&Stark effect (splitting up in electric field). potential energy(U) and total energy (E) have the follo wing relationships. Na+10 etc. L i+2. . the above relations do not .1 r or potential energyis not proportional to 1 r . In JEE problems. whileKre mains unchanged. this situation arises at two places. In som e problemsupposewe take zero potential energyin first orbit (U1 = 0). Thus. = R 2 2 1 1 2 3 . Itmeans that th is systemwill separate if 100 J of energyis supplied to this. . C30.. . . .0 nm Sol: min 1 .10 = 91. . = R 2 2 f i 1 1 n n . min 1 . positive or negative. .3 nm (d) 365. For instance. . Kinetic energyofa particle can t be negative. . Total energy of a closed systemis always negative and the modulus of this i s the binding energy of the system. The longest wavelength in this series corresp onds to the smallest energy difference between energy levels. .6 nm (b) 91.2 nm C31: Find the longest wavelength present in theBalmer series of hydrogen: Sol: In the Balmer series. = . 1 .Hence. . . Example 33:The value of series limit inLyman series is: (a) 121.1.. . binding energy of this systemis 100 J. . .097.. . Hence the initial statemust be ni = 3. . 1 .physicsashok. . C29. then themodulus of actual potentia l energyin first orbit (when reference point was at infinity) is added inUand E inall energy states.in 50 hold good. . = R . . 1 . suppose a systemhas a total energyof 100 J.while the potentialenergycan be zero.2 nm (c) 656.ATOMIC PHYSICS www. = R[1 0] . = R 1 1 4 9 . totalenergyof an open systemis either zero or greater than zero. It basically depends on the reference point where we have taken it zero. .min = 7 1 1. It is c ustomary to take zero potential energywhenthe electron is at infinite distance fromthe nucleus. nf = 2. in an atom(betw eennucleus and electron) and in solar system(between sun and planet). . . .10 = 6. Thus nf=1 . . . = R 2 i 1 1 n .5R 36 .5 nm. 1 . . Sol: The uv radiation 102. = 656 nm(near the red end of the visible spectrum) C32:Ahydrogen atomemits uv radiation of102.5.10 = 1 1 1. . Calculate the quantumnumbers of the states involved in the transition.1.56 × 10 7 m . . .5 nmlies inthe lyman region of spectrum. = 7 36 5. 2 i 1 n = 1 9 7 1 102.1.10.097.097.124 . 2 i 1 n = 1 1 .R . .. C34: Howmanydifferent wavelengthsmay be observed in the spectrumfroma hydrogen s ample if the atoms are excited to stateswith principal quantumnumber n ? Sol: The totalnumber of possible transitions is 1) + (n 2) + (n 3) + . Hence.. Ultraviolet light ofwavelength 800Åand 700Åwhen allowed to fallon hydrogen atom s in their ground state is found to liberate electronswith kinetic energy 1. . the unit of 1 R willbemetre i.10. = E0 + 1.I.in 51 2 i 1 n = 0. ..physicsashok.0 eVrespecti vely. . units ? Sol:We know that 1 . Find the value of the Planck constant. .. C36. we canwrite 1 .m.6 × 10 19 and 10 hc 700 .e. 3 Hence transition is from3 . C33:What is the unit of reciprocalofRydberg constant in S.6 × 10 19 . 1... the terminthe bracket is unitless. = R ..8 eVand 4..0 × 1.10. E . . 1 R = . Now. = R 2 2 f i 1 1 n n ..8 × 1. T . 10 hc 800 . = E0 + 4.ATOMIC PHYSICS www. . + 2 + 1 = (n n(n 1) 2 . hv = E0 + T where E0 = ground level energy and T = kinetic energy of electr on 0 hc . C35: Consider the following two statements: (A) Line spectra contain information about atoms only (B) Band spectra contain information aboutmolecules (a) BothAand B are wrong (b)Ais correct but B iswrong (c) B is correct butAis wrong (d) BothAand B are correct Sol: (c) Line spectra contain information about atoms andmolecules both. . Sol...1 2 i n = 10 ni . . .2 × 1.2 1.. . = 6. .Subtracting 8 hc 1 1 10.57 × 10 34 Js . 7 8 .6 × 10 19 or h = 27 8 2. = 2..6 10 56 3 10 . . . . . 6 eV). .in 52 C37: The excitation energy of a hydrogenlike ion in its first excited state is 4 0.ATOMIC PHYSICS www. Find the energy needed to remove the electronfromthe ion. .6 eV) × Z2 × 3 4 . . .6Z 1 = 4 × (13. V x. Sol: For E1.6Z n = 13. = 3x 4 V Example 34:Adoublyionised lithiumatomis hydrogen likewith atomic number 3.4 eV C38: The first ionization potentialof some hydrogen likeBohr atomis xV.6 9 1 . Find thewavelength of the radiation required to excite the electron inLi++ fromthe first to the thirdBohr orbit. . = 122.Then the value of the first excitation potential for this atomwillbe: (a) xV (b) x 2 V (c) 3 4 xV (d) 20 xV Sol: The value of first excitation potentialis given by x. . . . 40. .6 eV) Eion = 54.6Z2 2 2 1 1 1 2 . ionization energy= 2 2 13. n = 1 .physicsashok.8 eV. . Sol: The excitationenergy in the first excited state is E = 13. Z = 3. . = x 1 1 4 .8 = (13. E1 = 2 2 13. (Take ionization energy ofH-atomequal 13. . Z = 2 Now.4 eV E n = 3 3 E n = 2 2 E n = 1 1 . 8 eV . = hc .E = E3 E1 = 108. = 13. . = v .6eV We have . n = 3 . = 3 | v | . .E For E3. 1 | v | . e e e e v1 v2 v3 v4 . E3 = 2 2 13.E + E1 = E3 .8 Å = 114 Å Example 35: InBohr smodelof hydrogen atomwhenthe electron ismoving in one of the s tationaryorbits then: (a) velocityof the electron is fixed and no emission of energy takes place (b) velocitychanges continuouslybut no emission of energytakes place (c) energyis emitted but the velocitydoes not change (d) energyis emitted and the velocity also changes.6 3 3 .. Z = 3.E = 12400 108. Sol: (b) According to the second postulate ofBohr. = 4 | v | . = 2 | v | . Towhat energy level did it jump?How many spectral linesmay be emitted in the course of transition to lower energy levels? Calculate the shortest wavelength.physicsashok.25 eVan d 5. E = E0 2 2 Z n where E0 = 2. .888 or m= 3 Three lines are emitted.. .25 + 5.ATOMIC PHYSICS www. . = E3 E1 = E0 2 2 1 1 1 3 ..1 eV . .2 eV . An electron in an unexcited hydrogenatomacquired an energyof 12. min hc . . .0 .2 and 17.(1) Equation (1) and (2) gives E3 E2 = 17.6)(1/4 1/9) = 17. (lonization energyofHatom= 13. or .6 eV) Sol. .2 eV . Fromthe given conditions En E2 = (10.2 + 17) eV = 27.18 × 10 18 J = 1 rydberg . .E (energygap between unexcited state (n = 1) and an excited state (n= m)) 2 0 2 2 E Z 1 1 1 m . Sol. . . ..The excited atomcanmake a transition to the first excited state bysuccessivelyemitti ng two photons of energy 10.6 × 10 19 = 2. the atomfromthe same excited state c anmake a transition to the second excited state bysuccessivelyemitting two photons ofenergies 4.. = 1. .02 × 10 7 m = 1020 Å Example 37. .1 × 1. .min = 34 8 18 6.. . .6 10 3 10 9 2.. . 12.0 eVrespectively.in 53 Example 36. The shortestwavelength corresponds to the greatest ener gygap.95 eVrespectively.Alternatively..0 eV or Z2 (13.. . Ahydrogen like atom(atomic number Z) is in a higher excited state of quantumnumber n. Determine the values of n and Z.18 10 8 .18 × 10 18 × 12(1 1/m2) n = 3 n = 2 or 1 n = 1 2 1 m = 0.(1) and En E3 = (4.95) eV = 10. . Z2 (13.6) (5/36) = 17.0 . Z2 = 9 . Z = 3 Fromequation (1) Z2 (13.6) (1.4 1/n2) = 27.2 or (3)2(13.6) (1/4 1/n2) = 27.2 or 1/4 1/n2 = 0.222 or 1/n2 = 0.0278 or n2 = 36 . n = 6 ATOMIC EXCITATION WITH THE HELP OF COLLISION (a) An atomcanbe excited to an energyabove its ground state bya collisionwith an other particle inwhich part of their joint kinetic energy is absorbed by the atom. (b) An excited atomreturns to its ground state in an average of 10 8 s by emitting one ormore photons. ATOMIC PHYSICS www.physicsashok.in 54 (c) Energytransfer is amaximumwhenthe colliding particles have the samemass. (d) The energy used in this process will be of discrete nature. (.E = E2 E1 = hv = hc . ). NOTE: If the joint kinetic energy of colliding particles is less than 20.4 eV (c onsidering particles as Hatoms or one neutron and one H-(atom) then the nature of collision will be necessarily elastic. Example 38: Aneutronmovingwith speed vmakes a head-on collisionwith a hydrogen a tomin ground state kept at rest. Find the minimumkinetic energy of the neutron for which inelastic (completely or partially) collisionmay take place. The mass of neutron= mass of hydrogen = 1.67 × 10 27 kg. Sol: Let us suppose that neutron andH-atommove at speeds v1 and v2 after the col lision. Suppose an energy .E is used in thisway. On the basis of conservation of linearmomentumand energy,we canwrite mv = mv1 +mv2 .....(i) 1 2 mv2 = 2 1 1 mv 2 + 22 1 mv 2 + .E .....(ii) Fromequation (i) we have v2 = 2 1 v + 22 v + 2v1 v2 .....(iii) Now, fromequation (ii) v2 = 2 1 v + 22 v + 2 E m . .....(iv) . Fromequation(iii) and (iv) 2v1v2 = 2 E m . Hence, (v1 v2)2 = (v1 + v2)2 4v1v2 = v2 4 E m . Since v1 v2 is always real, v2 4 E m . . 0 mv2 . 4.E . 1 2 mv2 . 2.E Theminimumkinetic energyof the neutron needed for aninelastic collision correspo nds to transition from n = 1 to n = 2 . Kmin = 2 min 1 mv 2 = 2 × 10.2 eV . Kmin = 20.4 eV Example 39. Consider anexcited hydrogen atominstate nmovingwith a velocityv (v<< c). It emits a photon inthedirectionofitsmotionand changes its state to a lowr statem.Applymomentumand energyconservation principle to calculate the frequencyv of the emitted radiation. Compare thiswith the frequencyv0 emitted if the atomwere at rest. Sol. Let En and Em be the energies of electron in nth andmth states. Then ATOMIC PHYSICS www.physicsashok.in 55 En Em = hv0 ...(1) In the second casewhen the atomismovingwith a velocity v. Let v´ be the velocityof atomafter emitting the photon.Applyingconservationof linearmomentum, v v´ m m v mv = mv´ + hv c (m=mass of hydrogen atom) or v´ = v hv mc . . . .. .. ...(2) Applying conservationof energy En + 1 2 mv2 = Em + 1 2 mv´2 + hv or hv = (En Em) + 1 2 m(v2 v´2) hv = hv0 + 1 2 m 2 v2 v h mc . . . . . . . . . . . .. . . .. hv = hv0 + 1 2 m 2 2 2 2 2 2 v v h 2h v m c mc . . . . . . . . . . . hv = hv0 + 2 2 2 h v h c 2mc . . . Here the termis 2 2 2 h 2mc . is very small. So, can be neglected . hv = hv0 + h v h v c c . . . or v 1 v c . . . .. .. v0 or v = v0 1 1 v c . . . . .. .. ; v . v0 1 v c . . . .. .. as v < < c Example 40: Thewavelength ofD1 andD2 lines of sodiumare 5890Åand 5896Årespectively, if theirmean wavelength is 6000Åthen find the difference of excited energy states. Sol: E = hc . ....E = 2 hc . .. .E = 34 8 10 20 6.62 10 3 10 6 10 6000 6000 10 . . . . . . . . . . . .E = 3.31 × 10 22 J . .E = 22 19 3.31 10 1.6 10 . . . . , ........2 × 10 3 eV C39:Alithiumatomhas three electrons.Assume the following simple picture of the a tom.Two electronsmove ATOMIC PHYSICS www.physicsashok.in 56 close to the nucleusmaking up a spherical cloud aroundit and the thirdmoves outs ide this cloud in a circular orbit. Bohr smodel can be used for themotion of this third electron but n = 1 stat es are not available to it. Calculate the ionization energyof lithiumin ground state using the above picture . Sol: In this picture, the third electronmoves in the field of a total charge +3e 2e = +e. Thus, the energy are the same as that of hydrogen atoms. The lowest energy is E2 = 1 E 4 = 13.6eV 4 . = 3.4 eV Thus , the ionization energy of the atomin this picture is 3.4 eV. Example 41: Find the wavelengths in a hydrogen spectrumbetween the range 500 nmt o 700 nm. Sol: The energy of a photon ofwavelength 500 nmis hc . = 1242eV nm 500nm . = 2.44 eV The energy of a photon ofwavelength 700 nmis hc . = 1242eV nm 700nm . = 1.77 eV The energydifference between the states involved in the transition should, there fore, be between 1.77 eV and 2.44 eV. Figure shows some of the energies of hydrogen states. It is clear that onlythose transitionswhich and at n = 2mayemit photons of energybetween 1.77 eVand 2.44 eV. Out of these only n = 3 to n = 2 falls in the proper range. The energyof the photon emitted in the transition n = 3 to n= 2 is E . . = (3.4 1.5)eV= 1.9 eV.Thewavelength is . = hc .E = 1242eV nm 1.9eV . = 654 nm. Example 42: Calculate the (a) velocity, (b) energy, and (c) frequency of the ele ctron in first Bohr orbit of hydrogen atom. Sol. (a)We have, vn = 2 0 Ze 2. nh ; but here Z = 1 and n = 1 . v1 = 2 0 e 2. nh = 19 2 9 34 (1.6 10 ) 1 36 10 2 1 6.62 10 . . . . . .. . . . = 2.18 × 106 m/sec (b) We have, En = 4 2 2 2 2 0 me Z 8. n h Again here, z = 1 and n = 1 ATOMIC PHYSICS www.physicsashok.in 57 . E1 = 31 19 4 2 9 2 2 34 2 9.1 10 (1.6 10 ) 1 (4 9 10 ) 8 1 (6.62 10 ) . . . . . . . . .. . . . . = 21.758 × 10 19 joule = 19 19 21.758 10 1.6 10 . . . . = 13.6 eV (c) We have, v = 2 4 2 3 3 0 Z me 4. n h ; here also n = 1 and Z = 1 = 4 2 3 0 me 4. h = 6.57 × 1015 Hz. C40: Find out the radius ofthe hydrogen atomin ground state. Sol. We have, rn = 2 2 0 2 h n mZe . . ; here Z = 1 and n = 1 r1 = 34 2 9 19 2 31 (6.62 10 ) 1 4 9 10 1 (1.6 10 ) (9.1 10 ) . . . . . .. . . .. . . . . r1 = 0.53 Å Example 43: If thewavelength of the firstmember of the Balmer series of hydrogen spectrumis 6562Å, then calculate thewavelength of firstmember ofLymen series in the same spectrum. Sol. We have, for the first member of the Balmer series v1 = R 2 2 1 1 2 3 . . . . . . . = 5 36 R and for the first member ofLyman series, v2 = R 2 2 1 1 1 2 . . . . . . . = 3R 4 . 1 2 . . = 1 2 . . = 5R 36 × 4 3R = 5 27 . .2 = 1 5 27 . = 5 6552 27 . = 1215.18 Å Example 44.Thehydrogenatominits groundstate is excitedbymeans ofmonochromatic ra diationsofwavelength 975Å.Howmanydifferent lines are possible in the resulting spectrum?Calculate the l ongest wavgelength among them.Youmayassume the ionizationenergy for hydrogen atomto be 13.6 eV, the Planck constant = 6.63 × 10 34 Js. Sol. En= E0Z2/n2. The energyrequired to take the electronfromn = 1 to infinityis the ionization energyof the hydrogen atom. . 13.6 = E0(1/12 1/.) or E0 = 13.6 eV ATOMIC PHYSICS www.physicsashok.in 58 Therefore, for hydrogen En = 13.6/n2 eV The energyof the photon incident on hydrogen is 34 8 18 10 hc 6.63 10 3 10 E h 2.04 10 J 975 10 . . . . . . . . . . . . . Let the electron jump fromn= 1 to n=mafter absorbing the incident photon. Then .E = Em E1 = 13.6(1/12 1/m2)eV = 13.6(1 1/m2) × 1.6 × 10 19 J . 13.6(1 1/m2) × 1.6 × 10 19 = 2.04 × 10 18 . (1 1/m2) = 0.9375 or m= 4 The resulting transitions are shown in the figure. So there are six possible lin es. The longest wavelength corresponds to theminimumenergygap.Hence longest wavelengthcorresponds to transi tionfromm= 4 to m= 3 . 0 2 2 h E 1 1 3 4 . . . . . .. .. . 0 hc E 1 1 9 16 . . . . . .. .. E0 m = 4 m = 3 m = 2 m = 1 . 0 hc 144 7E . . . . 34 8 19 6.63 10 3 10 144 7 13.6 1.6 10 . . . . . . . . . . . . = 1.88 × 10 6 m = 18800 Å. Example 45: The energy of an excited hydrogen atomis 3.4 eV. Calculate the angula r momentumof the electron according to Bohr s theory. Ans: 2.11 × 10 2 joule × sec. Since the energy of an electron in nth level in hydrogen atomis En = 2 RCh n or, 3.4 = 2 13.6 n . RCh = 13.6 eV . n = 2 FromBohr s theory, L= n h 2. . L = 2 × 0.6 10 34 2 3.14 . . . = 2.11 × 10 3 joule sec. REDUCED MASS .In our earlier discussionwe have assumed that the nucleus (a protonin case of h ydrogenatom) remains at rest.With this assumption the values of the Rydberg constant R and the ionizatio n energy of hydrogen predicted byBohr s analysis arewithin 0.1%ofthemeasured values. Rather the proton and electron both revolve in circular orbits about their commo n centre ofmass.We can ATOMIC PHYSICS www.physicsashok.in 59 take themotion of the nucleus into account simplybyreplacing themass ofelectronm bythe reducedmass µ of the electron and nucleus. Here . = Mm M.m .....(i) where M= mass of nucleus. The reduced mass can also be written as, . = m 1 m M . Now, whenM> > m, m M . 0 or . . m For ordinary hydrogen we let M = 1836.2 m. Substituting in equation (i),we get µ = 0.99946mwhen this value is used instead of the electronmassmin the Bohr equations, the predicted values arewellwithin 0.1%of themeasured values. + cm m m v v Separation r Applying the Bohr model to positronium. The electron and the positron revolve about their common centre of mass, which is located midway between them because they have equal mass The concept of reduced mass has other applications.Apositron has the same rest mass as an electron but a charge +e. Apositroniumatomconsists of an electron and a positron, each withmassm, in orbit around their common centre ofmass. This structure lasts only about 10 6 s before two particles annihilate (combine) one another and disappear, but this is enough time to studythe positroniumspectrum. The reducedmass ism/2, so the energy levels and photon frequencies have exactlyhalf the values for the simpleBohrmodelwithinfinite protonmass. CM M m r2 r1 Now, let us provewhymis replaced bythe reducedmass .when motion of nucleus (proton) is also to be considered. In figure both the nucleus (mass =M, charge = e) and electron (mass =m, charge =e ) revolve about their centre ofmass (CM) with same angular velocity (.) but different linear speeds. Let r1 and r2 be the distance ofCMfromproton and electron. Let r be the distance between the proton and the electron. Then, Mr1 = mr2 .....(ii) r1 + r2 = r .....(iii) . r1 = mr M.m and r2 = Mr M.m .....(iv) Centripetal force to the electron is provided by the electrostatic force. So, mr2.2 = 2 2 0 1 e 4.. r or m Mr M r . . . . . . . .2 = 2 2 0 1 . e 4.. r or Mm M m . . . . . . . r3.2 = 2 0 e 4.. ATOMIC PHYSICS www.physicsashok.in 60 or .r3.2 = 2 0 e 4.. .....(v) where Mm M.m = . Moment of inertia of atomabout CM, I = 2 2 1 2 Mr .mr = Mm M M . . . . . . . r2 = .r2 .....(vi) According toBohr s theory, nh 2. = I... µr2..= nh 2. .....(vii) Solving equations (v) and (vii) for r,we get r = 2 2 0 2 n h e . . . .....(viii) Electrical potential energyof the system, U = 2 0 e 4 r . .. and kinetic energy,K = 1 2 I.2 = 1 2 µr2.2 Fromequation (v),.2 = 2 3 0 e 4.. .r , K = 2 0 e 8.. r .Total energyof the system, E = K + U = 2 0 e 8.. r Substituting value of r fromequation(viii),we have E = 4 2 2 2 0 e 8 n h . . .....(ix) The expression for Enwithout considering themotion of proton is En = 4 2 2 2 0 me 8. n h , i.e.,mis replaced byµ while considering themotionof proton. NOTE :(i) Variation of rn, vn and Enwithmass of election is as under, rn . 1 m , vn = independent ofmand En .m Sometimes the electron is replaced bysome another particlewhich has a charge e bu t mass different fromthemass of electron. Here, two cases are possible. Case 1: Let saymass of the replaced particle is x times themass of the electron and nucleus is still very heavycompared to the replaced particle, i.e., themotion of the nucleus is not to be considered. Inthis case Case 2: In this case motion of nucleus is also to be considered. Let say the reducedmass is ytime thema ss of the electron.e.. vn remains unchanged and En becomes y-times. (ii)Reduced massm= 1 2 1 2 m m m . .in 61 rnwill become 1 x times.ATOMIC PHYSICS www. C41:Apositroniumatomis a systemthat consist of a positron and an electron that o rbit each other. mass of t he replaced particle is comparable to themass of the nucleus. In this case themass of the electron is re placed bythe reducedmass of the nucleus and the replaced particle. m ofm1 and m2 is less than both the masses. As a result the wavelength inthe positroniumspectrallines are all twice thoseof the correspondin g lines in theH-spectrum. Sol: In this case reducedmasswillbe given by m. vn will remain unchanged and En becomes x times. the value of correct Rydb erg constant will be (consider hydrogen atom) (a) 4 H 2 3 .n = m' m . . Hence.physicsashok. i. Assuming the nucleus to be of finite mass MH. . rn will become 1 y times. Then. Example 46: Bohr s theoryassumes that nucleus is of infinitemass and so electron r otates round the stationary nucleus. 1 2 E n = 1 2 E 2n It means that the Rydberg constant for positroniumis half as large as it is for H-atom.M = m2 2m = m 2 wherem=mass of the electron. . Compare thewavelengths of the spectral lines of positroniumwith those of ordinaryhydroge n. the energy levels of a positroniumatomare E. . = mM m. Suppose that the radiiof the electron and nucleus orbits are re and rn respectivelythenbydefi nition of centre ofmass. . . . n e n r r . MHrn =mre whenmis themass of electron . (c) 4 H 2 3 H 0 mM e M m 8 ch . . . (d) 4 2 3 0 e 8. . . . . .M FromBohr s quantizationrule. . . . the centre ofmassOand the nucleus are always in a straight line. . sayO. .M and re = H H M r m. (b) 4 H 2 2 3 H 0 M m e mM 8 c h . . ch Sol: (c) In this case both electronand nucleuswill rotate about a common centre ofmass. . . . .M Let r = re + rn then rn = H mr m. r = H m m. . the electron. .H 0 mM e M m 8 ch .we have . r2 = nh 2. 4 2 3 0 e 8. . . . Find thewavelength of the radiationwhen the systemdeexcites fromits first excited state to the ground state. . .MH. . Sol. . The above equation can be comparedwithm. . .Apositromiumatomis a bound systemofan electron (e ) and its antiparticl e positron (e+) revolving about their centre ofmass. 2 4 e n 2 2 2 0 . .r2 = nh 2. .is the angular velocity about the centre ofmass] Putting thevaluesof rn and re.in 62 2 2 m. TheRydberg constantwill be given by R = H H mM m M . ch Obviously the reducedmass of the electron is H H mM m M . . . e e e e e m m m µ m m 2 .physicsashok. . . . . .re .ATOMIC PHYSICS www. [. . This problemcan be solved byBohr s theoryof the hydrogen atombyreplacing them ass of the electron by its reducedmass. . . .rn = nh 2. Example 47. . . .we get H H mM m M . . . e4me/8. Calculate the value of ionization energyof themuonic atom..Amuon can be captured by a nucleus to formamuonic atom. . . . 34 8 18 6. . .18 × 10 18 × 3/8 . E1 = 2. E1 = 2. . .18 × 10 18/8 . . .18 × 10 18(1/2 1/8) = 2. . 4 2 e n 2 2 2 0 e m Z E 8 h 2n . . ) is 207 times that of the electron and charge = 1. . . . . . . . . . . . .6 × 10 19 C.18 10 0.375 . = 1.375 . = 2.18 × 10 18(12/2n2) (. . . .. . . .18 × 10 18 × 0.E = E1 E2 = 2.E = hc/.18 × 10 18) When n = 1. . .Z e m E 2 8 h n .18 × 10 18/2 When n = 2. . .2165 × 10 7 m = 2433 Å Example 48:Themass ofmuon (.63 10 3 10 2. . .. . . En = 2.0 2h2 = 2. . 2 2 4 e 1 2 e m 2 m k e E m h . po tential is obtained froma step-up transformer T1whose output is converted into direct current by full-wave . .6 eV E1 = The ionizationenergy = E1 = 186 × 13. 186 × 13. .c. Th e electrons are accelerated to veryhigh speeds (upto 10%of velocity of light) by the d. . . This reducedmass is m = p p m m m m . 2 2 4 1 2 2 mk e E h .physicsashok. . = e e e e 1836m 207m 1836m 207m .c. .At present. it is well known that these rays are produced whenever fast moving electrons strike a high atomicweight solid like tungsten kept in vacuum.in 63 Sol: The ionization energyof themuonic atomis obtained by replacingme inH-atomfo rmula bythe reduced massmof the proton-muon system. 186me Thus the ground state energy is (n = 1. . .ATOMIC PHYSICS www.53 KeV PRODUCTION OF X-RAYS X-rayswere accidentallydiscovered byWilhelmRontgenin 1895 during the course of s ome experiments with a discharge tube. . These electrons are focussed on the targe t T with the help of a cylindrical shield Swhichsurrounds Fand ismaintained at a negative potential. .100 kV) applied between F and the anode (also called anticathode). This high d. . . potentialdifferenc e (of about 50 kV. (a) X-ray Tube: The essentialelements of amodernCoolidgeX-rayvacuumtubewhich iswidely used for c ommercial and medical purposes are shown in figure.6 eV = 2. . Z= 1). Electrons are produced thermionically from a tungsten filamentary cathode F which is heated to incandescence either by a storage battery or by a l ow-voltage alternating current froma stepdown transformer T2. amolybdenum .rectifier and a suitable filter. + 50 kV Electrons T A Cooling Tube Fins X-rays R F S B The targetTusuallyemployed inX-raytubes is amassive block oftungstenor inmanycas es. e.Most ofthemundergogla ncingcollisions . This number is determined by the temperature of the electron-emitting filament which itself is proportiona lto the heater current. Thatiswhymetalsliketungsten.mostmetalswillmelt. they give up their kinetic energy and thereby produce X-rays. ORIGIN OF X-RAYS X-rays are produced when high-speed electrons strike some material object.Undertheterrificbombardmentofthetar getbysomany electrons. The quality of X-rays is measured in terms of their penetrating power which is d ependent on the potential difference between filamentary cathode and the anode.e. the quality or penetrating po wer of X-rayscan be controlled by varying the potential difference between the cathode and anode. Howev er. T A Cooling Water Tube X-rays R F ST1 T2 Rectifier (b) Control of Intensity and Quality The intensity of X-rays depends on the number of electrons striking the target. Greater this accelerating voltage. of low frequency) as soft X-rays. Target materials of higher atomic weights yield a greater abundance of X-rays than those oflower atomic weights.ATOMIC PHYSICS ATOMIC PHYSICS plug embedded in the face of a solid copper anode. copper helps to conduct he at efficiently to the externalcooling finsorthewater-coolingsystem. Hence by controlling the filament current with the help of a rheostat R. It is customaryto refer to highly penetrating X-rays (i. majority of the electronsthatstrikeasolidtarget. higher the speed of the striking electrons and consequently. When the electrons strike the tungsten target. their abunda nce depends on the atomic weight ofthe target material.donothingspectacular atall. It has beenfound that apart fromthe intensity and quality ofX-rays. Obviously. it is pos sible to achieve separate control ofthe intensity and quality of X-rays independent of each other. more penetrating the X-rays produced .areused whichhave highmelting points and also have a highatomicweight (whichis essential for abundant production of X-rays). Being very good conductor of heat. It will be noticed from above explanation that in coolidge X-ray tube.platinumandmolybdenume tc. those possessing high frequency) as hard X-rays and to those less penetrating (i. The face of the copper anode is sloped at about 45º to the electron beam. thermionic em ission and hence intensity of X-rays can be controlled. 8 percen t of the energyof the electron beam goes into heating the target. It is found that nearly99. lose their energy a little bit at a time and thus mer ely increase the average kinetic energyofthe particles of the target material.physicsashok.in 64 .with the matter particles. But a smallnumber ofthe bombardingelectrons produce X-rays bylosing their kineti c energyinthe following two ways: (i) Someofthehigh-velocityelectronspenetratetheinterioroftheatomsofthetargetmate rialandareattracted www. physicsashok. .This spectrumhas a sharplydefined short-wavelength limit .in 65 bythe positive charge of their nuclei..min = 34 8 19 6. hvmax= hc/. In that case 1 2 mv2 = hvmax .602 × 10 19C h = 6. vmax = eV/h Now. = 0. then 1 2 mv2 = eV .62 × 10 34 J-s and c = 3 × 108 m/s..ATOMIC PHYSICS www.62 10 3 10 1. . 1 2 m(v2 v.min [. . c = v .. Thismust equal the energyof the X-rayphotons emitted. we get hvmax= eV. when v. X-Ray X-Ray + + + + X-Ray 1 v. .As an electronpasses close to the positive nucleus. it is deflected from its pathas shownin figure.. ..(i) If the electron is accelerated through a potential ofVvolts.min (or high-frequencylimit fmax)which corresponds to themaximumenergyofthe incident electron.602 10 V .min = eV or .(ii) From(i) and (ii). then its loss of energyis = ( 1 2 mv2 1 2 mv. The electronexperiences deacceleration duringits defl ectionin the strong field of the nucleus.min = hc/eV Substituting the values of e = 1.2 Continuous spectrum 1 2 mv2 v If. hc/.e. 2 mv..] . as shownin figure. the striking electronhas its velocityreduced fromvto v.. . d uring its passage through the atomofthe targetmaterial.. The energy lost during this de-acceleration is given off in the for mofX-rays of continuously varyingwavelength(and hence frequency).2). TheseX-rays produce continuous spectrumw hen analysed by Bragg spectrometer...2) = hv The highest ormaximumfrequency of the emitted X-rays corresponds to the case whe n the electron is completely stopped i. we get . (ii) Some ofthe high-velocityelectronswhile penetrating the interior ofthe atoms of the targetmaterial. as said earlier. TheseX-rays are independent ofthe na ture of the targetmaterial but are determined by the potentialdifference between the cathode and anode of t heX-ray tube. = 1.min = 12400 V Å [1Å = 10 10 m] SuchX-rays are veryaptly called braking radiations because they are due to braking or slowing down of high-velocity electrons is the positive field of a nucleus. knock .24 10 6 V . the continuous spectrumof theX-rays because they consist of a series of uninterrupte dwavelengths having a sharplydefined short-wavelength limit .min. m or . These radiations con stitute. .. and L. two linesK. During the jump anX-rayradiation is emittedwhose frequencyis given by E.When electrons fromouter orbits jump to fillup the vacancyso produce.. This gives the L-series o f theX-ray spectrumas shown byK. L.physicsashok. this vacancy inK-shell is filled up by an electron jumping fromM-sh ell. theX-rays emitted would be stillmore energetic andwould consequentlypossess stillhigher frequencyb ecause . the energydifference is g iven out in the formofXrays of definite wavelength (and frequency). L-shells etc) of the atoms. lines in figure (a). M (a) E = 0 2 20 . Usually. = hv Ek where Ek is the energyrequired to dislodge an electron formtheK-shell and El is that required for L-shell. K. KLMN O K. If. These wavelengths constitute the line sp ectrumwhich is characteristic of thematerial of the target. L. As shown in figure (b).in 66 off the tightlybound electrons in the innermost shells (likeK. it dislodges an electron fromthe L-orbit and an electron either fromM-orbit or other outer orbits takes i ts place so that X-rays of frequency lower than that of theK-series are produced. K L M e e e (a) K L M (b) X-Ray K -line . theX-rays emitted have verylarge energycontent and hence are highly penetrating. Figure (a) shows the case when the high-velocity incident electron knocks off on e electron fromtheKshell.ATOMIC PHYSICS www.E = (Ek EL). SuchX-rays arising frommillions of atoms produce the K -lines as shown in figure. Similarly. andK. however. Since this energy difference is comparativelyvery large. when the incident electron carries somewhat lesser amount of energy. L.E =(Ek Em) ismore than . of this series are detected although there ar emanymore. this vacancyinK-shellis filled by a nearby electron in t he L-shell. L. K. L. K L M N O . M.000 K. theseK. theX-rays produced byanX-ray tube consist of two parts: (i) one part consists ofa series of uninterruptedwavelengths having a short cutoffwavelength . LandMseries constitute the line spectra of theX-raysw hich are characteristic of thematerialused as target in theX-ray tube. L.200 2000 20. M. Hence. As stated earlier. L. SHELL Energy in eV (b) M Spectral lines ofM-series are produced ina similarway as shownin the energy-leve ldiagramof figure (b).min.This . 400 V Å (iii)The cut-offwavelength. L. These two are shownin figure.physicsashok. L.0 4. The va lue of . K. X-ray Spectrum As explained inX-ray spectrumconsists of (i) continues spectrumand (ii) line spectrum.400 V × 10 10 m or 12.0 3. K. L.min = 12.) X-ray Intensity Tungsten Target (b) (v) There is a shift of themaximumintensity position towards the short wavelength side as voltage is increased. (b) Line Spectrum . (a) (ii) It has a sharply-defined short wavelength limit given by . L. L.min decreases as this potentialdifference is increased.ATOMIC PHYSICS www.0 5 kV 10 kV 15 kV 20 kV K 25 kV . Intensity Continuous Spectrum .min .0 2.min is independent ofthenature ofthe targetmaterialbu t is inverselyproportional to the potentialdifference between the cathode and anode of anX-ray tube. K. L. L. L. L. Wave length(A. charged nucleus of an atomof the targetmaterial. (a) Continuous Spectrum (i)It is produced due to the de-acceleration of high-velocityelectronswhen theya re deflectedwhile passing near the positively.U. K. 0 1. f Continuous Spectrum K. (iv) The intensity of the continuous spectrum(given by the area enclosed bythe curve of figure (b) is found verynearlyproportional to the square of the applied voltage for a given target and to the atomic number of the target materialwhen a constant potential difference is applied.in 67 constitute the continuous spectrumand (ii) the other part consists of a number of distinct and discrets wavelengths wh ich constitute the line or discontinues spectrumof theX-rays. (iv) Line spectrumis characteristic of the targetmaterial used.andM-serie s formthe soft Xrays. In .(i) It is producedwhen electrons are dislodged fromthe innermost orbits ofthe atoms ofthe targetmaterialfollowed byelectronjumps fromouter orbits.K-series consists of those lines for which electron jumps end at K-level. (iii) K-series beingmost energetic constitute the hardX-rayswhereasL. (ii) It consists of discrete spectral lineswhichconstituteK-series. L-series andM-series etc. K. = 1 2 . line for a target material having an atomic n umber ofZ1 and v2 and Z2 are similar quantities for some different targetmaterial. 90 79 50 42 28 Cr 24 Cu Mo Sn Au Cf .ATOMIC PHYSICS www.4 kV and the current through it is 2 mA. is 2 1 . where v1 is the frequency of theK. Sol: . .physicsashok. = 2 1 2 2 (Z 1) (Z 1) . calculate: (i) the number of electrons striking the target per second (ii) the speedwithwhich theystrike it (iii) the shortest wavelength emitted Take e = 1. . as found byMoseley.2 Å C43: If the potential difference applied across anX-ray tube is 12. . C42:AnX-raytubeworks on 60. .000 = 0. K.000V. Mass No.000 V .min = 12. 400 V Å Here.in 68 fact.min= 12. (Z) (c) (v) There is a regular shift towards shorter wavelength in theKspectrumas the atomic number ofthe target is increased figure (c).What will be the wavelength ofX-rayemitted in it . The exact relationship. 400 60. V = 60. X-rays constituting the line spectrum are known as characteristicX-rays.1 × 10 31 kg .6 × 10 19 C and m = 9. The number of lines present in the spectrum depends both on the nature of target material and the excitation voltage. Calculate the maximumspeed of the . . 400 V = 12400 12400 = 1Å C44: Calculate theminimumapplied potentialrequired to produceX-rays of 1Åwavelengt h.93 105 V m . n = I e = 3 19 2 10 1. = 12400 1 = 12. V = min 12400 .25 × 1016 s 1 (ii). v 2eV 5. . v = 5. 400 = 6.4 kV C45:An X-ray tube passes 5 mAat a potential difference of 100 kV.min = 12400 V Å .6 × 107 m/s (iii) .6 10 . = 1.93 × 105 12. .Sol: (i) If n is the number of electrons striking the anode per second.min = 12. .93 × 105 v = 5. Sol: . . then I = ne . Calculate Planck s constant h if e = 1. Sol: . . Sol: .414 10 3 10 . Consider theK.63 × 10 34 J-s MOSELEY S LAW In 1913-14. . . . However. larger amount ofenergyis required to liberte an electron fromtheK. higher is the frequencyof theK.000 × 5 × 10 3 = 500W Power converted into heat = 99.414 Å. (Z b)2 or . B ecause there is greater positive charge onthe nucleus ofanelement of higher atomic number. .76 × 1011 C/kg and J = 4. . L andMshells of that element.min = 0. Take e/m= 1.By usingBragg s spectrometer for the purpose.88 × 108 m/s + 100 kv 5 mA Incident power = 100.5 4. . v 2eV 5. = 19 3 19 8 1.93 × 105 100.000 = 1.93 105 V m . there i s one very important difference.1 percent of the incident energy is converted into X-radiations. line produced byit.Moseleycarried out a systematic study of the characteristicX-ray spec tra ofvarious elements used as targets in anX-raytube. = a(Z b) . C46:AnX-raytube operated at 30 kVemits a continuousX-rayspectrumwith a short wav elength limit .602 10 30 10 0. It is foun d that higher the atomic number ofthe targetmaterial.ATOMIC PHYSICS www.physicsashok. the r emarkably similar to each other in the sense that each consists of K-L and M-series.min = ch eV .602 × 10 19 C and c = 3 × 108m/s. . . (Z b) or . The frequency of lines (in every series) produced froman element of higher atomic number is greater thanthat produced byanelement of lower atomic number.9%of 500 = 499.18 = 119 cal/s. .5W Heat produced / second = 499.18 jou les/cal. = 6.in 69 electrons striking the target and the rate of productionofheat at the target if only0. h = min eV c . The exactmathematical relationship betweenfrequency and atomic number is given by v . v = 5. It is due to the f act that binding energyof electrons increases aswe go fromone element to another ofhigher atomic number. line of the characteristicX-rayspectrumof any element. where Z is the atomic number of the element and a and b are constants for a part icular series but varyfrom one series to another i. lineswhich is . The constant b is known as nuclear screening constant. andK. Figure showsMoseleydiagramfor K. Its values for lines ofL-series ismore.e. their values for K-series are different fromthose for L-series etc. Itmaybe stated as follows: The frequency of a spectral line in the characteristic X-ray spectrumvaries dire ctly as the square of the atomic number ofthe element emitting it. b = 1. The above relation is known asMoseleylawfor the characteristic or lineX-ray spec trum. For lines ofK-series. . the frequencyof t he radiationgiven out is. technetium(43) andrhenium(75) etc. It provides the proper guideline that elementsmust be arranged in theperiodic table according to their atomic numbers and not their atomicweights. Accordingly. K. For example. Moseleylawhas led to thediscoveryof newelements like hafnium(72). whereR is Rydberg s constant. ifwe go by the atomic weight. . nickel (28Ni58. Atomic number Frequency An exact formofMoseley s lawis 1 . ButMoselylawdictates that as per their atomic numbers. .This lawhas beenalso h elpfulindetermining the atomic number of rare earths therebyfixing their position in the periodic ta ble. bytheindicationofgaps inMoselydiagram.Moseleylawhas been used to place elements in their proper sequence i n the periodic table in certain questionable cases. a correction factor and n1 and n2 the principal quantum numbers of the energy levels between which the transition occurs. It can be shown thatMoseleylawis in accordancewithBohr s theory of spectral emissi onfromatoms.physicsashok.in 70 obtained by plotting . v = 4 2 3 0 .As expected.9). 5 10 15 20 25 × 108 Al-13 10 Ca-20 Zn-30 Ze-40 Sn-50 K.)2 2 2 1 2 1 1 n n . . promethium(61) . .7) should precede cobalt (27C o58. .As shown inwhenan electron jumps froman orbit n2 to the orbit n1. = R(Z .ATOMIC PHYSICS www. potassium (19K39) should come before argon (18A40) and similarly. their order should be just opposite of the abo ve. Z the atomic number. This fact is further supported bythe chemical properties of these elements. the graph is linear. versus atomic number of different elements of the periodic table. Importance of Moseley law The great significance ofMoseley law lies in the fact that it proves for the fir st time that it is the atomic number and not the atomicweight of an elementwhichdetermines its characteristic properties (bothphysical and chemical). . . . .. . . Thismaybe put as v = 4 2 3 2 2 0 1 2 me 1 1 8 h n n .. . Z Bohr did not take into account the screening effect of electrons whereasMoseley did. . .and L-series of X-ray Spectrum The frequencies of the various lines in the K.Z2 2 2 1 2 1 1 n n . . .. (Z Formulae for K. . . . . .. Z2 or . 3 etc Here. . . h . the nuclear screening constant is unity. . K-series: The general formula is 1 .... . .and L-series of the X-ray line sp ectrumare given by the following empiricalformulae.me 8.. where n = 2. .. Z2 or v . . = R(Z 1)2 2 1 1 n . . . . That is why the expression becomes b) . physicsashok. . . = 1. = 8R 9 (Z 1)2 L-Series: The general formula is 1 . . line.4)2 (ii)For H. Take Rydberg constant = 10. . .4. screening constant is 7. . (i) For H. = R(Z 7. . = R(Z b)2 2 2 1 1 2 3 . . = R(Z 7.in 71 (i) For K. . 1 . 4 etc.43Åare emittedwhen an electron in a tungsten atom(Z = 74) is transferred fromtheM-level to L-level. . .4)2 C47: Find the nuclear screening constant for the L-series ofX-rays if it is know n that X-rayswith awavelength of . where n = 3. . Its wavelength as given byMoseley s lawis 1 . .. .. = 5R 36 (Z 7. . . = R(Z 1)2 1 1 4 . n = 3 . . = 3R 4 (Z 1)2 (ii)For K. . line is given out. 1 .e. .97 × 106m 1. . . n = 3 . n = 2. Here. . Sol:When electron jumps fromM to L-level.4)2 2 1 1 4 n . the first member of the L-series i. n = 4 1 .4)2 1 1 4 9 . = 3R 16 (Z 7. .4)2 1 1 4 16 .. = R(Z 1)2 1 1 9 .ATOMIC PHYSICS www. 1 . line. line. . = R(Z 7.. . L. . line. . 4).. absorption edges of copper occur at wavelengths 1. K-Series: 1 . = R 4 (Z 11)2 . . Sol: It shouldbe remembered that absorptionages are found inthe absorptionspectr umofX-rays. b = 6. .38 = (Z 3.3 (instead of1 for othermembers). . = 5R 36 (Z b)2 Substituting the givenvalues. . . (74 b) = 67.8. the value of screening constant forK. .Moreover. = R(Z 3..43 = 10.288 Å respectively.25.The absorption edge ofeachseries (ofline spectrum) represents the limit ofthat series.3)2 . .97 × 106 × 5 36 (74 b)2 . . . the short-wavelength limit of each series is called its absorption edge and iswritten as . 1010 1. . .3)2 L-Series: 1 . . Example 49: The K. . It maybe obtained by putting n= . .3)2 2 . = R(Z 11)2 2 1 1 4 . Calculate the atomic number of copper. in the formulae. . . . Corresponding value for L-series is 11(instead of 7. . forK-series. 1 . and L.1.member is 3.75 . .380 Å and 11. Inotherw ords.we have 1010 1. (74 b)2 = 4589.. = R(Z 3. . 3) (Z 11) .1. part of it is a .21Åand aweakK. Can you identify the impuritytaking the nuclear screening constant as unity. = 0. = 3 4 R(Z 1)2 Dividing one bythe other.537 0. = 11.1) . = R(Z 4 . impurity is nickelwhose atomicmass number is 28. = 1.987R 4 For impurity 10 1 1.3) (Z 11) . line is given bythe relation 1 1)2 1 1 . = 3 4 R(74 1)2 = 15. .537 Å. . Sol: Thewavelength for K.537.10. = 1. Z = 28 Obviously. line of .ATOMIC PHYSICS www.21 = 2 5329 (Z.38 = 2.in 72 .045 or (Z 3. 1010 11.288 = R 4 (Z 11)2 Dividing one bythe other. . . . line of .10. . we get 1. = 3 4 R(Z 1)2 For tungsten 10 1 0. for Ni = 28.43 or Z = 29 Example 50:An impure tungsten target emits a strongK. Absorption of X-rays When a narrowandmonochromatic beamofX-rays passes throughmatter. we get 2 2 (Z 3.Z= 74. .physicsashok. . Gi ven for tungsten.288 4.21. . Next.The strength of the ionization current is ameasure of the intensityof theX-ray.Absorption of x-rays can be studiedwith the help o f the apparatus shown in figure.bsorbed and the remaining part is transmitted. TheX-rays produced by anX-raytube are first made into awell defined narr owbeamby passing themthrough two fine slits S1 and S2 in the two lead plates. a sheet of the absorbing material is interposed in the path of the X-ray b eambefore it enters the ionization chamber. The ratio 0 I I of the two ionization currents can be used to measure the absorptioncoefficient of thematerial. The beamis thenmono chromatised byBragg reflectionfroma crystal(not showninthe figure) and allowedto enter the ionizatio nchamberwhichmeasures the ionization current. It is found that the ionization current and hence the intens ity ofX-rays is reduced by their passage through the absorber sheet. x .. .x + logeI0 or e 0 log I I = ..x + K ... dI I = . Now.ATOMIC PHYSICS www....x or 0 I I = e . dI / I = . Assuming that this rate is proportionalto the intensityI.. . I = I0. we have I0 x I O logeI0 = K Hence.dx .(i) Integrating both sides of the above equation. dx .physicsashok. If dI is the further decrease in intensityover a thickness dx of the absorber figure. logeI = .x or I = I0e . we get . then dI dx gives the rate of decrease of intensity with thickness.(iii) It is seen that intensity of the X-ray beamdecreases exponentiallywith the thick ness of the absorbing material as shown in figure. BRAGG S LAW .I where ..we have I0 I (I dI) x dx dI dx .(ii) The value ofthe integration constant Kcan be found fromthe known initial conditi onswhichare that when z = 0.in 73 L1 L2 S1 S2 Absorber Ionization Chamber Let I0 be the initial intensityof a homogeneousX-raybeamincident normallyon an a bsorber sheet and I the intensity after the beamhas travelled a thickness x of the absorber.... is a constant of proportionality and is called the linear absorption coe fficient of the absorber (it is also known asmacroscopic absorption coefficient or linear attenuation coefficien t). I or dI dx = . equation (ii) becomes logeI = . Substituting these values in equation (ii) above. .. Figure gives a .A P S Q E F M B G H N R Figure gives a 3-dimensional view of how a beam of monochromaticX-rays undergoesBragg s reflection from different planes in a NaCl crystal. + d sin. . where d is the interplanar spacing i. = 2d sin.in 74 2-dimensional viewof the same diagram. 2. it is obvious that ray no. refraction. Sin ce the two rays travel the same distance frompointsAandNonwards. 2 is reflected from atom B lying plane 2 immediately below atom A. The transparency depends on the density of the material.This equation is known as Bragg s Law. second order and third ordermaxima r espectively. 3 etc. The two reflected beamswillbe in phasewith eachother if this pathdifference equa ls an integralmultiple of la dnwill be antiphase if it equals an oddmultiple of . The beamis partically reflected at the successive layers rich in atoms Ray no. on a set of parallelplanes ofNaCl crystal. diffraction and polarisation etc.physicsashok. sheet of lead 1 cmthick can absorbX -rayswhereas aluminium sheet ofsame thickness cannot. 2 tr avels an extra distance = MB + BN Hence. 1 2 3 1 2 d d Plane 1 Plane 2 Plane 3 B = d sin. Hence. Whether two reflected rays will be in phase or antiphase with each other will depend on their path difference. the more penetrating the X-rays . where n= 1.ATOMIC PHYSICS www. It shows a beam ofmonochromaticX-rays incident at a glancing angle . N . (ii) They are not deflected by electric andmagnetic field. This pathdifference can be found bydrawing perpendicularsAMandANon rayNo. PROPERTIES OF X-RAYS Main properties ofX-raysmay be summarised as under: (i) Likevisible light. Higher the density of t he substance. the path difference between the two reflected beams is = MB + BN M. (iii) They posses high penetrating power and can pass throughmanysolidswhich are opaque to visible light. A . the less transparent it is to theX-rays.X-rays consist ofelectromagneticwavesofveryshortwavelength (or ofveryhighfrequency) and showreflection.1 is reflected fromatomAinplane1whereas rayno./2. vertical distance between two adjacent p lanes belonging to the same set.. = n. for the first order. The penetrating power ofX-rays depends upon (a) the voltage applied across the c athode and anode of theX-raytube and (b) the atomic number of thematerialof the cathode. interference. the conditionfor producingmaxima becomes 2d sin.e.Greater the accelerating potential of the X-ray tube and higher the atomic number of its target material. For example. 2. (iv) They ionize a gas and also eject electrons frommetals onwhich they fall.e. (v) Theycause fluorescence inmanysubstances like barium. (vii) They have a destructive effect on living tissue. (a) Industrial applications: Some of these applications are as under: . Exposure of humanbodyofXrays causes the reddening of skin and surface sores. Practical Applications of X-rays On account of their diverse and distinctive properties. tungstate and zinc sulphide etc. (vi) Theysuffer compton scattering.produced.medicine and research. in industry. cadmium.X-rays have beenput toman yuses in different fields of our dailylife i. malignant sores. (c) Medical Applications: These can be broadly divided into two classes. (iii) for identification ofchemical elements including determination oftheir ato mic numbers. In this way. cellulose and plastics. the presence of f oreign matter like bullets etc. porcelainand other insu lators. machine parts intended for withstanding highpressures. forgings and castings etc. In this way. one for diagnosis purposes (radiography) and the other for curative purposes (X-ray therapy). bef ore taking radiograph. Similarly.e. in its internal structure su ch as metals. (iii) to study the structure of rubber.ATOMIC PHYSICS ATOMIC PHYSICS (i) to detect and photograph defects within a body i. these powders settle in the gastrointestinal tract and if a radiograph is taken at that item. (iv) for analysing the structure of coraplex organic molecules by examining thei r X-ray diffraction patterns. cracks in wood. Since bones are more dense and hence more opaque to X-rays then flesh. (b) Applications for pure scientific research: (i) for investigating the structure of the atom. stones in kidneys and gall-bladders etc. barium or bismuth meal is given to the patient. a artificial means are adopted to create sufficient contrast between them. a contrasting radiogr aph ofhuman body can be obtained for leisurely study by interposing it between the X-ray tube and pho to film. the intestines stand out in sharp contrast to the surrounding tissues due to the fact that absorptioncoefficient o fbarium is greater on account ofitshigh atomic number. in the human body as well as diseased organs of the boy. Where the organs do not provide contrast as. This meal consists of milk to wh ich some amount of barium sulphate or bismuth carbonate has been added. alloys like cobalt-nickel. tumours. duraluminium.Afew hours after the meal has been taken. (ii) for studying the structure of the crytalline solids and alloys (X-ray cryst allography). radiographs are routinely used for the diag nosis of tuberculosis. tissues and metals. steel. It is due to diffe rential absorption of X-rays between bones. defectsindiamonds and other precious stones. cancers and tumours have been cure d bycontrolled exposure to X-rays of suitable quality. In such cases. in moulds. in the inte rnal organs ofthe human body can be accuratelylocated. (ii) X-ray or Rontgen Therapy Many types of skin diseases. bronze. The diffraction of X-rays bythese substances leads to valuable information about their molecular grouping. pepticulcers and ruptures etc. (ii) toanalysesthestructureofalloysandotherrcompositebodiesbydeterminingthecrystalfor minaningot with the help of diffraction of X-rays. for example. the intestines or other fleshy parts of the human body. Radiographs or X-ray photos are used for this purpose. artificial pearls and old paintings have been analysed. This curative power of X-rays is due to the fortu . (i) Radiography: X-rays are being widely used for detecting fractures. X-rays have been used for the identification of different types of cells and tis sues and for bringing about genetic mutations. www.in 75 .nate fact that diseased tissue is more susceptible to destructionthan the surrounding healthytissue.physicsashok. Unnecessar ylong exposure of human body to X-rays produces many injurious effects including the loss of white cells in the blood. sterility and harmfulgenetic changes. .. Parallel electric and magnetic fields act on on a charged particle moving pe rpendicular to these fields. slightlydiverging beamof charged particles focussed by amagnetic field ? 18. 6.Why a re the latter preferred in actual use ? 21.The tracks of the particles are labelled in the figure . Magneticmirror is a termfor the region of amagnetic field inwhich there is an intense concentration oflines ofinductionas shown inthe figure. Howdo you conclude that cathode rays are fast moving negativelycharged parti cles ? 16. Aneutralpion decays into two gamma photons.0 .What will happen to it ? 17. Does the speed of a charged particle changewhen (a) a magnetic field. . Suppose a charge d particle approaches amagneticmirror. Howis amonoenergetic. The occurrence ofa lower bound ofwavelengths ofX-rays produced inanX-raytube lends support to the quantumconcept ofradiation.Explainthis phenomenon. X-rays can be produced in cathode ray tubes and also in Coolidge tubes. Describe its subsequentmotion. What leads you do believe that X-rays are electromagneticwaves ? 2. Why is thewave nature ofmatter not apparent in our daily lives ? 22. Expl ainwhy. Themagnetic field is along the inward norma l to the place of the paper. Is this true or false ? 5. Why cannot a single photon be born?What conservation lawis in contradictionwith it ? . Water irradiatedwithX-rays is unsafe for drinking. 10. Describe its motion. (b) an electric field. IfYoung s experiment is repeatedwith electron beams interference is observed. a proton.Willtheyhave the samedeBro gliewavelength? 9. (b) later ? 13. 7.What happens to the electron ? 19. Achargedandanunchargedparticlehavethe samemomentum. Explain how. D oes thismean that an electron gets divided into twowhile passing through the slits ? 4. 15. Auniformelectric field acts normally on amoving charge. Aneutron. 3. Which tracks do the electron and alpha particle follow? 8. Can the charge be de flected through 90º ? 12.-rays pass throughi t.MODERN PHYSICS www. Auniformelectric field acts normally on amoving charge. acts on it for some time ? 14. X-rays are producedwhen a fast electron hits a proper target. Iswork done by the f ield on the charge (a) as it enters the field. an electron and an alpha particle enter a region of const ant magnetic fieldwith equal velocities. The electricalconductivityofa gas increaseswhenX-rays or . An electronmoves through a gas-filled region in the presence of a transversem agnetic field. Why does the target in anX-ray tube become hot ? 20.physicsashok. Fluerescence is produced by ultraviolet rays but never by infrared rays.. Why are tungstenor platinumwidely used as the target theX-raytubes ? 11.in 123 THINKING PROBLEMS X-rays 1. . Quarks inside protons and neutrons are thought to carryfractional charges 2 e.23. . . ... .Whyare theynot evidenced inMillikan s oildrop experiment ? .. 1 e 3 3 .... . Which level of the doubly-ionized lithium(Li++)ion has the same energy as th e ground state energyof the hydrogen atom? 17. the latter was discovered earlier. Rutherfor dmodel) but has a highly non-uniformdistributionin .-rays. 21.. Nuclear Fassion and Fusion 1. andtheBalm er series to the second orbit.-dcay. (Thomsonmodel.. Distinguish between excitation and ionization bycollision.. Cathode rays and ..-decay? 9.What inference did he drawfromthis result ? 25.e. the combination of . . spread over a small range ofwavelengths. How can Becquerel ray. When a nucleus undergoes . Can a spectral line belong to both the Lyman andBalmer series ? 19. Bohr s principle of quantization of angularmomentumis not a postulate but an e ssentialcondition.Why ? 20. is the product atomelectrically neutral ? In .physicsashok. If the a-decayofU238 is allowed fromthe point of viewof energy (the decaypro ducts have a totalmass less than themass ofU238) what preventsU238 fromdecaying all at once ?Why is its half -life so large ? 18. be separat ed ? 8. (Thomsonmodel. Explain this. Auraniumnucleus (atomic number 92.-rays emitted only in nuclear processes and not in orbital electron transitions ? 5.-particle and the resultant nucleus . i. of one particularwavelength each. he found that 1 in 2500 are deflected through verylarge angles.. Does a nucleus have to be bombardedwith fast or showneutrons in order for it to undergo fission ? 14.-. Explain how.-decay cause a change of element. called transmutation ? 10. Whyare . In practice.-particle tracks as observed in a cloud chambermuch shorter than .. mass number 238) emits an . When a radioactive substance emits an . Radioactivity. Can it be concluded from. 16. When Rutherford bombarded a thin foil of gold by .-p article tracks though they emerge froma radioactive samplewith almost the same speed ? 22.in 124 Atomic Structure.-particles are streams of electrons. 11.-decay that electrons exist inside the nucleus ? 4. An atomhas a continuous distribution ofmass in. Howis the radioactivityof an element affectedwhen it forms chemical compounds ? 7. 2. Do . Suggest a reason for this. 3..-particles. Why are .e. Experimental results inradioactivityshowsmallvariations fromthe results pred icted bytheory. the spectral lines froma glo wing gas should be sharp i. i.. What is a thermalneutron ? 13.-decay and .Rutherfordmodel). According to the theory of electron transitions.e. Does the relationE =mc2 suggest that mass can be converted to energy onlywhe n it is inmotion ? 12.and . theyare found to be somewhat diffu se .MODERN PHYSICS www. What are the principles that are obeyed in filling the orbits of an atom? 6. Although theLymanseries involves transitions to the ground level. Why has it not been possible so far to control the fusion process and obtain usable energy fromit ? 15.-particle its position in the periodi c table is lowered bytwo places. Inwhat respect do the ythen differ fromeach other? 23.... Is this true or false ? 24.. Explain the statement Themoderator ina nuclear reactor thermalizes the neutro ns.. What ismeant by the disposal of radioactivewaste in a nuclear reactor ? 29. Among .emits a . per unitmass ? 33.whatever be the half-life of the material. The isotope of hydrogen 3 1 H (tritium) is radioactive. .What would be its decay process and the product? 28. Is this true or false ? 31. Is this true or false? 30. Which yields greater energy per atom fission or fussion ? . What ismean by enrichment of uranium ? 32.-particles.-particles suffer equal and opposite deflections in an external elec tric field. It requires infinite time for all the atoms in a radioactive sample to decay ... protons and neutronswhich have the greatest penetrating power through matter andwhy ? 27.What are the atomic andmass numbers of the final nucleus ? 26. .and .-particle..-particles. . The electric fieldwillnot affect the initialvelocity. 13. (b)Work iis done.Why ? 37. D. i. The free elect ron and ionpairs produced can nowmove and conduct electricity. .-rays cause ionization bycollision inthe gas atoms.-ray photon.. 3. i. 2. 4. They are reflected .. Scientists say there is a change in the frequency of the photon as it moves awayfromthe star and call the difference infrequcney gravitational shift.. Alpha particles are heavy and positively charged. onlyonmomentum. (a) The speed does not change since the force is perpendicular to the displa cement at everypoint. (a) No work is done.A90º deflection requires that the initial ve locity be reduced to zero. as the displacement howhas a component parallel to the force.in 125 34.All these facts lead us to believe that X-rays are electroma gneticwaves. This is possiblewith ultraviolet rays as these have greate r photon energy than visible light. Can you explain this? SOLUTION OF THINKING PROBLEMS X-rays 1. 6. Aphoton is emitted by a dense star. False. refracted.. X-rays damage living cells and hence kill the bacteria present inwater . angularmomentum= nh/2. (Thomsonmodel. Yes. around the sun ? 39.e. is a basic lawin n ature. 5. If a nucleus emits only a . Infact electrons have associatedwaves. 10.. diffracted like ordinarylight waves.. the water actually becomes safer for drinking. No. Theyhave large atomic numbers and highmelting points. The phenomenon offluorescence consists of absorptionof higher energyphotons a nd re-radiation of lower energy visible light.e. When is a chain reaction said to be critical ? 35. The field does no work on the particle. Hence. Bohr s quantization principle. The energy and speed of the particle remain unchan ged. The quantumtheory predicts a lower bound (hv =Ve) ofwavelengths ofX-rays prod uced in anX-rays tube. The electrons are deflected to the right. 11. Aclassical atombased on . the gas becomesmore conducting. 7. It therefore des cribes a circlewith decreasing radius.MODERN PHYSICS www. Thus. No. This is found to be in good agreement withexperimentalobservations. and so they are defle cted the least to the left according to Fleming s left-hand rule. B. The de Brogliewavelength does not depend on charge.physicsashok. What is Bohr s correspondence principle ? 38. does itsmass number change?Does itsm ass change ? 36. it spirals inwards. X-rays or . Infrared rays have lower photon energy. Itwillonlyproduce a n additional acceleratedmotion perpendicular to the initial velociyt.Why dowe never speak of quantizationof the angularmomentumof a planet. Rutherfordmodel) is doomed to c ollapse. since the displacement due to the motion is perpendicul ar to the field. 12. 8. X-rays cannot be deflected by electric andmagnetic fields. The electronloses energydue to collisionswith the gas atoms. 9. So thi s supports the quantum theoryof light. so fluorescence cannot occur.At the accelerating voltage of the experiment thewave behaviour of electrons becomes quite prominent and interference fringes are observed. Due to themagnetic field.(b)When an electric field acts for some time. . a forcewill act along this axis. the particlewillmove in a circlewith the field as axis. The energyand speed of the particlewi ll change. Work is done by the field on the particle. 15. producing an acceleratedmotion perpendicular to t he plane of the circle. 14. The result willbe a helicalpath ofgradually increasing pitch.Due to the electric field. Their streaming character is shown by deflection in amagnetic field because a magnetic field can produce deflection only when charged particles are inmotion. the particlewill acquire a displac ement parallel to the force. Their negativelycharged character is shownbythe directionofdeflection inan e lectric field. . Let us resolve the velocityalong the field and perpendicular to it. that photon must also be at rest to conserve momentum. The resolved part along the field vII is called orbital velocity. i. caus .This difference depends onthe velocityofthe atom. the hardness and intensity of theX-rays can be controlled independently.which is also the anode of theX-raytube.6 × 10 34 Js. Th e balance is lost in inelastic collisions be between the electrons and the target atoms.mv)/Bq when . the observed frequency(and hencewavelengt h) becomes different fromthe emitted frequency. 19. 18.. 20.Themomentumofordinarymaterialb odies at ordinary speeds is verylarge and so the associatedwavelength is extremelysmallbecause of the verysmallvalue of h = 6.As a result. The charged particles followa helicalpath of periodT = 2.in 126 16. or become completelyfree fromthe attractive field ofthe nuc leus (ionization). Because they are held together by a strong force. small. Thus after a path of length p all the particles come down to the same point whatever be the angle of inclination of th eir initialmotionwith the field. The meson is at rest and so if a single photon is created. by the filament current. Asingle photoncannot be bornbecause the principle ofconservation ofmomentumw ould then be violated. one orbital electron in the atoma bsorbs part or allof the kinetic energyof the incident electron.e. 23. of higher energy (excitation). This iswhy two phot ons are bornwhichmove in opposite directions after creation. The charged particlewillmove along a helixwinding around the lines of induction. In aCoolide tube.. Fromthis place it begings tomove in the opposite direction. But it is not possible for a photon to be at rest. Atomic Structure. = (2. This energy appears as heat in the target. the electronreturns to the cathode via the external voltage circuit. They b ehave like fixed frequency sources inmotion. =h/p. This iswhy thewave nature ofmatter is not apparent in our daily lives. the charged partic le is strongly decelerated and so its drift velocitydecreases.m/Bqwhichis indepen dent of the velocityof the particles and pitch p = (2. becoming zero inthe case of a sufficientlyhigh fiel d gradient. Less than 1%of the incident electronenergy is actuallyconverted toX-rays. they are not exhibited sep arately. The intensity is controlled by the filamen t temperature.Due toDoppler effect. in conformitywith the principle of conser vation ofmomentum. Radioactivity. When amoving electron collideswith an atom. Thewavelengthof amatterwave is given by. 17. It is absorbed bythe target. i. Subsequ ently.physicsashok. The gas atoms emitting light due to electron transitions are inmotion. 21. The hardness is controlled by the applied voltage. Since the magneticmoment due to the orbitalmotion is opposite to themagnetic field is tends to push the charged particle out of the field.MODERN PHYSICS www.e. 22. Such independent control is not possible in the cathode ray tu be. Nuclear Fission and Fusion 1. the orbital electronmaymove to an outer orbit.m/Bq) v cos. 2. whereas radioactivit y is a nuclear process. The universalprinciple of stability ofa system. 5. hence the atomis left with t wo extra orbital electrons. . the atomis left with a ne t single positive charge. Chemical bonds involve onlyorbital electrons. i. that is. a systemlies instable equilibriumwhen its energy is at the lowest possibelvalue and Pauli s exclusion principle. The . It therefore has a double negative charge. No. Energies of thismagnitude oc cur innuclear processes but not inorbital electron transitions. no two electrons c anhave all their quantum numbers identical . The energyof a . In . 6. the atomic number decreases by2.e. In no way. 4. In . It cannot exist inside the nucleus as its de Broglie wavelength ismuch larger th an the dimensions of the nucleus. Bypassing themthrough transverse electric ormagnetic fields. 3.-rayphotonis of the order ofMeV.-decay.-particle. is actuallycreated at the instant o f . 7.ing the small spread of the spectral line.-decay.-decay and ejected at once.. No. although an electron. 8. . Hence. Thomsonmodel. that is.MODERN PHYSICS www.-decay.We have p = h/.. .. The lawof radioactive decay is statistical in nature. E . 15. 14. They bounce to and fro in the potentialwell bounded by the barrier befo re tunnelling through it. Thus finallywe find that I. 3/1 = n/1 . Z2 . m= 2. 12.physicsashok. Hence. 13. Hence the probability of escape is not the same for all the a-particles because all are not born inside the nucleus at the same time. Z.This is po ssible onlywith slowneutrons. Clearly. Because theBalmer series lies inthe visible region and the Lyman series in t he ultraviolet region.Moment ofmomentumis angular momentum.. For fission.Therefore. 20. .n . 16. innature allmoving bodies have an assoc iatedwave. In the stationary orbits thewaves associatedwiththe particlemust forma st ationarywave. theneutronmust be absorbed bythe fissionable nucleus. the element moves forward one place on the periodic table. 21.in 127 9.r)must be inmultiples ofwavelength. h/p or pr = nh/2. He nce pr is the angularmomentum(L) of the electron. n = 2.. .-decay. Bohr s principle of quantizationof angularmomentumis seen to be an essential c onditionwhile considering de Broglie smatterwave principle. .The averages over a large number of experimental results conformexact lywith the theory.When E is constant. Z2 / n2 . . n2 .. 2. (hv)max = E0/4 and (hv)min = 5E0 /36 or 5E0/36h < v < E0/4h. In . No. For the Lyman series n = 1. Fusion occurs onlyat temperatures of the order of 106K. In .. Spectral lines fromhydrogen arise fromthe relation hv = E0(1/n2 1/m2) withm> n. 10. 11. or 2... Yes.. . The emission of a-particles is caused by quantummechanical tunnelling throug h the repulsive Coulomb barrier. a spectral line cannot belong to both series. If r is the radius ofthe stationaryorbit.Rutherfordmodel.-particles have greater ionizing power than . This follows fromcomparisonwith the law of equipart ition of energy as applied to gasmolecules.. the same value of v cannot satisfy both the ser ies. This is a neutronwithenergyof the order of(3/2) kT. . 19.= nh/2. (hv)max = E0 and (hv)min = (3/4) E0 or 3 4 E0/h < v < E0/h For the Balmer series. Thismakes it extreme ly difficult to control fusion processes. .r = n .-particles and so they lose th eir energymuch earlier than . .. m= 3.. the element moves back two places on the periodic table. its circumference (equalto 2. Here c2 appears only as a constant and does not suggest motion. Thermalisation of a neutron brings down its energy froma high value of about (3/2) kT. individial expe rimental resultswillshowslight variations.whereTis the absolute tem perature of the surroundings and k is Boltzmann s constant. 18. n = 3 17.r = n. Otherwise the atomas awhole in empty. their emissionlowers the atomic numb er by2 andmass number by 4. 92 and 234. . 92U234 . So emittion of . Theydiffer in respect of their origin. 24.-particles lowers the position of the element by two places in the periodic table. whereas electrons in cathode rays are orbital electrons. 25. 23.. 22. 92U238 2He4 2 × 1e0 .-particles. This iswhy their track lengths are shorter th an those of . Yes.-particles are heliumatoms. An atomconsists of a smallcentralizedmass containing positivelycharged parti cles. Elements are arranged in the periodic table according to their atomic number.-particles originate in the nucleusw hen a neutron is converted into a proton.-particles because of collisions. Since . while the former is fissionable. These are highly radioactive. They hav e to be disposed of in sealed containerswhich can contain their radioactive emissions. .physicsashok. This happens in a controlled chain reaction. Fusion yields greater energyper unit mass. Neutrons. 36. Themass is reduced.-decay. 34. e. ina nuclear reactor.This is called thermalization. 38. Because large n corresponds to a verylarge value at which classical and quan tumresults are identical by Bohr s correspondence principle. True. two nuclideswith Z of the order of 40 to 50 are created. Thomson smodel. 32. For large quantumnumbers.-particles are deflectedmore due to their larger specific charge.g. Naturaluraniumcontains less than1%ofU235mixedwithU238. . 35. 31. quantummechanical results reduce to classical resu lts.in 128 26.MODERN PHYSICS www. As the photonmoves out against strong gravitational attraction its energydec reases and so its frequencyis expected to decrease. 39.This is gravitational shift. 28. 37. These are called radioactivewaste . In nuclear fission. This is called enrichment of uranium. and is performed bythemoderator. The latter is not fis sionable. Themass number does not change. Fission yields greater energyper atom. 29. of the several produced by the fission of one nucl eus.t. False. 27. The proportion ofU235must be increased artificiallyfor the urani umto be used in fission.. because theyare electricallyneutral and so they do not interactwit hmatter electrically. with half-lives or thousands of years. In N = N0e . The only possible decay process is . for N = 0. 30. t = . When exactly one neutron. The decayproduct would be 2He3. 33.This follows fromthe exponentialnature ofthe decay. is permitted to cause further fission. The neutrons emitted in fissionmust be slowed down inorder to cause further fission in other nuclei. 07] . energy of . Statement-2 : The binding energy per nucleonwith increase in atomic number first increases and then decreases.in 129 ASSERTION-REASON (A) Statement-1 is True. Statement-2 : In pair production. 2. Statement-1 : The nuclear energycan be obtained bythe nuclear fissionof heavi er nuclei aswell as nuclear fusion oflighter nuclei. Statement-1 : The ratio of rate of production (R) of neutrons to the rate of leakage of neutron froma spherical body of 92U235 is directlyproportional to radius (r) Statement-2 : Rate of production of neutron is directly proportional to volume b ut rate of leakage of neutrons is directlyproportional to area. Statement-1 : Asmallmetal ball is suspended in a uniformelectric fieldwith an insulated thread. Statement-1 : Ionisation energyof atomic hydrogen is greater than atomic deut erium.2 eV. 7. Statement-2 :When an electron beamstrikes the target in anX-ray tube. Statement-2 is True. Statement-2 : Heavy elements exhibit radioactivity. 10. Statement-1 : The difference in the frequencies ofseries limit ofLymanseries andBalmer series is equal to the frequency of the first line of the Lyman series. Statement-2 :Wavelengthof photon is directlyproportional to the energyof emitted photon 3. the electron is not emitted. Statement-2 is True. If high energyX-ray beamfalls on the ball. If two photons of eachof en ergy 2.02MeV. energy is converted into mass. Statement-2 : In photoelectric effect. 8. 6.physicsashok. [JEE. the ballwill be deflected in the direction of electric field. thewave lengthsofthe characteristic X-rays do not change.5 eVstrike on an electron ofaluminium. 9. X-ra y for the same material. X-raymust be greater then the wavelength ofK. Statement-1 :Work function of aluminiumis 4. Statement-2 is NOT a correct expla nation for Statement-1 (C) Statment-1 is True. Statement-2 is a correct explanati on for Statement-1 (B) Statement-1 is True. Statement-1 : In a hydrogen atomenergy of emitted photon corresponding to tra nsition fromn = 2 to n = 1 ismuch greater as compared to transition fromn = . ray is greater than 1. electron is emitted onlyif energyofeach o f incident photonis greater than thework function. 5. Statement-1 : Electron capture occursmore often than positron emission in hea vy elements. Statement-2 :Wavelength of L. part of th e kinetic energy is converted intoX-rayenergy.MODERN PHYSICS www. Statement-1 : For pair production. Statement-2 : Ionisation energy is directly proportional to reducedmass 4. Statement-1 : Ifthe acceleratingpotentialinanX-raytube is increased. Statement-2 :Difference inenergyof two atomic levels is proportionalto the energ yofemitted or absorbed photon. Statement-2 is True 1. Statement-2 is False (D) Statement-1 is False. to n = 2. . 6 Z2 2 2 1 2 1 1 n n . Column I Column II (A) Binding energy per nucleon for middle order (P) Optical Model or elements is (B) Nuclear force depends on (Q) Shell model (C) For nuclear fission Z2 A is (R) 8.physicsashok. dN/dt is (P) Greater than Half life proportional to (B) Mean life of radioactive substance is (Q) Less than Half life (C) Intensity I of . Column I Column II (A) Radius of orbit depend on principal quantum (P) Increase number as (B) Due to orbital motion of electron. .5 eV explained by (T) Charges of Nucleons (U) Spin of Nucleons (V) Greater than 15 (W) Less than 15 5. 28.x (U) Temperature. volume 4. 20. pressure. ray of initial intensity I0 after (R) Number of atoms of pa rent radioactive transversing the thickness of x of the absorber is substance still undecayed at time t (related as) (D) The radioactive decay rate is not affected by (S) I = I0/x (T) I = I0e .MODERN PHYSICS www.e. 8. Column I Column II (A) Radius of orbit is related with atomic number (Z) (P) is proportional to Z . . Column I Column II (A) Work function of copper is 4 eV. If two (P) 13. 126 (S) 2. 50. .5 eV strike an electron of copper emission of electrons (B) Cathode rays get deflected by (Q) 13. eV (C) Ionisation energy of H like atom is (R) both electric and magnetic field (S) 1H1 (D) Greater wavelength in transition from n = 2 (T) 1H3 to n = 1 is for (U) Not possible (V) Possible 2.8 MeV (D) Magic numbers are 2. then total energy of (T) is proportional to 1/n5 electron will 3. i. 82. as (R) is proportional to 1/n2 (C) If electron is going from lower energy level to higher energy level then velocity of electron will (S) is proportional to n2 (D) If electron is going from lower energy level to higher energy level. Column I Column II (A) Rate of disintegration. . (Q) decrease Magnetic field arises at the centre of Nucleus is proportional to principal quantum no. .in 130 Match the Column 1.6 Z2/n2eV photons each of energy 2. . (D) Velocity of an electron related with atomic (S) is proportional to Z3 number (Z) .(B) Current associated due to orbital motion (Q) is inversely proportion to Z electron with atomic number (Z) (C) Magnetic field at the centre due to orbital (R) is proportional to Z2 motion of electron related with Z. 5 eV (C) Binding energyper nucleon (R) 3 eV (D) Photoelectric threshold ofmetal (S) 20 eV (T) 10 keV (U) 8MeV 8.Match the statements inColumn I to the appropriate process (es) fromColumn-II. aga inst the number (i). Choose the appropriate value of energy fromcolumn II for each of the physical quantities in column I andwrite the corresponding letterA. etc. column I lists some physical quantities&the column II gives approx. Cetc. 09] (A) The energy ofthe systemis increased (P) System: Acapacitor. initiallyat rest Process :The nucleus fissions into two fragments of nearly equalmasses and some neutrons are emitted. (iii).MODERN PHYSICS www.025 eV (B) EnergyofX-rays (Q) 0.-decay energyof electrons. Column. of the physical quantityin the answer book. In your answer. (ii). B. 06] Column I Column II (A) Nuclear fusion (P) Converts somematter into energy (B) Nuclear fission (Q) Generallyoccurs for nucleiwith lowatomic number (C) . 97] (A) Energyofthermalneutrons (P) 0. 07 ] (B) electron emissionfromamaterial (Q) Photoelectric effect (C) Mosley s law (R) Hydrogen spectrum (D) Change of photonenergyinto kinetic (S) . Process : It is connected to a b attery which is converted into energy of random (Q) System: Agas inan adiabatic contain er fitted motion ofits parts with an adiabatic piston (C) Internal energyof the systemis converted into Process :The gas is compressed by pushing itsmechanical energy.in 131 6. initiallyuncharg ed (B) Mechanical energyis provided to the system.II gives certainsystemsundergoing aprocess. the piston (D) Mass of the systemis decreased (R) System: Agas in a girid container Process :The gas gets cooled due to colder atmosphere surrounding it (S) System: Aheavynucleus. Match the followingColumns [JEE.Column-I suggests changes insomeofthe parameters related to the system. Column I Column II (A) Transition between two atomic energylevels (P) CharacteristicX-rays [JEE. 9. energy values associated with some of them.-decay (R) Generallyoccurs for nucleiwith higher atomic number (D) Exothermic nuclear reaction (S) Essentially proceeds byweek nuclear forces 7. Column I Column II [JEE.Match thesewiththe physicalphenomena giveninColumn II. In the following. Column I Column II [JEE. the sequ ence of column I should be maintained. (T) System:Aresistivewire loop Process :The loop is placed in a time varrying magnetic fieldperpendicular to its plane . Some laws/processes are given inColumnI.physicsashok. n2 = 2 (c) n1 = 8. 1 2 . (given in terms of the Rydberg constant R for the hydrogen atom ) equal to (a) 9/5R (b) 36/5R (c) 18/5R (d) 4/R 8. (b) min .n2 where n1 and n2 are the principal quantum numbers of the two states. to .2 MeV (c) 59 keV (c) 136 eV 3. h aving non zero velocities.MODERN PHYSICS www. . 5. A particle of mass M at rest decays into two particles of masses m1 and m2.8 (d) 79. ray .0 (d) 1 2 m m 6. is (a) 2 1 m m (b) 1 2 m m (c) 1. = 0. potential and total energies decrease.51 MeV (b) 1. The electron in a hydrogen atom makes a transition n1 . (d) min . potential energy increases and its total energ y remains the same. n2 = 3 4. The longest wavelength photon that will be emitted has wavelength .2 (c) 51. The energy difference between K and L levels in this atom is about (a) 0. where max . The K X . Electrons with energy 80 keV are incident on the tungsten target of an X-ray . . The ratio of the de Broglie wavelengths of the particles. (d) Its kinetic. where 0 < min . > 0 (c) 0 to max . Apply the Bohr atom model and consider a ll possible transitions of this hypothetical particle to the first excited level. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. The time period of the electron in the initial state is 8 times that in the final state.2 (b) 49. < max . (b) Its kinetic energy decreases. The energy (in eV) required to remove both the electron from a neutral helium atom is (a) 38. where min . (c) Its kinetic and total energies decrease and its potential energy increases. < . 7. The electron in a hydrogen atom makes a transition from an excited state to t he ground state.0 2. X-rays are produced in an X-ray tube operating at a given accelerating voltag e. Which of the following statements is true ? (a) Its kinetic energy increases and its potential and total energies decrease. n2 = 1 (d) n1 = 6.021 nm. The possible values of n1 and n2 are (a) n1 = 4. Assume the Bohr model to be valid. to max .in 132 Level 1 1.6 eV is required to remove one of the electrons from the neut ral helium atom. n2 = 1 (b) n1 = 8. An energy of 24. emission line of tungsten occurs at . The wavelength of the continuous X-rays has values from (a) 0 to .physicsashok. < . For a photoelectric cell. the graph in Figure.tube. and the characteristics X-rays spectrum of tungsten. K shell electrons of tungsten have 72.155A.5 keV energy. X-rays emitted by the tube contain (a) a continuous X-ray spectrum (Bresmsstrahlung) with a minimum wavelength of a bout 0. showing the variation of the c ut-off voltage V0 with frequency (v) of incident light is O v V0 (a) O v V0 (b) O v V0 (c) O v V0 (d) .155A. (b) a continuous X-ray spectrum (Bremsstrahlung) with all wave-lengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of ab out 0. 9. (b) 4 . 13. (b) . . (c) . 1 v2 v1 e V .v2 v1. (c) . . the stopping p otential is V/3. e V .in 133 10. the stopping potential is V0. The threshold frequency for photoelectric emission is (a) 3 2v (b) 2 3v (c) 5 3v (d) 3 5v 12. 1 v1 v2 e V . (d) . . h . Monochromatic light of frequency v1 irradiates a photocell and the stopping potential is found to be V1. . the stopping potential is V. . radius R0 contracts to radius R.p (d) 2 p.MODERN PHYSICS www. From this statement one may conclude that the wave velocity of light is eq ual to (a) 3 x 108 ms 1 (b) . 14. the stopping potential for photoelectric current is V0/2. . . . When the same surface is illuminated by light of wavelength 2 . (b) . h . When a centimeter thick surface is illuminated with light of wavelength . . When a certain photosensitive surface is illuminated with monochromatic ligh t of frequency v.. .0 c. 1 v1 v2 e V . A star of mass M0. (c) 6 . . The threshold wavelength for the surface is (a) 3 4 . . . . . . h . 11..physicsashok. is p = h / . . . h . . (d) 3 8 . When the same surface is illuminate d by monochromatic high of frequency v/2. . Energy radiated by the s tar assuming uniform density in each case while temperature remains unchanged is (a) . What is the new stopping potential of the cell if it is irradiated by monochroma tic light of frequency v2? (a) 1 . The energy of a photon of frequency v is E = hv and the momentum of a photon of wavelength . is (l >> d) (a) 2 2 hc N P d t . . . . . (d) 2 2 16hc N P d t .. . (b) 2 2 hc N 4P d t . 3 0 0 R c 1 R 15. . . . . . . . . . . . . (c) 2 2 4hc N P d t . . . 0 0 R c 1 R (d) . . 2 0 0 R c 1 R (c) . . .. The sensor has an opening that is 4d in diameter. . . . . t he number of photons entering the sensor if the wavelength of light is .. . . . Assuming all energy of the lamp is given off as light. .. . . .. .. . . . . A sensor is exposed for time t to a lamp of power P placed at a distance . . . .. . 2 1 . . . An electron is lying initially in the n = 4 excited state.n 1. 6 1 . . (b) 41 43 31 . .n 1.2n 1. (c) . . . .2n 1. . (a) . . If we take into account the recoil of the atom when the photon is emitted. 3 1 . . 17. . Due to a n electron transition from n = 2 to n = 1 in a hydrogen atom. is emitted. If elements of quantum number greater than n were not allowed.2 eV corresponds to light of wavelength 0 . then (a) 41 43 31 . (d) the data is not sufficient to reach a co nclusion. If the same electron first de-exci tes to n = 3 state by emitting a photon of frequency v43 and then goes from n = 3 to n = 1 state by emitting a photon of frequency v31 . . (b) 0 .41 . . . . (d) n. (c) 0 . .43 . . 2 2 n n 1 .31 (d) Data Insu fficient 18.n 1. . . (c) n. . . . 2. = 0 . 16. the electron de-e xcites itself to go to n = 1 state directly emitting a photon of frequency v41 . (b) .. . . . the number of possible elements in nature would have been (a) n . light of wavelength . . A photon of energy 10. 2 then : (a) . Binding energy per nucleon of 1H2 and 2He4 are 1.252 MeV (d) 23.1 = . Energy released in the process 1H2 + 1H2 = 2He4 is : (a) 20.1 > .1 and .9 MeV 21. line (c) Q and P represents K.1 > . It emi ts a photon of wavelength .1 < .2 (b) .2 24.26 MeV (b) 7 MeV (c) 15. and K. After sometime if th e de Broglie wavelength of the two are .2 (c) . When an electron moving at a high speed strikes a metal surface.0 MeV respective ly.6 MeV 23. while returning to the ground state. lines respectively (d) Position of K.physicsashok. mass of oxygen nucleus = 15. The value of n is (a) 1 (b) 2 (c) 3 (d) 4 26.2 MeV (d) 23. In a characteristic X-ray spectra of some atom superimposed on a continuous X-ray spectra : P Q . (d) The electron may undergo elastic collision with the metal surface. particles is : (a) E n(E) Y X (b) E n(E) Y X (c) E n(E) Y X (d) E n(E) 22.in 134 19. Two electrons are moving with the same speed V. depend on the particular atom 25. (c) The entire energy of the electron may get converted to heat.6 MeV (c) 25. and K. 20. line (b) Q represents K. The value of n is : 1 . The graph showing the energy spectrum of . A hydrogen atom is in an excited state of principal quantum number n.2 (d) . Relation intensity (a) P represents K. Difference between nth and (n + 1)th Bohr s radius of H atom is equal to its (n )th Bohr s radius.0026 amu. The en ergy released per oxygen nuclei is (given mass of nucleus = 4.MODERN PHYSICS www.994 a mu) (a) 7. One electron enters a region of uniform electric field while the other enters a region of uniform magnetic field .1 > . which of th e following are possible? (a) The entire energy of the electron may be converted into an X-ray photon. (b) Any fraction of the energy of the electron may be converted into an X-ray ph oton.2 or . A star converts all of its 2He4 nuclei completely into oxygen nuclei.8 MeV (b) 16.1 eV and 7.min . .R . Initially they have the same number of nuclei. Two atoms of X fuse to give one atom of Y and an energy Q is released. The binding energies of nuclei X and Y are E1 and E2 respectively. and . (c) 11/10. .1) (b) ( R 1) R . . . (d) 1/9. The ratio of the number of nuclei x1 to that of x2 will b e 1/e after time : (a) 1/10. (c) R R 1 . . respectiv ely.(a) .R(. 27. (b) 1/11. Two radioactive materials x1 and x2 have decay constant 10 . Then : (a) Q = 2E1 E2 (b) Q = E2 E1 (c) Q < 2E1 E2 (d) Q > E2 2E1 28. . (d) (R 1) . It can emit a maximum energy photon of 204. If the hal f life of neutrons is 700 s.MODERN PHYSICS www. Assuming that about 200 MeV energy is released per fission of 92U235 nuclei. It makes a transition to quantum state n.6 eV be elastic and more th an 27.physicsashok. (d) Density of nucleus is directly proportional to mass number. P and r. the magnitude of linear momentum and orbital radius of an electr on in a hydrogen atom corresponding to the quantum number n are E.3 . There is a stream of neutrons with a kinetic energy of 0. An electron is excited from a lower energy state to a higher energy state in a hydrogen atom. The count rate for 10 gram radioactive material was measured time (hrs) at different times and this had been shown in figure scale given. The wavelength of first Balmer line for 1H1 . The fraction of neutrons will decay before they travel a distance 10 m. according to the Bohr s theo ry of hydrogen atom: (a) EPr is proportional to 1/n (b) P/E is proportional to n0 (c) Er is not constant for all orbits (d) Pr is proportional to n. 34. A hydrogen like atom of atomic number Z is an excited state of quantum numbe r 2n. then (a) Z = 2 (b) Z = 4 (c) n = 1 (d) n = 2 37. a ph oton of energy 40. What would be mass of 92U235 consumed per day in the fission of reactor of power 1 MW approximately ? (a) 10 kg (b) 100 kg (c) 1 gram (d) 10 2 gm 33.2 and .96 × 10 5 (b) 3.1. The energy. and 9000 approximately (b) 3 hrs and 14100 approximately (c) 3 hrs and 235 approximately 75 50 25 3 6 9 100 12. Which of the following quantities decrease in the excitation.2 eV be inelastic.675 × 1 0 27 kg : (a) 3. 32. (a) Potential enrgy (b) Angular speed (c) Kinetic energy (d) Angular momentum 35. The half life of material and the total count in the first half value period respectively are : (a) 4 hrs.5 (d) 10 hrs and 157 approximately time in hr. An electron with kinetic energy varying from 5 eV to 50 eV is incident on a hydrogen atom in its ground state. 38. 1H2 and 2He4 and . eV. 36.in 135 29.8 eV is emitted. At t = 0 activity of radioactive substance is 1600 Bq and t = 8 sec activity remains 100 Bq. The correct statement is l are : (a) density of nucleus is independent of mass number (A) (b) Radius of nucleus increases with mass number (A) (c) Mass of nucleus is directly proportional to mass number (A).0327 eV. The collision : (a) may be elastic (b) may be partially elastic (c) must be completely inelastic (d) from zero to 13.96 × 10 6 (c) 2. . The activity at t = 2 sec is : (a) 200 Bq (b) 400 Bq (c) 600 Bq (d) 800 Bq 30.96 × 10 4 (d) None 31. mn = 1. 1 ..1 (b) .2 > . The correct statements are (a) .2 (c) . (c) Wavelength of continuous spectrum is dependent on the potential difference a cross tube (d) None of these ..2 .....3 . Regarding X ray spectrum which of the following statement (a) The characteristic X ray spectrum is emitted due to excitation of inner elec trons of atom (b) Wavelength of characteristic spectrum depend on the potential difference acr oss the tube.2 (d) .3 39.respectively...1 . (b) Radius of an orbit is double.MODERN PHYSICS www.1 .25 × 10 5 kg 47. then : 1 2 3 C B A .85 × 10 2 gram (b) 0.25 × 10 4 kg (b) 12. which of the following statement are consisten t with Bohr s theory : (a) Energy of a state is double. . The energy of .9949 a.3 = . 43.3 and .in 136 40. .m.3 (a) .13 MeV (c) Energy needed to remove one proton from 8O16 is 15. When Z is doubled in an atom.1 + . If the wavelength of light in an experiment on photo electric effect is doub led : (a) The photoelectric emission will not take place. .5 × 10 4 kg (c) 2.0001 a. Then : (a) Binding energy per nucleon in 8O16 is 7.2 . (a) 6.physicsashok. What mass 210Po is ne eded to power a thermoelectric cell of 13 watt output.61 MeV.2 (c) 1 2 3 1 2 .85 × 10 4 gram 45.u. (d) Binding energy per nucleon is minimum for middle order elements 42.97 MeV (b) Energy is needed to remove one proton 8O16 is 12. What would be power output after 1 year : (The half life of 210Po is 138 days) (a) 8.Wavelength is more than work function of metal.2.3 MeV.2. Photons of wavelength 6620 Å are incident normally on a perfectly reflecting s creen. and 15. 8O15 and 8O16 are respectively 15. . (c) Velocity of electrons in an orbit is doubled (d) Radius of orbit is halved.m.5 × 108 (d) None of these 44.u. .3 = .1. (b) The photoemission may or may not take place.0030 a. Calculate the number of photons per second falling on the screen as total power of photons suc h that the exerted force is 1N : (a) 5 × 1026 (b) 5 × 1025 (c) 1.1. ..2 (b) . Their weights are in the ratio of 1 : 2. .m. particles emitted by 210Po is 5. 41. The atomic masses of 7N15. The wavelength and frequency of photons in transition 1. . A sample contains 10 2 kg each of two substances A and B with half lives 4 sec and 8 sec respectively.159 watt (c) 0. (c) The stopping potential will increase (d) The stopping potential will decrease under the condition that energy of phot on of doubled. The amounts of a and B after an interva l of 16 sec.u. (d) All the above 46.179 watt (d) 8. the binding energy per nucleon is : (a) Maximum for middle order element (b) Minimum for lighter elements (c) Binding energy per nucleon suddenly increases for some mass number called ma gic numbers.5 × 10 3 kg (d) 1.3. 15 .1 + .2 and 3 for H like a tom are . .. . the ratio of U235 to U238 is 1 : 40 (b) Critical mass of ura nium is 10 kg (c) 92U235 : 92U238 = 1 : 4 (d) All the above . . . Which of the following pair constitute very similar radiations ? (a) Hard U. . . 48. The correct option are : (a) In uranium ore.V.V. (b) Soft U. ray and soft X ray. . ray and hard X ray (c) Very hard X ray and low frequency Y ray (d) Soft X ray and Y ray 49. (d) 1 2 3 1 2 . . Fast neutrons can easily be slowed down by : (A) the use of lead shielding (B) passing themthroughwater (C) elastic collisionswith heavynuclei (D) applying a strong electric field (ii). 94] (A) . .M1 themass of a 20 10 Ne nucleus&M2 themass of a 40 20Ca nucleus.-rays. 98] (i) Let mp be themass of a proton. (C) Innuclear fusion. ..86 54. If the average power radiated bythe star is 1016W.Yand Zare four nuclei indicated on the curve.X. The binding energy per nucleon of 16O is 7..5 5. 1H3 + p & 1H2 + 1H3 . 53. each having an energy of 0. 97] (A) 2 (B) 4 (C) 6 (D) 10 55. energyis released by fragmentation of a very heavy nucle us. Consider . Which ofthe following statement(s) is (are) correct ? [JEE..64 (C) 4. Themaximumkinetic energyofphotoelectrons emitted froma surfacewhenphotons of energy6 eVfallon it is 4 eV. particles & . 95] (A) 3. Given a sample of 131I at time t = 0.physicsashok. (B) . Then : (A) M2 = 2M1 (B) M2 > 2M1 (C) M2 < 2M1 (D) M1 < 10(mn + mp) (ii) The half-life of 131I is 8 days. (D) In nuclear fission. Binding energyper nucleon vs.52 (B) 3.. (B) The rest mass of a stable nucleus is greater than the sumof the rest masses of its separated nucleons.in 137 50. . (D) . 2He4 + n.23 (D) 7. Increasing order of penetrating powers. It produces energy via.97 MeV & that of 17O is 7. .. . 52. 99] 30 60 90 120 Z Y X W 8.mass number curve for nuclei is shownin the figure. . the processes 1H 2 + 1H2 . The energy in MeV required to remove a neutron from17Ois : [JEE. .0 Mass Number of Nuclei in MeV ..mn themass of a neutron. . A star initially has 1040 deutrons. 94] (A) The rest mass of a stable nucleus is less than the sumof the rest masses of its separated nucleons. .75 MeV.5 MeV. Select the correct alternative(s). the stopping potential inVolts is : [JEE.5 8. the radiations are : [JEE.. we c an assert that : (A) no nucleuswill decay before t = 4 days (B) no nucleuswill decay before t = 8 days (C) all nucleiwill decay before t = 16 days (D) a given nucleusmaydecayat anytim e after t = 0 56(a). (C) . the d euteron supplyof the star is exhausted in a time of the order of : [JEE. energy is released byfusion two nuclei ofmediummass (appro ximately100 amu)..W...MODERN PHYSICS www. The process that would release energyis [JEE. 93] (A) 106 sec (B) 108 sec (C) 1012 sec (D) 1016 sec 51(i).. [JEE.0 7.. particles. Y+ Z . 2Y (D) X . 2Z (B)W. X + Z (C) W.Binding Energy/nucleon (A)Y. 2000 Scr. [mp = 1. Find the probability. after absorbing energy.1/.Then (A) X&Yhave the same decay rate initially (B) X &Ydecay at the same rate always. Initiallyboth of themhave the same number ofatoms. then [JEE.. Aparticle ofmassMat rest decays into two particles ofmassesm1 andm2. . (D) Its kinetic. (C)Ywilldecay at a faster rate thanX (D) X willdecay at a faster rate thanY 57. 06] (A) 1 (B) 1/2 (C) 3/4 (D) 1/4 .physicsashok. (C) Its kinetic and total energies decrease and its potential energy increases. and EH and ELi their respective energies. 2000 Scr.0 (D) 2 1 m / m 58. having non-zero velocities. (C)Alpha particles are singlyionized heliumatoms (D) Protons and neutrons have exactlythe samemass (E)None (e) The half-life period of a radioactive elementXis same as themean-life time o fanother radioactive element Y.Whichof the following statements is true ? [JEE.Apply theBohr atommodel and consider allp ossible transitions of this hypotheticalparticle to the first excited level.-particles and an unk nown nucleus. [JEE. The radio of the de-Brogliewavelengths of the particles.] (A) lH > lLi and |EH| > |ELi| (B) lH = lLi and |EH| < |ELi| (C) lH = lLi and |EH| > |ELi| (D) lH < lLi and |EH| < |ELi| 61. potential and total energies decrease. 02 Scr. The longest wavelength pho ton that willbe emitted has wavelength . The potentialdifference applied to anX-ray tube is 5 kV and the current thro ugh it is 3. a nucleus disintegrateswithin 2 half lives.in 138 (b) Order ofmagnitude of density ofUraniumnucleus is. potential energyincreases and its total energyr emains the same.2. 02 Scr.The unknown nucleus is (A) nitrogen (B) carbon (C) boron (D) oxygen (d) Which of the following is a correct statement ? (A) Beta rays are same as cathode rays (B) Gamma rays are high energyneutrons.] (A) Its kinetic energy increases and its potential and total energies decrease.2mA. is [JEE. 99] (A)m1/m2 (B)m2/m1 (C) 1. decays into two .] (A) 2 × 1016 (B) 5 × 1016 (C) 1 × 1017 (D) 4 × 1015 60.MODERN PHYSICS www.(a) Imagine an atommade up of a proton and a hypotherical particle of double themass of the electron but having the same charge as the electron.] (A) 9/(5R) (B) 36/(5R) (C) 18/(5R) (D) 4/R (b). If lH and lLi are their respective electronic angularmomenta. Then the number of electrons striking the target per second is [JEE.(given in terms ofthe Rydberg constant R for the hydrogen atom) equ al to [JEE. (B) Its kinetic energydecreases. Givena sample ofRadium-226 having half-life of 4 days. A Hydrogen atom and Li++ ion are both in the second excited state.67 × 10 27 kg] (A) 1020 kg/m3 (B) 1017 kg/m3 (C) 1014 kg/m3 (D) 1011 kg/m3 (c) 22Ne nucleus. The electronin a hydrogen atommakes a transition froman excited state to th e ground state. 59. The cut-offwavelength of the emittedX-rays is [JEE. 2E(n) (D) . . 92 53 39 E U .2 : . Electronswith de-Brogliewavelength .001 0.The smallestwavelength in the infrared region of the hydrogen spectrum(to the nearest integer) is [JEE. 2E(n) (C) . 236 .140 .236 .94 .97 . [JEE.3 = 1 : 2 : 4 (B) Ratio ofwork functions .1 : . (D) . 2E(n) (B) . let E denote the rest mass energy of a nucleus a nd n a neutron.1 : . .004 1/ nm 1 metal 1 metal 2 metal 3 (A) Ratio ofwork functions . E I .3 is anexperiment of photoelectric effect is plotted as shown in the figure. ..3 = 4 : 2 : 1 (C) tan . E Kr . E Y . 07] (A) . 236 . E I . (B) 0 2h mc .physicsashok. Which of the following statement(s) is/are correct? [Here .MODERN PHYSICS www. In the options given below.94 . is thewavelength of the incident ray]. (C) 2 2 2 0 2 2m c h . and stopping potential (V) of three metals havingwork functions . 0 6] 63. 92 53 39 E U . . 66. .97 . . . . The correct option is : [JEE. V 0.140 . . . . .236 . . E Ba . fallon the target in anX-ray tube.in 139 62.is directly proportional to hc/e. where h is Planck s constant and c is the speed of light (D) The violet colour light can eject photoelectrons frommetals 2 and 3. [JEE.137 . 08] (C) Intensityof the characteristicX-rays depends on the electricalpower given to theX-rays tube. . 92 56 36 E U . The graph between 1/.0 = .1. 92 56 36 E U . E Ba . E Kr . E Y .2 and . 07] (A) 2 0 2mc h . 07] (A) 802 nm (B) 823 nm (C) 1882 nm (D) 1648 nm 65.002 0. 2E(n) 64. .2 : .137 . (D) Cut-offwavelengthof the continuousX-rays depends on the energyof the electro ns intheX-rays tube. Which one of the following statements isWRONGin the context ofX-rays generat ed fromaX-raytube? (A)Wavelength of characteristicX-rays decreaseswhen the atomic number ofthe targ et increases (B) Cut-offwavelengthof the continuousX-rays depends on the atomic number of the target. The largestwavelengthinthe ultraviolet regionofthe hydrogenspectrumis 122 nm . 67.Use this plot to choose the correct choice(s) given below: Figure 100 200 A 0 2 4 6 8 B/A (A) Fusionof two nucleiwithmass numbers lying in the range of 1 <A< 50 will release energy. The quantumnumber n of the state finally populated inHe+ ions is [JEE. 08] (A) 2 (B) 3 (C) 4 (D) 5 . Assume that the nuclear binding energyper nucleon(B/A) versusmass number (A) is as shown inthe figure. (B) Fusionof two nucleiwithmass numbers lying inthe range of 51 <A< 100 will release energy. (C) Fission of a nucleus lying inthemass range of100 <A< 200will release energyw hen broken into two equalfragments. [JEE. (D) Fission ofa nucleus lying inthemass range of 200 <A< 260will release energyw henbroken into two equalfragments. 08] 68. 08] (A) 20 years and 5 years.0 eV. 08] (A) 1 4 (B) 1 2 (C) 1 (D) 2 71.r = 3. Due to the speed at which these galaxies are travelling.in 140 69. q and r havingwork functions .p = 2. 450 nmand 350 nmwith equal intensities illuminates each of the plates. The half lives of S1 and S2 can be : [JEE.physicsashok. 09] (A) I p q r V (B) I p q r V (C) I p q r V (D) I r q p V Passage PASSAGE : 1 Figure 1 below depicts three hypothetical atoms.5 × 10 7 m (B) 5. respectively (B) 20 years and 10 years. 08] (A) 6. Thewavelength oflight emitted in the visible region byHe+ ions after collisi onswithHatoms is [JEE.0 eV.5 eVand . The ratio of the kinetic energy of the n = 2 electron for theHatomto that of He+ ion is [JEE.MODERN PHYSICS www.6 × 10 7 m (C) 4. Aradioactive sampel S1 having an activity of 5µCi has twice the number of nucl ei as another sample S2 which has an activity of 10 µCi. respectively. but their pattern remains the same. The distance between the segments is representative of the energy difference between the various levels .8 × 10 7 m (D) 4. these lines are shifted.Alight beamconta ining wavelengths of 550 nm. Energy levels are represented a s horizontal segments.0 × 10 7 m 70. respectively (C) 10 years each (D) 5 years each 72. . Photoelectric effect experiments are performed using three different metal p lates p. All possible transitions between energy levels are indicated by arrows Atom #1 Atom # 2 Atom # 3 Scientists can observe the spectral lines of atoms that are dominant in far-away galaxies. This allows researchers to use the spectral pattern to determine which atoms they are seeing . T he correct I Vgraph for the experiment is [JEE.q = 2. Which is proportional to the energy. Table 1 below shows spectroscopic measurements made by researchers trying to determine the atomic ma keup of a particular faraway galaxy. but rather is determined from mea suring the frequency of light. Light energy is not measured directly.. . (D) four distinct spectral lines emanating from six different electrons. because it contains four different energy levels (B) Atom 3. Scientist observing an actual atom similar to hypothetical Atom 1 in the figu re might observe (A) three spectral lines close together and two other spectral lines close toget her (B) light blinking at six different frequencies (C) a much brighter light emanating from one electron than from any other. Certain transitions from one energy level to another result in the emission of a photon of radiation . 4. The laws of atomic physics prohibit electron movements between certain energy states. E4. In which of the three hypothetical atoms depicted in Figure 1 does the energy of the light released by the atom vary the least (A) Atom 1 (B) Atom 2 (C) Atom 3 (D) It is impossible to tell 3...in 141 Table 1 Frequencies Measured 868440 880570 879910 856390 1. Atom 2 and Atom 3). or E2 and E3 or E3 and E4 e tc. Figure 1 de picts the (A) number of electrons and the amount of energy the atom contains (B) distance an electron travels from one part of the atom to another (C) energy released by the atom as an electron as it moves from one energy state to another (D) frequency with which the atom s electrons move from one energy state to anothe r 2. The energy levels in a newly discovered gas are expressed as: . Based on Figure 1.. whereas others can only take place if a photon is absorbed. because there is a comparatively small difference between exactly tw o of the four frequencies listed in Table 1 5. In atomic physics. and why ? (A) Atom 2... w hich of the following is most accurate ? (A) Atom 2 has the same number of forbidden transitions than Atom 3 (B) Atom 2 has more forbidden transitions than atom 3 (C) Atom 3 has the same number of forbidden transitions as Atom 1 (D) Atom 1 has fewer forbidden transitions than Atom 2 PASSAGE : 2 In quantum mechanics. For each of three hypothetical atoms (Atom 1. which of the atoms in Figure 1 (Atom 1. (D) Atom 3.. Based on the spectroscopic measurements shown in Table 1. E2.. these prohibitions are called forbidden transitions. or Atom 3) is most similar to the one the scientists were observing. Energy can only take certain values E1. One of these quantities is the energy. some quantities are discrete and cannot be continuous. which are called energy levels. because it contains four different energy levels (C) Atom 1. Atom2.MODERN PHYSICS www. because the frequencies listed in Table 1 indicate a high level of a tomic activity. The energy cannot take any values between E1 and E2. E3.physicsashok. The results obtained are as follows : n En(eV) 2 144 3 64 4 36 6. Take z = 1 for simplicity. in which E1z2 is the ground state energy. The ground state energy is (A) 144 eV (B) + 144 eV (C) 244 eV (D) none of the above . . An experiment is designed to measure the energy as a functions of the level. but do not a ssume that the gas is hydrogen.2 1 n 2 E E z n . The ionization energy of the gas must be (A) 244 eV (B) 576 eV (C) 144 eV (D) +13.6 eV 7. a cathode emits electrons when kinetic energy of Emax will not be able to reach the anode therefore the current will stop altogether. in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1. For example.2 kW/m2. electrons travel to the anode. A transition from the n = 2 state to the n = 3 state results : (A) in emission of a photon of energy 144 eV (B) in emission of a photon of ener gy 80 eV (C) in emission of an ultraviolet photon 10. Light intensity is defined as the energy flowing per unit area per unit time for an area perpendicular to the direction of energy flow. The most efficient modern photovoltaic cells can covert the Sun s energy into electrical energy with an efficiency of 35%. In an experiment. is 0.6 eV (D) only accomplished if a photon is absorbed. This solar energy can be converted directly into electrical energy via the photoelectric effect. in the photoelectrical cathode (emitter) incidentlight Anode (collector) photocell A Ammeter cell shown in figure 1. averaged over 24 hours. Which of the following shapes is most likely to represent the graph of En ver sus 1/n2 ? (A) (B) (C) (D) 9.2 kW/m2.5 m2 (B) 12 m2 (C) 285 m2 (D) 6850 m2 12. Approximately what area would have to be covered by such cell s in order to supply a household with 20 kW-hourse of electrical energy per day (A) 0. averaged over 24 hours. a cathode emits electrons when PASSAGE : 3 The power per unit area reaching the Earth s surface from the sun. An electron w . is 0.PASSAGE : 3 The power per unit area reaching the Earth s surface from the sun. 11. The value of this stopping voltage is dependent only on Emax . For example. and a small electric current flows. A transition from the n = 4 state to the n = 3 state results : (A) in emission of a photon of energy 28 eV (C) in emission of an infrared photon (D) only accomplished if a photon is absorbed (B) in emission of a photon of energy 13. This solar energy can be converted directly into electrical energy via the photoelectric effect. the frequency of light incident on the cathode of a MODERN PHYSICS 8. www.physicsashok. but the intensity is varied. even an electron ejected from the cathode with a photoelectric cell is held constant. As the intensi ty of the incident light is increased. The behavior of light is sometimes explained in terms of particles and sometimes in terms of waves. The energy of each incident photon is given by Ep = hf. then no electrons will be ejected from the cathode at all. The ejected :E =E W max p A voltage source can be connected across the photoelectric cell to oppose the cu rrent flow. (A) increases.in 142 . called the stopping voltage. (D) The angle through which light is refracted when it moves from one medium to another is a function of frequency. because more electrons are ejected from the cathode as the number of photons striking (B) remains the same. 13. At a critical applied voltage. (C) increases. An individual photonincident on the cathode collides with an electron and is absorbed. Which of the following CANNOT be explained by a theory that refers to light in t erms of waves alone (A) Current flow in a photoelectric cell can be stopped by reducing the intensit y of the incident light while maintaining the same frequency. because each incident photon shares its energy between sev eral electrons in the cathode. and is a constant intrinsic to the material of which the cathode is composed. The maximum kinetic energy Emax of an electron liberated from the cathode is given by illuminated by light of a high enough frequency. transfe rring all of its energy to the electron. the stopping voltage. This minimum energy is known as the work function. where f is the frequency of incident light and h is Planck s constant. because the energy supplied to one electron depends only o n the energy of an individual photon. W.ithin the cathode requires a minimum energy to break free from the cathode surface. rather than intensity. because the electrons in the cathode absorb more energy per unit time (D) remains the same. If Ep is less than W. (C) Current flow in a photoelectric cell can be stopped by reducing the frequenc y of the incident light while maintaining the same intensity. (B) An electron requires energy to escape from the surface of a photosensitive c athode. it increases. penetrates through the outer electron cloud and is abruptly accelerated by the large positive charge on the nucleus of a heavy atom. PASSAGE : 4 An x. This minimum wavelength corresponds to : (A) the smallest number of emitted photons (B) the highest energy photons emitte d (C) the type of cathode used (D) the type of anode material used. This increased current increases the number of electrons striking the target increasing the overall intensity. The K. a heated filament cathode. fills the vacancy in the n = 1 shell. (C) The wavelength of incident light is long. 15.000 V. The production of x-rays increases with increasing atomic number but is typically no more than 1% efficient. Figure 1 is a ske tch of intensity versus wavelength for a molybdenum target with an accelerating voltage of 35.MODERN PHYSICS www. peak in figure 1 corresponds to an electron transition from sta te n = 2 to n = 1. because (A) it is due to a higher atomic number target metal (B) K.min . and the work function of the catho de is low. transition. transition is more probable than the K. say from n = 2. corresponds to a transition from state n = 3 to n = 1. is the more energeti c transition (C) the K. (B) The wavelength of the incident light is short. line peak is highe r than the K. (D) The wavelength of the incident light is long. emitting an x-ray photon whose energy corresponds to the energy difference betwe en the n = 2 and n = 1 shells. The photon energy E = hf = hc/. so its intensity is higher (D) the K. and an anode containing the metal target sealed under high vaccum in a glass envelope. The spectral line is produced when an electron. 16. and the work function of the c athode material is high. The intensity of x-rays is proportional to the number of photons created. No shorter wavelengths are emitted from the tube. The current to the heated filament in the cathode is increased while the acc elerating voltage is kept constant. where h is Planck s constant and c is the speed of light. such as n = 1 shell electron. the remaining energy appearing as heat in the target metal. The sharp line spectra that can be seen at higher voltages occur when the incident electrons eject an inner shell e lectron. The continuous or bremsstrahlung spectrum that is always present is produced when the electron Kp Ka I . The sharp K. Under which of the following conditions will the stopping voltage across a p hotoelectric cell be greatest(A) The wavelength of the incident light is short. Figure shows that the continuous x-ray spectrum has a minimum (cut-off) wave length.physicsashok.ray tube consists of two metal electrodes.in 143 14. Two different types of x-ray spectra may be seen. What effect does this have on the minimum wavelength value ? . transition occurs in the valence shells instead of the inner shells 17. where K. and the work function of the cathode material is high. The heated filament in the cathode emits electrons which are accelerated by a high DC voltage and collide with the positive anode target. and the work function of the cathode material is low. whereas l ead is never used. (B) It will move to longer wavelength values (C) There will no longer be a cutoff wavelength. 19. . tungsten (also used in ordinary light bulb filaments) is the most common target meal. 18. If the accelerating voltage. this is because(A) lead has too many electrons (B) lead becomes radioactive (C) lead would melt. V0. (C) They may diappear because all energies may exceed those of the n = 3 to n = 1 transition (D) The K. occur at longer wavelengths but the K. (D) The minimum wavelength will remain the same. it would seem that lead (Z = 82) would be a better target material than tungsten (Z = 74). How ever. Because the efficiency of x-ray production increases with increasing atomic number.(A) The minimum value will move to shorter wavelength values. whereas tungsten has a very high melting point (D) lead will not get hot enough to produce and x-rays. occur at shorter wavelengths. increased while keeping the filament curren t constant. the overall intensity will also increase. What effect will this have on the wavelength position where the two peaks are observed? (A) They will occur at the same wavelengths (B) The peaks occur at shorter wavelengths due to the higher wavelengths due to the higher energies available. Z. 00 A.in 144 PASSAGE : 5 Student in a medical physics class are given the assignment of planning a nuclea r medicine facility.10 (B) 0. Z and mass number A. there is a biological half-life of 12 hour s (for the biological excretion of the carrier).25 days ? (A) 2.0 hours (B) 7. The most common isotope used in diagnostic work is Technicium. One advantage of the Thallium isotope is the low whole body absorbed dose per millicurie . what would be the whole body dose in millirads f or a 70 kg person ? (A) 34 mrad (B) 49 mrad (C) 72 mrad (D) 134 mrad 27. When it is t agged onto a polyphosphate carrier used for a certain procedure. 99 21. (A) 100 mCi (B) 150 mCi (C) 300 mCi (D) 400 mCi 24. about what fraction is lef t after 5. ZTcA + 1e0 What are the atomic number . If the Mo99 has a physical half-life of 67 hours.45 22. the mass number.5 days ? (A) 0. 203 25. radiation protection and safety.MODERN PHYSICS www. 98 (D) 43. They not only design the rooms and choose the major equipment they also will hav e to solve elementary problems dealing with treatment. ZHgA + Is (A) 80. half-thickness for shielding and the decay schemes of rep resentative isotopes. What a re the atomic number Z. 201 (B) 80. of the Tc isotope ? (A) 41. If the recommended amount to be injected for a heart scan is 10 microcurie per kg of body mass.M.5 mCi (B) 5 mCi (C) 10 mCi (D) 20 mCi 26. which for Tl-201 is 70 millirads/millicurie. doses. If the sample has an activity of 80 millicuries initially. What was the activity of the Technic ium at 8. 99 (B) 42.25 (C) 0. The Thallium -201 half-life is 74 hours. The cow was milked at 8. 99 (C) 43.M. This isotope of Technicium has a physical half life of 6 hours. It is furnish ed from a generator or cow in which the negative beta decay of Molybdenum -99 produces the desirablemetas table isotope of Technicium according to the following decay scheme ? 42Mo99 . At 2. and the general principles of physics of typical isotopes that might be used in diagnostic nuclear medicine. 202 (C) 81.00 PM the activity of the milked sample i s measured by a technician and found to be 200 millicuries. of the Mercury isotope produced in the decay of th e Thallium -201 ? 81Tl201 + 1e0 . Another feature that makes Technicium a desirable isotope for diagnostic nuc lear medicine use is that it is a pure gamma emitter.00 A.40 (D) 0. A. what will be the activity after 9. 201 (D) 81. 20. They are required to be familiar with concepts of half life.physicsashok.5 hours (C) 10 hours (D) 14 hours 23. What is the effective half life in this case ? (A) 4. It deca ys by electron capture when the nucleus captures one of the atom s own orbital electrons (converting a proton in the nucleus into an uncharged neutron). Thallium -201 is used for myocardial prefusion studies of the heart. What is the meaning of the terminology pure gamma emitt . with the emission of gamma rays used for the imaging. 4. 28. (B) Particles emitted cannot escape tissue while the gama radiation escapes (C) The isotope decay emits only penetrating gamma radiation that can escape fro m tissue and be detected.er ? (A) The gamma radiation stays in the patient s tissue while the electrons are dete cted.4 cm of lead. Th e half life of a radioactive isotope is the amount of time necessary for one-half of the initial amount of it s nuclei to decay. what was the doese received by the patient in rem units ? (A) 4 mrems (B) 400 mrems (C) 1200 mrems (D) 4000 mrems 29. (D) The isotope decay emits energetic electrons that act like gamma rays. How many half thicknesses are necessary to reduce the radiation penetra tion to less than 1% and how thick would the lead be ? (A) 2 half .2 cm (C) 7 half-thicknesses. The half-thickness of lead for the absorption of the gamma radiation from a particular isotope is 0.4 cm PASSAGE : 6 Medical researchers and technicians can track the characteristic radiation patte rns emitted by certain inherently unstable isotopes as they spontaneously decay into other elements. the initial number of nuclei. 3. A patient is injected with Technicium and is estimated to have received a wh ole body dose of 400 millirads. 1. the number nuclei remaining after a given period to N0.2 cm (B) 4 half-thicknesses. If the Quality Factor is 1 for gamma radiation and 3 for low energy neu trons. 2. .8 cm (D) 11 half-thicknesses .thickness. the decay curves of isotopes 39Y90 and 39Y91 are graphed below as functions of the ratio o f N. physicsashok.7 days if the i nitial samples of the two isotopes contain equal numbers of nuclei ? (A) 1 : 1 (B) 1 : 2 (C) 2 : 1 (D) 10 : 1 32. The half-life of 39Y90 is approximately : (A) 2.2 0. Which of the following conclusions is / are supported by the information giv en in the passage ? . if there are 1. 39Y90 accumulates in the gastrointestinal tract. which of the following situations will exist three days after a patient inhales these substances. If the total amount of each isotope inhaled goes to the specified are a.8 0.4 0. What will the approximate ratio of 39Y90 to 39Y91 be after 2. assum ing none of the isotopes leave the assuming none of the isotopes leave the specified areas due to physiol ogical factors ? (A) The amount of 39Y91 in the gastrointestinal tract will be approximately equa l to the total amount inhaled.0 1 2 3 4 5 6 Time (days) N/N0 39Y90 0.5 0.1 0.8 0. whe reas 39Y91 accumulates in the bones.3 0.000 nuclei to begin with ? (A) 50 (B) 125 (C) 250 (D) 500 34. 33. Approximately how many 39Y91 nuclei will exist after three half -lives have passed.6 0. (B) The amount of 39Y90 in the bones will be approximately one-half of the total amount inhaled (C) The amount of 39Y90 in the gastrointestinal tract will be approximately onehalf of the total amount inhaled (D) None of the 39Y91 inhaled will be left in the bones.1 0.9 1.7 0.2 0. When inhaled by humans.6 0.7 0.3 0.in 145 0.4 days (C) 27 days (D) 48 days 31.7 days (B) 5.9 1.MODERN PHYSICS www.5 0.0 30 60 90 120 150 180 Time (days) N/N0 39Y91 30.4 0. . In the core of fusion reactor. a reactor is termed successful ifLawson number is greater t han 5 × 1014s/cm3. Special techniques are used which confine the plasma for a time t0 befor e the particles fly away fromthe core.I. Usually. a gas of heavyhydrogen is fullyi onized into deuteron nuclei and electrons. 09] .44 × 10 9 eVm. the product nt0 is calledLawsonnumber. 3 2He + n+ energy. Ifn is the density(number/volume) ofdeuterons. can be thought of as a candidate for fusion reactor. III. Only one-quarter of the original amount of 39Y90 will remain after 116 days. Nucleiof heavy hy drogen.6 × 10 5 eV/K. It may be halpful to use the following : Boltzmann constant k = 8. This collection of 2 1 H nuclei and electrons is known as plasma. In one of the criteria. 39Y90 is less stable than 39Y91 II. [JEE. The D D reaction is 2 1 H + 2 1 H . known as deuteron and denoted by D. 39Y90 and 39Y91 are both radioactive (A) I only (B) III only (C) I and II only (D) I and III only PASSAGE : 7 Scientists are working hard to develop nuclear fusion reactor. = 1. 2 1 H . 2 0 e 4. the temperatures in the reactor core are too high and no materialwall c an be used to confine the plasma. The nuclei move randomlyin the reactor core and occasionally come close enough for nuclear fusion to take place. Also neglect any interaction from other particles in the core.0 × 1014 cm 3.in 146 .0 × 109 K (D) 4.5 kT.0 × 109 K < T < 5.0 × 109 K < T < 2. Assume that two deuteron nuclei in the core of fusion reactor at temperature T a re moving towards each other. confinement time = 4.0 × 10 12 s www.0 × 109 K < T < 4.0 × 10 1 s (C) deuteron density = 4.0 × 1024 cm 3.0 × 1012 cm 3. (A) 1.MODERN PHYSICS MODERN PHYSICS 35.0 × 1023 cm 3.0 × 109 K (C) 3. confinement time = 9. The minimum temperature T required for them to reach a separation of 4 × 10 15 m is in the range. confinement time = 5.0 × 10 11 s (D) deuteron density = 1. 36. Which of these is most promising based on Lawson criterian ? (A) deuteron density = 2.physicsashok.0 × 109 K 37. the gas becomes plasma because of (A) strongnuclearforceactingbetweenthedeuterons (B) Coulomb force acting between the deuterons (C) Coulomb force acting between deuteron-electron pairs (D) the high temperature maintained inside the reactor core.0 × 109 K (B) 2. confinement time = 1. Results of calculations for four different designs of a fusion reactor using D D reaction are given below. whenthe separationbetweenthemis large enou ghto neglect Coulomb potential energy.0 × 10 3 s (B) deuteron density = 8. each withkinetic energy1. In the core of nuclear fusion reactor.0 × 109 K < T < 3. ) Assuming the Bohr model of the atom to be applicable to this syste m.m on a surface perpendicular to the beam and possessing a reflection coefficient .097 x 107 m 1 ) 3. (i) derive an expression for the radius of the nth Bohr-orbit. .5 6. When a surface is irridated with light of . The binding energy of an electron in the ground state of He atom is equal to E0 = 24.13 ms and energy E = 10 J. A short light pulse of energy E = 7. A particle of charge equal to that of an electron and mass 208 times the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge +3e.in 147 Level 2 1. 7. what is the work function of the surface and the wavelength of the second source ? If the photo-electrons (after emission from the source) are subjected to a magnetic field of 10 tesla what changes will be observed in the a bove two retarding potentials? 2. 8. Find the energy required to remove both electrons from the atom.6V is applied across the photo tube. You may assume the ionization energy for hydrogen atom as 13.1% of the energy of cathode rays is converted into X-ray radiation and heat is generated in the target at the rate o f 120 calorie per second. Only 0.physicsashok. = 4950A. 4. How many different lines are possible in the resulting spectrum ? Calculate the longest wavelength amongst them. and (iii) find the wavelength of the radiation emitted when the . . When a different s ource of light is used. That photon liberated a photoelectron from a stationary hydrogen atom in the ground state. (Rydberg s constant = 1. it is found that the critical retarding potential is changed to 1. = 0.5 J falls in the form of a narrow and alm ost parallel beam on a mirror plate whose reflection coefficient is .000 V. An X-ray tube works at a potential difference of 100.6 eV. (ii) find the value of n for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom. The angle of incidence is 30°. (Take the ma ss of the nucleus to be infinite. meson jumps from the third orbit to the first orbit. A laser emits a light pulse of duration . In terms of the corpuscular theory find the momentum transferred to the plate. Hydrogen atom in its ground state is excited by means of monochromatic radiat ion of wavelength 975A. A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. 0. .6 eV.1 V. a photo current appears which vanishes if a retarding potential greater then 0. What current does the tube pass and what is the energy and velocity of an electr on when it reaches the target? 5. F ind the velocity of the photoelectron. Find the mean pressure exerted by such a light pulse when it is focussed into a spot of diameter d = 10 .60.MODERN PHYSICS www. = 0. . and the wavelength of the first line of the Lyman series. obtained without taking into account the motion of the nucleus. differ from the more accurate corresponding values of these quantities? 10. if suc h a system is (a) a mesonic hydrogen atom whose nucleus is a proton ( in a mesonic atom an ele ctron is replaced by a meson whose charge is the same and mass is 207 that of an electron). Taking into account the motion of the nucleus of a hydrogen atom. Calculate the separation between the particles of a system in the ground sta te. How much (in pe r cent) do the binding energy and the Rydberg constant.9. the corresponding binding energy. . find the ex pressions for the electron s binding energy in the ground state and for the Rydberg constant. (b) a positronium consisting of an electron and a positron revolving around thei r common centre of masses. 14. Calculate f . Earth-Sun distance = 1. is a small angle between the dir ection of the original velocity of the electron and the direction of final velocity of the electron arr iving at the screen. Suppose th at the Sun and the Earth are point particles with their existing masses and that Bohr s quantization rule for a ngular momentum is valid in the case of gravitation. mass of the Sun = 2.3 cm ( L >. . (c) Show that two plane waves representing deflected beams in the upper and lowe r parts give interference pattern on the screen.physicsashok.in 148 11.0 x 1030 kg. Calculate the number of bright bands in the interfence pattern. (b) Calculate the angle of deflection of electron beam which has the collision p arameter b and the electron does not collide with the wire. Assume a to be directed permanently towards the nucleus. the distance of the electrons closest b L Electron Beam Screen approach to the wire if uncharged is represented by bmax = 10 4 m.MODERN PHYSICS www. Estimate the time during which an electron moving in a hydrogen atom along a circular orbit of radius r = 50 pm would have fallen onto the nucle us. The wire carries uniform positive charge with charge density . A parallel electron beam is obtained from the application of accelerating voltage V0 = 2 x 104 V.5 x 1011 m. 12. a . (b) What is the value of the principal quantum number n for the present radius? Mass of the Earth = 6.4 x 10 11 c/m. The electrons after passing the charged copper wire land on the screen located at distance L = 0. . At the beginning of the experiment the electron beam is confined within the normal distance to the wire of b ( collision parameter) (see Figure). The Earth revolves round the Sun due to gravitational attraction.0 x 1024 kg.b) from the wire. (a) Calculate the minimum radius the Earth can have for its orbit. These electrons are sent travelling in the direction normal to an infinitely long straight copper wire of radius r0 = 10 8 m as shown in figure. 13. If f . To stop the flow of photoelectrons produced by electromagnetic radiation inc ident on a certain metal. . In accordance with classical electrodynamics an electron moving with an acce leration a loses its energy due to radiation as : 2 3 2 a 3c 2e dt d . (a) Determine electric field E due to the charged copper wire and sketch a graph of electric field E as a function of distance measured from the axis of the wire to inside as well as o utside the wire. = 4. Find : . . each of intensity 1. 18. and all surfaces of truck are perfectly black. The beam falls normally on an area 10 4 m2 of a clean metallic surface of work function 1. If the photoelectric threshold of t he metal is 1500A. . what is the frequency of the incident radiation? Is this radiation visible? 15. where Z i s a constant and e is the magnitude of the electric charge. 4800A.2 eV to excite the electronfrom the second Bohrorbit tothe third Bohr orbit. Neglect fraction. . = 4900 A°) incident normally upon a surface p roduces a pressure of 5 x 10 7 N/m2 on it. 17. A single electron orbits around a stationay nucleus of charge +Ze. A beam of light consists of four wavelength 4000A. .9 eV.5 x 10 3 wm 2. Amonochromatic beam of light ( . 6000A. . find the number of photons falling per second on a unit area of thin surface.Assuming no loss of light energy calculate the number of photo electrons libe rated per second. Assuming that 25% of the light incident is reflected and the re st absorbed. If intensity of sun rays is . calculate tension in thread used to keep truck stationary. and 7000A.negative potential of 300 volts is required. A toy truck has dimensions as shown in Figure and its width normal to the plane of this paper is d. It requires 47. Thread b a h 16. Sun rays are incident on it as shown in figure. Radiations from hydrogenic C gas corresponding to 2nd line of Lyman series f alls on a hydrogenic atom where rotating particle s mass and charge are unknown. All the wavelengths are equal or smaller than that of the absorbed photon. 20. Th e work fraction for sodium is 1. (iii) The wavelength of the electromagnetic radiation required tomove the electr on from. Assume that the positive cloud has a char ge +2e uniformly distributed over the volume of a sphere of radius 0. 19. Nucleus contains one proton only. Consequently. A gas of hydrogen like atoms can absorb radiations of 68 eV.82 eV. what will be the wavelength of K line? 21. 22.E. (b) Identify the gas atoms. t he atoms emit radiations of only three different wavelengths. (a) Determine the initial state of the gas atoms. Stopping potentials of 24. 110 and 115 kV are measured for photoelectro ns emitted from a certain element when it is radiated with monochromatic x-ray. The kinetic energy of the fastest photo-electrons emitted form sodius is 0. Light from a discharge tube containing hydrogen atoms falls on the surface o f a piece of sodium. (iv) Thekinetic energy. (b) Find the maximum K. Find : (i) the energy of the photons causing the photoelectric emission. (a) Find the equilibrium position of the two electrons. within which two electrons sit at equlibrium positions.6 eV). (iii) the charge in the angular momentum of the eletron in the hydrogen atom in the above transition.50 A.of the electron ejected due to aforesaid radiation fall ing on a hydrogenic potassium having some of the atoms in the ground energy level & some in the 4th energy level. (Ionization potential of hydrogen is 13. According to the Thomson model. If this element is used as a target in an x-ray tube. (iv) the recoil speed of the emitting atom assuming it to be at rest before the transition.MODERN PHYSICS www. (v) the radius of the first Bohr orbit.in 149 (i) the value of Z (ii) the energy required to excite the electron from the third to the fourth Boh r orbit. (c) Find the minimum wavelength of the emitted radiation. . a helium atom consists of a cloud of positiv e charge. (a) Find a relation for valid values of mass and charge of the rotating particle s. the potentialenergy and the angular momentumof the elect ron in the first Bohr ortib. (d) Find the ionization energy and the respective wavelength for the gas atoms. (b) What is the frequency of small radial oscillations of the electrons about th . 100. Assume that the electron s are symmetrically placed with respect to the centre. 23. (ii) the quantum numbers of the two levels involved in the emission of these pho tons.physicsashok.73 eV. This atom is excited to 4th excited energy level. Assume that the electrons move symmetrically with identical amplitudes.284% respectively. 0. 24. 7. A sample contains 10 2 kg each of two substances A and B with half lives 4 sec ond and 8 second respectively. 25. Find the ratio of the mass of the proton to that of the electron in terms of R1. The half-lives of these isotopes are 2. 92U235 and 92U238 present in the ratio of 0. Find the amounts of A and B after an interval of 16 second. The Rydberg constants for hydrogen and singly ionized helium are R1 and R2 r espectively. Calculate the contribution to activity (in %) of each isotope in this sample.5 x 105 y ears. .5 x 109 years respectively. 26.1 x 108 years and 4.2.006%. A sample of uranium is a mixture of three isotopes 92U234. Their weights are in the ratio of 1. .71% and 99. R2 and R.eir equilibrium positions. 94] 2..Mass of the heliumatom= 4.2 x 1 07 J of electric energy per day at the end of 693 days.00260 a. Determine the total value of blood i n the bodyof the person. -particle? 29.6 days.0141 u.physicsashok. Pb206 = 205. Consider the following reaction . Assuming an efficiency of 10% for the thermoelectric machine.72 x 1010 . Calculate the power output froma sample of 1020 Cmatoms.. Po210 has half life of 138. 244 94 Pu = 244.u.Aft er 10 sec the number of undecayed nuclei remains to 12. Polonium ( 84Po210 ) emits 2He4 particles and is converted into lead (82Pb20 6). the formerwith a probabilityof 8%and the laterwitha probabi lityof92%. what is the average energy of the . 10 3 kg of radioactive isotope (atomic mass 226) emits 3. Calculate the half-life of the isotope.2 x 102 joule is released in one hour in this pro cess. 28. This reaction is used for producing electric power in a space mission. 96] (i)mean-life ofthe nuclei and (ii)The time inwhich the number ofundecayed nuclearwill further reduce to 6. -particle s in a second.97440 a. the ratio of U238 to Pb206 nuclei is 3. Calculate : [JEE.98264 a.072220 u.in 150 27. If 4. Level 3 1. 4. 2He4 = 4.Themasses involved in . In an ore containing uranium. Eachfission releases 200MeVof energy. At a given instant there are 25%undecayed radio-active nuclei in a sample.u. Take t he half-life of U238 to be 4.5%. Asmall quantityof solution containing 24Na radionuclide (half life 15 hours) or activity1. 30. 2H1 + 2H1 = 4He2 + Q. The element Curium 248 96 Cmhas a mean life of 1013 seconds..m.8 x 108 disintegrations per second.m.MODERN PHYSICS www. Also find the initial activity of the material. Assume that the radioactive solutionmixes uniformly in the blood of the person. Assuming that one 7Be radioactive nuclei is prod uced by bombarding 1000 protons. A7 Li target is bombarded with a proton beam current of 10 4 A for 1 hour to p roduce 7Be of activity 1. Its primary decaymodes are spontaneous fission and .25%o fthe reduced number. 97] . (Given : masses of nuclei Po210 = 209. 3.m. howmuch Po210 is required to produce 1. decayare as follows : 248 96 Cm= 248. determine its half-life.0microcurie is injected into the blood of a person. assuming that all the lead present in the ore is the final stable product of U238.0024 u This is a nuclear _______ reaction inwhich the energyQis released is _______MeV. decay.002603 u. (1 Curie = 3.u.m. [JEE. 96] Mass of the deuteriumatom= 2. Calcul ate the age of the ore. (Iu = 931MeV/C2) [JEE.u.Asample of the blood of value 1 cm3 taken af ter 5 hours shows an activity of 296 disintegrations per minute.064100 u & 42 He = 4. 1 a. = 931 MeV and Avogadro number = 6x1023/mol.7 × 1010 disintegrations per second) [JEE.5 x 109 year. . [JEE.. = 2N0.5. there areN0 nucleiof the element. the element has a decay constant .At time t = 0... 98] (a) Calculate the number Nof nuclei ofAat time t. calculate the number of nucleiofAafter one halflife ofA&also th e limiting value ofNas t .. Nuclei of a radioactive elementAare being produced at a constant rate . . (b) If .. Frequencyof a photon emitted due to transition ofelectronofa certain elemrnt fromLtoKshellis found to be 4.Also assume that all the emitted photoelectrons are collected by plateB and thework dunction of plateAremains constant at the value 2 eV. Ground state energy of hydrogen atomis 13. When a beamof 10.with photons of ener gy 5 eVeach. then plot a graph b etween anode potential and current. 0. 2000] 8.MODERN PHYSICS www. Z and the ground state energy(in eV) for this atom. (Neglect the time taken by photoelectron to reach plate B) [JEE. 03] 11. a phot on ofenergy40.He+ ions thus formed are intheir fourth excited state. UsingMoseley s law.9 eV.53%ofthe incident photons eject photoelectrons. It can emit a maximumenergyphoton of 204 eV. originating from all possible transitionbetween a group of levels. 02] 10.-particles. If itmakes a transition to quantumstate n. (b) themagnitude of the electric field between the platesAand B at t = 10 s and (c) the kinetic energy of the most energetic photoelectron emitted at t = 10 s w hen it reaches plate B. These photoelectrons pass througha regioncontaining .2 × 1018 Hz. photons of energy 5 eVfalls on the cat hode havingwork function 3 eV.8 eV is emitted. emittinga single photoninthis process. These levels have energies bet ween 0. 9. [Take. Photoelectrons are emittedwhen 400 nmradiationis incident on a surface ofwork function 1.7 × 10 12 C. (a) if the saturation current is iA = 4µA for intensity10 5W/m2.6 eV.physicsashok.6 eV. Plate B carries a positive charge of 33.14 × 10 15 eV s] [JEE.6 eV photon of intensity 2W/m2 falls on a platinumsurface of area 1 × 104 m2 and work function5. Determine (a) the number of photoelectrons emitted up to t = 10 sec.in 151 6. [JEE. Ahydrogenlike atom(described bytheBohrmodel) is observed to emit sixwavelengt hs. that are likelyto be emitted during and after the combination. Find the number ofphotoelectrons emitted per sec and theirminimumandmaximumenergies in eV. cal culate theminimumenergy (in eV) that can be emitted bythis atomduring de-excitation. [JEE. Find the energies in eVofthe photons.Amonochromatic beamof light.-particle to formaHe+ ion. 03] . 99] 7(a). 2000] (b).85 eVand 0. Two metallic platesAand B each of area 5 × 10 4m2. Ahydrogen-like atomof atomic number Z is in an excited state of quantumnum ber 2 n. given t hat the Rydberg s constant R = 1. h = 4. assume that one photoelectron is emitted for every106 incident photons.1 × 107 m 1.544 eV(including both these values) [JEE. (b) Calculate the smallest wavelength emitted in these transitions. find the atomic number of the element. (b)Also drawa graph for intensity of incident ratiation of 2 × 10 5W/m2 ? [JEE.Also. are placed at a separation of 1 cm. Find n. starts falling onplateAat t = 0 so that 1016 photons fall on it per squaremeter per second. lying in the 2 to 4eVrange. In a photoelectric experiment set up. 02] (a) Find the atomic number of the atom. [JEE.Amaximumenergyelectron combineswith an . what is themean life of the sample ? [JEE. Aradioactive samle emits n .75 n .12.-particles in 2 sec. 03] . In next 2 sec it emits 0.-particles. find . (P). line of the X-ray produced. [JEE. 05] (a) atomic number of the nucleus (b) the frequency ofK. If the total energy of particle is 2E0. Find [JEE. (C) . (D) .MODERN PHYSICS www. (P) 3.2. (D) . (Q) 5. (B) . equalto th ede-Broglie swavelength of electron in the level fromwhich it originated. The potential energy of a particle ofmassmis given by 0 E 0 x 1 V(x) 0 x 1 . (C) .in 152 13. 1 and x > 1 respectively. (C) . (R). The ratio of radii of nucleus to that of heliumnucleus is (14)1/3. (C) . . (B) . (P) 6. (A) . when 0 . (R) 8. (R).1/. B C D B A A B A A B Match the Column 1. (QS). (S) . (Q) 9. (D) . . .53Å and Rydberg constant (R) = 1. (PS). 05] 14. . (Q).2 are are the de-Brogliewavelengths of the particle. (PR). (C) . (P). (B) . . (P). (S). (D) . (V). (D) . (PQT). (R). 06] Answer Key Assertion-Reason Q. (R). (D) . (T). (D) . (B) . (B) . Highly energetic electrons are bombarded on a target of an element containin g 30 neutrons. (C) . (S). (D) . (PQR) 7. (A) .1 and . .What is the value of n ? [Take :Bohr radius (r0) = 0. (U). Inhydrogen-like atom(z=11). (A) .physicsashok. (B) . (A) .1 × 107 m 1 and c = 3 × 108 m/s) 15. (A) . (A) . (C) . (P). (A) . (C) . (B) . nth line ofLymanseries haswavelength. (B) . (C) . (T). (U) 4. (A) . (T). 1 2 3 4 5 6 7 8 9 10 Ans.1 × 107m 1] [JEE. (PR). (S). (B) . . (A) . . . (Q). (U). (D) . (R = 1. (Q). (PQ). (S) 2. . x . (U). E/c. If the motion of the nucleu s is not taken into account. 2 . 66 67 68 69 70 71 72 Ans.s 7. 52 53 54 Ans. . . cos 2. 31 32 33 34 35 36 37 38 39 40 Ans. (ii-A) (i-CD). . 35 nN. . . D C D B C A C D D A Q. 57 59 60 61 62 63 64 65 Ans.R/m.106 m/s.. (ii-D) 55 Passage Q. . . (b-A) 51 (i-B). (d-E). .3. these values (in the case of a hydrogen atom) are greater by m/M .1. B B C A B C D A C D Q. . (b-B). 50.. 31 32 33 34 35 36 37 Ans. where m and M are the masses of an electron and a proton. 50 atm 6. 11 12 13 14 15 16 17 18 19 20 Ans.05 5%. b . is the reduced mass of the system. . C A B C AC A B A Q. 21 22 23 24 25 26 27 28 29 30 Ans.MODERN PHYSICS www. n = 4. 41 42 43 44 45 46 47 48 49 50 Ans. 1. . 11 12 13 14 15 16 17 18 19 20 Ans. 4. 2. 8. (p) . C B A D B B D A D A Q. . .physicsashok. ABC CD AC AB AB AC AD AC AB C Q. where . . (c-B). 2.in 153 Level 1 Q. AD C B Q. 9. B B B D A D A C ABC C Q. B C B D D A B Level 2 1.426 x 10 14 n2 (ii) 25 (iii) 5. . (i) 8. e4 / 2 3 . B C AD BC ABC BD ABD AB AC BD Q. where m is the mass of the electron. 1. P .478 x 10 11 m 3.A 5. 4. .d2c. . (e-C) 58 (a-C).2 .1. 1 2 3 4 5 6 7 8 9 10 Ans. = 4125A. 21 22 23 24 25 26 27 28 29 30 Ans. . 0 . = 18800A. C D D C D C B D D B Q.6 . .9 eV 0 . B BD C C A A A 56 (a-C). 0. . 1 2 3 4 5 6 7 8 9 10 Ans. B A D A C B C B C A Q. . 4 R 79 eV. 88 eV(5 .45% 27.298 µW 5.72 x 1011 18.Min.57 x 1021 per day. (a) 1. (a) n = 2 (b) z = 6 (c) 28. . (a) 0. electron jumps from 4th to 2nd orbits. particle = 1 9. 3) 7.(1.868 x 109 year 30.6 eV. . .36 A . for r > r0 = 0.63 eV(4 . ) .25×1019per sec. 0. energy = 10. . 3) &2. 2.3 nm 9. . T½ = 1573 year.5 MeV 29.41%. . e. 0. 1. (a) 0.13%.. NB = 2. 2. (a) 2.8144 m/s 22. T½ = 8. . during combination = 3. no 15. .6 x 10 17 2 2 0 . A0 = 4. .65 nm. .53 keV. (i) x = 5 . ps 13 e 4 / r c m t 2 3 2 2e . 11..11 x 10 34 Js 0.93 x 1020 17. .285 pm. (b) 2000 N/C. 11. 12.32 A. (b) 6.8 eV.S. .E. . z = 4..55 eV..76 x 1016 Hz 24.056 x 10 34 kg m2 s 1 (v) 1. 6. (a) 5 × 107. 33. . 2.. G. v = 7. 2 4 4.in 154 10. Fusion. . 20. (ii) 40 seconds 3.(a) n = 2.163 A 21. 46. . 2.365 eV..243 . 1 [.. Energy per .3 x 10 138 m (b) 2.43 s. (a) 3. (a) t t 0 N .5 x 10 3 26.63 x 106 s. (b) 0 0 f 4 V q .MODERN PHYSICS www. 6 litre 2. 1 R2 1 R R 1 0. (b) 106 pm. 2N 2 6.53 eV (iii) 36. ] . (b) 0 0 3N . Level 3 1. tmeans = 14. (b) 4052.45 x 1016 cycles/s.58eV. (c) 23 eV . NA = 6. 320 gm 28.25 A (b) 1. 23. . (a) 2 r E q L 0 . h (b) The radiation connot eject electron.75 25. (i) t1/2 = 10 sec. .physicsashok.5 A.06 x 10 11 m 19. 217. (iv) 1. 0.N e. 2.5 x 1074 13. . (ii) 16. (d) 25. . 0. . b cos acos hsin sin c d 16. m. 51.. after combination = 3. (c) 500 14. 5 eV 8. . . .25 x 10 4 kg. 2 ln(4 / 3) 13. z = 42 11. 2 14.95 sec.546 × 1018 Hz 15. 8µA I 4µA 2V VP I=2×10 5W/m2 I = 10 5W/m2 12. n = 24 X X X X .75n = N0(1 e 4.). 6.10. 1. v = 1. PH NO. TIRUPATI.NUCLEAR PHYSICS DHANALAKSHMI NAGAR NEAR ANNAMAIAH CIRCLE. 9440025125 . (ii) Isobars: Nuclides (nucleiof different elements) having the sameAvalue but d ifferent Z andNvalues are called isobars. eg. Types of Nuclei : Nuclei are of three types: (i) Isotopes: Nuclei having the same Z value but different NandAvalues are calle d isotopes of the element.g. 1H3 and 2He4 are isotones.-particleswere made to strike a thin gold foil. protons and neutrons. e. Thus. The neutron numberN: This is equal to the number of neutrons in the nucleus. (d) Aneutron has amass of 1. (f) Nucleus has a positive charge.physicsashok. R0 ~.These particles are collectivelycalled nucleons. (c) This structure ofatomwas revealed by the experiments ofRutherford inwhich a beamof . (iii) Isotones: Nuclides having the same Nvalue but differentAand Z values are k nown as isotones. Most nucleus are almost sphericalwith an average radius Rgiven by R = R0A1/3 whereAis the mass number and R0 is a constant.6 × 10 19 C). (b) The onlyexceptionis the ordinary hydrogen nucleuswhich is just a single prot on.A= Z + N Symbolically a nucleusXshall be represented as A ZX . Themass numberA: This is equal to the number of nucleons (protons + neutrons) in the nucleus. 1. 1H2 and 1H3 are isotopes of hydrogen.2 × 10 15 m .NUCLEAR PHYSICS www. 1H1. (b) The electrons surround this nucleus to formthe atom.2 fm = 1.in 1 RADIO ACTIVITY INTRODUCTION OF ATOMIC NUCLEUS (a) The atomic nucleus consists oftwo types of elementaryparticles. e. Radius of atomic nucleus: The size of the nucleus is of the order of fermi (fm). PROPERTIES OF AN ATOMIC NUCLEUS Composition: (a) All nucleiare composed of two types of particles protons and neutrons.6726 × 10 27 kg and charge +e (= 1. (e) Most of themass of the atomis concentrated in the nucleus. viz.g. 1H3 and 2He3 are isobars. (c) A proton has a mass of 1. The atomic number Z: This is equal to the number of protons in the nucleus.6750 × 10 27 kg and is electrically neutral. (d) Nucleus can be regarded as a small spherical volume situated at the centre o f atom. 1 fm = 10 15. energy equivalent of 1 a.u.u. is equal to 931.. NA of at oms.66056 × 10 27 kg) × (3 × 108 m/s)2 = 1.NUCLEAR PHYSICS www.66056 × 10 27 kg (b) FromEinstein relation. kg or 1 a.m.66056 × 10 27 kg (e) Matter can be viewed as a condense formof energy. 1 a.5MeV/c2 Density: Mass of a nucleus can be taken approximatelyAm. = 931. fromEinstein relation. Mass =Am Also. (c) This unit is such that mass of the carbon isotope 6C12 is exactly 12 amu.5MeV. i. E = 931.in 2 Mass : (a) Nuclearmasses have been accurately determinedwith the help of themass spectr ometer. (h) Energy equivalent of 1 a.u.u.whichmeans.e.m. 1 a.m. [. (g) Rest mass energyis E =mc2.m.m. (f) The energy corresponding to themass of a particlewhen it is at rest is calle d rest mass energy. = 1. . = 1.m. wheremis the mass of proton or a neutron andAis mass number .022055 10 . Bydefinition eachC12 atomhas a mass of 12 a.u.m. Sol: (a) Onemole of C12 has a mass of 12 g and containsAvogadro number.6 × 10 19 J. Thus.m. fromAvogadro s number.R3 = 3 0 4 R A 3 . . = A 1g N = 3 23 10 6.physicsashok.u. assuming the nucleus to be a sphere of radius R.5MeV Hence.u. R = R0A1/3] The nuclear densityis thus given by. (b) It is expressed in a. 1 a.e. 1 eV = 1. Example 51: (a) Calculate the value of 1 a.m.u = 931.m.) × c2 = (1.u. .u. = mass volume = 3 0 Am .u. 12 g corresponds to (12NA) a. its volume is V = 4 3 .m. atomicmass unit. .4924 × 10 10 J Since.5MeV /c2. (b) Determine the energy equivalent of 1 a.m. (d) 1 a.u. restmass energy E = mc2 Hence.u. E = (1 a.m. i. (a) Nuclear density . = 3 0 3m 4. = 2.3 × 1017 kg/m3. Putting R0 = 1. This is almost 1014 times the densityofwater.2 × 10 15mand m= 1.2 fm = 1. we get.3 × 1017 kg/m3 (b) It is nearly the same for all nuclei.67 × 10 27 kg. .4 R A 3 .R It is thus independent of themass numberAand is therefore nearly the same for al lnuclei. = 2. . physicsashok.16 × 10 43 m3 Mass of oxygen atoms (A= 16) is approximately 16 a.16.16 10 m . (c) These are short ranged forces (effective upto 10 fm).(1. d . . (b) The Coulomb s repulsive force between two protons in a nucleus is about 1036 t imes as large as the gravitational force between them.5 × 10 15mor less. (b) These are about 100 times stronger than the Coulomb s force.5 fm . .2)3 × 16 × 10 45 m3.3 × 1017 kg/m3 NUCLEAR STABILITY (a) High densityof the nucleus suggests a very tight packing of protons and neut rons in it. 10 fm (g) Let Fpp.R3 = 3 0 4 R A 3 . protonwith neutrons.66 10 1.5 fmor 0. Properties of Nuclear forces : (a) These are strong attractive forces.2 × 10 15.NUCLEAR PHYSICS www.in 3 C48: Find themass density of the oxygen nucleus 8O16. . Nuclear force : It arises due to interaction betweenprotons. (d) They contain a small component of repulsive force which is effective up to a distance of the order of 0. It is therefore surprising that a nucleus shou ld be so stable. R = R0A1/3] = 4 3 . (e) These forces are charge independent. [.10. . Sol: Volume = 4 3 . Fpn and Fnn denote themagnitude of the nuclear force by a proton on .This force is essentially a very strong attractive force and overcomes the electrostatic repul sion between the proton inside the nucleus. (c) Nuclei are stable because of the presence of another force.u. R0 = 1.m. kg/m3 = 2. Therefore densityis . [. and neutronwit h neutrons. = mass volume = 43 3 16 amu 1. A= 16] = 1. m = 27 43 3 16 1. called the nucle ar force. This repulsive component prevents the collapse of the nucleus. then 0. (f) Let d be the range of the effectiveness of nuclear forces. 6. by a proton on a neutron and bya neutron ona neutron respectively. 1 . N = Z line N v/s Z curve N (b) It must be greater than or equal to unity and less than 1. 1. Fpp = Fpn = Fnn. N Z .6 Z i.e. Then for a separation of 1 fm. .a proton. N/Z ratio : (a) N/Zratio inside a nucleus is responsible for stability of a nucleus. c2 If . = [M {Z mp + (A Z)mn}] × 931.52878 u) (931MeV/u) = 492 MeV Negative sign indicates boundedness of nucleons.m× 931.E. (d) When excess of neutrons or protons in a nuclide is there then the nuclide . B.) : The binding energy is equal to thework thatmust be done to split the nucleus int o particles constituting it.E.E.E.This repulsion becomes so great in nucleiwithmore than 1 0 protons or so that an excess of neutrons. Mass of proton = mp . = . B.m. = [M {Zmp + (A Z)mn}]c2 The quantity.9349 u]c2 = (0. Mass of neutron =mn. = Energyofeach nucleon individually.E.Mass of neutron = 1. and. = . = 492MeVinmagnitude.52878 u)c2 = (0. represents boundedness of nucleons inside the nucleu s. Mc2 + B.m=M {Z mp + (A Z)mn} . which produce onlyattractive nuclear forces. Rest mass energyof nucleus =Mc2. .E. Binding energy per nucleon : Binding energyper nucleonis obtainedbydividing the binding energyof the nucleus bythe numberA of nucleons in the nucleus. ThusN/Z ratio increaseswith increase in Z.u.physicsashok.00867 u.M {Zmp + (A Z)mn} is called mass-defect (.decayor . .m) . A 0 50 100 150 200 250 2 4 6 8 10 .9349 u and that of 1His 1.Atomicmass of 56Fe is 55. Z)mn c2 Thus. Sol: Z = The number of protons in 56 26Fe = 26 and the number of neutrons. per nucleon = B.00783 u + 30 × 1.mis in a. . C49: Find the binding energy of 56 26Fe . = Z mpc2 + (A .in 4 (c) Reason forN/Zratio to be greater than unityis due to the fact that protons a re positivelycharged and repel on another electrically.e. Energyof nucleus+B. B.E.E. then B.E.. Let Mass of nucleus =M. Hence. Mass defect = .A Z = 56 26 = 30 Then binding energy of 56 26Fe = [M {Zmp + (A Z))mn}]c2 = [26 × 1.5MeV or B. Rest mass energyof proton =mpc2. Binding Energy (B.m.-decayto achieve the requiredN/Zratio for stability.5MeV NOTE: Negative sign of B.00783 u.E.E. i.NUCLEAR PHYSICS www.This causes radioactive disintegrations of nuclid es. is required for stability.00867 u 55. B. Rest mass energyof neutron =mnc2. A . MeV Mass number.56 26Fe 4He 6Li Binding energy per nucleon. Z = number of proton = 6 A Z = number of neutrons = 12 6 = 6 .physicsashok.22 12 MeV= 7.in 5 (a) The adjacent figure shows the dependence of the B.099 × 931.u.m.00759 + 6 × 1 × 1. Binding energy= 92. A = 92. Sol: M = mass of C12 atom = 12 a.m.7MeVand gradually decreaseswith increasingA.u. per nucleon.u. Y m3 K3 X m4 K4 After collision Initial energy: Ei = m1c2 + m2c2 + K1 + K2 Final energy: Ef = m3c2 + m4c2 + K3 + K4 .E. Mass of proton = 1. b)Y.00898 a. (ii) The conservationof energy.E./Aonthemass number Aofthe nucleus. Mass of neutron = 1.5MeV = 0.E. TheB.009) a.E. (iii) The conservationofmomentum.00898 a. Q-Value or Energy of a reaction : Let m2.m.22MeV (Negative sign indicates boundedness of the nucle on) Binding energy per nucleon = B.NUCLEAR PHYSICS www. .m.5 MeV = 92.00898) in a. .u. per nucleon is highest for 56 26Fe .m× 931.u.00759 a. Mass-defect.m. and (iv) The conservationof angularmomentum.00759 a. mn = mass of neutron = 1.m = M {Zmp + (A Z)mn} = 12 (6 × 1.68MeV NUCLEAR COLLISIONS (a) Anuclear reaction in which a collision between particle a and nucleus X prod ucesY and particle b is represented as a + X Y+ b (b) The reaction is sometimes expressed inthe shorthand notationX(a.22MeV Hence. mp = mass of proton = 1. Example 52: Find the binding energy of 12 6C ?Also find the binding energy per nucleon. (c) B. B. (b) Nucleons in nucleiwithmass number from50 to 60 have the highest B. /Afor these nuclei amounts to 8.m.u.E.m3 are nuclearmasses ofXandYrespectively.m. (c) Reaction are subjected to the restrictions imposed by (i) The conservationof charge. a m1 K1 X m2 K2 Before collision .E. = (12 12.m.E. = .u = 0. B.u.u.m.099 a. Mass of 6C12 atom = 12 a. 01410 + 6. Q = . Calculate the energy prod uced in this reaction in MeV.01 513 amu. Speed of light = 3 × 108m/s.17 × 10 13 J = 13 19 1. 6. Sol: 0n1 1H1 + 1e0 Mass defect. Becquerel discovered accidentally that uraniumsalt crystals emit an invisible radiationwhich affected a photographic plate even though it was properly covered.00783 = 8. 8.00549 amu = 0. Q = [(m1 + m2) (m3 + m4)]c2 J or. Sol. that is released or absorbed in a nuclear re action is called the QValue or disintegration energyof the reaction.00549 × 931 MeV (.6747 × 10 27 Kg. when masses are in a. Find theQ-value of the reaction 1H2 + 3Li6 .1 MeV INTRODUCTION OF RADIOACTIVITY (a) In 1896.physicsashok. Mass defect: The quantity [(m1 +m2) (m3 +m4)] is called themass defect of the reaction and is given by .m.02384 amu .0013 × 10 27 kg . (b) In 1898. 1 amu = 931MeV) Q = 5. Ei = Ef. . [(m1 + m2) (m3 + m4)]c2 = (K3 + K4) (K1 + K2) The energy. Mass of proton = 1. Mass of neutron = 1.73MeV Example 54. Q = .6725 × 10 27 kg.m = [mass of neutron (mass of proton +mass of 1e0)] = 1.3 × 10 30 × (3 × 108)2 kg m2/s2 = 1. [(m1 + m2) (m3 +m4)]c2.Marie and PierreCurie and otherworkers showed thatmanyother substanc es also emitted similar radiations.02933 = 8.01601 amu and 1. respectively. 931. (fromenergyconservation) . . Q = [(m1 + m2) (m3 + m4)] 931. and 1H1are.in 6 Since.6747 × 10 27 kg (1.02384 + Q or Q = 0.u.. 2.01601 + 1.6725 + 0.6 10 . Energyreleased.01513 = 8.c2 = 1. .5 MeV Example 53:Aneutron breaks into a proton and electron. Suppose 1H2 + 3Li6 . Mass of an electron = 9 × 10 31 Kg. eV = 0. Totalmass of left-hand side = 2.u.02933 amu Totalmass on right-hand side = 7.0009) × 10 27 kg = 0. 3Li6. 3Li7 + 1H1 The rest masses of 1H2. .00783 amu.NUCLEAR PHYSICS www. 3Li7 + 1H1 + Q.5MeV. Hence.m.01410 amu. 3Li7. . 7.m.m.17 10 1.m= (m1 + m2) (m3 + m4) ina. C50:Uraniumsalt crystals emit (a) visible radiation (b) invisible radiation (c) anytype of electromagnetic radiation (d) soundwaves Sol: This is according to discovery ofradioactivityofBecquerel. (d) It is statisticalprocess that obeys the laws of chance.in 7 (c) The propertyof spontaneous emission ofradiation fromthe substance is called radioactivityand such type of substance is called radioactive substance.many nuclides are unstable and spontaneouslychange into other nuclides b yradioactive decay. (i) No single phenomenon has played so significant role in the development of nu clear physics as radioactivity. Alpha decay (a) In this type of decay. unlike the case of atomic radiation. (j) It is not influenced byexternal parameters such as pressure. (d) Radioactivityis due to the decay or disintegrationof unstable nuclei. not anatomicphenomenon. (e) The decayof nucleus takes place to achieve the stable end products. Th.Whichwere eventuallyidentifiedas 42 He nucleus.NUCLEAR PHYSICS www. (c) The energyliberatedduring radioactive decaycomes fromwithinindividualnucleiw ithout externalexcitation.) and gamma)(. (h) Radioactivity is explained on the basis of quantummechanics. (f) Some examples of radioactive substances are:U. beta(. C51: The radioactivityis a/an (a) optical phenomenon (b)Atomic phenomenon (c) nuclear phenomenon (d) photoelectric phenomenon Sol: Nuclear radiations are obtained fromthe nuclei hence it is a nuclear phenom enon. (e) The radiations arebeing emitted fromthenucleihence it is anuclear phenomenon .) . Kinds of Decay There are five kinds of radioactive decays. electronand highenergyphotonrespectively. (b) It reduces the proton number Z of the nucleus by 2. temperature. When radioactivitywas discovered. . Po and Np. (c) It reduces the neutron number Nof the nucleus by 2. onlythree kinds of radioactive decays alpha(. the unstable nucleus emits an alpha particle. (b) Despite the strength of the forces that hold nucleons (protons and neutrons) together to forman atomic nucleus. ch emical reaction (combination) or phase ofmatter.) were known. RADIOACTIVE DECAY (a) The decayof radioactive substancemeans disintegration of nucleiofthe substan ce byemissionof different radiations. Ra. Later two more kinds of radioactive decays namely positron emission and electron capturewere added.physicsashok. (g) Electronic configuration ofatomdoes not have anyrelationshipwith radioactivi ty. 7 × 10 27 kg and energy 8. Tl. Aradon nucleus Rn86 222. (a) The atomic number will be reduced by 2 and the mass number by 4.physicsashok.8 × 10 13 J Now m1v1 =m2v2 . (i) In this decay.in 8 (d) It reduces themass number (i. AZ X represent atomic mass of the particle X. Q-Value for .6 × 10 25 kg. (j) On emission of . with E = 8. Z +N) of the nucleus by 4.. .7 × 10 27 = 3. (b) mass of resulting nucleus = m1 = 3. the nucleus decreases itsmass number tomove towards stability . + 42 He . . . . . (g) The nucleus before the decay is called the parent nucleus and that obtained after the decay is called the daughter nucleus.. the binding energyper nucleon increases and the r esidualnucleus tends towards stability.-particle. A = 222 4 = 218 and Z = 86 2 = 84 The resulting nucleus is 84Po218. Example 55.6 × 10 25 6. (e) It changes the element itself hence the chemical symbol of the residualnucle us is different fromthat of the originalnucleus. the parent nucleus is bismuth (Bi) and the daug hter nucleus is thallium (Tl). (a)What is the resulting nucleus? (b) Find the velocity of recoilof the nucleus.-decay. 4 .-decay: If . . The . undergoes .-particle = m2 = 6.-particle hasmass 6. Y m He . Y . c2 NOTE: The quantity m. A 4 . .. then Q-value = . . Y He .8 × 10 31 J.-decay process is given by AZ X A 4 Z 2 . . (h) Alpha decay occurs in all nucleiwithmassA> 210. 1/ 2 27 13 1/ 2 .-decay. He In the above example of . (f) The alpha decay processmay be represented as A A 4 4 Z Z 2 2 X . ofmass 3. Ex: Let us consider the example given below: 212 208 4 83 81 2 Bi . .-particle = v2 = 1/2 2 2E m . ..NUCLEAR PHYSICS www. Sol.7 ×10 27 kg Velocity of . . Z Z 2 2 m X m .e.533 × 10 25 kg Let v1 = its velocity of recoil Mass of . A . . 533 10 .2 1 25 1 (2m E) (2 6. . . v1 = 3. . . . .8 10 ) v m 3. .7 10 8.1 × 10+5 ms 1 . . . . c2 Final restmass energy. .physicsashok. (h) Astreamof beta particles coming frombulk of unstable nuclei is called beta r ay. (k) An example of beta decay is 6C14 7N14 + e + . Thus.. + . ..NUCLEAR PHYSICS www. Zm . . ... an i solated neutron decays to a proton. the energyQis shared by the antineutrino and the beta particle. (j) The beta decay processmay be represented by AZ A A Z 1Y . (b) It increases the atomic number (Z) of nucleus by 1. REi = . the kinetic energyof the beta particle can be anything between zer o and amaximumvalueQ. n p + e + . + . .+ decay) (a) .-decay: (a) . A . A . . ) AZ X A Z 1Y . the residual nucleus A Z 1Y . an electron and a newparticle nam ed antineutrino are created and emitted fromthe nucleus. . p n + e+ + v (b) It reduces the atomic number (Z) of nucleus by 1. Z Z 1 m X m Y .. It is chargeless and has quantumnumber ±½. (. Positron emission (. -decay process is given by AZ X A Z 1Y . a neutronwill convert itself into a proton tomove towards stability.+ decay is a process in which a proton is converted into a neutron with emi ssion of positron (e+) and neutrino (v). or 1e0. A . On the other-hand. .. .in 9 Beta Decay (a) Beta decay is a process inwhich a neutron is converted into a proton. (d) An isolated proton does not beta decay to a neutron. REf= . + e + . n p + e + . (c) It does not alter themass number (A). + .. + . Z 1 e e m Y Zm (Z 1) m . (d) If a nucleus is formedwithmore number of neutrons than needed for stability. . (c) It does not alter themass number (A). Depending on the fra ction taken away by the antineutrino. (g) The electron emitted fromthe nucleus is called a beta particle and is denote d bythe symbol . . (f) The antineutrino is denoted bythe symbol . It is supposed to have zero rest mass like photon. (i) It is also called betaminus decayas negatively charged beta particles are em itted. Q = REi REf = . c2 . . decay: If . (antineutrino) Q-value for .m X . then Initial rest mass energy. (e) When a neutron is converted into a proton. AZ e . c2 Because ofthe largemass. does not share appreciable kinetic energy. . c2 .decay or position . a positron and a neutrino are cre ated and emitted fromthe nucleus. a proton c onverts itselfinto a neutron. + .emission : If the . The positron so emitted is called a beta plus particle.. .+ + v . ..NUCLEAR PHYSICS www. . AZ e .. ..(v) This process is called beta plus decay.When a proton in a nucleus converts itself into a neutron.. n + e+ + v] AZ X A Z 1Y . (k) Neutrino and antineutrino are antiparticles of each other. It is charge-less particle. p . a prot on converts itselfinto a neutron. . Similarly. a positron and a neutrino are created and emitted fromthe nucleu s.+ + v If the unstable nucleus has excess protons than needed for stability.. In the process.. p n + e+ + v . Zm . + e+ + v or AZ X A Z 1Y .. both the particles are destroyed and energy is made available. then R. A . . (h) The positron (e+) has a positive electric charge equalinmagnitude to the cha rge on an electrons and has a mass equal to themass of an electron. A . A . Q = R. . . .+-decay or position-emission is given by AZ X A Z 1Y . (j) When an electron and a positron collide. A . the decay process is represented as AZ X A Z 1Y . (g) The neutrino is denoted by the symbol v.m X .. A .When an electron and a positron collide.. c2 R. Z 1 e e m Y (Z 1)m m .Ei = . c2 Q = .Ei Z Z 1 e m X m Y 2m .+ decay process is represented as AZ X A Z 1Y . (f) When a proton is converted into a neutron.+ + v.Ef = . (l) The ... . (i) Positron is called the antiparticle of electron. Positron is called the antiparticle of electro n. c2 R. Z Z 1 e m X m Y 2m . .Ef = . + .in 10 (e) If the unstable nucleus has excess protons than needed for stability. neutrino and antineutrino are antiparticles of each other.. + .(iv) The positron e+ has a positive electric charge equalinmagnitude to the chargeon an electron and has amass equal to the mass of an electron. . . .. . both the particles are destroyed an d energyismade available... .(vi) Can an isolated proton decayto a neutron emitting a positron and a neutrino as s uggested byequation (iv)? . + e+ + v [.. + .. .physicsashok.. . an isolated neutron decays to a proton as suggested by equation (i).On the other h and. (n) An example of .Themass ofa neutron is larger than themass ofa proton and hence theQ-value of su ch a processwould be negative.+ decay is 29Cu64 28Ni64 + e+ + v . an isolated proton does not beta decay to a neutron. (m) The positronis also called beta plus particle. So. 990 432 a.25Mg = 24.in 11 Asimilar process. × (931.m.u 13me) (24.. a vacancy is created in the atomic shel l and X-rays are emitted following the capture. p + e n + v (d) In this process.u.254MeV . p + e n + v] (i) An example of electron capture is 29Cu64 + e 28Ni64 + neutrino Q-Value of K-capture process: IfK-capture process is given by AZ X + e A Z 1Y .) 2 × 0.u.5MeV/a.511 2 MeV c . -decay is Q = [m (19O) m(19F)]c2 = [19. .] (931.physicsashok.m.m.003576 a. .NUCLEAR PHYSICS www. Ap roton in the nucleus combineswith this electron and coverts itselfinto a neutron.m.m.998403 a. . Electron capture (a) When the nucleus has toomanyprotons relative to the number ofneutrons. 18.m.. A . In this process. known as electron capture. 19F = 18.) = 4. [.276MeV 1. -decay] (b) 25Al 25Mg + e+ + v. 12 me) me]c2 = [(0.m. .+ -decay is Q = [(mass of 25Al nucleus) (mass of 25Mg nucleus) (mass of positron)]c2 = [(24. (c) Aneutrino is created in this process and emitted fromthe nucleus.u.u. the nucleus captures one of the atomic electrons (most like an electron fromthe K shell). Z Z 1 m X m Y . takes place in certain nuclides.u. (f) When an atomic electron is captured. [.004593 a.. 25Al = 24.5MeV/a.985939 a.m. A .Aneutrino is created in the process and emitted fromthe nucleus. (b) Aproton inthe nucleus combineswith this electron and converts itselfinto a n eutron. atomic number (Z) of the nucleus decreases by 1.022 MeV = 3. Sol: (a) The Q-value of .m.u. c2 Example 56: Calculate theQ-value in the following decays(a) 19O 19F + e + .m.+-decay] The atomic masses of 19O = 19.004593 a.u. + v [.990432 a.003576 a.u.m.c2 = 4.u.m.u.819MeV (b) The Q-value of .u. (g) This process is also calledK-capture.985939 a.998403 a. (e) This process does not alter the value ofmass number (A). then Q = . + v.) 2me]c2 = 0. (h) The processmay be represented as AZ X + e A Z 1Y . the n ucleus captures one of the atomic electrons (most likely an electron fromtheK-shell). .NUCLEAR PHYSICS www. v.-particles is relatively low: c/15). the nucleus maybe excited to higher energies. (m) Thewavelength ofthis radiation is given bythe common relation. we can observe a streamof . 122 keV photons and 14 keVphotons coming fromthe 57Co source. = e . (k) In this decay. the quantumstates of the nucleons vary. the daughter nucleus is generally formed in one of its excited states. atomic number (Z) aswell asmass number (A) of the nucleus rem ain constant. . (b) This decayprocess is related to the transitions between two nuclear energy l evels. = (c/30 Mass m. (d) In the ground state. . (j) In this decay. The daughter nucleus in an excited state eventuallycomes to its ground s tate byemitting one photon ormore than one photon of electromagnetic radiation. the nucleons occupy those quantumstates whichminimise t he total energy of the nucleus.+ 136 keV 14 keV 0 keV 2nd excited state 1st excited state Ground state 57Fe When 57Co is taken in bulk.+ particles. and . Comparison among the kinds of decay (a) The velocityof . (f) The energydifferences in the allowed energylevels of a nucleus are generally large (in the order ofMeV). < c m. (g) It is difficult to excite the nucleus to higher energylevels byusualmethods of supplying energylike heating etc. NOTE: The . rays are collectively called nuclear radiation. = +2e For . .in 12 Gamma Decay (a) Nucleus has also energylevels like atoms have.-particles: 0 . 57Co . =mass of an electron q.. = 4 amu charge q. (e) The higher energystates are also available to the nucleons and if appropriat e energyis supplied. . 136 keV phot ons. (l) The electromagnetic radiationemitted in nuclear transitions is called gamma ray. (h) When an alpha or a beta decay takes place. (c) The protons and neutrons inside a nucleusmove in discrete quantumstateswith definite energies. (i) The process of a nucleus coming down to a lower energylevelbyemitting a phot on is called gamma decay.when c is the velocity of light v.physicsashok. (n) An example of gamma decayis shown in figure below: . = hc E where E = energy of the photon. in 13 For ..NUCLEAR PHYSICS www. a beamof .-particles . 87 * 38Sr 87 38Sr + . In amagnetic field.-rays are undeviated and . The penetrability of . + v 64 29Cu + e 64 28Ni + v Gamma decay A * Z X AZ X + .-rays also exceed s the penetrabilityof x-rays.-rays.-decay AZ X A 4 4 Z 2 2 . This particle is a/an .-decay AZ X A Z 1Y .-rays are electromagneticwaves.. . . 238 92U 234 4 90 2 Th .e. The *denotes anexcited nuclear estate. d enotes neutrino and antineutrino particles respectively. = c Rest mass.-rays is 0 . . .-rays Magnetic field (d) Table for various decay directed into paper Decay Transformation Example .m. = 0(like photon) q. Y He . v. He . + e+ + v 64 29Cu 64 28Ni + e+ + v Electroncapture AZ X + e A Z 1Y . (c) In a magnetic field. . Carboard Al Lead . C52:An a-particle is bombarded on 14N.-rays: Since . . + e + .As a result.-particles aremost deviated .10000 times higher than the penetrability of . Even a thick slab of lead may not stop all the -rays.-denotes a gamma-rayphoton and v and . Positron emission AZ X A Z 1Y . and . hence theypropagatewith the speed of ligh t. .-particles . 14 6C 14 7N + e + . a 17O nucleus is formed and a particle is emitted.100 times higher than the penetrability o f b-rays and 1000 . i. -particles penetrate the cardboard but are stopped by a sheet of aluminium.physicsashok. . = 0 (like photon) (b) The penetrability of .particles from radioactive materials are stopped by a piece of cardboard.-rays splits into three parts. (a) neutron (b) proton (c) electron (d) positron . calculate the value of wavelength of resulting photonwhen the nucleus of 27 13Al reaches the ground state (Eg = 0) from the state in which Eex.. (.-particle) Sol: (i) e+(Positron) (ii) . onlyquantumstates of the nucleons vary.NUCLEAR PHYSICS www.Which of the following particles is emitted in the decay ? (a) Proton (b) Neutron (c) ..E(eV) = 6 12400 (1. (Antineutrino) (iii) 90Th234(Thorium) C58: Following the origin of gamma decay.... .in 14 Sol: 14 4 7 N .physicsashok...-particle) (Protactinium) (iii) 92U238 2He4 + .... 1.. and on ..015 .-radiation..... ..... (Uranium) (.The related figure is shown adjacent... .... neither the atomic number nor themass number change s.. Inthis decay. 2He 17 1 8 1 O .. N=mass number Atomic number = 24 11 = 12 C57: Fill up the blanks (i) hv e + . Each decay is an independent event.-photon) (electron) (ii) 90Th234 1B0 + 91Pa234 + . protons and neutrons are there in a nucleus of atomic num ber 11 andmass number 24 ? Sol: Number of electrons or protons. = 1..10 = 0.. C55: Give an equation representing the decay ofa free neutron.. . p 11 p is equivalent to 11 H ...015 MeV 27Mg 12 0 MeV 27Al Sol: . C53: In a radioactive decay... (Thorium) (..(in Å) = 13 12400 ..015MeV. C54: . Sol: 1 0 n 1 0 1 1 H e v.-rays emitted bya radioactivematerial are (a) electromagneticwaves (b) electrons orbiting around the nucleus (c) charged particles emitted by the nucleus (d) neutral particles Sol: Aneutronin the nucleus decays emitting an electron. C56: Howmanyelectrons.particle (d) photon Sol: Photon is equivalent to ..012217 Å LAW OF RADIOACTIVE DECAY When the radioactive substance is only disintegrating: Radioactive decay is a randomprocess. 0). . Z= 11 and number of neutrons..... it is transformed into another nuclidewhichmay .e cannot tellwhen a particular nucleuswilldecay.When a given nucleus decays. (c) Other half lives range fromabout 10 20s to 1016 years. dN dt . This equationmay be expressed in the form dN dt = . Tm = 0 0 N 0 tdN .693 . Mean life(Tm) : Mean life of a radioactive sample is defined as the average of the lives of all nuclei. is proportional to the number of nuclei. .T . N .5N0 = N0e . dN dt .t Definition of decay constant : The probability of decay per second for a particular process for a sample is cal led the decay constant for that process for that sample. t 0 . . .t. ln 0 N N = . that are present i. . Radioactivity lawof decay gives. the exponential decay equation gives. (b) The half-life for the decayof the free neutron is 12. (a)It takes one half-life to drop to 50%of anystarting value. . where N0 is the initial number of parent nuclei at t = 0. Half life : The time period for the number of parent nuclei to fall to 50%is called the half -life. = dN Ndt .NUCLEAR PHYSICS www. 0. is called the decay constant. . to yield.8min..N where .693 . . dt .e.dt and integrated to get 0 N N dN N . N.e.physicsashok. . T = 0. The number that survive at time t is therefore N = N0e . T. andmay be related to . the rate of decay i.in 15 or may not be radioactive. dN dt = . .T = ln|2| = 0. When there is a very large number of nuclei in a samp le. ifwe put N =N0/2 at t = T. = . .t. . .693 Tm NOTE : (i) Radioactive decay equation. . = T 0. . where T is half-life of the sample.N . Tm = 1 .303 t log 0 N N . = 1 . = 2. .693 . can also be written as N = N0 t /T 1 2 . . (ii) The decay constant is also given as. N = N0e . and T = 0. in 16 Activity of radioactive substance: Since the number of atoms is not directlymeasurable. The activityis characterized by the sa me half-life.3600s .t) = .(N0e .693 . A = . N = 3.7.7 0.9 × 10 9 s 1) × 3. × 6 × 1023 atoms . but the curie (Ci) is oft en used in practice.N = (2. whereNis the number of nuclei present in 1mg of 198Au. (c) Activity is given as. 1 becquerel (Bq) = 1 disintegrations per second (dps) 1 curie (Ci) = 3. = 2.693 days = 3.693 T = 0. whereA0 = . = 0. (b) the average life and (c) the activity of 1.9 days.N =A0e .24.NUCLEAR PHYSICS www.7 days.03 × 1018 atoms Thus. .693 2. A = . N = 10 3 198 .7 × 1010 dps 1 rutherford = 106 dps (b) Rate of decayof 1 gmsubstance is called specific activity. Example 57: The half-life of 198Au is 2. BothNandAdecrease exponentiallywith time.N. Take atomicweight of 198Au to be 198 g/mol. Calculate (a) the decayconstant.N .00mg of 198Au. Units of activity: (a) The SI unit for the activityis the becquerel (Bq).693 = 2.t A = dN dt = .7 days = 0.9 × 10 6 s 1. = 12 10 . Atomic mass of 198Au = 198 g .8 × 1012 disintegrations per sec.03 × 1018 (atoms) = 8. but N = N0e .A= .physicsashok. N = 1 mg 198g ×Avogadro no . . . (c) Activity of 1 gmRa226 is 1Ci. (b) Tm= T 0. wemeasure the decay rate or activity (A) A= dN dt .693 2. Sol: (a) The half-life and the decay constant are related as T = 0.t.N0 is the initial activity. 7 10 . .8 10 3. Ci = 240 Ci .8. We assume that only one emission takes place per atom. . N e.Howmany of . . no. of atoms in 1 cm3.0 . N e.. . Sol.36 × 105 × 107 = 3. Estimate the number of atoms in1 cm3 ofthe element. . Over a period of 140 days.e. (C) 0 1 10. . Two radioactive materialA1 andA2 have decayconstants of 10 . of emissions = 140 × 12 × 1012 . .9.0t 1 = 9. Hence (B) is correct.2 × 1016 active nuclei at certain instant.. .. 9 0t 1 e e .36 × 1022 C60:Aradioactive sample has 3. . In 140 days. .NUCLEAR PHYSICS www. .-emissions per day is found to be 12 × 1012. of atoms in 1 µg = 28 × 12 × 1013 . i. 0 0 10 t 1 0 t 2 0 N N e N N e . 10g = 3.physicsashok. e. no. . 1 9 0t 2 N e N . N e. . of atoms present = 2 × 140 × 12 × 1012 (since 140 days is the half-life) . no. 1t 10 0t 1 0 0 N . (D) 1 Sol.. 2t 0t 2 0 0 N . from a sample of i nitial mass 1 µg.1 . Acertain element has a density of 10 g cm 3 and half-life of 140 days.. (B) 0 1 9. initial no.0t 0 t 1 9 . N e. the average number of . If initially they have same number ofnuclei.in 17 Example 58. the ratio of number of their undecayed nucleiwill be (1/e) after a time (A) 0 1 .0 and . e. C59. . . . Thus in four half-lives the number is reduced to 1 2 .2 × 1016 = 2 × 1015 nuclei . 1 2 . . . 1 2 . . The number ofremaining active nuclei is. . = 1 16 × 3. . . i. .these nucleiwill stillbe in the same active state after four half-lives ? Sol: In one half-life the number of active nuclei reduces to half the original n umber.e. . 1 16 th of the original number. . . . . . . . 1 2 . . . . t (1000 s 1) = (1200 s 1) e . t1/ 2 t1/ 2 t1/ 2 t1/ 2 100%. Calculate its half life.25% ofthe reduced number.Also find the total energy released during th is period if the energy released per disintegration is 8 × 10 13 J.t . no. In 260 days. activity = . A=A0e .NUCLEAR PHYSICS www. its half life is t1/2 = 10 s Relation betweenhalf life andmean life is 1/ 2 mean t t 10 s ln (2) 0.12..5%.25% . 86400 × . Sol:We have. Therefore. T = ln 2 .6. Calculate the activityof the sample after 260 days have elapsed. i. half-life = 228min C62.25%. t = 4 t1/2 = 4(10) s = 40 s Example 59. . tmean = 14.Aft er 10 second themumber of undecayed nuclei reduces to 12.. the number of undisintegrated atomswill reduc e to 1/4th.25%. Sol.t .in 18 C61: The activityof a radioactive sample falls from1200 s 1 to 1000 s 1 in 60minutes . Sol.Asample initiallycontains 1020 radioactive atoms of half-life 130days . number of nuclei has been reduced to half (25%to 12.50%.At a given instant there are 25%undecayed radioactive nuclei in a sample.5%)..43 s (ii) Frominitial100%to reductiontill6. . = ln(6 / 5) t = ln(6 / 5) 60min but.. it takes four half lives.693 0. two half-lives. (i) In 10 second. disintegration constant = ..N = 0. ln 2 T = ln(6 / 5) 60min T = ln 2 ln(6 / 5) × 60 min = 0.693 130 .5%Calculate : (i)mean life of the nuclei (ii) the time inwhich the number of undecayed nucleiwill further reduce to 6.182 × 60 min . . .physicsashok...t = ln(6/5) .e. of atoms present after 260 days = N = 1 4 × 1020 Also.. 5/6 = e . = 0..6931/T where T = 130 days = 130 × 86400 s Now.693 . total energy released = 3 4 × 1020 × 8 × 10 13 = 6 × 107 J . number of disintegrated atoms = 3/4 × 1020 Energy per disintegration = 8 × 10 13 J .1 4 × 1020 = 1.54 × 1012 s Number of atoms present initially= 1020 Number of atoms present after 260 days = 1 4 × 1020 . It means A.. . A = 4.dt or v t 0 0 .-particles are emitted per second.e .. Volume of this space = 4. .r2 (v dt) .-pa rticles strike the plate normallyand come to rest.. Itmeans during an elemental time intervaldt a number (A. . dt) of . dv .t Example 61. (in forward direction) or m dv dt ... dt or v = u. Sol. A= . If a spherical surface of radius rwith centre at position of nuclide be considered thenA . Concentration of .t or dv = u.-particles are in a space having shape of a spherical shell of radius r and radial thickness (v dt) as shown in figure.-particles of a verage kinetic energyE = 11.NUCLEAR PHYSICS www. u. Aradio nuclidewith half life T = 69. Find the velocityof it after time t. .-particle = 9 × 10 31 kg)] Sol. Themass of t he body is decreasing exponentiallywith disintegration constant . .-particles cross this surface (v dt) r per second. [Mass of . .t and thrust force on the body is Ft = ur dm dt .-particles be v then above calculated (A.physicsashok. N = . If velocity of . r = 2mfromnucli de is n= 3 × 1013 perm3.Assuming that themass is ejected bac kwardswith a relative velocity u.-particles at distance r fromnuclide is 2 n A dt 4 r (v dt) .t So dm dt .25 eV. .t) (ur = u) or (m0e . . = m0. Abodyofmassm0 is placed on a smooth horizontal surface. Mass of the bodyleft after time t is m = m0e .-particles at distance. Let activity (rate of decay) of the nuclide beAnuclie per second. dt) . or activityofthe nuclide.. Initially the bodywas at rest. (i) Calculate number ofnuclie in the nuclide at that instant.-particles cross this surface.in 19 Example 60.. .At an instant concentration of . (ii) If a small circular plate is placed at distance r fromnuclide suchthat ... .e .-p articles. calculate pressure experienced bythe plate due to collision of .e .NwhereNis number of nuclei Hence. = u(m0.r2 vn But activity.31 second emits . .t) dv dt = m0 u. . .4..(1) . .6931 .. .r2vn . 4 r2vnT N 0.= log 2 T . but decay constant . + . . anti neutrino] (a) High energyparticles for outer space. is used to estimate the age oforg anic samples. v 2E m . Force. .p = mv Due to transfer ofmomentum.6931NS 4.14. .6931 NS 4 r 4 r 4 r T .physicsashok. of particles striking per second or F = mv × 2 0.6 1022 0. . .r2) . also called carbon dating. Let area of small circular plate be S. beta particles.-particles cross unit area per seco nd. . E 1 mv2 2 . The technique is based on the . . . F = . .6931N 4. P = F S = 2 0. 14 6 C 147 N + .r T = 1.r T mv but v = 2E m .p × no. .. P = 2 0. . then number of .r T Pressure. (b) The carbon dioxidemolecules of the earth s atmosphere have a constant ratio (~ . (ii) At distance r fromthe nuclide (A/4.-particles striking the plate per second 2 2 2 A S N S 0. . Momentumtransferred to plate due to collision is 0 = mv .6931N 2mE 4.08 × 10 4 Nm 2 RADIOACTIVE DATING OR CARBON DATING Radiocarbon dating. P = Force per unit area . the plate experiences a forcewhichis equal to rate o ftransfer ofmomentum. 2 N 4 r nT 2E 9. induce nuclear rea ctions in the upper atmosphere and create carbon . substituting this value in equation(1).NUCLEAR PHYSICS www. .6931 m .-activity of the radioactive isotope C14.-particles.in 20 Kinetic energy of . . called cosmic rays. 1. .4 × 10 12) ofC14 to C12 . . . Momentumof eachparticle just before collisionismv andafter collisionparticles co me to rest ormomentum becomes zero. .-decay of C14. (e) Thus bymeasuring the . (f) Half life of 14C is 5739 y. it is possible to estimate the age of amaterial. (c) All living organisms also showthe same ratio as theycontinuously exchange CO 2withtheir surroundings. an organismcan no longer absorbCO2 and the ratio C1 4/C12 decreases due to the .-activityper unit mass. (d) However after its death.isotopes. . be the decayconstant of 14C-activity . the sample is t year old. = . using radioactive decay law. . A=A0e . t = T 0.t = ln 15.18 that of a contemporary specimen..218 .A ) . A=A0e .3 12.. the sample is 1805 y old. = 0. then (14C-activity of sample) = (14C-activity of living tree )e .218 0.3 disintegrations of 14C per gram per minute.18 .we get 12. Asample from an ancient piece of charcoal shows 14C activity to be 12.NUCLEAR PHYSICS www.. t = 1 .3 .3 disintegrations per gramperminute. 0 A A = 0. [as. is decayconstant of 14C-activity Putting values ofAandA0 in equation (i). t year ago fire occurred and let .t i.3 dis.physicsashok..3 = 15. t = 1805 y Thus.693 T ] .t ..We have..A0 = 15. per min.Howlong ago did the fire occur ?Half-life o f 14C-activity is 5730 y. ratio between the 14C-activityof burnt charcoal and that of a living tree is given.e. Howold is this sample ?Half life of 14C is 57 30 y.t = ln(1.693 ln 0 A A . = 0. . 14 14 0 C activity of charcoal (say. [. Example 63: The relative radiocarbon activityin a piece of charcoal fromthe rema ins of an ancient camp fire is 0. t = 0.218 . permin. 14C-activity of the old sample.e. Sol: 14C-activity of a living tree.. A= 12.t = 0 A A .18 Suppose. Sol: Here..24) = 0..693 T. .t .693 × 5730 y. [T = half-life of 14C -activity= 5730 y] . = 0. Suppose.218 0.t = ln 0 A A .A) C activity of a living tree (say.3 e .3 dis.t .in 21 Example 62:When charcoalis prepared froma living tree it shows a disintegration rate of 15. .. t = 0.(i) where . ln 0 A A . 56| = 1.3 ln|5.4 × 104 y Thus.18 = 8268.693 T ] = 5730 0.t (whichis applicable onlywhenthe radioactive substance is onlydisintegrati ng).0. N=N0e . but the radioactive lawchanges for various types ofdisintegrationofthe substance.4 × 104 years ago.Let us first enlist these types ofdisintegration as: .693 ln 1 0. fire occurred 1. Radioactivity law for different types of disintegration of the radioactive subst ance It is seenthat radioactive disintegration of a radioactive substance is not only ruled bythe radioactive law. ) 1 A (4) Simultaneous disintegrations of parent nuclei: It dealswith the casewhen a parent nucleusmaydisintegrate in a number ofways int o different products. decays B (.in 22 (1) Radioactive substance only disintegrates: For this radioactive law is dN dt = . It deals with the case when production and the decay of the radioactive substanc e are taking place simultaneously.Herewewilldealwiththe net ra te ofdisintegration of the preparation. q = constant rate of formation ofA. Let us discuss these types of disintegrations.) Formation (q) A Let. then decayr ate. after a period of time.1 and . A .N or N = N0e .N . B 1 . . and simultaneously the p roduction ofAis taking place at a constant rate q. (6) Disintegration of isotopes: Apreparationmayhave a number ofradioactive isotopes. Radioactive substance only disintegrates : Suppose disintegration ofAinto Bis taking placewithdecay constant . dN dt = ..2 C Parent nucleusAmay decay in B or Cwith decay constants . successive daughter nuclei deca y at the same rate as it is formed. (5) Radioactive equilibrium: In a radioactive series. N = N0e . Nis the number of nuclei ofApresent at time t.) B (3) Successive disintegrations of the products: It dealswith the casewhen a substanceAdecays into a substance Band Bsuccessively decays into a third substance Cwith the same or different decay rates.t Disintegration with continuous production: Suppose a substanceAdecays into B with decay constant . Disintegration ratewillbe given by dN dt = q + . ) 2 decays C (. Formation A decays (.t (2) Disintegration with continuous production of the radioactive substance.2 respectively. B decays (.physicsashok. This situationis called radioactive equilibrium.N.NUCLEAR PHYSICS www. where .N is the rate of decay ofA. we get.Rearranging. . . N = 1 .1 e... . . . . .. (1 e . Then Rate of formation = N Rate of decay = N .2 E t 1 e ms . Q . . . . . . Example 65.t] Example 64.. Nuclei of a radioactive element Aare being produced at a constant ra . t . number of nucleidisintegrated in time t t . . .dt . Therefore. ..physicsashok.N = dt.N or N t 0 0 dN dt N . is produced in a reactor at a constant rate . Integration gives. . 0 0. .q . find the increase in temperature ofmmass ofwater in time t. or N = . .t . . So . .q .= increase in temperature ofwater = Q ms . . energy released till time .t) Number of nuclei formed in time t = . Aradionuclidewith disintegration constant .2 E t 1 e.. .t and number of nuclei left after time t .DuringeachdecayenergyE0 isreleased. 0 t E t 1 e. . t . .. . . .. . Sol. . Let Nbe the number of nuclei at any time t.1 e. .. of nuclei ofAinitially present. .. . net rate of formationof nuclei at time t is dN dt = . Specific he at ofwater is s. . . . nuclei per second. . 0 0.. . .NUCLEAR PHYSICS www. . . 20%ofthis energyisutilised inincr easingthe temperature ofwater.N0 q)e . . . ..t . But only20%of it is used in raising the temperature ofwater. . Finally..in 23 dN . . t . . where Q = ms. .N = t 0 ..whereN0 is the no. . [q + (. .Assume that there is no loss of energythroughwater surface. 0 N N dN . . . . . . . . . =2N0. calculate the number of nucleiofAafter one half life ofA..te .. The element has a decay constant ... there areN0 nucleiof the element. .. (a) Calculate the number Nof nuclei ofAat time t (b) If ..At time t = 0 . and al so the limiting value ofN as t . To calculate rate of release of energy at time t and total energy released upto time t. . . (ii)Substituting . .NUCLEAR PHYSICS www. in equation (1). we get N = .. their decay starts as soon as their production is started. . . in equation (1).in 24 Sol. dN dt = ..we get .physicsashok. . .N e. Since.N = dt or 0 N t N 0 dN dt N . or e dN qT Nlog 2 dt T .. Aradionuclide with half life T is produced in a reactor at a constan t rate q nuclei per second. . is decay constant which is equal to e log 2 T . . .. Sol. . (a) Let at time t . = 2. . t 0 N . nucleiproduced are radioactive. During each decay. . . . . . 1 .. . If production of radionuclide is starte d at t = 0. (i) rate of release of energyas function of time t and (ii) total energy released upto time t.. number of radioactive nuclei areN. .Nwhere . = 2N0 or N = 2N0 Example 66. . Let at some instant number of nuclei in the radionuclide beN.N0 and t . at instant t. Then rate ofits de cay= . = 2. .. calculate. Rate of formation = A Rate of decay = N t = t N = N Solving this equation. rate of decayat that instant and totalnumber of decays upto that instantmust be known. Net rate of formation of nucleiofA. . net r ate of increase of nuclei e dN q N q Nlog 2 dt T .N or dN . energyE0 is released. therefore.. therefore. . . rate of production is q nuclei per second. . Since.N0 and t = t1/2 = ln (2) . . . we get 0 N 3 N 2 .(1) (b) (i) Substituting . . . . N= 0 and at t. . . . .N = ? N t 0 0 e dN dt qT N log 2 T .. . . e dN dt qT N log 2 T .. .(1) Integrating above equationwith limits at t = 0. . . t loge 2 T e qT N 1 e log 2 . . . .. . 1 andB succes sivelydecays into another stable product Cwith a decay constant . .(ii) Rate of change of number of nuclei ofC is 3 dN dt = (rate of decay ofB) = .in 25 Hence. Successive disintegration : Suppose a radioactive substanceAdecays into B with decay constant .(iii) Solving (i). . .. . rate of release of energy at time t loge 2 T 0 0 t AE qE 1 e.2N2 . 3 dN dt = ..2N2 hence....1t N2 = 0 1 2 1 N . . N1. .2N2 . . . 2 dN dt = .1 A decays Let N0 be the number of nuclei ofApresent at t = 0.1N1 ..2N2 . N1 = N0e . . Decay rate ofAis given by dN dt . Rate of decay ofA= . therefore.. . . decays C (stable product) .NUCLEAR PHYSICS www. .(i) Rate of change of no of nucleiofB is 2 dN dt = (Rate of decay ofA) (Rate of decay of B) But. = . . N3 be the number of nuclei ofA. . .physicsashok. . t But the number of nuclei remaining undecayed at that instant isN. total energy released upto this time = (qt N)E0 t loge 2 0 T 0 e qTE qtE 1 e log 2 .2.1t ..1N1 . B and C respectively at anyinstant t. .1N1 and rate of decay of B = . . Therefore.. . . . . rate of decay. . . t loge 2 A N q 1 e T .. . N2. Total number of nuclei produced upto time t = q . Since..we get N1 = N0e . . . . tot al number of nucleiwhich decayed upto time t = (qt N) Hence.2 B . (ii) and (iii). energyE0 releases during each decay. .. ....2t] N3 = N0 2t 1t 1 2 2 1 e e 1 . . . . . . . . . .. . . .1t e . [e . . . . N dt but. (e. .1t Rate of change of the number of nucleiofB is 2 dN dt = . Set up the rate equations for the populations ofA. N2 = 0 1 1t 2t 2 1 N (e e ) ( ) . . N3 = .NUCLEAR PHYSICS www. N3 3 0 . . )dt . N dt .2N2 3 dN dt = . 1 2 . .. N2 and N3 be the number ofA.2 t 2 0 . e. .. . . = 0 1 2 2 1 N . . . . . N2 is given as. . . .2. .2N2 where .in 26 NOTE : In this case total number of nuclei remains constant. Decay rate forAnucleiwill be 1 dN dt = . . .The population ofBnucleus as functi on of time is given by N2(t) = 0 1 1t 2t 2 1 N (e e ) ( ) . Initiallythere are onlyAnuclei and their number isN0.BandC. after integrating. .2N2 = decay rate of B and .1N1 . . N1 + N2 + N3 = N0 at any time.physicsashok. . B further decays to a stable nucleusCwith a decayconstant . .. . dN = t 2 2 0 . . . . N3 = 1 2 t 0 1 2 t t 2 1 0 N .1. hence. . Example 67:Aradioactive nucleusAdecays to a nucleus B with a decay constant .. . B and C nuclei respectively present at a time t.. . N1 = N0e . Calculate the population ofCas a function of time t. . . .2N2 . .. .1N1 = decay rate ofA Rate of change of the number of nuclei ofC is 3 dN dt = (Rate of decay of B) = . . Sol: Let N1. . .1N1 . . . For example a nucleusA may either a-decay to a nucleusB or . . .. e. . . . 1 1 . . e.. Simultaneous disintegration: Aradioactive nucleus can decay bytwo different processes.. .. e. 1t 2t 1 2 1 2 . N3 = N0 2t 1t 1 2 2 1 e e 1 .2 be the decay constants for these two decay processes. . . . N3 = 0 2 1 N . . B 1 . . . .1 and .. . A . . . . . .. .. = 0 1 2 2 1 N . e. .... .. . . . . e. . . 2 t 1t 1 2 [. . ... . . .2 C Let .. . . .. ] + N0 . .-decay to nucleus C. . e.t t t 1 2 0 . . . . .. . ..1dt + . Thus.eff = . . Showthat the effective half-life t of the nucleus is given 1 t = 1 1 t + 2 1 t ..NUCLEAR PHYSICS www. . . ..2dt.1dt + .1dt .2dt .eff = .1 + .where . Thus.eff = .2dt ..in 27 The probability that an active nucleus decays by the first process in a time int erval dt is .2 = 2 ln 2 t Probability of decay by the first process = .eff.effdt = . ...1 = 1 ln 2 t For the second process..2dt. The half li fe for the first process is t1 and that for the second process is t2.physicsashok.1dt Probability of decay by the second process = . . Example 68:Aradioactive nucleus can decay bytwo different processes. ln 2 t = 1 ln 2 t + 2 ln 2 t . If the effective decay constant is .1 + .1 dt + .effdt.effdt = .1 + .2 . the probability the it either decays by the first process or by the secon d process is . this probability is also equal to .1dt + . the probability that it decays by the second process is .As decay constant. is defined as the probabilityofdecayper second for a particular proc ess for a sample.. Hence...2 + .eff is the effective decay constan t.2 For a number of different process for decay.effd t. 1 t = 1 1 .2dt This probability also equals to .. Sol: The decay constant for the first process is .2dt Probability that it either decay by the first process or the second process = . Similarly. .1 > >. Anumber N0 of atoms of a radio active element are placed inside a cl osed volume. number of increase ofdaughter nuclei are dN2 = . N0 so that N2 = N0 e.2N2dt or dN2 = .2 i.1t dt .2. Consider two limiting cases ..1N1dt .t + 2 1 t Proved.. Sol.1<< .The radiactive decayconstant for the nucleus of this element is .1t .1N0 e. Example 69..2 and .2N2dt (N1 = N0 e.2N0 e.1.(1) Case-1 : When . The daughter nucleus that f ormas a result of the decay process are assumed to be radioactive toowitha radioactive decayconstant .1>> .. (t1/2)1 < < (t1/2)2 (t1/2 = half life) We can assume that N20 .e.2.1t ) or 2 dN dt + .2t (N20 = number of daughter atoms at time t = 0) .Dete rmine the time variation of the number of such nucleus..2N2 = .. In time intervaldt. . therefore.-particle. Total activity of radionuclide at t = 1620 sec. . That of second isotope.A t t = 0.e. . particles fromthe radionu clide are equal. .1N0) or N2 t 2 0 1 0 2 2 0 dN dt N N . . Aradio nuclide consists of two isotopes. Case-2 : When .. respectively.and . Activity of first isotope at t = 1620 sec. . .in 28 Physicallythismeans that parent nucleipractivallyinstantlytransforminto daughter nuclei.2N2 . . and . Let it beA0.particles fromthe radionuclide are equal.30103. probabilities of getting .-emissionwith half livesT1 = 405 second and T2 = 1620 second.-emission and the other by . N0(1 e.physicsashok.2. Since. If at t = 0. . (t1/2)1 > > (t1/2)2 In this case number of parent nuclei can be assumed to remain constant over a si zable time interval and is equal to N0. total number of nuclie in the ratio-nuclide areN0. . t /T1 0 1 0 1 A A A 2 16 . . log10 13 = 1. This transforms equation (1) into 2 dN dt = (.which thendecay according to the lawof radioactive decaywith decay constant .11394 Sol..2t ) Example 70.2 i.. .1 < <. at t = 0.A=A1 +A2 = 9 16 A0 . Given log10 2 = 0. ..NUCLEAR PHYSICS www. Probabilityofgetting . Calculate their res pective probabilities at t = 1620 second.. Which after integration gives N2 = 1 2 .. . ..One of the isotopes decays b y . initial activities of two isotopes are equal. . calculate time t when total number of nuclie remained undecayed becomes equal to 0 N 2 . probabilities of getting . . t /T2 0 2 0 1 A A A 2 2 .. 1 1 . respectively. That of second isotope. . Let at t = 0. number of nuclei of two isotopes beN01 andN02.P A 1 A 9 . . 2 2 P A 8 A 9 . . 2 02 2 02 2 A N N log 2 T . . . and that of getting . .-particle. 1 01 1 01 1 A N N log 2 T . Initial activityof first isotope. . .NUCLEAR PHYSICS www. t /T2 t /1620 2 02 0 N N 1 4 N 1 2 5 2 . . A1 =A2. .. . . . t /T1 t / 405 1 01 0 N N 1 1 N 1 2 5 2 . number of nucleiof first isotope that remain undecayed. . . . . . . .. . .. N0 and 02 N 4 5 .. .. . .. . That of second isotope. therefore 01 02 1 2 N N T T . .. total number of nuclei. . . N0 At time t. ..physicsashok. . . . But it is equal to 0 N 2 . . = 4 t /1620 0 N 1 4 1 5 2 2 .. .. . . ... . . . ... . . .in 29 Since. or 01 1 02 2 N T 1 N T 4 . . N = N1 + N2 = t / 405 0 N 1 5 2 . . . . . 01 N 1 5 . . . . . + t /1620 0 4 N 1 5 2 . or t /1620 1 8 2 13 . 4 t /1620 0 0 N 1 4 1 N 5 2 2 2 .. Initially.. . . Total number of nuclei remaining undecayed at time t. N0 =N01 +N02 . . . ... ... .. . .. . . The intermediatemembers of each decayseries havemuch shorter half-lives than their parent nuclide. decays at the same rate as it is formed. .... 1620 t s log 2 . RB. . . are all equal at equilibrium. .. .. ... . af ter a period of time an equilibriumsituationwill come about inwhich each successive daughter B. C. log 2 = log 8 log 13 or ....... . Radioactive equilibrium : Decayof 238 92 U into a stable end product 206 82 Pb is aradioactive serieswhichcontains anumber ofintermediate members. t 1620 .... .log13 3log 2. Taking log. . .. .Nwe have. or .. 1620 t s 1134 s log 2 . Thus the activities RA. and sinceR = . ifwe start with a sample NA nuclei of a parent nuclideA... RC.. ..As a result.log13 log8. 1N1 = .BNB = .2 and .1. .1 = ln 2 T .Hence activities of the two willbe equal.8 × 104. . B and C respectively then.A2.2 are decay constant of 238U and 234Urespectively. . . N1 and N2 are number of atoms of 238U and 234U respectively. The half-life of 234U is 2.10 (year) 1 .. This situation is called radioactive equilibrium. .ANA = .2N2. Example 71: The atomic ratio between the uraniumisotope 238Uand 234Uin amineral sample is found to be 1. = 5 ln 2 2. . A = .. half life of 238U = 4. .5 × 105 year Disintegration of isotopes : Suppose a sample is a mixture of three radioactive isotopesA. where T is half life of 238U. B and C respectively at any instant. IfA1. × 2.A3 are decay rates ofA..5.2 = 5 ln 2 2.2N2 + .. . . where.NUCLEAR PHYSICS www..1N1 + .8 × 104 × 2. the net decay rate is Anet = A1 + A2 + A3 but.physicsashok. . . 1 2 N N = 1.1N1 = .2N2 .. Now.5 × 105 year. . .3N3 . Sol: The two isotopes are in radioactive equilibrium. ..net = 1 1 2 2 3 3 N N N .. .in 30 .. Let .. The above equation can be used to establish the decay constant (or half-life) of anymember ofthe series if the decayconstant of anothe rmember and their relative proportions in a sample are known. B and C. . .5 × 105year .8 × 104. B and C respectively.3 be the decay constant ofA. N2 and N3 be the number of nuclei of isotopesA.N . T = 1 2 N N . ln 2 T 1 2 N N . we have .1. Find the half-life of 238U.CNC = .10 (year) 1..5 × 105 year = 1..5. N1.net N = . Thus. .. Example 72:Asample of uraniumis amixture of three isotopes 234 92 U. .N . . .006%. . .5 × 105 years. .The half life of these isotopes are 2.284%respectively. . .net = 1 1 2 2 3 3 1 2 3 N N N N N N . .5 × 109 years respectively. 7. N = N1 + N2 + N3 hence net decay constant.71%and 99. 235 92U and 238 92U present in atomic ratio of 0.1 × 108 years and 4. . . 0. Calculate the contribution of activity (in %) of each isotope in this sample. Nis the total no. . of nuclei in the sample at any instant . 71 7.5.2.284 .71 : 99. we haveN 1 : N2 : N3 = 0.5 : 1 : 99.006 : 0. T1. Comparing (i) and (ii) we get A1 :A2 :A3 = 1 0.006 T : 2 0.NUCLEAR PHYSICS www. [as ...71 T : 3 99.5 = 24 : 1 : 22.06 = 24 100 24 1 22.3 are decay constants of these isotopes and as activityA= .10 : 8 0.284 4.1N1 : .1. .in 31 Sol: Let N1.284 4.284 T = 5 0. A1 :A2 :A3 = 1 1 N T : 2 2 N T = 3 3 N T .. T2 and T3 are half lives of the isotopes.N. N2 and N3 be the number of the three isotopes in the sample...06 A1 :A2 :A3 = 24 : 1 : 22..5.physicsashok.3N3 A1 :A2 :A3 = 1 1 (ln 2)N T : 2 2 (ln 2)N T : 3 3 (ln 2)N T .. = ln 2 T ] .(ii) where.10 : 9 99. .006 2.10 = 60 2.(i) If .1. we hav e A1 :A2 :A3 = ..06 .2N2 : . and Contribution of 238 92 U = 46.18 10 9 t .06 100 24 1 22.00%: 2.t 128 e . . . Example 73. ln 2.5 × 109 years and 7.13 × 108 years. Sol. ln 2. (ln 2/T1)t 1 0 N . Contribution of 235 92 U = 2. .12%: 46.1/ 7. Let N0 be the initial number of atoms. = 2400 47. N e.88%in activityof the sample.06 %: 2206 47.1/T2 1/T1.t.06 = 51. .18 .13 108 1/ 4. .t 1 2 N / N e .06 . . . . or t 7 1.06 %: 100 47. × 109 = 5. 128 eln 2 1. . Then fromN= N0e .88% Activityratio = 51%: 2.88% . The isotopes of uraniumU238 andU235 occur in nature in the ratio 128 : 1. or 27 eln 2 1..12%. .make an estimate ofthe age of the earth. : 22. . . . . . respectively.5 109 .12%: 46. . Contribution of 234 92 U = 51%. . . : 1 100 24 1 22. N e.The half-lives of U238 and U235 are 4.06 .Assuming that at the time ofthe earth s formationtheywere in equal ratio. and (ln 2/ T2 )t 2 0 N .18 10 9 t .9 × 109 years . . (iii) In the .. . Neglect any recoilof the residualnucleus. .NUCLEAR PHYSICS www. . restmass energyis converted to ki neticmass energy(K3. . the energyQis shared bythe antineutrinos and the bet a particle. decay process..m. y 1 m m . .002603 a. . + Ky = Q & p. = Q . . . = py p2 2m . Note: (i) The kinetic energyKE. * K.-particle emitted in the decay 238Pu 234U+ . .m.in 32 Types of Nuclear Collision: Exoergic collision / reaction: IfQis positive.04095 a. of 4He = 4. and the reactionis called exoergic.-particle.m. . .u. . . . + 2 y y p 2m = Q p2 2m . of the emitted . . . The atomic masses of 238Pu = 238. .. = Q K.-part icle emerges (since momentummust be conserved). . .u.Q C63: Find the kinetic energy of the . The kinetic energy of the beta particle can be anything between zero andmaximumvalue ofQ. and hence most ofthe disintegration energy appears as theKE of the .04955 a.-particle is never quite equal t o the disintegration energyQ because the nucleus recoils with a small amount of kinetic energywhen the . * KE A 4 Q A . Sol: Using energyconservation. .Ais themass number of the parent nucleus.u.physicsashok. radiation energyor both. . 1 4 A 4 . . = A 4 A . of 234U = 234. . K. (ii) Themass numbers of nearlyallalpha-emitters exceed 210. . .K4 etc).. . canhave.u.04955 a.m(238Pu)c2 = m(234U)c2 + m(4He)c2 + K or K = [m(238Pu) m(234U) m(4He)]c2 = (238.u.0059970 a.5)MeV/a. Find theminimumandmaximumkinetic energy that the beta particle 01 e .m. × (931.59MeV C64: Neon-23 beta decay in the followingway: 23 10 Ne 23 11 Na + 01 e .u.9898 u. respectively.m.m.u. 234. + . The atomicmasses of 23Ne and 23Na are 22.002603 a.04095 a.u. Sol: Reactant Product .]c2 = 0.m. = 5.9945 u and 22. 4.m. 4.9945 10me 23 11 Na 22.0047 u . .physicsashok.m2 =mass of the target nucleus. me Total 22. .-particle and neutrino share this energy. the reaction is endoergic. . .4MeV The . . Q = .NUCLEAR PHYSICS www. Endoergic collisions (a) IfQis negative. wherem1 =mass of the bombarding particle.4MeVwhenthe antineutrino does not get anyshare.0047 u) × 931. is called threshold energyEth. particle can range from0 to 4. m1c2 + m2c2 + K1 = (m3 + m4)c2 + K3 + K4 & p1 = p3 + p4. (c) Threshold energyEth: Theminimumamount of energy that a bombarding particle m ust have in order to initiate an endoergic reaction.9945 10 me Total 22.5MeV/u) = 4. (b) For endoergic reaction to take place aminimumenergy has to be supplied.4MeV.we get Eth = Q 1 2 m 1 m .m c2 = (0.Themaximumkinetic energyofa beta p article inthis decayis.9945 u 22. . . Q + K1 = K3 + K4 Q + 2 1 1 p 2m = 2 3 3 p 2m + 24 4 p 2m = 2 3 3 p 2m + 2 1 3 4 (p p ) 2m . therefore.9898 u = 0.in 33 23 10 Ne 22. . Usingmomentumconservationalso.9898 11me 01 e .m= 22. Energyof .9898 10 me Mass defect. . . . . 1 3 4 2p p m + 2 1 4 p m 2 3 3 4 p 1 1 m m . . . 2Q = 0 2 1 24 4p m 4 3 4 1 1 m m . . . . . . . . . 1 3 4 2p p m + 2 1 p 4 1 1 1 m m . . .. . . . . . . . . . . . . 2 1 4 1 p 1 1 m m . . . . . . . . . . . 2Q + 2 1 1 p 2m = 2 3 3 4 p 1 1 m m . 7. .0783 u.01600 u. 1. . Q 3 4 3 4 1 m m m m m . . the appr opriate number of electron . .0866 u respectively. C65: Howmuch energymust a bombarding proton possess to cause the reaction 7 3 Li + 11 H 74 Be + 1 0 n atomicmasses of 7Li. . 1H. Sol: Since themass of an atominclude themasses of the atomic electrons. Q 3 4 3 4 1 m m m m m . .2Q . . 0 2 1 1 p 2m . K1 . 7Be and 1 0 n are 7.01693 u and 1. If m1 + m2 ~ Q 1 2 1 m m . . . . . Q + T..00176 u × (931. = O O 2m T . = TO + Tp . ( 1. so that momentumis conserved. p) O17.65MeV Negative sign ofQindicates endoergic reaction. TO Tp p X . + p p 2m T .cos.5MeV/c2) = 1...89 MeV. if the kinetic energy of the incoming .02383 8.0 MeV.NUCLEAR PHYSICS www.09MeV.. . Energymust be supplied for this r eaction to take place. .00300 amu = 2. 7N14 + 2He4 8O17 + 1p1 Let Q . .Making use ofthe table of atomicmasses find the energies of the following re achings Li7(.value of reactionQ.cos. theminimumkineti c energyof the incident can be foundwiththe formula.Withmomentumconservationtakeninto account.Q = 1 1 7 .=60º to themotiondirection of the alpha-particle has a kinetic energyTp = 2.09MeV Let TO is kinetic energyafter collisionof oxygen.79 MeV Example 74: Find the energyof the reactionN14(.. T... Y Sol: T. Q-value. = O O 2m T .physicsashok.0MeV and protonoutgoingat anangle ..0124 + 1.00260) (10. .65 MeV) = 1. . .(ii) . = 60º .00867) Q = 0. p p 2m T .= 4. Eth = 1 m M .01601 + 4.c2 = 0. .02559)u. . The energyis supplied as kinetic energy of the bombarding proton.(i) Momentumconservationalong x-directiongives. . N .in 34 massesmust be subtracted fromthe given values to getmasses of nuclei.. Tp = 2. The incident p rotonmust havemore than this energy because the system must posses some kinetic energy even after t he reaction. . . = 4.. 2m T . . C66.cos. ..cos.. . Energy conservation gives. . n) B10 Sol. 2m T .. Reaction is.-particle is T. 5B10 + 0n1 Q = (7. . . = [m( 7 3 Li ) 3me+m(11 H ) me]c2 [m(74 Be ) 4me + m(1 0 n )]c2 = [m( 7 3 Li ) +m(11 H ) m(74 Be ) m(1 0n )]c2 = (8. 3Li7 + 2He4 . ..sin.(v) Putting the value TO fromequation (iv) in equation (v).. p p 2m T .(iii) Squaring the equation (ii) and (iii) on both sides and adding the result. ( 2m T . 2 p p (2m T )(2m T ) .. Q = TO + Tp T. 2m. T.= O O 2m T .(iv) Fromequation (i).cos. .. + mpTp 2 p p m m T T .cos.we get...) = 2mOTO .. + 2mpTp(cos2.T. ..sin. TO = O 1 m [m.cos..)2+ 2mpTp sin2..] . . . p p 2m T . . + sin2. ... . = 2mOTO .we get..Momentumconservationalong y-direction gives. 09 MeV 4.in 35 Q = Tp T.NUCLEAR PHYSICS www.09 2 4. a uraniumnucleus after capturing neutronhad become so unstable that instead ofdisintegrating byej ecting oneor two particles. to produce transuranic elements by bombarding uraniumwith neutrons.e. It was discovered in 1939 that the heavyunstable uraniumnucleuswhen bombarded by neutrons splits into two almost equalfragmentswhich flyapart with great speed and the amount of energ yreleased per fission is about 200MeV. the fi ssion process itself was discovered in 1939 by German radio chemists Otto Hahn and his two associates Meitner and Strassmann.0 MeV + 1 16 [1 × 4 + 1 × 2.1.After bombarding uraniumwith neutrons. both natural and artificial.09 . . To their great surprise. Q = 1.physicsashok.-rays Unstable . the nucleus isonly chipped off rather than broken and accordingly. + mp Tp 2 p p m m T T .14MeV NUCLEAR FISSION Continuous research or artificialtransmuttionand especiallythe studyof inducedra dioactivity. . This division of a nucleus into two approximately equalparts as called nuclear f ission. they found that the atoms produc ed bythe bombardment of uraniumbelonged to elementswhich lie near the centre ofthe periodic table. Discovery of fission The starting point in the discovery of nuclear fission can be traced to the atte mpts of Fermi in 1934. + O 1 m [m. cos.] = 2. it had split up into two parts.4. 0n1 92U236 92p 143n 92U236 . T. culminated in the discovery of nuclear fissionwhich is accompanied by the release of enormous amounts of energy. However. theyperformed a series of chem ical separations to identify the products.-rays Unstable Antimony Nuclide 51Sb133 51p 82n . The actualfissionprocess canbeunderstoodwiththe helpoffigurewhich shows a uraniu mnucleus capturing a neutron.Obviou sly. the amount of energyreleased is comparatively less i. fromabo ut 10 to 23MeV.2. In ordinarynuclear disintegrations.co60º] MeV . It amy. behaving like a liquid drop. So great is the release of energy that the two fission fragments fly apart in opposite di rections with tremendous speeds. Inbreaking up. . i. however. ne utron and . be noted that not all uraniumnuclei break into Sb and N b as shown in figure. the uraniumnucleus.Niobium Nuclide 41p 48n 41Nb99 (a) (b) (c) The newly-formed nucleus of figure (b) isunstable and starts breaking up into tw o parts.-rays. splashes out smalldroplets.e. However.-part icles. lead.-rays etc.-emissions in which neutrons are converted into protons in the nucleus.NUCLEAR PHYSICS www.U235. deuterons and . thallium. s oon after itwas found that other elementsof high atomicweight could also bemade to undergo fissionand that particles other than neutron could be equallyeffective inthis respect. In this process. Fission can also be produced in uraniumand thoriumby high-energy .physicsashok. protons. a large number of nuclides of intermediate charge andmass are found. In 1947. nucleus divides in the ground statewithout bombardment byparticles fromoutside. Out ofall the neutrons ejected during the fission of uranium.1 × 108 Y U234 0. Their study is a promising source of information about the mechanismof the fission process i tself and also offers the .With bismuth (Z = 83) fission was detected with 50M eV deuterons whereas tantalum(Z= 73) required . Similarly.Two other nuclideswhich do not occur in nature but have proved to be fissionalbe by neutrons of all energies are 92U233 and 94Pu239.51 × 109 Y U235 0. some heavynucleihave beenfound to undergo spontaneous fission. Pu239 are important inthe large-scale application of nuclear energy.Th232 and Pa231 undergo fissionw henbombardedwith fast neutrons.in 36 There are at least 30 different ways inwhich a fissile nuclide can divide itself .-particles of 400MeVenergy. Mass distribution of Fission Products During uraniumfission. In general.48 × 105 Y It is found that slowneutrons cause fission ofU235 but not ofmore abundant isoto peU238whichrequires fast neutronswith energies exceeding 1MeV. Itmaybe noted that divisionofa fissile nucleus into three fragments ofcomparable sizes (ternaryfission) has been observed although it is a rare event. platinumand tantalumwas achieved inUSAbymeans of . occurring about 5 times permillionbin aryfissions.mercury. protons. gold. The delayed neutrons originate fro munstable fragments that decay by neutron emission on theirway to becoming stable nuclei. uraniumwas the first element to undergo fission. about 99 per cent are ejected inan extremely short interval oftime and are called prompt neutrons. deutrons and neutrons of 100 MeV and more..-particles.28% 4. Types of Fission Reactions Historicallyspeaking. fission fragments are unstable nucl ei containing an excess number of neutrons. Naturaluraniumcontains three principal isotopeswith the following relative abund ance: U238 99. The experimental evidence seems to favour pairs of fissionfragments of unequalmasses (asymmetrical fission ) accompanied byone to five or some time more neutrons. Finally. a stable nuclide results. It isworthnothing that onlythree fissilematerialsU233. The remaining one per cent of neutrons are emitted a little later and are called delayed neutrons. success ful fission of bismuth.006% 2.After a series of .714% 7. . (i) light groupwithmass numbers from85 to 104 and (ii) heavygroupwithmass numbers from130 to 149. Investigations of the fiss ion products ofU235 have shown that the range of their mass numbers is from72 to 158. Themost probable type of fissionwhich occurs in about 7%of the total cases.About 97%of theU235 nuclei undergoing fission give fragmentswhich fall into two groups as shown inthe fission yield cu rve of figure . give s fission productswith mass numbers 95 and 139.possibility of discovering hitherto unknown nuclides. 54 79 54Xe133 5d Unstable Xenon O . Energy Distribution of Fission Products Energy distribution among the fission products can be found bymeasuring their kinetic energywith the help of suitable ionization chambers. Consequently. The results of such studyonU235 fissionhave shown that the energy distribution curve is not uniform. fission fragments have toomany neutrons in their nuclei for stability.One suchfission decay chain is shownin figure which startswith one of the unstable fragments of the fission ofU235 nucleus.NUCLEAR PHYSICS www. rather it is a doublepeaked . most ofthemdecay byelectron emission.-particles. These series are called fission decay series and chain has threemembers on the average althoughlonger and shorter chains occur frequently.in 37 60 80 100 120 140 160 180 10 100 1000 10000 95 139 Mass Number (A) Number of Fragments Asmentioned earlier. 55 78 55Cs133 Stable Cesium O . Each fragment starts a short radio-active series involvingmany emission of . 51 82 51Sb133 5m Unstable Antimony 52 81 52Te133 60m Unstable Tellurium O . 53 80 53I133 5d Unstable Iodine O .physicsashok. curvewithmaxima at 67MeVand 100MeV. It is seen that while the greatest probability is for a fragment of 100MeV, the areas under the two peaks which represent the total number of particles in the two groups are approximatelyequal. 0 100 200 300 400 40 60 80 100 120 Energy (MeV) Number of frangments 67 100 Neutron Emission in Nuclear Fission One ofthe notable features ofnuclear fission is thatwhile it is initiated byneut rons it is also accompanied by the emissionof fast-moving neutrons. The number of neutrons released depends on themode of fission and on the energyof the neutronswhichinduce fission. The average values for the numb er of neutrons emitted per thermalneutron absorbed bythe three important fissilematerials are given bel ow: U235 2.43 U233 2.50 Pu239 2.89 These neutrons are emitted bythe fission fragments and not bythe compound nucleu s. NUCLEAR PHYSICS www.physicsashok.in 38 The neutrons emitted as a result of fission process (i.e. fissionneutrons) can b e divided into two groups: (i) Prompt Neutrons: Thesemake up about 99.36%of the total fission neutrons and are ejected by the product nucleiwithin 10 14 second of the fission process. Prompt .-rays are also e mitted at the same time. (ii) DelayedNeutrons: These constitute about 0.64%of the totalneutrons fromthe f ission ofU235.These are emittedwithgraduallydecreasing intensityfor severalminutes after actual fiss ionprocess.Although the number ofdelayed neutrons is small, theyhave a strong influence on the time-depe ndent behaviour of chainreacting systems based on fission and play an important role in the controlofnuclear-fiss ion reactors. Fissile and Fissionable Nuclides Elements likeU235, U233 and Pu239 undergo fission by neutrons of energy fromalmo st zero upwards. Such nuclei are referred to as fissile nuclides. On the other hand, U238 and Th232 nu clei which have a fission threshold at 1MeV are said to be fissionable nuclides. In general, fissile nuclides have either an even number of protons and an odd nu mber of neutrons or odd numbers ofboth. Fissionalbe nuclides, onthe other hand, have either even number ofprotons and neutrons or an odd number of protons and an even number of neutrons. Fission Energy One ofthe striking features ofthe fissionprocess is themagnitude of the energyre leasedwhich is about 200 MeV per fission ofU235 nuclide. Before 1939, the largest known nuclear reaction energywas 22.2MeV associatedwith Li6 (d, .) He4 reaction. The amount ofenergyreleased per fissionofU235 nuclidemaybe calculated bythe foll owing threemethods: (i) Binding-energymethod:Asmentioned above all stable fission products havemass numbers in the range 72 to 158where the average binding energyper nucleon is about 8.5MeV.However, in the neighbourhood of uranium, its value is 7.6MeV. Hence, average binding energy per nucleon is (8 .5 7.6) = 0.9MeV greater inthe fission products thanin the compound nucleus ofU235. The excess en ergyis released as fission energy. Its value is 235 × 0.9 ~. 200MeVper fission ofU235 nuclide (which has 235 nucleons). (ii) MassDefectMethod: The energy released per fission can also be estimated byc omparing themass of the interacting particles and the final fission products. As mentioned, U235 splits in manyways and the nuclei obtained in the greatest yi eld in fission by slow neutrons havemass numbers of 95 and 139. The fissionproducts being initiallyradi oactive, undergo many .-emissions to formultimatelystable nuclides. Ifmolybdenum-95 and lanthanum-139 are takenas pair of stable products fromfission ofU235, the fission reaction can bewritten as 92U235 + 0n1 62Mo95 + 57La139 + 20n1 Comparingmasses on both sides of the above equationwe get, mass ofU235 nuclide = 235.124 amu mass of one neutron = 1.009 amu Total = 236.133 amu mass ofMo95 nuclide = 94.946 amu mass of La139 nuclide = 138.955 amu mass oftwo neutrons = 2.018 amu Total = 235.919 amu mass of defect = 236.133 235.919 = 0.214 amu Therefore, energy released per fission ofU235 nucleus = 0.214 × 931 ~. 200 MeV (iii) Kinetic energymeasurementmethod: The total amount ofenergyreleased per fis sionis equal to the sum ofthe following energies: (a) the kinetic energy of fission fragments.As seenfromfigure the average value of this energy for U235 is 167MeV. NUCLEAR PHYSICS www.physicsashok.in 39 (b) the kinetic energyof fissionneutrons. Since the average number of neutrons e mitted per fission ofU235 is 2.43 or say 2.5 and the average kinetic energyof these neutrons is 2MeV, tota l kinetic energyof fission neutrons is 2.5 × 2 = 5MeV. (c) the kinetic energy of prompt .-rays. Its value is about 7MeV. (d) the total energy of the decay process in the fission decay chains. This includes the energy carried away byradiations like .-rays, .-rays and neutr ons. Its value is nearly 21 MeV. The totalof all the above energies is = 167 + 5 + 7 + 21 = 200 MeV C67:AU235 nucleus is fissioned bya thermalneutron and two fission fragments and two neutrons are produced. Compute the fissionenergyreleased if the average binding energyper nucleon is 7. 8MeVin fissionedU235 nucleus and 8.6MeVin the fission fragments. Sol: Greater binding energyof the fissionfragments indicates that there has been release ofenergyduring fission of low-binding energynucleusU235. 236 × 7.8) Fission energy released is = (234 × 8.6 = 171.6 MeV Theory of Fission Process The first attempt to explain themechanismoffission processwasmade byBohr andWhee lerwho accounted formanyof the properties offission on the basis of the liquid-dropmodelof the nu cleus. 1 2 3 4 5 6 The shape of the drop depends on a balance betweenthe surface tension forces and Coulombic repulsive forces. The excitation energy given to the drop during the capture of the slow o r thermalneutron sets up oscillationswithin the drop. These oscillations tend to distort the spherical sh ape so that the drop becomes ellipsoid in shape.The surface tension forces trytomake the dropreturn to its or iginalspherical shapewhile the excitation energytends to distort the shape stillfurther. Ifthe excitation e nergyandhence oscillations are sufficientlylarge, the drop pattains the dumbbell shape as shownin figure. TheCoul ombic repulsive forces thenpush the two bells further apart untilthe dumbbell splits into two similar dro ps eachofwhich assumes a spherical shape. The sequence of steps leading to fission is shown in figure. However, if the excitation energy is not large enough, the ellipsoidwillreturn t o the sphericalshape. In that case, the excitation energyis givenout in the formof .-rays and the process beco mes a radioactive capture process rather than fission process. LECTURE 5 Nuclear Reactors : (a) Anuclear rectoris a systemdesignedto controlthe chainreactionoffissionwithco ntinuous energyproduction. (b) Ausefulfactor for describing the levelof operation of a reactor is the repro duction constant K. It is defined as the average of neutrons available fromeach fission that will cause another fi ssion. For a controlled or self sustained chain reactionKmust bemaintained close to unit. i.e. K. 1 for controlled chain reaction. NUCLEAR PHYSICS www.physicsashok.in 40 (c) Fuel: This is the fissionablematerial.Commonly usedmaterials areU238 enriched inU235 and plutonium(Pu239). (d) Moderator:Fastmoving neutrons cannot triggerthe fission ofU235 and have a very high chance of being captured by U238 which is not fissionable. It is therefore necessary to usemoderators to slowdown the neutrons. Control rods Shield Moderator Fuel elements Nuclear Reactor (e) Coolant: Air, water or CO2 are used as a coolant to remove the heat released inside the reactor. (f) Control Rods: Cd(cadmium) which is a good absorber of neutrons is used to controlthe rate of fissionand also to shut down the reactor in case of emergency. (g) Types of reactors: (i) Thermal reactors: In these reactors fission is produced by slowneutrons ro thermalneutrons. (ii) Breeder reactor: Breeder reactors generally use fast neutrons in these reac torsU238 is converted into Pu239 by capture of neutrons. Pu239 is fissionable. Thus such reactors also prod uce fuel in addition to the energyreleased through fission. (h) Critical mass: For a fuel there is a criticalmass belowwhich the fissionable material is completely safe. But for amass above the criticalmassmore neutrons are produced than are lost so that the chain reaction builds up rapidlyand the systemexplodes. The atomic bombare therefore stocked as subcriticalmass such that the combinedma ss is greater than the criticalmass resulting in a spontaneous explosion. Example 75: In a nuclear reactor, fission is produced in 1 g for U235 (235.0439u ) by a slowneutron (1.0087 u). Assume that 35Kr92 (91.8973 u) and 56Ba141 (140.9139 u) are produced inall react ion and no energy is lost. (a) Write the complete reaction, (b) Calculate the energy released per fission, (c) Calculate the total energy produced in kilowatt hour.Given 1 u = 931.5MeV/c2 . Sol: (a) The nuclear fission reaction is 92U235 + 0n1 56Ba141 + 36K92 + 3 0n1 (mBa + mKr + 3mn) (b) Mass defect, .m= [(mu + mn) .m= 256.0526 u 235.8373 u = 0.2153 u Energyreleased per fission, Q = 0.2153 u × 931.5MeV/u = 200.6MeV (c) Number of atoms in 1g = 6.02 1023 235 . = 2.56 × 1021 Energy released in fission of 1 g ofU235 = 200.6 × 2.56 × 1021 MeV = 5.14 × 1023 MeV = (5.14 × 1023) × (1.6 × 10 13)J = 8.2 × 1010 J = 8.2 × 1010 W-s = 10 6 8.2 10 3.6 10 . . KWh = 2.28 × 104 KWh NUCLEAR PHYSICS www.physicsashok.in 41 Example 76: In neutron-induced binary fission of 92U235 (235.044) two stable end -products usuallyfound are 42Mo98 (97.905) and 54Xe136 (135.917).Assuming that these isotopes have come fro mthe original fission process, find (i) what elementary particles are released (ii)mass defect of the reaction (iii) the equivalent energy released. Sol: (i) The reaction can bewritten as 0n1 + 93U235 = 43Mo98 + 51Xe136 It is seen that the totalZ-value of the two stable fission products is (42 + 54) =96. It is 4 unitsmore than that of 92U235. For balance, the original unstable fission products must have go t Z = 92. Obviously, the originalunstable productsmust have emitted 4 .-particles before becoming stable. Now,mass number on right-hand side is 2 units less than on the left-hand side. It means that towfis sion neutronsmust have been produced.Hence, the fission reaction canbe represented by the following equation : 0n1 + 92U235 = 42Mo98 + 54Xe136 + 4 1e0 + 20n1 (ii) .m= LHS mass RHS mass LHSmass = (1.009 + 235.055) = 236.053 amu RHSmass = (97.905 + 135.917 + 4 × 0.0055 + 2 × 1.009) = 235.842 amu . .m = 236.053 235.842 = 0.211 amu (iii) Energy released = 0.211 × 931 = 196MeV Example 77. About 180MeVenergy is releasedwhen one nucleus of 92U235 undergoes f ission. estimate the energy released from1 kg ofU235, assuming that each nucleus undergoes fission. Sol. 1 kg of U235 = 1000 g = 1000 235 mole . number of atoms = 1000 235 × 6.02 × 1023 . total energy released = 6.02 235 × 1026 × 180MeV = 6.02 18 235 . × 1027 × 106 × 1.6 × 10 19 J = 7.37 × 1013 J C68. Calculate the energy released in slowneutron capture by Pu239.Mass of Pu239 = 239.127 amu, Pu240 = 240.1291 amu, 0n1 = 1.008665 amu. Sol. Energy released =mass defect in energy units = (239.127 + 1.008665 2401) amu = 0.006565 × 931MeV = 6.1 MeV Example 78.Anuclear reactor generates P =20MWpower at efficiency. =60%bynuclear fission of a radionuclidewhose half life isT= 2.2 years. If each fission releases energyE = 200MeV, calculate t ime during which µ = 10mole of the radionuclidewill be consumed completely. (Avogadro number, N = 6 × 1023, loge 2 = 0.693, 1 year = 3.15 × 107 s) NUCLEAR PHYSICS www.physicsashok.in 42 Sol. To operate the nuclear reactor, let the number of fissions required per sec ond be n0. Thenenergy released per second byfission reactions = n0E Since, efficiencyofthe reactor is ., therefore, power output fromthe reactor = . n0E. But it is equal to P therefore, P = .n0E or 0 n P E . . Let at an instant number of nuclei of radionuclide be n then rate of decay= .nwh ere . is decay constant which is equal to e log 2 T . Hence, net rate of decrease of nuclei 0 dn n n dt . . . . . . .. .. or e e dn n log 2 P n Elog 2 PT dt T E ET . . . . . .. . . . . . . . . . e dn dt PT n E log 2 Et . . . . . ...(1) At t = 0, number of nuclei are n = µNand time t is to be calculated when all the n uclei are consumed or when n = 0, t = ? Integrating equation (1)withthese limits, 0 t µN 0 e dn dt PT n E log 2 ET . . . . . . . . e 8 . . e e e T µN Elog 2 t log 1 10 log 1.0576 s log 2 PT . . . . . . . . . . Example 79. The element Curium 248 96 Cmhas a mean life of . = 1013 second. Its primary decaymodes are spontaneous fission and .-decay, the former with a probabilityof P1 = 8%and latt er with a probability of P2 = 92%. each fission releasesE = 200MeVenergy. The masses involved in .-decay are as follows 248 96 Cm= 248.072220 u, 244 94 Pu = 244.06400 u and 42 He = 4.002603 u. Calculate the power output froma sample ofN = 1020 Cmatoms. (1 u = 931MeV/c2) Sol. Decayconstant, 1 10 13 sec 1 mean life ( ) . . . . . . Rate of decay froma sample of N atom,A= .N = 107 sec 1 Since, probabilities of fission and .-decay are P1 and P2 respectively, therefor e, rate of decay due to fission, A1 = P1A or A1 = 8 × 105 sec 1 and rate of decay due to .-emission,A2 = P2A= 9.2 × 106 sec 1. Since, each fission releases energyE, therefore, rate of release of energydue to fission =A1 . E NUCLEAR PHYSICS www.physicsashok.in 43 Equation of .-decay is 248 244 4 96 94 2 Cm... Pu . He Massmost during each .-decay, ..m= [248.072220 (244.06400 + 4.002603)] u = 5.617 × 10 3 u . Energy released during each .-decay, E´ = 5.617 ×10 3 × 931 MeV E´ = 5.23MeV . Rate of release of energy due to .-decay =A2 . E´ . Total rate of release of energy =A1E +A2E´ But totalrate of release of energy is equal to power output. Therefore, power output, P = A1E + A2E´ = 3.33 × 10 5W Nuclear fusion It is the process of combining or fusing two lighter nuclei into a stable and he avier nuclide. Inthis case also, large amount of energyis released becausemass of the product nucleus is less tha n themasses oftwo nuclei which are fused. Many reactions between nuclei of lowmass numbers have been brought about by acce lerating one or the other nucleus in a suitable manner. These are often fusion processes accompanied by release of energy. However, reactions involving artificially-accelerated particles cannot be regard ed as ofmuch significance for the utilizationof nuclear energy.To have practicalvalue, fusion reactionsmus t occur in suchamanner as tomake themself-sustaining, i.e.more energymust be released thanis consumed in i nitiating the reaction. It is thought that the energyliberated inthe Sun and other stars of themainsequence type is due to the nuclear fusion reactions occurring at the very high stellar temperature of 30million ºK. S uchprocesses are called thermonuclear reactions because they are temperature-dependent. Steller Thermonuclear Reactions: Following two sets of thermonuclear reactions have been proposed as sources of e nergy in the Sun and other stars of themain sequence: (i) proton-proton (p - p) chain and (ii) carbon-nitrogen (C-N) cycle. At lowtemperatures corresponding to those in the Sunwhenit was first formed, the proton-proton chain was predominant. In the present state of the Sun with its higher central tempera ture and larger He4 concentration, the C-Ncycle is supposed to be the main source of its energy. Proton-Proton Chain It is so called because the step involves the combination of two protons.When tw o protons fuse, they produce a deuteronnucleus, a positron and a neutrino thus: 1H1 + 1H1 = 1D2 + 1e0 + v × 2 The deutron thencombineswith another proton to yield helium-3. 1D2 + 1H1 = 2He2 + . × 2 The two helium-3 nuclei fuse to produce helium-4 thus 2He3 + 2He3 = 2He4 + 21H1 + 1e0 It should be noted that for the third reaction to occur, each of the first two r eactionsmust occur twice. The net effect ofthe reactions is 41H1 = 2He4 + 21e0 + 2. + 2v NUCLEAR PHYSICS www.physicsashok.in 44 Obviously, four hydrogen atoms are fused to produce one heliumatomwith a total e nergyrelease of about 26.7 MeV.When the kinetic energy of neutrions is substracted, the energy is 26.2 MeV. The emitted positrons are annihilated by free electronswith the production of .-rays. Carbon-Nitrogen Cycle It was proposed byH.A. Bethe to account for the energy production in the Sun and other stars ofmain sequence. In this cycle, carbon acts as a nuclear catalyst. The cycle startswhen a proton (hydrogen atom) first interactswith carbon-12 nucleuswiththe release of fusionenergythus 6C12 + 1H1 = 7N13 + . The product N13 is known to be radioactive, emitting a positronwith a half-life of 10minutes. Hence, it decays in a very short time according to the relation 7N13 = 6C13 + 1e0 + v The stable C13 nucleus reactswith another proton, therebyliberatingmore energy 6C13 + 1H1 = 7N14 + . The stable product N14 combineswith third proton thus 7N14 + 1H1 = 8O15 + . TheO15 nucleus is a positron emitterwith a half-life of 2.06minuteswhich decays by the process 8O15 = 7N15 + 1e0 + v Finally, the resultingN15 nucleus interactswiththe fourthproton thus: 7N15 + 1H1 = 6C12 + 2He4 By adding up the above six equations and cancelling out those nucleiwhich appear on both sides, it is seen that four hydrogen atoms are consumed and, in return, 2 positrons, 3 .-rays and one heliumnucleus are created. In otherwords, hydrogen is burned and heliumis created. The overall processmay bewritten as 41H1 = 2He4 + 21e0 + 24.7 MeV The annihilation of positrons supplies an additional energy of 2MeVso that total energyreleased is 26.7 MeV. The fusion energy releasedmayalso be found by the loss ofmass during the above r eaction: 41H1 = 4 × 1.008144 = 4.032576 amu 2He4 = 4.003873 amu 21e0 = 2 × 0.000558 = 0.001115 amu . mass loss = 4.032576 (4.003873 + 0.001115) or .m = 0.028857 amu . energy released = 931 × 0.028857 = 26.7MeV It is worthnoting that the above energy release is less than that in nuclear fis sion. However, its value is 26.7/4 = 6.7MeVper nucleon as compared to less than 1MeVper nucleon in fission p rocess. Controlled Thermonuclear Reactions The fact that nuclear fusion reactions release large amounts of energy, as in st ars, has attracted much attention and continuous search is beingmade for finding practicalmeans or contr olled release of such energy. It has however, been found that reactions ofC-Ncycle and proton-proton c hain occur too slowly to be of any practical use.Other thermonuclear reactionswhich occurmuchmore rapi dlyand depend on aboundant hydrogen isotopes like deuteron (1D2 or 1H2) and tritium(1T3 or 1H3) a nd hence seemmore practicalproposition, are as under: (i) 1H2 + 1H2 = 2He3 + 0n1 + 3.3 MeV NUCLEAR PHYSICS NUCLEAR PHYSICS or 1D2+ 1D2 = 2He3+ 0n1+3.3MeV (ii) 1H2+ 1H2 = 1H3+ 1H1+4MeV or 1D2+ 1D2 = 1T3+ 1H1+4MeV (iii) 1H2 + 1H3 = 2He4 + 0n1 + 17.6 MeV or 1D2+ 1T3 = 2He4+ 0n1+17.6MeV The most important aspect of the above nuclear fusion reactions is that deuteriu m is available easily and aboundantly. It occurs in nature with an aboundance of one part in six thousands of hydrogen and can be separated fromthe lighter isotope quite cheaply. Five litres of water contain ab out 1/8 gram of deuterium but its energy content if it could be used as a fuel in a thermo-nuclear reactor , would be equivalent to 130 litres of petrol! The more than 5 × 1019 kg of water present in the oceans could t hus supply world s power requirement for severalmillion years at negligible cost ifthe deuteriumcould be utilized to provide energyby fusion reactions. However, as discussed below, there are some difficult problems to be solved before manmade controlled fusion reactors can become a reality. Condition for Controlled Fusion In order to provide useful energy, the fusion process must be self-sustaining. O nce the temperature of deuterium (or a mixture of D2 and T3) has been raised to the point at which fusi on occurs at an appreciable rate, the energyreleased must be sufficient, at least, to maintain that temperat ure. The minimumtemperature is known as critical ignition temperature and may be defined as that temperature above which the rate of energy production by fusion exceeds the rate of energy loss. Its value is about 5 keV (i.e. 50 million ºK) for a D-D-reaction. At these temperatures, the atoms are entirely stripped of their electrons. The result is a completely ionized gas or plasma consisting of atomic nuclei (like deuterons, tr itons and protons) and electrons in rapid random motion. It is practically impossible to contain such p lasma in walls of ordinary materials. In the Sun, the fusion reactions are contained bya tremendous gravita tional pressure. Such a high pressure is yet not available for controlled thermo-nuclear reactions on earth a lthough the plasma can be contained in a magnetic field. Hence, main problem is to devise an apparatus in which plasma can be obtained by meansofa magnetic field at the kinetic temperaturesrequired for the fusionreactions to proceed. Another necessary condition for a self-sustaining thermonuclear system is known as Lawson criterion. It is based on the requirement that in the operation of a fusion reactor, the total us efulrecoverable energyshould be at least sufficient to maintain the temperature of te reacting nuclei. Lawson criterion can be expressed in terms of the product nt where nis the number of reacting nuclei per m3 and t is th e time inseconds in which the thermonuclear reaction takes place. The minimum value of nt for D-T system is 7 × 1019 and D-D system is 2 × 1021. In controlled thermonuclear reactions, t is taken as the time during which the high- temperature plasma can be confined. It will be seen from above that both the critical ignition temperature and Lawso n criterion are much more favourablefor D-TsystemthanforD-Dsystem. Buttheformersystemhasthedraw-backthat i trequires tritium which has to be obtained by nuclear reactions (because it does not occur in nature). Tritium can, however, bemadebybombardinglithiumwithslowneutronsinareactorthus: 3Li6+ 0n1= 1H3+ 2He4+4.8MeV Thisreactioncanbemadetoservetwousefulpurposes.InthethermonuclearreactorusingD-T reaction, the escaping neutrons carry off much of the energy (about 14 MeV per neutron). T his energy can be converted into heat byslowing down these fast neutrons in a blanket of beryllium surrounding the reactor in whichplasmaisproduced. Theslowneutronsarecapturedbylithiumwhichproducestritium.T heblanket couldthusconsistofmoderator (i.e.beryllium),coolantandlithium.Theheatgeneratedby themoderation and absorption of neutrons could thus be transferred by coolant to external heat exchangers and then to turbines whichcould run alternators. www.physicsashok.in 45 00386 amu res pectively. provides the high temperature and pressure nece ssaryfor the successful working of the hydrogen bomb (fusion bomb). it is seen that 4 hydrogen atoms fuse to produce o .The essentialconditions for the operationof the hydrog en bombare extremely high temperatures and pressures required for the fusion to start.35 kW/m2 anthe earth i s 1. Thus the amount of fusiblematerialin a hydrogen bombis not limited.5 × 1011mfromthe Sun. the radioactive cobalt is pulver ised and converted into a gigantic radioactive cloudwhich canspread over thousands ofkilometres killing everything living in that area. If the activematerial in an atomic bombexceeds the critical size.e. t he fusionitselfmaintains the temperatures to keep the process going. Howmuch hydrogenmust be converted to heliumif solar constant is 1. involves combining oftwo lighter nuclei into on heavynucleus.Once started.Hydrogen bomb cannot explode unless ignited i. on the other hand. (ii) It hasno limitation of a critical size of the fusiblematerialunlike anatomi c bomb. it has no upper lim it. (i) Fission involves breaking up of a heavy nucleus into lighter nuclei. (ii) The links of the fission process are neutronswhile the links of a fusion pr ocess are protons. Fusion proceeds best withthermalparticleswhere thermalmeans temperatures ofmillions of ºK. Afusion bomb is superior to a fission bomb because of the following reasons: (i) The energy release in a hydrogen bomb is open-ended i. spontaneous explosion results.When the hydrogenbomb explodes.During explosion.00813 amu and 4. But there aremanydifferences inthemechanisms of the two processes.NUCLEAR PHYSICS www.e. byfirst exploding.Hydrogen bomb is based on the fusion if the hydrogen atoms into heavier ones by the therm onuclear reactionswith release ofenormous energy. It depends on howmuch fusiblematerial is present in the bomb.physicsashok. it gives off neutrons which act on the cobalt cover and render it intensivelyradioactive due to the fo rmation ofCo60 that is 300 timesmore powerfulthan radium. Fusion. the atombomb (fission bomb) is used as a primer which.in 46 Hydrogen bomb This bomb is 1000 timesmore powerfulthan the atomic bombwhich is based onnuclear fission. For this purpose. heated to critical ignition temperature and anyamount of fusi blematerial is safe until ignited. Example 80: the masses of 1H1 and 2He4 atoms are 1. (iii) Fission proceeds best with thermalneutronswhere thermalmeans roomtemperatu re. inits essentials as 41H1 = 2He4 + 1e0 Neglecting the two positrons.. Cobalt Bomb It consists of a hydrogen bomb which is encased in a sheath of metallic cobalt a nd is more lethal and destructive thana simple uncased hydrogenbomb. Sol: This thermo-nuclear reactionmay bewritten. Fission and Fusion: One thing common between the two nuclear processes si that they release very lar ge amounts of energy. 67 × 1.67 MeV = 6.002866 amu energy produced = 0. = 26.42 × 1014 J .02 × 1026 × 10.67 × 10 13 = 6.68 MeV This is the energy released when four hydrogen atoms fuse.03252 amu Mass ofone heliumatom = 4.03252 4. Mass of 4 hydrogen atoms = 4 × 1.00386 = 0. energy produce d by one hydrogen atom.00813 = 4. . Hence.02866 × 931 = 26.m = 4.68 4 = 6.6 × 10 13 = 10.ne atomof helium.67 × 10 13J = 6.00386 amu Decrease inmass. 01478 a. energy released per kg = 1. and mass of 2He4 = 4. Find also the ra te of consumption of heliumto maintain the radiative power of the star at 4 × 1021MW. Mass of hydrogen consumed is = 25 14 38.9 × 10 27 kg.015 amu. = 5. 3. Energy released per fusion =mass defect in amu = (3 × 4.Estimate the total energyreleased if 1 kg of deuteriumundergoes complete fusion. 2.724MeV No. Sun 1.00388 amu.27 1. deuteron.0000 amu. 2He3 + 0n1.27 MeV .004 × 931MeV = 3. Sol.017 + 1.02568 amu = 0.5 × 1011m.00388 + Q . In some stars.004 amu .NUCLEAR PHYSICS www. × (1.009) = 0.physicsashok.35 kW/m2 = 1. Surface area of the sphere = 4. .724MeV Energy released per deuteron = 1 2 × 3. three 2He4 nuclides fuse together to form6C12 ofmass 12.01 × 1026 MeV = 5.18 10 6.95 × 108 tones/second C69: Calculate the energyliberatedwhen aHeliumnucleus is formed bythe fusion of two deuteriumnuclei.009 amu.28 × 1022 = 38. . Sol. Sol: The reactionmaybewritten as 1H2 + 1H2 = 2He4 + Q . respectively. 9 × 1 C71.5 × 1011)2 = 28.m. Howmuch energy is released per fusion of 6C12 ? Rest mass of 2He4 = 4. Q = 0. heliumand the neut ron havemasses 2.18 × 1025 J/s It also represents the energy emitted by the Sun per second.42 10 .in 47 The solar constant represents the amount of energy received per second by 1m2 area held perpendicular to the Sun s rays at a distance equal to the mean distance of the earth fromthe Sun.6 × 10 19 .5 × 1011 Given value of solar constant = 1.862 × 3.9 MeV C70.u.28 × 1022 m2 .59 × 1011 kg/s = 5.02568 × 931 = 23.02 1026 2 .017 amu and 1.R2 = 4. Mass difference = 2 × 2. The mass 1H2 = 2.00263 12.002603 amu. energy received by this surface area per second is = 1350 × 28.6 10 19 106 n J .015 (3.35 × 1000 W/m3 = 1350 J/s/m2 Total energyemitted bythe Sunis equal to the energyreceived bythe inner surface of the imaginary sphere drawnwith Sun as centre and radius = 1. energy released by n atoms 7.mass of 2He4 atom= 6.0000) amu = 0. . energy released = 0.6 × 1026 × 106 × 1. of deuterons in 1 kg = 6.01478 = 4. In the fusion reaction 1H2 + 1H2 .007809 × 931MeV = 7.01478 + 2. 9 × 10 27 kg = 7. . n = 1.03163 × 1040 × 6.3 . . 4 × 1021 × 106 (given) .1 × 1013 kg .103163 × 1040 mass of heliumatoms burnt per second 1. = . . = . .