Solution Manual of Physics by Arthur Beiser

Inha University Department of PhysicsChapter 1. Problem Solutions 1. If the speed of light were smaller than it is, would relativistic phenomena be more or less conspicuous than they are now? 3. An athlete has learned enough physics to know that if he measures from the earth a time interval on a moving spacecraft, what he finds will be greater than what somebody on the spacecraft would measure. He therefore proposes to set a world record for the 100-m dash by having his time taken by an observer on a moving spacecraft. Is this a good idea? ¡ ¼Sol¡ ½ All else being the same, including the rates of the chemical reactions that govern our brains and bodies, relativisitic phenomena would be more conspicuous if the speed of light were smaller. If we could attain the absolute speeds obtainable to us in the universe as it is, but with the speed of light being smaller, we would be able to move at speeds that would correspond to larger fractions of the speed of light, and in such instances relativistic effects would be more conspicuous. ¡ ¼Sol¡ ½ Even if the judges would allow it, the observers in the moving spaceship would measure a longer time, since they would see the runners being timed by clocks that appear to run slowly compared to the ship's clocks. Actually, when the effects of length contraction are included (discussed in Section 1.4 and Appendix 1), the runner's speed may be greater than, less than, or the same as that measured by an observer on the ground. Inha University Department of Physics 5. Two observers, A on earth and B in a spacecraft whose speed is 2.00 x 10 8 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A's reckoning before the watches differ by 1.00 s? (b) To A, B's watch seems to run slow. To B, does A's watch seem to run fast, run slow, or keep the same time as his own watch? ¡ ¼Sol¡ ½ Note that the nonrelativistic approximation is not valid, as v/c = 2/3. (a) See Example 1.1. In Equation (1.3), with t representing both the time measured by A and the time as measured in A's frame for the clock in B's frame to advance by to, we need from which t = 3.93 s. (b) A moving clock always seems to run slower. In this problem, the time t is the time that observer A measures as the time that B's clock takes to record a time change of t o . s 00 1 255 0 3 2 1 1 1 1 2 2 2 0 . . · × · , _ ¸ ¸ , _ ¸ ¸ − − · , _ ¸ ¸ − − · − t t c v t t t Inha University Department of Physics 7. How fast must a spacecraft travel relative to the earth for each day on the spacecraft to correspond to 2 d on the earth? 9. A certain particle has a lifetime of 1.00 x10 -7 s when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created? ¡ ¼Sol¡ ½ From Equation (1.3), for the time t on the earth to correspond to twice the time t 0 elapsed on the ship’ s clock, ¡ ¼Sol¡ ½ The lifetime of the particle is t 0 , and the distance the particle will travel is, from Equation (1.3), m/s, 10 60 2 2 3 so 2 1 1 8 2 2 × · · · − . , c v c v relating three significant figures. m 210 99 0 1 s 10 00 1 m/s 10 0 3 99 0 1 2 7 8 2 2 0 · − × × · − · − ) . ( ) . )( . )( . ( / c v vt vt to two significant figures. Inha University Department of Physics 11. A galaxy in the constellation Ursa Major is receding from the earth at 15,000 km/s. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm, what is the corresponding wavelength measured by astronomers on the earth? ¡ ¼Sol¡ ½ See Example 1.3; for the intermediate calculations, note that , / / c v c v c c o o o + − · · · 1 1 λ ν ν ν ν λ where the sign convention for v is that of Equation (1.8), which v positive for an approaching source and v negative for a receding source. For this problem, , . . . 050 0 m/s 10 0 3 km/s 10 50 1 8 7 − · × × − · c v so that nm 578 050 0 1 050 0 1 nm 550 1 1 · − + · + − · . . ) ( / / c v c v o λ λ Inha University Department of Physics 13. A spacecraft receding from the earth emits radio waves at a constant frequency of 10 9 Hz. If the receiver on earth can measure frequencies to the nearest hertz, at what spacecraft speed can the difference between the relativistic and classical Doppler effects be detected? For the classical effect, assume the earth is stationary. ¡ ¼Sol¡ ½ This problem may be done in several ways, all of which need to use the fact that when the frequencies due to the classical and relativistic effects are found, those frequencies, while differing by 1 Hz, will both be sufficiently close to v o = 10 9 Hz so that v o could be used for an approximation to either. In Equation (1.4), we have v = 0 and V = -u, where u is the speed of the spacecraft, moving away from the earth (V < 0). In Equation (1.6), we have v = u (or v = -u in Equation (1.8)). The classical and relativistic frequencies, v c and v r respectively, are ) / ( ) / ( ) / ( ) / ( , ) / ( c u c u c u c u c u o o r c + − · + − · + · 1 1 1 1 1 2 0 ν ν ν ν ν The last expression for v o , is motivated by the derivation of Equation (1.6), which essentially incorporates the classical result (counting the number of ticks), and allows expression of the ratio . ) / ( 2 1 1 c u r c − · ν ν Inha University Department of Physics Use of the above forms for the frequencies allows the calculation of the ratio 9 9 2 10 Hz 10 Hz 1 1 1 1 − · · + − − · − · ∆ ) / ( ) / ( c u c u o r c o ν ν ν ν ν Attempts to solve this equation exactly are not likely to be met with success, and even numerical solutions would require a higher precision than is commonly available. However, recognizing that the numerator is of the form that can be approximated using the methods outlined at the beginning of this chapter, we can use . The denominator will be indistinguishable from 1 at low speed, with the result 2 1 1 ) / ( c u − − 2 2 2 1 1 1 ) / )( / ( ) / ( c u c u ≈ − − , 9 2 2 10 2 1 − · c u which is solved for km/s. 13.4 m/s 10 34 1 10 2 4 9 · × · × · − . c u Inha University Department of Physics 15. If the angle between the direction of motion of a light source of frequency v o and the direction from it to an observer is 0, the frequency v the observer finds is given by where v is the relative speed of the source. Show that this formula includes Eqs. (1.5) to (1.7) as special cases. ¡ ¼Sol¡ ½ The transverse Doppler effect corresponds to a direction of motion of the light source that is perpendicular to the direction from it to the observer; the angle θ = tπ/2 (or t90 o ), so cos θ = 0, and which is Equation (1.5). For a receding source, θ = π (or 180 o ), and cos θ = 1. The given expression becomes , / 2 2 1 c v o − · ν ν , / / / / c v c v c v c v o o + − · + − · 1 1 1 1 2 2 ν ν ν which is Equation (1.8). For an approaching source, θ = 0, cos θ = 1, and the given expression becomes , / / / / c v c v c v c v o o − + · − − · 1 1 1 1 2 2 ν ν ν which is Equation (1.8). θ ν ν cos ) / ( / c v c v o − − · 1 1 2 2 Inha University Department of Physics 17. An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.90c relative to the earth. What is his height as measured by an observer in the same spacecraft? By an observer on the earth? 19. How much time does a meter stick moving at 0.100c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion. ¡ ¼Sol¡ ½ The astronaut’ s proper length (height) is 6 ft, and this is what any observer in the spacecraft will measure. From Equation (1.9), an observer on the earth would measure ¡ ¼Sol¡ ½ The time will be the length as measured by the observer divided by the speed, or ft 6 2 90 0 1 ft 6 1 2 2 2 . ) . ( ) ( / · − · − · c v L L o s 10 32 3 m/s 10 0 3 100 0 100 0 1 m 00 1 1 8 8 2 2 2 − × · × − · − · · . ) . )( . ( ) . ( ) . ( / v c v L v L t o Inha University Department of Physics 21. A spacecraft antenna is at an angle of 10 o relative to the axis of the spacecraft. If the spacecraft moves away from the earth at a speed of 0.70c, what is the angle of the antenna as seen from the earth? ¡ ¼Sol¡ ½ If the antenna has a length L' as measured by an observer on the spacecraft (L' is not either L or L O in Equation (1.9)), the projection of the antenna onto the spacecraft will have a length L'cos(10 o ), and the projection onto an axis perpendicular to the spacecraft's axis will have a length L'sin(10 o ). To an observer on the earth, the length in the direction of the spacecraft's axis will be contracted as described by Equation (1.9), while the length perpendicular to the spacecraft's motion will appear unchanged. The angle as seen from the earth will then be . ) . ( ) t a n ( a r ct a n / ) cos( ) s in ( a r ct a n o o o o c v L L 14 70 0 1 10 1 10 10 2 2 2 · 1 1 ] 1 ¸ − · 1 1 ] 1 ¸ − ′ ′ The generalization of the above is that if the angle is 00 as measured by an observer on the spacecraft, an observer on the earth would measure an angle θ given by 2 2 1 c v o / t a n t a n − · θ θ Inha University Department of Physics 23. A woman leaves the earth in a spacecraft that makes a round trip to the nearest star, 4 light- years distant, at a speed of 0.9c. ¡ ¼Sol¡ ½ The age difference will be the difference in the times that each measures the round trip to take, or ( ) ( ) yr. 5 9 0 1 1 9 0 yr 4 2 1 1 2 2 2 2 · − − · − − · ∆ . . / c v v L t o 25. All definitions are arbitrary, but some are more useful than others. What is the objection to defining linear momentum as p = mv instead of the more complicated p = γmv? ¡ ¼Sol¡ ½ It is convenient to maintain the relationship from Newtonian mechanics, in that a force on an object changes the object's momentum; symbolically, F = dp/d t should still be valid. In the absence of forces, momentum should be conserved in any inertial frame, and the conserved quantity is p = -γmv, not mv 27. Dynamite liberates about 5.4 x 10 6 J/kg when it explodes. What fraction of its total energy content is this? ¡ ¼Sol¡ ½ For a given mass M, the ratio of the mass liberated to the mass energy is . . ) . ( ) . ( 11 2 8 6 10 0 6 m/s 10 0 3 J/kg 10 4 5 − × · × × × × M M Inha University Department of Physics 29. At what speed does the kinetic energy of a particle equal its rest energy? ¡ ¼Sol¡ ½ If the kinetic energy K = E o = mc 2 , then E = 2mc 2 and Equation (1.23) reduces to 2 1 1 2 2 · − c v / (γ = 2 in the notation of Section 1.7). Solving for v, m/s 10 60 2 2 3 8 × · · . c v 31. An electron has a kinetic energy of 0.100 MeV. Find its speed according to classical and relativistic mechanics. ¡ ¼Sol¡ ½ Classically, m/s. 10 88 1 kg 10 11 9 J/eV 10 60 1 MeV 200 0 2 2 8 31 19 × · × × × × · · − − . . . . e m K v Relativistically, solving Equation (1.23) for v as a function of K, . ) / ( 2 2 2 2 2 2 2 1 1 1 1 1 , _ ¸ ¸ + − · , _ ¸ ¸ + − · , _ ¸ ¸ − · c m K c K c m c m c E c m c v e e e e Inha University Department of Physics With K/(m e c 2 ) = (0.100 MeV)/(0.511 MeV) = 0.100/0.511, m/s. 10 64 1 511 0 100 0 1 1 1 m/s 10 0 3 8 2 8 × · , _ ¸ ¸ + − × × · . ) . / ( ) . ( . v The two speeds are comparable, but not the same; for larger values of the ratio of the kinetic and rest energies, larger discrepancies would be found. 33. A particle has a kinetic energy 20 times its rest energy. Find the speed of the particle in terms of c. ¡¼Sol¡ ½ Using Equation (1.22) in Equation (1.23) and solving for v/c, 2 1 , _ ¸ ¸ − · E E c v o With E = 21E o , that is, E = E o + 20E o , . . c c v 9989 0 21 1 1 2 · , _ ¸ ¸ − · Inha University Department of Physics 35. How much work (in MeV) must be done to increase the speed of an electron from 1.2 x 10 8 m/s to 2.4 X 10 8 m/s? ¡ ¼Sol¡ ½ The difference in energies will be, from Equation (1.23), MeV 294 0 0 3 2 1 1 1 0 3 4 2 1 1 MeV 511 0 1 1 1 1 2 2 2 2 1 2 2 2 2 . ) . / . ( ) . / . ( ) . ( / / · 1 1 ] 1 ¸ − − − · 1 1 ] 1 ¸ − − − c v c v c m e 37. Prove that ½γmv 2 , does not equal the kinetic energy of a particle moving at relativistic speeds. ¡ ¼Sol¡ ½ Using the expression in Equation (1.20) for the kinetic energy, the ratio of the two quantities is . / 1 1 ] 1 ¸ − − · , _ ¸ ¸ − · 2 2 2 2 2 2 2 2 1 1 1 1 2 1 1 2 1 c v c v c v K mv γ γ γ Inha University Department of Physics 39. An alternative derivation of the mass-energy formula E O = mc 2 , also given by Einstein, is based on the principle that the location of the center of mass (CM) of an isolated system cannot be changed by any process that occurs inside the system. Figure 1.27 shows a rigid box of length L that rests on a frictionless surface; the mass M of the box is equally divided between its two ends. A burst of electromagnetic radiation of energy E o is emitted by one end of the box. According to classical physics, the radiation has the momentum p = E o /c, and when it is emitted, the box recoils with the speed v ≈ E 0 1Mc so that the total momentum of the system remains zero. After a time t ≈ L/c the radiation reaches the other end of the box and is absorbed there, which brings the box to a stop after having moved the distance S. If the CM of the box is to remain in its original place, the radiation must have transferred mass from one end to the other. Show that this amount of mass is m = E O 1c 2 . ¡¼Sol¡ ½ Measured from the original center of the box, so that the original position of the center of mass is 0, the final position of the center of mass is . 0 2 2 2 2 · , _ ¸ ¸ − , _ ¸ ¸ + − , _ ¸ ¸ + , _ ¸ ¸ − S L m M S L m M Expanding the products and canceling similar terms [(M/2)(L/2), mS], the result MS = mL is obtained. The distance 5 is the product vt, where, as shown in the problem statement, v ≈ E/Mc (approximate in the nonrelativistic limit M >> Elc 2 ) and t ≈ L/c. Then, . 2 c E c L Mc E L M L MS m · · · Inha University Department of Physics 41. In its own frame of reference, a proton takes 5 min to cross the Milky Way galaxy, which is about 10 5 light-years in diameter. (a) What is the approximate energy of the proton in electronvolts?. (b) About how long would the proton take to cross the galaxy as measured by an observer in the galaxy's reference frame? ¡ ¼Sol¡ ½ To cross the galaxy in a matter of minutes, the proton must be highly relativistic, with v ≈ c (but v < c, of course). The energy of the proton will be E = E o γ, where E O is the proton's rest energy and . However, γ, from Equation (1.9), is the same as the ratio L O /L, where L is the diameter of the galaxy in the proton's frame of reference, and for the highly-relativistic proton L ≈ ct, where t is the time in the proton's frame that it takes to cross the galaxy. Combining, 2 2 1 1 c v / / − · γ eV 10 s/yr 10 3 s 300 ly 10 eV 10 19 7 5 9 · × × ≈ ≈ · · ) ( ) ( ) ( c ct L E L L E E E o o o o o γ 43. Find the momentum (in MeV/c) of an electron whose speed is 0.600c. ¡ ¼Sol¡ ½ Taking magnitudes in Equation (1.16), c c c c v v m p e / . ) . ( ) . )( / . ( / MeV 383 0 600 0 1 600 0 MeV 511 0 1 2 2 2 2 · − · − · Inha University Department of Physics 45. Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV ¡ ¼Sol¡ ½ When the kinetic energy of an electron is equal to its rest energy, the total energy is twice the rest energy, and Equation (1.24) becomes c c c c m p c p c m c m e e e / . ) / ( / ) ( , GeV 94 1 keV 511 3 3 or 4 2 2 2 4 4 4 4 · · · + · The result of Problem 1-29 could be used directly; γ = 2, v = ( /2)c, and Equation (1.17) gives p = m e c, as above. 3 3 47. Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.500 GeV ¡ ¼Sol¡ ½ Solving Equation (1.23) for the speed v in terms of the rest energy E O and the total energy E, c c E E c v o 963 0 500 3 938 0 1 1 2 . ) . / . ( ) / ( · − · − · numerically 2.888 x 10 8 m/s. (The result of Problem 1-32 does not give an answer accurate to three significant figures.) The value of the speed may be substituted into Equation (1.16) (or the result of Problem 1-46), or Equation (1.24) may be solved for the magnitude of the momentum, c c c c E c E p o / . ) / . ( ) / . ( ) / ( ) / ( GeV 37 3 GeV 938 0 GeV 500 3 2 2 2 2 · − · − · Inha University Department of Physics 49. A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. Find its mass (in MeV/c 2 ) and speed (as a fraction of c). ¡ ¼Sol¡ ½ From E = mc 2 + K and Equation (1.24), Expanding the binomial, cancelling the m 2 c 4 term, and solving for m, ( ) 2 2 4 2 2 2 c p c m K mc + · + . / ) ( ) ( ) ( ) ( 2 2 2 2 2 2 2 MeV 874 MeV 62 2 MeV 62 MeV 335 2 c c K c K pc m · − · − · The particle's speed may be found any number of ways; a very convenient result is that of Problem 1-46, giving . . c c K mc pc c E p c v 36 0 MeV 62 MeV 874 MeV 335 2 2 · + · + · · Inha University Department of Physics 51. An observer detects two explosions, one that occurs near her at a certain time and another that occurs 2.00 ms later 100 km away. Another observer finds that the two explosions occur at the, same place. What time interval separates the explosions to the second observer? ¡ ¼Sol¡ ½ The given observation that the two explosions occur at the same place to the second observer means that x' = 0 in Equation (1.41), and so the second observer is moving at a speed m/s 10 00 5 s 10 00 2 m 10 00 1 7 3 5 × · × × · · − . . . t x v with respect to the first observer. Inserting this into Equation (1.44), ms. 97 1 m/s) 10 (2.998 m/s 10 00 5 1 ms 00 2 1 1 1 1 2 8 2 7 2 2 2 2 2 2 2 2 2 2 2 . ) . ( ) . ( ) / ( / ) / ( · × × − · − · − − · − − · ′ c t x t t c x t c x t ct x tc x t t (For this calculation, the approximation is valid to three significant figures.) An equally valid method, and a good cheek, is to note that when the relative speed of the observers (5.00 x 10 7 m/s) has been determined, the time interval that the second observer measures should be that given by Equation (1.3) (but be careful of which time it t, which is to). Algebraically and numerically, the different methods give the same result. ) / ( ) / ( 2 2 2 2 2 1 1 t c x ct x − ≈ − Inha University Department of Physics 53. A spacecraft moving in the +x direction receives a light signal from a source in the xy plane. In the reference frame of the fixed stars, the speed of the spacecraft is v and the signal arrives at an angle θ to the axis of the spacecraft. (a) With the help of the Lorentz transformation find the angle θ ' at which the signal arrives in the reference frame of the spacecraft. (b) What would you conclude from this result about the view of the stars from a porthole on the side of the spacecraft? ¡ ¼Sol¡ ½ (a) A convenient choice for the origins of both the unprimed and primed coordinate systems is the point, in both space and time, where the ship receives the signal. Then, in the unprimed frame (given here as the frame of the fixed stars, one of which may be the source), the signal was sent at a time t = -r/c, where r is the distance from the source to the place where the ship receives the signal, and the minus sign merely indicates that the signal was sent before it was received. Take the direction of the ship's motion (assumed parallel to its axis) to be the positive x-direction, so that in the frame of the fixed stars (the unprimed frame), the signal arrives at an angle 0 with respect to the positive x-direction. In the unprimed frame, x = r cos θ and y = r sin θ . From Equation (1.41), , / ) / ( cos / ) / ( cos / 2 2 2 2 2 2 1 1 1 c v c v r c v c r r c v vt x x − + · − − − · − − · ′ θ θ and y’ = y = r sin θ. Then, Inha University Department of Physics 55. A man on the moon sees two spacecraft, A and B, coming toward him from opposite directions at the respective speeds of 0.800c and 0.900c. (a) What does a man on A measure for the speed with which he is approaching the moon? For the speed with which he is approaching B? (b) What does a man on B measure for the speed with which he is approaching the moon? For the speed with which he is approaching A ? ¡ ¼Sol¡ ½ (a) If the man on the moon sees A approaching with speed v = 0.800 c, then the observer on A will see the man in the moon approaching with speed v = 0.800c. The relative velocities will have opposite directions, but the relative speeds will be the same. The speed with which B is seen to approach A, to an observer in A, is then . ) / ( cos / , / / )) / ( (cos s in t a n 1 1 ] 1 ¸ + − · ′ − + · ′ ′ · ′ c v c v c v c v x y θ θ θ θ θ θ 2 2 2 2 1 sin arctan and 1 (b) From the form of the result of part (a), it can be seen that the numerator of the term in square brackets is less than sinθ , and the denominator is greater than cosθ , and so tan θ and θ’ < θ when v ≠ 0. Looking out of a porthole, the sources, including the stars, will appear to be in the directions close to the direction of the ship’ s motion than they would for a ship with v = 0. As v àc, θ’ à0, and all stars appear to be almost on the ship’ s axis(farther forward in the field of view). . . ) . )( . ( . . / c c c V v v V V x x x 988 0 900 0 800 0 1 900 0 800 0 1 2 · + + · ′ + + ′ · Inha University Department of Physics (b) Similarly, the observer on B will see the man on the moon approaching with speed 0.900 c, and the apparent speed of A, to an observer on B, will be . . ) . )( . ( . . c c 988 0 800 0 900 0 1 800 0 900 0 · + + (Note that Equation (1.49) is unchanged if V x ’ and v are interchanged.) B A O’ V x ’ v S’ (moon) S Inha University Department of Physics Chapter 2 Problem Solutions 1. If Planck's constant were smaller than it is, would quantum phenomena be more or less conspicuous than they are now? 3. Is it correct to say that the maximum photoelectron energy KE max is proportional to the frequency ν of the incident light? If not, what would a correct statement of the relationship between KE max and ν be? ¡ ¼Sol¡ ½ Planck’ s constant gives a measure of the energy at which quantum effects are observed. If Planck’ s constant had a smaller value, while all other physical quantities, such as the speed of light, remained the same, quantum effects would be seen for phenomena that occur at higher frequencies or shorter wavelengths. That is, quantum phenomena would be less conspicuous than they are now. ¡ ¼Sol¡ ½ No: the relation is given in Equation (2.8) and Equation (2.9), ), ( max o h h KE ν ν φ ν − · − · So that while KE max is a linear function of the frequency ν of the incident light, KE max is not proportional to the frequency. Inha University Department of Physics 5. Find the energy of a 700-nm photon. 7. A 1.00-kW radio transmitter operates at a frequency of 880 kHz. How many photons per second does it emit? ¡ ¼Sol¡ ½ The number of photons per unit time is the total energy per unit time(the power) divided by the energy per photon, or ¡ ¼Sol¡ ½ From Equation (2.11), eV. 77 1 m 10 700 m eV 10 24 1 9 - 6 . . · × ⋅ × · − E Or, in terms of joules, J 10 84 2 m 10 700 m/s) 10 s)(3.0 J 10 63 6 19 9 8 34 − − − × · × × ⋅ × · . . ( E . photons/s 10 72 1 Hz) 10 s)(880 J 10 63 6 J/s 10 00 1 30 3 34 3 × · × ⋅ × × · · − . . ( . ν h P E P Inha University Department of Physics 9. Light from the sun arrives at the earth, an average of 1.5 x 10 11 m away, at the rate of 1.4 x 10 3 W/m 2 of area perpendicular to the direction of the light. Assume that sunlight is monochromatic with a frequency of 5.0 x 10 14 Hz. (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? (b) What is the power output of the sun, and how many photons per second does it emit? (c) How many photons per cubic meter are there near the earth? ¡ ¼Sol¡ ½ (a) The number of photons per unit time per unit are will be the energy per unit time per unit area (the power per unit area, P/A), divided by the energy per photon, or ). . . / 2 21 14 34 - 2 3 m s photons/( 10 2 4 Hz) 10 s)(5.0 J 10 (6.63 W/m 10 4 1 ⋅ × · × ⋅ × × · ν h A P (b) With the reasonable assumption that the sun radiates uniformly in all directions, all points at the same distance from the sun should have the same flux of energy, even if there is no surface to absorb the energy. The total power is then, , . ) . ( ) . ( ) / ( W 10 0 4 m 10 5 1 4 W/m 10 4 1 4 26 2 11 2 3 2 × · × × · − π π S E R A P where R E-S is the mean Earth-Sun distance, commonly abbreviated as “1 AU,” for “astronomical unit.” The number of photons emitted per second is this power divided by the energy per photon, or . photons/s 10 2 1 Hz 10 0 5 s J 10 (6.63 J/s 10 0 4 45 14 34 - 26 × · × ⋅ × × . ) . )( . Inha University Department of Physics 11. The maximum wavelength for photoelectric emission in tungsten is 230 nm. What wavelength of light must be used in order for electrons with a maximum energy of 1.5 eV to be ejected? ¡ ¼Sol¡ ½ Expressing Equation (2.9) in terms of λ = c/ν and λ 0 = c/ν 0 , and performing the needed algebraic manipulations, (c) The photons are all moving at the same speed c, and in the same direction (spreading is not significant on the scale of the earth), and so the number of photons per unit time per unit area is the product of the number per unit volume and the speed. Using the result from part (a), . . . ) . 3 13 8 2 21 photons/m 10 4 1 m/s 10 0 3 m s photons/( 10 2 4 × · × ⋅ × nm. 180 m eV 10 24 1 m 10 230 eV 5 1 1 nm) 230 1 1 6 9 1 0 · 1 ] 1 ¸ ⋅ × × + · 1 ] 1 ¸ + · + · − − − − . ) )( . ( ( ) / ( max max hc K K hc hc o o λ λ λ λ Inha University Department of Physics 13. What is the maximum wavelength of light that will cause photoelectrons to be emitted from sodium? What will the maximum kinetic energy of the photoelectrons be if 200-nm light falls on a sodium surface? 15. 1.5 mWof 400-nm light is directed at a photoelectric cell. If 0.10 percent of the incident photons produce photoelectrons, find the current in the cell. ¡ ¼Sol¡ ½ The maximum wavelength would correspond to the least energy that would allow an electron to be emitted, so the incident energy would be equal to the work function, and ¡ ¼Sol¡ ½ Because only 0.10% of the light creates photoelectrons, the available power is (1.0x10 -3 )(1.5x10 -3 W) = 1.5x10 -6 W. the current will be the product of the number of photoelectrons per unit time and the electron charge, or nm 539 eV 3 2 m eV 10 24 1 6 · ⋅ × · · − . . max φ λ hc where the value of φ for sodium is taken from Table 2.1. From Equation (2.8), eV. 3.9 eV 3 2 m 10 200 m eV 10 24 1 9 6 · − × ⋅ × · − · − · − − . . max φ λ φ ν hc h K A 48 0 m eV 10 1.24 m 10 400 J/s 10 5 1 1 6 - 9 6 µ λ λ . ) )( . ( ) ( / · ⋅ × × × · · · · − − e hc P e hc P e E P e I Inha University Department of Physics 17. A metal surface illuminated by 8.5 x 10 14 Hz light emits electrons whose maximum energy is 0.52 eV The same surface illuminated by 12.0 x 10 14 Hz light emits electrons whose maximum energy is 1.97 eV From these data find Planck's constant and the work function of the surface . ¡ ¼Sol¡ ½ Denoting the two energies and frequencies with subscripts 1 and 2, . , max, max, φ ν φ ν − · − · 2 2 1 1 h K h K Subtracting to eliminate the work function φ and dividing by ν 1 - ν 2 , s eV 10 1 4 Hz 10 5 8 Hz 10 0 12 eV 52 0 eV 7 19 15 14 14 1 2 1 2 ⋅ × · × − × − · − − · − . . . . . max, max, ν ν K K h to the allowed two significant figures. Keeping an extra figure gives s J 10 6.64 s eV 10 14 4 34 - 15 ⋅ × · ⋅ × · − . h The work function φ may be obtained by substituting the above result into either of the above expressions relating the frequencies and the energies, yielding φ = 3.0 eV to the same two significant figures, or the equations may be solved by rewriting them as , , max, max, 1 1 2 1 2 2 2 1 2 1 φν ν ν ν φν ν ν ν − · − · h K h K subtracting to eliminate the product hν 1 ν 2 and dividing by ν 1 - ν 2 to obtain eV 0 3 Hz) 10 5 8 Hz 10 (12.0 Hz) 10 eV)(12.0 52 0 Hz) 10 eV)(8.5 7 19 14 14 14 14 1 2 2 1 1 2 . . . ( . ( max, max, · × − × × − × · − − · ν ν ν ν φ K K (This last calculation, while possibly more cumbersome than direct substitution, reflects the result of solving the system of equations using a symbolic-manipulation program; using such a program for this problem is, of course, a case of "swatting a fly with a sledgehammer".) Inha University Department of Physics 19. Show that it is impossible for a photon to give up all its energy and momentum to a free electron. This is the reason why the photoelectric effect can take place only when photons strike bound electrons. ¡ ¼Sol¡ ½ Consider the proposed interaction in the frame of the electron initially at rest. The photon's initial momentum is p o = E o /c, and if the electron were to attain all of the photon's momentum and energy, the final momentum of the electron must be p e = p o = p, the final electron kinetic energy must be KE = E o = pc, and so the final electron energy is E e = pc + m e c 2 . However, for any electron we must have E e 2 = (pc) 2 + (m e c 2 ) 2 . Equating the two expressions for E e 2 ( ) ( ) ( ) ( ) , ) ( ) ( ) ( 2 2 2 2 2 2 2 2 2 2 2 c m c m pc pc c m pc c m pc E e e e e e + + · + · + · ( ). ) ( 2 2 0 c m pc e · or This is only possible if p = 0, in which case the photon had no initial momentum and no initial energy, and hence could not have existed. To see the same result without using as much algebra, the electron's final kinetic energy is pc c m c m c p e e ≠ − + 2 4 2 2 2 for nonzero p. An easier alternative is to consider the interaction in the frame where the electron is at rest after absorbing the photon. In this frame, the final energy is the rest energy of the electron, m e c 2 , but before the interaction, the electron would have been moving (to conserve momentum), and hence would have had more energy than after the interaction, and the photon would have had positive energy, so energy could not be conserved. Inha University Department of Physics 21. Electrons are accelerated in television tubes through potential differences of about 10 kV. Find the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube. What kind of waves are these? ¡ ¼Sol¡ ½ For the highest frequency, the electrons will acquire all of their kinetic energy from the accelerating voltage, and this energy will appear as the electromagnetic radiation emitted when these electrons strike the screen. The frequency of this radiation will be Hz 10 4 2 s eV 10 14 4 V) 10 10 1 18 15 3 × · ⋅ × × · · · − . . )( ( e h eV h E ν which corresponds to x-rays. 23. The distance between adjacent atomic planes in calcite (CaCO 3 ) is 0.300 nm. Find the smallest angle of Bragg scattering for 0.030-nm x-rays. ¡ ¼Sol¡ ½ Solving Equation (2.13) for θ with n = 1, o d 9 2 nm 0.300 2 nm 030 0 2 . . ar cs in ar cs in · , _ ¸ ¸ × · , _ ¸ ¸ · λ θ Inha University Department of Physics 25. What is the frequency of an x-ray photon whose momentum is 1.1 x 10 -23 kg m/s? ¡ ¼Sol¡ ½ From Equation (2.15), Hz 10 0 5 s J 10 63 6 m/s) kg 10 m/s)(1.1 10 0 3 18 34 23 - 8 × · ⋅ × ⋅ × × · · − . . . ( h cp ν 27. In See. 2.7 the x-rays scattered by a crystal were assumed to undergo no change in wavelength. Show that this assumption is reasonable by calculating the Compton wavelength of a Na atom and comparing it with the typical x-ray wavelength of 0.1 nm. ¡ ¼Sol¡ ½ Following the steps that led to Equation (2.22), but with a sodium atom instead of an electron, m, 10 8 5 kg) 10 m/s)(3.82 10 (3.0 s J 10 63 6 17 26 - 8 34 − − × · × × ⋅ × · · . . , Na Na C cM h λ or 5.8 x 10 -8 nm, which is much less than o.1 nm. (Here, the rest mass M Na =3.82 x 10 -26 kg was taken from Problem 2-24.) Inha University Department of Physics 29. A beam of x-rays is scattered by a target. At 45 o from the beam direction the scattered x-rays have a wavelength of 2.2 pm. What is the wavelength of the x-rays in the direct beam? ¡ ¼Sol¡ ½ Solving Equation (2.23) for λ, the wavelength of the x-rays in the direct beam, pm 5 1 45 pm)(1 426 2 pm 2 2 1 . ) cos . ( . ) cos ( · − − · − − ′ · o C φ λ λ λ to the given two significant figures. 31. An x-ray photon of initial frequency 3.0 x 10 19 Hz collides with an electron and is scattered through 90 o . Find its new frequency. ¡ ¼Sol¡ ½ Rewriting Equation (2.23) in terms of frequencies, with λ = c/ν and λ’ = c/ν’ , and with cos 90 o = 0, C c c λ ν ν + · ′ and solving for ν’ gives Hz 10 4 2 m/s 10 0 3 m 10 43 2 Hz 10 0 3 1 1 19 1 8 12 19 1 × · 1 ] 1 ¸ × × + × · 1 ] 1 ¸ + · ′ − − − . . . . c C λ ν ν The above method avoids the intermediate calculation of wavelengths. Inha University Department of Physics 33. At what scattering angle will incident 100-keV x-rays leave a target with an energy of 90 keV? ¡ ¼Sol¡ ½ Solving Equation (2.23) for cos φ, 432 0 keV 90 keV 511 keV 100 keV 511 1 1 1 2 2 . cos · , _ ¸ ¸ − + · , _ ¸ ¸ ′ − + · ′ − + · E mc E mc C C λ λ λ λ φ from which φ = 64 o to two significant figures. 35. A photon of frequency ν is scattered by an electron initially at rest. Verify that the maximum kinetic energy of the recoil electron is KE max = (2h 2 ν 2 /mc 2 )/(1 + 2hν/mc 2 ). ¡ ¼Sol¡ ½ For the electron to have the maximum recoil energy, the scattering angle must be 180 0 , and Equation (2.20) becomes mc 2 KE max = 2 (hv) (h v'), where KE max = (hv - hv') has been used. To simplify the algebra somewhat, consider , ) / ( ) / ( ) / ( c C C νλ ν λ λ ν λ λ ν λ λ ν ν 2 1 2 1 1 + · + · ∆ + · ′ · ′ where ∆λ = 2λ C for φ = 180 o . With this expression, . ) / ( ) / ( ) ( ) )( ( max c mc h mc h h KE C νλ ν ν ν 2 1 2 2 2 2 2 + · ′ · Using λ C = h/(mc) (which is Equation (2.22)) gives the desired result. Inha University Department of Physics 37. A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron. If the electron moves off at an angle of 40 o with the original photon direction, what is the energy of the scattered photon? ¡ ¼Sol¡ ½ As presented in the text, the energy of the scattered photon is known in terms of the scattered angle, not the recoil angle of the scattering electron. Consider the expression for the recoil angle as given preceding the solution to Problem 2-25: . ) cos ( s in ) cos ( ) cos )( / ( s in ) cos ( ) / ( s in t a n φ λ λ φ φ φ λ λ φ φ λ λ φ θ − , _ ¸ ¸ + · − + − · − + ∆ · 1 1 1 1 1 C C For the given problem, with E = mc 2 , λ = hc/E = h/(mc) = λ C , so the above expression reduces to . ) cos ( s in t a n φ φ θ − · 1 2 At this point, there are many ways to proceed; a numerical solution with θ = 40 o gives φ = 61.6 0 to three significant figures. For an analytic solution which avoids the intermediate calculation of the scattering angle φ, one method is to square both sides of the above relation and use the trigonometric identity sin 2 φ = 1 - cos 2 φ = (1 + cos φ)(1 – cos φ) to obtain φ φ θ cos cos t a n − + · 1 1 4 2 (the factor 1 - cos φ may be divided, as cos φ = 1, φ = 0, represents an undeflected photon, and hence no interaction). This may be re-expressed as Inha University Department of Physics or 1 2 1 4 1 2 ), cos ( cos ) t a n )( cos ( φ φ θ φ − − · + · − . t a n t a n cos , t a n cos θ θ φ θ φ 2 2 2 4 1 4 3 2 4 1 2 1 + + · − + · − Then with λ’ = λ + λ C (1 – cos φ) = λ C (2 – cos φ), eV 335 40 4 3 40 4 1 keV) 511 4 3 4 1 2 2 2 2 · + + · + + · ′ · ′ ) ( t a n ) ( t a n ( t a n t a n o o E E E θ θ λ λ An equivalent but slightly more cumbersome method is to use the trigonometric identities 2 2 1 2 2 2 2 φ φ φ φ φ s in cos , cos s in s in · − · in the expression for tan θ to obtain , _ ¸ ¸ · · θ φ φ θ t a n ar ct an , cot t a n 2 1 2 2 2 1 yielding the result θ = 61.6 o more readily. Inha University Department of Physics 39. A positron collides head on with an electron and both are annihilated. Each particle had a kinetic energy of 1.00 MeV Find the wavelength of the resulting photons. 41. Show that, regardless of its initial energy, a photon cannot undergo Compton scattering through an angle of more than 60 o and still be able to produce an electron-positron pair. (Hint: Start by expressing the Compton wavelength of the electron in terms of the maximum photon wavelength needed for pair production.) ¡ ¼Sol¡ ½ The energy of each photon will he the sum of one particle's rest and kinetic energies, 1.511 MeV (keeping an extra significant figure). The wavelength of each photon will be ¡ ¼Sol¡ ½ Following the hint, pm 0.821 m 10 21 8 eV 10 51 1 m eV 10 24 1 13 6 6 · × · × ⋅ × · · − − . . . E hc λ , min E hc mc hc mc h C 2 2 2 2 · · · λ where E min = 2mc 2 is the minimum photon energy needed for pair production. The scattered wave- length (a maximum) corresponding to this minimum energy is λ’ max = (h/E min ), so λ C = 2λ’ max . At this point, it is possible to say that for the most energetic incoming photons, λ ~ 0, and so 1 - cos φ = ½ for λ ' = λ C /2, from which cos φ = ½and φ = 60 o . As an alternative, the angle at which the scattered photons will have wavelength λ’ max can m be found as a function of the incoming photon energy E; solving Equation (2.23) with λ ' = λ' max ) Inha University Department of Physics . / cos max max E mc E hc C C C 2 2 1 1 1 + · + ′ − · − ′ − · λ λ λ λ λ λ φ This expression shows that for E >> mc 2 , cos φ = ½and so φ = 60 o , but it also shows that, because cos φ must always be less than 1, for pair production at any angle, E must be greater than 2mc 2 , which we know to be the case. 43. (a) Show that the thickness x 1/2, of an absorber required to reduce the intensity of a beam of radiation by a factor of 2 is given by x 1/2 = 0.693/µ. (b) Find the absorber thickness needed to produce an intensity reduction of a factor of 10. ¡ ¼Sol¡ ½ (a) The most direct way to get this result is to use Equation (2.26) with I o /I = 2, so that . . ln / µ µ µ 693 0 2 2 1 · · ⇒ · − x e I I x o (b) Similarly, with I o /I = 10, . . ln / µ µ 30 2 10 10 1 · · x Inha University Department of Physics 45. The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m -1 . find the thickness of lead required to reduce by half the intensity of a beam of such gamma rays. ¡ ¼Sol¡ ½ From either Equation (2.26) or Problem 2-43 above, mm 9 8 m 78 693 0 2 1 - 2 1 . . ln / · · · µ x 47. The linear absorption coefficients for 2.0-MeV gamma rays are 4.9 m -1 in water and 52 in -1 in lead. What thickness of water would give the same shielding for such gamma rays as 10 mmof lead? ¡ ¼Sol¡ ½ Rather than calculating the actual intensity ratios, Equation (2.26) indicates that the ratios will be the same when the distances in water and lead are related by m 106 0 m 9 4 m 52 m 10 10 or 1 - 1 - 3 O H Pb Pb O H Pb Pb O H O H 2 2 2 2 . . ) ( , · × · · · − µ µ µ µ x x x x or 11 cm two significant figures. Inha University Department of Physics 49. What thickness of copper is needed to reduce the intensity of the beam in Exercise 48 by half. ¡ ¼Sol¡ ½ Either a direct application of Equation (2.26) or use of the result of Problem 2-43 gives m, 10 47 1 m 10 7 4 2 5 1 - 4 2 1 − × · × · . . ln / x which is 0.015 mm to two significant figures. 51. The sun's mass is 2.0 x 10 30 kg and its radius is 7.0 x 10 8 m. Find the approximate gravitational red shift in light of wavelength 500 nm emitted by the sun. ¡ ¼Sol¡ ½ In Equation (2.29), the ratio 6 1 - 4 2 8 30 2 11 2 10 12 2 m 10 0 7 m/s) 10 0 3 kg) 10 0 2 kg m N 10 67 6 − − × · × × × ⋅ × · . ) . ( . ( . )( / . ( R c GM (keeping an extra significant figure) is so small that for an “approximate” red shift, the ratio ∆λ/λ will be the same as ∆ν/ν, and pm. 1.06 m 10 1.06 ) 10 m)(2.12 10 500 12 - 6 - 9 2 · × · × × · · ∆ − ( R c GM λ λ Inha University Department of Physics 53. As discussed in Chap. 12, certain atomic nuclei emit photons in undergoing transitions from "excited" energy states to their “ground” or normal states. These photons constitute gamma rays. When a nucleus emits a photon, it recoils in the opposite direction. (a) The nucleus decays by K capture to , which then emits a photon in losing 14.4 keV to reach its ground state. The mass of a atom is 9.5 x 10 -26 kg. By how much is the photon energy reduced from the full 14.4 keV available as a result of having to share energy and momentum with the recoiling atom? (b) In certain crystals the atoms are so tightly bound that the entire crystal recoils when a gamma-ray photon is emitted, instead of the individual atom. This phenomenon is known as the Mössbauer effect. By how much is the photon energy reduced in this situation if the ex- cited 2576Fe nucleus is part of a 1.0-g crystal? (c) The essentially recoil-free emission of gamma rays in situations like that of b means that it is possible to construct a source of virtually mono- energetic and hence monochromatic photons. Such a source was used in the experiment described in See. 2.9. What is the original frequency and the change in frequency of a 14.4-keV gamma-ray photon after it has fallen 20 m near the earth's surface? Co 57 27 Fe 57 26 Fe 57 26 ¡ ¼Sol¡ ½ (a) The most convenient way to do this problem, for computational purposes, is to realize that the nucleus will be moving nonrelativistically after the emission of the photon, and that the energy of the photon will be very close to E ∞ = 14.4 keV, the energy that the photon would have if the nucleus had been infinitely massive. So, if the photon has an energy E, the recoil momentum of the nucleus is E/c, and its kinetic energy is , here M is the rest mass of the nucleus. Then, conservation of energy implies ) / ( / 2 2 2 2 2 Mc E M p · Inha University Department of Physics . ∞ · + E E Mc E 2 2 2 This is a quadratic in E, and solution might be attempted by standard methods, but to find the change in energy due to the finite mass of the nucleus, and recognizing that E will be very close to E ∞ , the above relation may be expressed as eV. 0 1 1.9 keV 10 9 1 m/s) 10 kg)(3.0 10 2(9.5 J/keV) 10 60 1 keV) 4 14 2 2 3 6 2 8 26 - 16 2 2 2 2 2 × · × · × × × · ≈ · − − ∞ ∞ . . ( . ( Mc E Mc E E E If the approximation E ≈ E ∞ , is not made, the resulting quadratic is , 0 2 2 2 2 2 · − + ∞ E Mc E Mc E which is solved for . 1 ] 1 ¸ − + · ∞ 1 2 1 2 2 Mc E Mc E However, the dimensionless quantity E ∞ /(Mc 2 ) is so small that standard calculators are not able to determine the difference between E and E ∞ . The square root must be expanded, using (1 + x) 1/2 ≈ 1 + (x/2) - (x 2 /8), and two terms must be kept to find the difference between E and E ∞ . This approximation gives the previous result. Inha University Department of Physics It so happens that a relativistic treatment of the recoiling nucleus gives the same numerical result, but without intermediate approximations or solution of a quadratic equation. The relativistic form expressing conservation of energy is, with pc = E and before, . ) ( , ) ( E E Mc Mc E E Mc E Mc E − + · + + · + + ∞ ∞ 2 2 2 2 2 2 2 2 or Squaring both sides, canceling E 2 and (Mc 2 ) 2 , and then solving for E, . )) / ( ( )) / ( ( ) ( , _ ¸ ¸ + + · + + · ∞ ∞ ∞ ∞ ∞ ∞ 2 2 2 2 2 1 2 1 2 2 Mc E Mc E E E Mc E Mc E E From this form, , ) / ( 2 2 2 1 1 2 Mc E Mc E E E ∞ ∞ ∞ + , _ ¸ ¸ · − givin g t h e s a me r es u lt . (b) For t h is s it u a t ion , t h e a bove r es u lt a pplies , bu t t h e n on r ela t ivis t ic a ppr oxima t ion is by fa r t h e ea s ies t for ca lcu la t ion ; eV. 10 8 1 m/s) 10 kg)(3.0 10 2(1.0 J/eV) 10 6 1 eV 10 4 14 2 25 2 8 3 - 19 2 3 2 2 − − ∞ ∞ × · × × × × · · − . . ( ) . ( Mc E E E (c) Th e or igin a l fr equ en cy is Hz. 10 48 3 s eV 10 14 4 eV 10 4 14 18 15 3 × · ⋅ × × · · − ∞ . . . h E ν From Equation (2.28), the change in frequency is Hz. 6 7 Hz) 10 48 3 m/s) 10 (3.0 m 20 m/s 8 9 18 2 8 2 2 . . ( ) )( . ( · × × · , _ ¸ ¸ · − ′ · ∆ ν ν ν ν c gH Inha University Department of Physics 55. The gravitational potential energy U relative to infinity of a body of mass m at a distance R from the center of a body of mass M is U = -GmM/ R. (a) If R is the radius of the body of mass M, find the escape speed v, of the body, which is the minimum speed needed to leave it permanently. (b) Obtain a formula for the Schwarzschild radius of the body by setting v c = c, the speed of light, and solving for R. (Of course, a relativistic calculation is correct here, but it is interesting to see what a classical calculation produces.) ¡ ¼Sol¡ ½ (a) To leave the body of mass M permanently, the body of mass m must have enough kinetic energy so that there is no radius at which its energy is positive. That is, its total energy must be nonnegative. The escape velocity v e is the speed (for a given radius, and assuming M >> m) that the body of mass m would have for a total energy of zero; . , R GM v R GMm mv e e 2 or 0 2 1 2 · · − (b) Solving the above expression for R in terms of v e , , 2 2 e v GM R · and if v e = c, Equation (2.30) is obtained. Inha University Department of Physics Chapter 3. Problem Solutions 1. A photon and a particle have the same wavelength. Can anything be said about how their linear momenta compare? About how the photon's energy compares with the particle's total energy? About how the photon’ s energy compares with the particle's kinetic energy? ¡ ¼Sol¡ ½ From Equation (3.1), any particle’ s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. With a common momentum p, the photon’ s energy is pc, and the particle’ s energy is , which is necessarily greater than pc for a massive particle. The particle’ s kinetic energy is 2 2 2 ) ( ) ( mc pc + ( ) ( ) 2 2 2 2 2 mc mc pc mc E K − + · − · For low values of p (p<<mc for a nonrelativistic massive particle), the kinetic energy is K ≈ p 2 /2m, which is necessarily less than pc. For a relativistic massive particle, K ≈ pc – mc 2 , and K is less than the photon energy. The kinetic energy of a massive particle will always be less than pc, as can be seen by using E = (pc) 2 + (mc 2 ) 2 to obtain . ) ( 2 2 2 2Kmc K pc · − Inha University Department of Physics Chapter 3. Problem Solutions 3. Find the de Broglie wavelength of a 1.0-mg grain of sand blown by the wind at a speed of 20 m/s. ¡ ¼Sol¡ ½ For this nonrelativistic case, m; 10 3 3 m/s) kg)(20 10 0 1 s J 10 63 6 29 6 34 − − − × · × ⋅ × · · . . ( . mv h λ quantum effects certainly would not be noticed for such an object. 5. By what percentage will a nonrelativistle calculation of the de Broglie wavelength of a 100-keV electron be in error? ¡ ¼Sol¡ ½ Because the de Broglie wavelength depends only on the electron's momentum, the percentage error in the wavelength will be the same as the percentage error in the reciprocal of the momentum, with the nonrelativistic calculation giving the higher wavelength due to a lower calculated momentum. The nonrelativistic momentum is s, m kg 10 71 1 J/eV) 10 eV)(1.6 10 kg)(100 10 1 9 2 2 22 19 - 3 31 / . . ( ⋅ × · × × × · · − − mK p nr and the relativistic momentum is ( ) ( ) m/s, kg 10 79 1 MeV 511 0 100 0 1 22 2 2 2 2 2 ⋅ × · + · − + · − . / ) . ( . ( c mc mc K c p r Inha University Department of Physics Chapter 3. Problem Solutions 7. The atomic spacing in rock salt, NaCl, is 0.282 nm. Find the kinetic energy (in eV) of a neutron with a de Broglie wavelength of 0.282 nm. Is a relativistic calculation needed? Such neutrons can be used to study crystal structure. ¡ ¼Sol¡ ½ A nonrelativistic calculation gives keeping extra figures in the intermediate calculations. The percentage error in the computed de Broglie wavelength is then % . . . . . / ) / ( ) / ( 8 4 71 1 71 1 79 1 · − · − · − nr nr r r r nr p p p p h p h p h ( ) eV 10 03 1 m) 10 eV)(0.282 10 6 939 2 m) eV 10 24 1 2 2 2 3 2 9 - 6 2 6 2 2 2 2 2 2 − − × · × × ⋅ × · · · · . . ( . ( ) ( / λ λ mc hc mc hc m p K (Note that in the above calculation, multiplication of numerator and denominator by c 2 and use of the product hc in terms of electronvolts avoided further unit conversion.) This energy is much less than the neutron's rest energy, and so the nonrelativistic calculation is completely valid. Inha University Department of Physics Chapter 3. Problem Solutions 9. Green light has a wavelength of about 550 nm. Through what potential difference must an electron be accelerated to have this wavelength? 11. Show that if the total energy of a moving particle greatly exceeds its rest energy, its de Broglie wavelength is nearly the same as the wavelength of a photon with the same total energy. ¡ ¼Sol¡ ½ If E 2 = (pc) 2 + (mc 2 ) 2 >> (mc 2 ) 2 , then pc >> mc 2 and E ≈ pc. For a photon with the same energy, E = pc, so the momentum of such a particle would be nearly the same as a photon with the same energy, and so the de Broglie wavelengths would be the same. ¡ ¼Sol¡ ½ A nonrelativistic calculation gives eV, 10 0 5 m) 10 eV)(550 10 511 2 m) eV 10 24 1 2 2 2 6 2 9 - 3 2 6 2 2 2 2 2 2 − − × · × × ⋅ × · · · · . ( . ( ) ( ) ( ) / ( λ λ mc hc mc hc m p K so the electron would have to be accelerated through a potential difference of 5.0 x 10 -6 V = 5.0 µV. Note that the kinetic energy is very small compared to the electron rest energy, so the nonrelativistic calculation is valid. (In the above calculation, multiplication of numerator and denominator by c 2 and use of the product he in terms of electronvolts avoided further unit conversion.) Inha University Department of Physics Chapter 3. Problem Solutions 13. An electron and a proton have the same velocity Compare the wavelengths and the phase and group velocities of their de Broglie waves. 15. Verify the statement in the text that, if the phase velocity is the same for all wavelengths of a certain wave phenomenon (that is, there is no dispersion), the group and phase velocities are the same. ¡ ¼Sol¡ ½ Suppose that the phase velocity is independent of wavelength, and hence independent of the wave number k ; then, from Equation (3.3), the phase velocity v p = (ω/k ) = u, a constant. It follows that because ω = uk , ¡ ¼Sol¡ ½ For massive particles of the same speed, relativistic or nonrelativistic, the momentum will be proportional to the mass, and so the de Broglie wavelength will be inversely proportional to the mass; the electron will have the longer wavelength by a factor of (m p /m e ) = 1838. From Equation (3.3) the particles have the same phase velocity and from Equation (3.16) they have the same group velocity. . p g v u dk d v · · · ω Inha University Department of Physics Chapter 3. Problem Solutions 17. The phase velocity of ocean waves is , where g is the acceleration of gravity. Find the group velocity of ocean waves π λ 2 / g ¡¼Sol¡ ½ The phase velocity may be expressed in terms of the wave number k = 2π/λ as . , gk gk k g k v p · · · · 2 or or ω ω ω Finding the group velocity by differentiating ω(k ) with respect to k , . p g v k k g k g dk d v 2 1 2 1 2 1 1 2 1 · · · · · ω ω Using implicit differentiation in the formula for ω 2 (k ), , g v dk d g · · ω ω ω 2 2 so that , p g v k k k gk g v 2 1 2 2 2 2 2 · · · · · ω ω ω ω ω the same result. For those more comfortable with calculus, the dispersion relation may be expressed as ), ln ( ) ln ( ) ln ( g k + · ω 2 from which . , p g v k v k dk d 2 1 2 1 and 2 · · · ω ω ω Inha University Department of Physics Chapter 3. Problem Solutions 21. (a) Show that the phase velocity of the de Broglie waves of a particle of mass m and de Broglie wavelength λ is given by 2 1 , _ ¸ ¸ + · h mc c v p λ (b) Compare the phase and group velocities of an electron whose de Broglie wavelength is exactly 1 x 10 -13 m. ¡ ¼Sol¡ ½ (a) Two equivalent methods will be presented here. Both will assume the validity of Equation (3.16), in that v g = v. First: Express the wavelength x in terms of v g , 19. Find the phase and group velocities of the de Broglie waves of an electron whose kinetic energy is 500 keV. ¡ ¼Sol¡ ½ For a kinetic energy of 500 keV, . . / 978 1 511 511 500 1 1 2 2 2 2 · + · + · − · mc mc K c v γ Solving for v, , . ) . / ( ) / ( c c c v 863 0 978 1 1 1 1 1 2 2 · − · − · γ and from Equation (3.16), v g = v = 0.863c. The phase velocity is then v p = c 2 /v g = 1.16 c. . 2 2 1 c v mv h mv h p h g g g − · · · γ λ Inha University Department of Physics Multiplying by mv g , squaring and solving for v g 2 gives . ) / ( ) ( 1 2 2 2 2 2 2 2 1 − 1 1 ] 1 ¸ , _ ¸ ¸ + · + · h c m c c h m h v g λ λ Taking the square root and using Equation (3.3), v p = c 2 /v g , gives the desired result. Second: Consider the particle energy in terms of v p = c 2 lv g ; ( ) ( ) ( ) ( ) . / ) ( 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 mc hc v c mc mc mc pc E p + , _ ¸ ¸ · − · + · λ γ Dividing by (mc 2 ) 2 leads to that so 1 1 1 2 2 2 2 , ) / ( λ mc h v c p + · − , / ) ( ) ( ) ( ) / ( 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 h mc mc h mc h mc h v c p λ λ λ λ + · + · + · − which is an equivalent statement of the desired result. It should be noted that in the first method presented above could be used to find λ in terms of v p directly, and in the second method the energy could be found in terms of v g . The final result is, or course, the same. Inha University Department of Physics (b) Using the result of part (a), , . . . ( c c v p 00085 1 s J 10 63 6 m) 10 m/s)(1.0 10 kg)(3.0 10 1 9 1 2 34 13 - 8 31 · , _ ¸ ¸ ⋅ × × × × + · − − and v g = c 2 /v p = 0.99915c. For a calculational shortcut, write the result of part (a) as . . . ( c c hc mc c v p 00085 1 m eV 10 24 1 m) 10 eV)(1.00 10 511 1 1 2 6 13 - 3 2 2 · , _ ¸ ¸ ⋅ × × × + · , _ ¸ ¸ + · − λ In both of the above answers, the statement that the de Broglie wavelength is “exactly” 10 -13 m means that the answers can be given to any desired precision. 23. What effect on the scattering angle in the Davisson-Germer experiment does increasing the electron energy have? ¡ ¼Sol¡ ½ Increasing the electron energy increases the electron's momentum, and hence decreases the electron's de Broglie wavelength. From Equation (2.13), a smaller de Broglie wavelength results in a smaller scattering angle. Inha University Department of Physics Chapter 3. Problem Solutions 25. In Sec. 3.5 it was mentioned that the energy of an electron entering a crystal increase, which reduces its de Broglie wavelength. Consider a beam of 54-eV electrons directed at a nickel target. The potential energy of an electron that enters the target changes by 26 eV. (a) Compare the electron speeds outside and inside the target. (b) Compare the respective de Broglie wavelengths. ¡ ¼Sol¡ ½ (a) For the given energies, a nonrelativistic calculation is sufficient; m/s 36 4 kg 10 1 9 J/eV) 10 eV)(1.60 54 2 2 31 19 - . . ( · × × · · − m K v outside the crystal, and (from a similar calculation, with K = 80 eV), v = 5.30 x 10 6 m/s inside the crystal (keeping an extra significant figure in both calculations). (b) With the speeds found in part (a), the de Brogile wavelengths are found from or 0.167 nm outside the crystal, with a similar calculation giving 0.137 nm inside the crystal. m, 10 67 1 m/s) 10 kg)(4.36 10 11 9 s J 10 63 6 10 6 31 34 − − − × · × × ⋅ × · · · . . ( . mv h p h λ Inha University Department of Physics Chapter 3. Problem Solutions 27. Obtain an expression for the energy levels (in MeV) of a neutron confined to a one-dimensional box 1.00 x 10 -14 m wide. What is the neutron's minimum energy? (The diameter of an atomic nucleus is of this order of magnitude.) 29. A proton in a one-dimensional box has an energy of 400 keV in its first excited state. How wide is the box? ¡ ¼Sol¡ ½ The first excited state corresponds to n = 2 in Equation (3.18). Solving for the width L, ¡ ¼Sol¡ ½ From Equation (3.18), MeV. 5 20 J 10 28 3 m) 10 kg)(1.00 10 67 1 8 s) J 10 63 6 8 2 13 2 2 14 - 27 2 34 2 2 2 2 . . . ( . ( n n n mL h n E n · × · × × ⋅ × · · − − − The minimum energy, corresponding to n = 1, is 20.5 MeV fm. 45.3 m 10 53 4 J/eV) 10 eV)(1.60 10 kg)(400 10 67 1 8 s) J 10 63 6 2 8 14 19 - 3 27 2 34 2 2 · × · × × × ⋅ × · · − − − . . ( . ( mE h n L Inha University Department of Physics Chapter 3. Problem Solutions 31. The atoms in a solid possess a certain minimum zero-point energy even at 0 K, while no such restriction holds for the molecules in an ideal gas. Use the uncertainty principle to explain these statements. 33. The position and momentum of a 1.00-keV electron are simultaneously determined. If its position is located to within 0.100 nm, what is the percentage of uncertainty in its momentum? ¡ ¼Sol¡ ½ Each atom in a solid is limited to a certain definite region of space - otherwise the assembly of atoms would not be a solid. The uncertainty in position of each atom is therefore finite, and its momentum and hence energy cannot be zero. The position of an ideal-gas molecule is not restricted, so the uncertainty in its position is effectively infinite and its momentum and hence energy can be zero. ¡ ¼Sol¡ ½ The percentage uncertainty in the electron's momentum will be at least %. 1 3 10 1 3 eV) 10 eV)(1.00 10 2(511 m) 10 00 1 4 m) eV 10 24 1 2 4 2 4 4 2 3 3 10 6 2 . . . ( . ( ) ( · × · × × × ⋅ × · ∆ · ∆ · ∆ · ∆ − − − π π π π K mc x hc mK x h x p h p p Note that in the above calculation, conversion of the mass of the electron into its energy equivalent in electronvolts is purely optional; converting the kinetic energy into joules and using h = 6.626 x 10 -34 J· s will of course give the same percentage uncertainty. Inha University Department of Physics Chapter 3. Problem Solutions 35. How accurately can the position of a proton with v << c be determined without giving it more than 1.00 keV of kinetic energy? ¡ ¼Sol¡ ½ The proton will need to move a minimum distance , E h v t v ∆ ≥ ∆ π 4 where v can be taken to be that so 2 2 , m E m K v ∆ · · pm. 0.144 m 10 44 1 eV) 10 MeV)(1.00 10 938 2 2 m eV 10 24 1 2 2 2 2 4 2 13 3 6 6 2 · × · × × ⋅ × · · · ∆ · ∆ − − . ( . ) ( π π π π K mc hc mK h E h m K t v (See note to the solution to Problem 3-33 above). The result for the product v∆t may be recognized as v∆t ≥ h/ 2πp; this is not inconsistent with Equation (3.21), ∆x ∆p ≥ h/ 4π . In the current problem, ∆E was taken to be the (maximum) kinetic energy of the proton. In such a situation, ( ) , p v p m p m p E ∆ · ∆ · ∆ · ∆ 2 2 2 which is consistent with the previous result. Inha University Department of Physics 37. A marine radar operating at a frequency of 9400 MHz emits groups of electromagnetic waves 0.0800 µs in duration. The time needed for the reflections of these groups to return indicates the distance to a target. (a) Find the length of each group and the number of waves it contains. (b) What is the approximate minimum bandwidth (that is, spread of frequencies) the radar receiver must be able to process? ¡ ¼Sol¡ ½ (a) The length of each group is m. 24 s) 10 m/s)(8.0 10 0 3 5 - 8 · × × · ∆ . ( t c The number of waves in each group is the pulse duration divided by the wave period, which is the pulse duration multiplied by the frequency, waves. 752 Hz) 10 s)(4900 10 0 8 6 8 · × × − . ( (b) The bandwidth is the reciprocal of the pulse duration, ( ) MHz. 5 12 s 10 0 8 1 - 8 . . · × − Inha University Department of Physics Chapter 3. Problem Solutions . / / π ν 2 m C · 39. The frequency of oscillation of a harmonic oscillator of mass m and spring constant C is The energy of the oscillator is E = p 2 /2m + Cx 2 /2, where p is its momentum when its displacement from the equilibrium position is x. In classical physics the minimum energy of the oscillator is E min = 0. Use the uncertainty principle to find an expression for E in terms of x only and show that the minimum energy is actually E min = hν/2 by setting d E/d x = 0 and solving for E min . ¡ ¼Sol¡ ½ To use the uncertainty principle, make the identification of p with ∆p and x with ∆x, so that p = h/ (4πx), and . ) ( 2 2 2 2 2 1 8 x C x m h x E E , _ ¸ ¸ + , _ ¸ ¸ · · π Differentiating with respect to x and setting , 0 · E dx d , 0 1 4 3 2 2 · + , _ ¸ ¸ − Cx x m h π which is solved for . mC h x π 2 2 · Substution of this value into E(x) gives . m in 2 2 2 2 2 8 2 2 ν π π π π h m C h mC h C h mC m h E · · , _ ¸ ¸ , _ ¸ ¸ + , _ ¸ ¸ , _ ¸ ¸ · Inha University Department of Physics Chapter 4. Problem Solutions 1. The great majority of alpha particles pass through gases and thin metal foils with no deflections. To what conclusion about atomic structure does this observation lead? 3. Determine the distance of closest approach of 1.00-MeV protons incident on gold nuclei. ¡ ¼Sol¡ ½ The fact that most particles pass through undetected means that there is not much to deflect these particles; most of the volume of an atom is empty space, and gases and metals are overall electrically neutral. ¡ ¼Sol¡ ½ For a "closest approach", the incident proton must be directed "head-on" to the nucleus, with no angular momentum with respect to the nucleus (an "Impact parameter" of zero; see the Appendix to Chapter 4). In this case, at the point of closest approach the proton will have no kinetic energy, and so the potential energy at closest approach will be the initial kinetic energy, taking the potential energy to be zero in the limit of very large separation. Equating these energies, m. 10 14 1 J 10 60 1 C) 10 60 1 79 C m N 10 99 8 4 1 or 4 13 13 2 19 2 2 9 initial 2 2 initial − − − × · × × ⋅ × · , _ ¸ ¸ · · . . . )( ( ) / . ( , m in m in K Ze r r Ze K o o πε πε Inha University Department of Physics 5. What is the shortest wavelength present in the Brackett series of spectral lines? 7. In the Bohr model, the electron is in constant motion. How can such an electron have a negative amount of energy? ¡ ¼Sol¡ ½ The wavelengths in the Brackett series are given in Equation (4.9); the shortest wavelength (highest energy) corresponds to the largest value of n . For n →∞, ¡ ¼Sol¡ ½ While the kinetic energy of any particle is positive, the potential energy of any pair of particles that are mutually attracted is negative. For the system to be bound, the total energy, the sum of the positive kinetic energy and the total negative potential energy, must be negative. For a classical particle subject to an inverse-square attractive force (such as two oppositely charged particles or two uniform spheres subject to gravitational attraction in a circular orbit, the potential energy is twice the negative of the kinetic energy. m 1.46 m 10 46 1 m 10 097 1 16 16 6 1 - 7 µ λ · × · × · → − . . R Inha University Department of Physics 9. The fine structure constant is defined as α = e 2 /2ε o hc. This quantity got its name because it first appeared in a theory by the German physicist Arnold Sommerfeld that tried to explain the fine structure in spectral lines (multiple lines close together instead of single lines) by assuming that elliptical as well as circular orbits are possible in the Bohr model. Sommerfeld's approach was on the wrong track, but α has nevertheless turned out to be a useful quantity in atomic physics. (a) Show that α = v 1 /c, where v, is the velocity of the electron in the ground state of the Bohr atom. (b) Show that the value of α is very close to 1/137 and is a pure number with no dimensions. Because the magnetic behavior of a moving charge depends on its velocity, the small value of α is representative of the relative magnitudes of the magnetic and electric aspects of electron behavior in an atom. (c) Show that αa o = λ c /2π, where a o is the radius of the ground-state Bohr orbit and λ c is the Compton wavelength of the electron. ¡ ¼Sol¡ ½ (a) The velocity v, is given by Equation (4.4), with r = r 1 = a o . Combining to find v 1 2 , . , α ε ε π ε πε πε · · · , _ ¸ ¸ · · c h e c v h e me h m e ma e v o o o o o o 1 2 so 4 4 4 2 1 2 2 4 2 2 2 2 2 1 (b) From the above, ( ) ( )( )( ) , . . . / . . 3 8 34 2 2 12 2 19 10 30 7 m/s 10 00 3 s J 10 63 6 m N C 10 85 8 2 C 10 60 1 − − − − × · × ⋅ × ⋅ × × · α Inha University Department of Physics so that 1/α = 137.1 to four significant figures. A close cheek of the units is worthwhile; treating the units as algebraic quantities the units as given in the above calculation are . ] ][ [ ] ] 1 [J] m] [N [s] [m] s J ] [N][m [C [C 2 2 2 · ⋅ · Thus, α is a dimensionless quantity, and will have the same numerical value in any system of units. The most accurate (November, 2001) value of 1/α is , .03599976 137 1 · α accurate to better than 4 parts per billion. (c) Using the above expression for α and Equation (4.13) with n = 1 for a o , , π λ π π ε ε α 2 2 1 2 2 2 2 C o o o mc h me h hc e a · · · where the Compton wavelength λ C is given by Equation (2.22). Inha University Department of Physics 11. Find the quantum number that characterizes the earth's orbit around the sun. The earth's mass is 6.0 x 10 24 kg, its orbital radius is 1.5 x 10 11 m, and its orbital speed is 3.0 x 10 4 m/s. 13. Compare the uncertainty in the momentum of an electron confined to a region of linear dimension a o with the momentum of an electron in a ground-state Bohr orbit. ¡ ¼Sol¡ ½ With the mass, orbital speed and orbital radius of the earth known, the earth's orbital angular momentum is known, and the quantum number that would characterize the earth's orbit about the sun would be this angular momentum divided by ; ¡ ¼Sol¡ ½ The uncertainty in position of an electron confined to such a region is, from Equation (3.22), ∆p > /2a o , while the magnitude of the linear momentum of an electron in the first Bohr orbit is . . . . ( 74 34 11 4 24 10 6 2 s J 10 06 1 m) 10 m/s)(1.5 10 kg)(3.0 10 0 6 × · ⋅ × × × × · · · − h h mvR L n (The number of significant figures not of concern.) ; o o a a h h p h · · · π λ 2 the value of ∆p found from Equation (3.13) is half of this momentum. Inha University Department of Physics 15. What effect would you expect the rapid random motion of the atoms of an excited gas to have on the spectral lines they produce? ¡ ¼Sol¡ ½ The Doppler effect shifts the frequencies of the emitted light to both higher and lower frequencies to produce wider lines than atoms at rest would give rise to. 17. A proton and an electron, both at rest initially, combine to form a hydrogen atom in the ground state. A single photon is emitted in this process. What is its wavelength? ¡ ¼Sol¡ ½ It must assumed that the initial electrostatic potential energy is negligible, so that the final energy of the hydrogen atom is E 1 = -13.6 eV. The energy of the photon emitted is then -E l , and the wavelength is nm, 91.2 m 10 12 9 eV 6 13 m eV 10 24 1 8 6 1 · × · ⋅ × · − · − − . . . E hc λ in the ultraviolet part of the spectrum (see, for instance, the back endpapers of the text). Inha University Department of Physics 19. Find the wavelength of the spectral line that corresponds to a transition in hydrogen from the n = 10 state to the ground state. In what part of the spectrum is this? ¡ ¼Sol¡ ½ From either Equation (4.7) with n = 10 or Equation (4.18) with n f = 1 and n i = 10, nm, 92.1 m 10 21 9 m 10 097 1 1 99 100 1 99 100 8 1 - 7 · × · × · · − . . R λ which is in the ultraviolet part of the spectrum (see, for instance, the back endpapers of the text). 21. A beam of electrons bombards a sample of hydrogen. Through what potential difference must the electrons have been accelerated if the first line of the Balmer series is to be emitted? ¡ ¼Sol¡ ½ The electrons’ energy must be at least the difference between the n = 1 and n = 3 levels, eV 1 12 9 8 eV) 6 13 9 1 1 1 1 3 . . ( · · , _ ¸ ¸ − − · − · ∆ E E E E (this assumes that few or none of the hydrogen atoms had electrons in the n = 2 level). A potential difference of 12.1 eV is necessary to accelerate the electrons to this energy. Inha University Department of Physics 23. The longest wavelength in the Lyman series is 121.5 nm and the shortest wavelength in the Balmer series is 364.6 nm. Use the figures to find the longest wavelength of light that could ionize hydrogen. ¡ ¼Sol¡ ½ The energy needed to ionize hydrogen will be the energy needed to raise the energy from the ground state to the first excited state plus the energy needed to ionize an atom in the second excited state; these are the energies that correspond to the longest wavelength (least energetic photon) in the Lyman series and the shortest wavelength (most energetic photon) in the Balmer series. The energies are proportional to the reciprocals of the wavelengths, and so the wavelength of the photon needed to ionize hydrogen is nm. 13 91 nm 6 364 1 nm 5 121 1 1 1 1 1 2 1 2 . . . · , _ ¸ ¸ + · , _ ¸ ¸ + · − − → ∞ → λ λ λ As a check, note that this wavelength is R -1 . 25. An excited hydrogen atom emits a photon of wavelength λ in returning to the ground state. (a) Derive a formula that gives the quantum number of the initial excited state in terms of λ and R. (b) Use this formula to find n i for a 102.55-nm photon. ¡ ¼Sol¡ ½ (a) From Equation (4.7) with n = n i , , , _ ¸ ¸ − · 2 1 1 1 i n R λ which is solved for Inha University Department of Physics . / 1 1 1 2 1 − · , _ ¸ ¸ − · − R R R n i λ λ λ (b) Either of the above forms gives n very close (four place) to 3; specifically, with the product λR = (102.55x10 -9 m)(1.097x10 7 m -1 ) = 1.125 rounded to four places as 9/8, n = 3 exactly. 27. When an excited atom emits a photon, the linear momentum of the photon must be balanced by the recoil momentum of the atom. As a result, some of the excitation energy of the atom goes into the kinetic energy of its recoil. (a) Modify Eq. (4.16) to include this effect. (b) Find the ratio between the recoil energy and the photon energy for the n = 3 → n = 2 transition in hydrogen, for which E f - E i = 1.9 eV. Is the effect a major one? A nonrelativistic calculation is sufficient here. ¡ ¼Sol¡ ½ (a) A relativistic calculation would necessarily involve the change in mass of the atom due to the change in energy of the system. The fact that this mass change is too small to measure (that is, the change is measured indirectly by measuring the energies of the emitted photons) means that a nonrelativistic calculation should suffice. In this situation, the kinetic energy of the recoiling atom is , ) / ( M c h M p K 2 2 2 2 ν · · where m is the ftequency of the emitted photon and p = h/λ = hν/ c is the magnitude of the momentum of both the photon and the recoiling atom. Equation (4.16) is then Inha University Department of Physics . ) ( , _ ¸ ¸ + · + · + · − 2 2 2 2 1 2 Mc h h Mc h h K h E E f i ν ν ν ν ν This result is equivalent to that of Problem 2-53, where h ν = E ∞ . and the term p 2 /(2M) corresponds to E ∞ - E in that problem. As in Problem 2-53, a relativistic calculation is manageable; the result would be , , _ ¸ ¸ , _ ¸ ¸ + + · − −1 2 1 2 1 1 ν ν h Mc h E E i f a form not often useful; see part (b). (b) As indicated above and in the problem statement, a nonrelativistle calculation is sufficient. As in part (a), ( ) ( ) , . . , / 9 6 2 2 2 10 01 1 eV 10 939 2 eV 9 1 2 and 2 2 − × · × · ∆ · ∆ ∆ · · Mc E E K M c E M p K or 1.0 x 10 -9 to two significant figures. In the above, the rest energy of the hydrogen atom is from the front endpapers. Inha University Department of Physics 29. Show that the frequency of the photon emitted by a hydrogen atomin going from the level n + 1 to the level n is always intermediate between the frequencies of revolution of the electron in the respective orbits. ¡ ¼Sol¡ ½ There are many equivalent algebraic methods that may be used to derive Equation (4.19), and that result will be cited here; . 3 1 1 2 n h E f n − · The frequency v of the photon emitted in going from the level n + 1 to the level n is obtained from Equation (4.17) with n i = n + 1 and n f = n; . ) ( ) ( 1 ] 1 ¸ + + − · 1 ] 1 ¸ − + · ∆ · 2 2 2 1 1 2 2 1 2 1 1 1 n n n h E n n h E ν This can be seen to be equivalent to the expression for v in terms of n and p that was found in the derivation of Equation (4.20), but with n replaced by n + 1 and p = 1. Note that in this form, ν is positive because E l is negative. From this expression , n n f n n n n f n n n n hn E < 1 1 ] 1 ¸ + + + · 1 1 ] 1 ¸ + + + − · 1 2 1 2 2 2 2 1 2 2 2 1 2 3 1 ν as the term in brackets is less than 1. Similarly, Inha University Department of Physics , ) )( ( ) )( ( ) ( 1 2 2 1 1 2 2 1 3 1 1 1 1 2 + + > 1 ] 1 ¸ + + · 1 ] 1 ¸ + + + − · n n f n n n f n n n n h E ν as the term in brackets is greater than 1. 31. A µ − muon is in the n = 2 state of a muonic atom whose nucleus is a proton. Find the wavelength of the photon emitted when the muonic atom drops to its ground state. In what part of the spectrum is this wavelength? ¡ ¼Sol¡ ½ For a muonic atom, the Rydberg constant is multiplied by the ratio of the reduced masses of the muoninc atom and the hydrogen atom, R' = R (m'/m e ) = 186R, as in Example 4.7; from Equation (4.7), nm, 0.653 m 10 53 6 m 10 097 1 186 3 4 3 4 10 1 - 7 · × · × · ′ · − . ) . ( / / R λ in the x-ray range. e p e m m m m 1836 207 · · , µ e p p m m m m m m 186 · + · ′ µ µ Inha University Department of Physics 33. A mixture of ordinary hydrogen and tritium, a hydrogen isotope whose nucleus is approximately 3 times more massive than ordinary hydrogen, is excited and its spectrum observed. How far apart in wavelength will the H α lines of the two kinds of hydrogen be? ¡ ¼Sol¡ ½ The H α lines, corresponding to n = 3 in Equation (4.6), have wavelengths of λ = (36/5) (1/R). For a tritium atom, the wavelength would be λ T = (36/5) (1/RT), where RT is the Rydberg constant evaluated with the reduced mass of the tritium atom replacing the reduced mass of the hydrogen atom. The difference between the wavelengths would then be The values of R and RT are proportional to the respective reduced masses, and their ratio is . 1 ] 1 ¸ − · 1 ] 1 ¸ − · − · ∆ T T T R R 1 1 λ λ λ λ λ λ λ . ) ( ) ( ) / ( ) / ( H e T T e H T e T e H e H e T m m m m m m m m m m m m m m R R + + · + + · Using this in the above expression for ∆λ, , ) ( ) ( H e H e e H T e m m m m m m m m 3 2 λ λ λ ≈ 1 ] 1 ¸ + − · ∆ where the approximations m e + rn H ≈ m H and m T ≈ 3m H have been used. Inserting numerical values, nm. 0.238 m 10 38 2 kg) 10 67 1 3 kg) 10 11 9 2 m 10 097 1 5 36 10 27 31 1 - 7 · × · × × × · ∆ − − − . . ( . ( ) . ( ) / ( λ Inha University Department of Physics 35. (a) Derive a formula for the energy levels of a hydrogenic atom, which is an ion such as He + or Li 2+ whose nuclear charge is +Ze and which contains a single electron. (b) Sketch the energy levels of the He' ion and compare them with the energy levels of the H atom. (c) An electron joins a bare helium nucleus to form a He + ion. Find the wavelength of the photon emitted in this process if the electron is assumed to have had no kinetic energy when it combined with the nucleus. ¡ ¼Sol¡ ½ (a) The steps leading to Equation (4.15) are repeated, with Ze 2 instead of e 2 and Z 2 e 4 instead of e 4 , giving , 2 2 2 4 2 1 8 n h e Z m E o n πε ′ − · where the reduced mass m' will depend on the mass of the nucleus. (b) A plot of the energy levels is given below. The scale is close, but not exact, and of course there are many more levels corresponding to higher n. In the approximation that the reduced masses are the same, for He + , with Z = 2, the n = 2 level is the same as the n = 1 level for Hydrogen, and the n = 4 level is the same as the n = 2 level for hydrogen. Inha University Department of Physics The energy levels for H and He + : (c) When the electron joins the Helium nucleus, the electron-nucleus system loses energy; the emitted photon will have lost energy ∆E = 4 (-13.6 eV) = -54.4 eV, where the result of part (a) has been used. The emitted photon's wavelength is nm. 22.8 m 10 28 2 eV 4 54 m eV 10 24 1 8 6 · × · ⋅ × · ∆ − · − − . . . E hc λ Inha University Department of Physics 39. The Rutherford scattering formula fails to agree with the data at very small scattering angles. Can you think of a reason? ¡ ¼Sol¡ ½ Small angles correspond to particles that are not scattered much at all, and the structure of the atom does not affect these particles. To these nonpenetrating particles, the nucleus is either partially or completely screened by the atom's electron cloud, and the scattering analysis, based on a pointlike positively charged nucleus, is not applicable. 37. A certain ruby laser emits 1.00-J pulses of light whose wavelength is 694 nm. What is the minimum number of Cr 3+ ions in the ruby? ¡ ¼Sol¡ ½ The minimum number of Cr 3+ ions will he the minimum number of photons, which is the total energy of the pulse divided by the energy of each photon, ions. 10 49 3 m/s) 10 s)(3.0 J 10 63 6 m) 10 J)(694 00 1 18 8 34 9 - × · × ⋅ × × · · − . . ( . ( / hc E hc E λ λ Inha University Department of Physics 43. What fraction of a beam of 7.7-MeV alpha particles incident upon a gold foil 3.0 x 10 -7 m thick is scattered by less than 1 o ? ¡ ¼Sol¡ ½ The fraction scattered by less than 1 o is 1 - f, with f given in Equation (4.31); 41. A 5.0-MeV alpha particle approaches a gold nucleus with an impact parameter of 2.6 x 10 -13 m. Through what angle will it be scattered? ¡ ¼Sol¡ ½ From Equation (4.29), using the value for 1/4πε o given in the front endpapers, 11.43, m) 10 6 2 C) 10 60 1 79 C m N 10 (8.99 J/MeV) 10 eV)(1.60 0 5 2 13 2 19 2 2 9 13 - · × × ⋅ × × · − − . ( . )( )( / . ( cot θ keeping extra significant figures. The scattering angle is then . . t a n ) . ( cot o 10 43 11 1 2 43 11 2 1 1 · , _ ¸ ¸ · · − − θ Inha University Department of Physics 45. Show that twice as many alpha particles are scattered by a foil through angles between 60 o and 90 o as are scattered through angles of 90 o or more. ¡ ¼Sol¡ ½ Regarding f as a function of 0 in Equation (4.31), the number of particles scattered between 60 o and 90 o is f (60 o ) - f (90 o ), and the number scattered through angles greater than 90 o is just f (90 o ), and , ) ( cot ) ( cot ) ( cot ) ( ) ( ) ( 2 1 1 3 45 45 30 90 90 60 2 2 2 · − · − · − o o o o o o f f f so twice as many particles are scattered between 60 o and 90 o than are scattered through angles greater than 90 o . , . ) . ( cot . ( . )( ( ) / . )( . ( cot cot 16 0 5 0 J/MeV) 10 MeV)(1.6 7 7 C) 10 6 1 79 C m N 10 m)(9.0 10 0 3 m 10 90 5 2 4 1 2 4 2 2 13 - 2 19 2 2 2 9 7 3 - 28 2 2 2 2 2 2 2 · , _ ¸ ¸ × × × ⋅ × × × · , _ ¸ ¸ , _ ¸ ¸ · , _ ¸ ¸ · − − o o o K Ze nt K Ze nt f π θ πε π θ πε π where n, the number of gold atoms per unit volume, is from Example 4.8. The fraction scattered by less than 1 o is 1 - f = 0.84. Inha University Department of Physics 47. In special relativity, a photon can be thought of as having a “mass” of m = E ν /c 2 . This suggests that we can treat a photon that passes near the sun in the same way as Rutherford treated an alpha particle that passes near a nucleus, with an attractive gravitational force replacing the repulsive electrical force. Adapt Eq. (4.29) to this situation and find the angle of deflection θ for a photon that passes b = R sun from the center of the sun. The mass and radius of the sun are respectively 2.0 x 10 30 kg and 7.0 x 10 8 m. In fact, general relativity shows that this result is exactly half the actual deflection, a conclusion supported by observations made during solar clipses as mentioned in Sec. 1.10. ¡ ¼Sol¡ ½ If gravity acted on photons as if they were massive objects with mass m = E v /c 2 , the magnitude of the force F in Equation (4.28) would be ; 2 r m GM F s un · the factors of r 2 would cancel, as they do for the Coulomb force, and the result is , cot cos s in s un s un GM b c m GM b mc 2 2 2 and 2 2 2 2 · · θ θ θ a result that is independent of the photon’ s energy. Using b = R sun , . . deg . ) . )( / . ( t a n t a n 7 8 0 10 43 2 m) 10 m/s)(7.0 10 (3.0 kg 10 0 2 kg m N 10 67 6 2 2 4 8 8 30 2 2 11 1 2 1 ′ ′ · × · , _ ¸ ¸ × × × ⋅ × · , _ ¸ ¸ · − − − − s un s un R c GM θ Inha University Department of Physics Chapter 5 Problem Solutions 1. Which of the wave functions in Fig. 5.15 cannot have physical significance in the interval shown? Why not? 3. Which of the following wave functions cannot be solutions of Schrödinger's equation for all values of x? Why not? (a) ψ =A sec x; (b) ψ = A tan x; (c) ψ = A exp(x 2 ); (d) ψ = A exp(-x 2 ). ¡¼Sol¡ ½ Figure (b) is double valued, and is not a function at all, and cannot have physical significance. Figure (c) has discontinuous derivative in the shown interval. Figure (d) is finite everywhere in the shown interval. Figure (f) is discontinuous in the shown interval. ¡ ¼Sol¡ ½ The functions (a) and (b) are both infinite when cos x = 0, at x = ±π/2, ±3π/2, …±(2n+1)π/2 for any integer n, neither ψ = A sec x or ψ = A tan x could be a solution of Schrödinger's equation for all values of x. The function (c) diverges as x → ±∞, and cannot be a solution of Schrödinger's equation for all values of x. Inha University Department of Physics 5. The wave function of a certain particle is ψ = A cos 2 x for -π/2 < x < π /2. (a) Find the value of A. (b) Find the probability that the particle be found between x = 0 and x = π/4. ¡ ¼Sol¡ ½ Both parts involve the integral ∫cos 4 xdx , evaluated between different limits for the two parts. Of the many ways to find this integral, including consulting tables and using symbolic- manipulation programs, a direct algebraic reduction gives [ ] [ ] [ ] , cos cos ) cos ( cos ) ( cos cos ) cos ( ) (cos cos x x x x x x x x x 4 2 4 1 2 2 1 2 2 2 1 2 1 8 1 2 1 8 3 2 1 4 1 2 4 1 2 2 1 2 2 4 + + · + + + · + + · + · · where the identity cos 2 θ = ½(1+cos 2θ) has been used twice. (a) The needed normalization condition is [ ] 1 4 2 2 2 2 2 2 2 8 1 2 1 8 3 2 2 2 4 2 2 2 · + + · · ∫ ∫ ∫ ∫ ∫ + − + − + − + − + − ∗ / / / / / / / / / / cos cos cos π π π π π π π π π π ψ ψ xdx xdx dx A xdx A dx The integrals 2 2 4 1 2 2 2 2 2 1 2 2 4 4 and 2 2 / / / / / / / / s in cos s in cos π π π π π π π π + − + − + − + − · · ∫ ∫ x dx x x dx x are seen to be vanish, and the normalization condition reduces to . , π π 3 8 or 8 3 1 2 · , _ ¸ ¸ · A A Inha University Department of Physics (b) Evaluating the same integral between the different limits, [ ] , s in s in cos / / 4 1 32 3 4 2 4 0 32 1 4 1 8 3 4 0 4 + · + + · ∫ π π π x x x dx x The probability of the particle being found between x = 0 and x = π/4 is the product of this integral and A 2 , or ( ) ( ) 46 0 4 1 32 3 3 8 4 1 32 3 2 . · + · + π π π A 7. As mentioned in Sec. 5.1, in order to give physically meaningful results in calculations a wave function and its partial derivatives must be finite, continuous, and single-valued, and in addition must be normalizable. Equation (5.9) gives the wave function of a particle moving freely (that is, with no forces acting on it) in the +x direction as ) )( / ( pc Et i Ae − − · Ψ h where E is the particle's total energy and p is its momentum. Does this wave function meet all the above requirements? If not, could a linear superposition of such wave functions meet these requirements? What is the significance of such a superposition of wave functions? ¡ ¼Sol¡ ½ The given wave function satisfies the continuity condition, and is differentiable to all orders with respect to both t and x, but is not normalizable; specifically, Ψ ∗ Ψ = A * A is constant in both space and time, and if the particle is to move freely, there can be no limit to its range, and so the integral of Ψ ∗ Ψ over an infinite region cannot be finite if A ≠ 0. Inha University Department of Physics A linear superposition of such waves could give a normalizable wave function, corresponding to a real particle. Such a superposition would necessarily have a non- zero ∆p, and hence a finite ∆x; at the expense of normalizing the wave function, the wave function is composed of different momentum states, and is localized. 9. Show that the expectation values <px> and <xp>) are related by <px> - <xp> = / i This result is described by saying that p and x do not commute, and it is intimately related to the uncertainty principle. ¡ ¼Sol¡ ½ It's crucial to realize that the expectation value <p x> is found from the combined operator , which, when operating on the wave function Ψ(x, t), corresponds to "multiply by x, differentiate with respect to x and multiply by /i," whereas the operator corresponds to "differentiate with respect to x, multiply by /i and multiply by x." Using these operators, x p ˆ ˆ p x ˆ ˆ , ) ( ) ˆ ( ˆ ) ˆ ˆ ( 1 ] 1 ¸ Ψ ∂ ∂ + Ψ · Ψ ∂ ∂ · Ψ · Ψ x x i x x i x p x p h h where the product rule for partial differentiation has been used. Also, . ) ˆ ( ˆ ) ˆ ˆ ( 1 ] 1 ¸ Ψ ∂ ∂ · , _ ¸ ¸ Ψ ∂ ∂ · Ψ · Ψ x x i x i x p x p x h h Inha University Department of Physics Thus Ψ · Ψ − i p x x p h ) ˆ ˆ ˆ ˆ ( and i dx i dx i xp px h h h · Ψ Ψ · Ψ Ψ >· − < ∫ ∫ ∞ ∞ − ∞ ∞ − * * for Ψ(x, t) normalized. 11. Obtain Schrödinger’ s steady-state equation from Eq.(3.5) with the help of de Broglie’ s relation- ship λ = h/ mv by letting y = ψ and finding ∂ 2 ψ/ ∂x 2 . ¡¼Sol¡ ½ Using λν = v p in Equation (3.5), and using ψ instead of y , . cos cos , _ ¸ ¸ − · , _ ¸ ¸ , _ ¸ ¸ − · λ π πν π ψ x t A v x t A p 2 2 2 Differentiating twice with respect to x using the chain rule for partial differentiation (similar to Example 5.1), , s in s in , _ ¸ ¸ − · , _ ¸ ¸ − , _ ¸ ¸ − − · ∂ ∂ λ π πν λ π λ π λ π πν ψ x t A x t A x 2 2 2 2 2 2 ψ λ π λ π πν λ π λ π λ π πν λ π ψ 2 2 2 2 2 2 2 2 2 2 2 2 , _ ¸ ¸ − · , _ ¸ ¸ − , _ ¸ ¸ · , _ ¸ ¸ − , _ ¸ ¸ − · ∂ ∂ x t A x t A x cos cos Inha University Department of Physics The kinetic energy of a nonrelativistic particle is ) ( , U E h m m h m p U E KE − · , _ ¸ ¸ · · − · 2 2 2 2 2 1 that so 2 1 2 λ λ Substituting the above expression relating ψ λ ψ 2 2 2 1 and x ∂ ∂ , ) ( ) ( ψ ψ π ψ λ π ψ U E m U E h m x − − · − − · , _ ¸ ¸ − · ∂ ∂ 2 2 2 2 2 2 2 8 2 h which is Equation (5.32) Inha University Department of Physics 13. One of the possible wave functions of a particle in the potential well of Fig. 5.17 is sketched there. Explain why the wavelength and amplitude of &P vary as they do. ¡ ¼Sol¡ ½ The wave function must vanish at x = 0, where V →∞. As the potential energy increases with x, the particle's kinetic energy must decrease, and so the wavelength increases. The amplitude increases as the wavelength increases because a larger wavelength means a smaller momentum (indicated as well by the lower kinetic energy), and the particle is more likely to be found where the momentum has a lower magnitude. The wave function vanishes again where the potential V →∞; this condition would determine the allowed energies. Inha University Department of Physics 15. An important property of the eigenfunctions of a system is that they are orthogonal to one another, which means that m n dV m n ≠ · ∫ ∞ + ∞ − 0 ψ ψ Verify this relationship for the eigenfunctions of a particle in a one-dimensional box given by Eq. (5.46). ¡ ¼Sol¡ ½ The necessary integrals are of the form dx L x m L x n L dx L m n ∫ ∫ · ∞ + ∞ − 0 2 π π ψ ψ s in s in for integers n, m, with n ≠ m and n ≠ -m. (A more general orthogonality relation would involve the integral of ψ n * ψ m , but as the eigenfunctions in this problem are real, the distinction need not be made.) To do the integrals directly, a convenient identity to use is )], cos ( ) [cos ( s in s in β α β α β α + − − · 2 1 as may be verified by expanding the cosines of the sum and difference of α and β. To show orthogonality, the stipulation n ≠ m means that α ≠ β and α ≠ -β and the integrals are of the form Inha University Department of Physics , ) ( s in ) ( ) ( s in ) ( ) ( cos ) ( cos 0 1 0 · 1 ] 1 ¸ + + − − − · 1 ] 1 ¸ + − − · ∫ ∫ ∞ + ∞ − L o L m n L x m n m n L L x m n m n L dx L x m n L x m n L dx π π π π π π ψ ψ where sin(n - m)π = sin(n - m)π = sin 0 = 0 has been used. 17. As shown in the text, the expectation value <x> of a particle trapped in a box L wide is L/2, which means that its average position is the middle of the box. Find the expectation value <x 2 >. ¡ ¼Sol¡ ½ Using Equation (5.46), the expectation value <x 2 > is . s in dx L x n x L x L n ∫ , _ ¸ ¸ · > < 0 2 2 2 2 π See the end of this chapter for an alternate analytic technique for evaluating this integral using Leibniz’ s Rule. From either a table or repeated integration by parts, the indefinite integral is . s in cos s in s in s in 1 ] 1 ¸ + − − , _ ¸ ¸ · , _ ¸ ¸ · ∫ ∫ u u u u u u n L du u u n L dx L x n x 2 8 1 2 4 2 4 6 2 3 3 3 3 2 2 π π π where the substitution u = (nπ/L)x has been made. Inha University Department of Physics This form makes evaluation of the definite integral a bit simpler; when x = 0 u = 0, and when x = L u = nπ. Each of the terms in the integral vanish at u = 0, and the terms with sin 2u vanish at u = nπ, cos 2u = cos 2nπ = 1, and so the result is As a check, note that wh ich is t h e expect a t ion va lu e of <x 2 > in t h e cla s s ica l limit , for wh ich t h e pr oba bilit y dis t r ibu t ion is in depen den t of pos it ion in t h e box. . ) ( 1 ] 1 ¸ − · 1 ] 1 ¸ − , _ ¸ ¸ · > < 2 2 2 3 3 2 2 1 3 1 4 6 2 π π π π n L n n n L L x n , lim 3 2 2 L x n n · > < ∞ → 19. Find the probability that a particle in a box L wide can be found between x = 0 and x = L/ n when it is in the nth state. ¡ ¼Sol¡ ½ This is a special case of the probability that such a particle is between x 1 and x 2 , as found in Example 5.4. With x 1 = 0 and x 2 = L, . s in n L x n n L x P L L 1 2 2 1 0 0 · 1 ] 1 ¸ − · π π Inha University Department of Physics 21. A particle is in a cubic box with infinitely hard walls whose edges are L long (Fig. 5. 18). The wave functions of the particle are given by K K K 3, 2, 1 3, 2, 1 3, 2, 1 , , , s in s in s in · · · · z y x z z x n n n L z n L y n L x n A π π π ψ Find the value of the normalization constant A. ¡ ¼Sol¡ ½ The normalization constant, assuming A to be real, is given by . s in s in s in * * , _ ¸ ¸ , _ ¸ ¸ , _ ¸ ¸ · · · ∫ ∫ ∫ ∫ ∫ dz L z n dy L y n dx L x n A dxdy dz dV L z L y L x 0 2 0 2 0 2 2 1 π π π ψ ψ ψ ψ Each integral above is equal to L/2 (from calculations identical to Equation (5.43)). The result is 2 3 3 2 2 or 1 2 / , _ ¸ ¸ · · , _ ¸ ¸ L A L A Inha University Department of Physics 23. (a) Find the possible energies of the particle in the box of Exercise 21 by substituting its wave function ψ in Schrödinger's equation and solving for E. (Hint: inside the box U = 0.) (b) Compare the ground-state energy of a particle in a one-dimensional box of length L with that of a particle in the three-dimensional box. ¡ ¼Sol¡ ½ (a) For the wave function of Problem 5-21, Equation (5.33) must be used to find the energy. Before substitution into Equation (5.33), it is convenient and useful to note that for this wave function . , , ψ π ψ ψ π ψ ψ π ψ 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 L n z L n y L n x z y x − · ∂ ∂ − · ∂ ∂ − · ∂ ∂ Then, substitution into Equation (5.33) gives , ) ( 0 2 2 2 2 2 2 2 · + + + − ψ ψ π E m n n n L z y x h and so the energies are ). ( , , 2 2 2 2 2 2 2 z y x n n n n n n mL E z y x + + · h π (b) The lowest energy occurs when n x = n y = n z = 1. None of the integers n x , n y , or n z can be zero, as that would mean ψ = 0 identically. The minimum energy is then , min 2 2 2 2 3 mL E h π · which is three times the ground-state energy of a particle in a one-dimensional box of length L. Inha University Department of Physics 25. A beam of electrons is incident on a barrier 6.00 eV high and 0.200 nm wide. Use Eq. (5.60) to find the energy they should have if 1.00 percent of them are to get through the barrier. ¡ ¼Sol¡ ½ Solving equation (5.60) for k 2 , 1 - 10 9 2 m 10 15 1 100 m 10 200 0 2 1 1 2 1 × · × · · − . ) ln ( ) . ( ln T L k Equation (5.86), from the appendix, may be solved for the energy E, but a more direct expression is ( ) eV 95 0 J/eV 10 6 1 kg 10 1 9 2 m 10 15 1 s J 10 05 1 eV 00 6 2 2 19 31 2 1 10 34 2 2 2 . ) . )( . ( ) . )( . ( . ) ( · × × × ⋅ × − · − · − · − · − − − − m k U m p U KE U E h 27. What bearing would you think the uncertainty principle has on the existence of the zero-point energy of a harmonic oscillator? ¡ ¼Sol¡ ½ If a particle in a harmonic-oscillator potential had zero energy, the particle would have to be at rest at the position of the potential minimum. The uncertainty principle dictates that such a particle would have an infinite uncertainty in momentum, and hence an infinite uncertainty in energy. This contradiction implies that the zero-point energy of a harmonic oscillator cannot be zero. Inha University Department of Physics 29. Show that for the n = 0 state of a harmonic oscillator whose classical amplitude of motion is A, y = 1 at x = A, where y is the quantity defined by Eq. (5.67). ¡ ¼Sol¡ ½ When the classical amplitude of motion is A, the energy of the oscillator is . , k h A h k A ν ν · · so 2 1 2 1 2 Using this for x in Equation (5.67) gives , 1 2 2 2 · · · k m k h m y ν π ν ν π h where Equation (5.64) has been used to relate ν, m and k . 31. Find the expectation values <x> and <x 2 > for the first two states of a harmonic oscillator. ¡ ¼Sol¡ ½ Th e expect a t ion va lu es will be of t h e for ms dx x dx x ∫ ∫ ∞ ∞ − ∞ ∞ − ψ ψ ψ ψ * * 2 and It is far more convenient to use the dimensionless variable y as defined in Equation (5.67). The necessary integrals will be proportional to , , , , dy e y dy e y dy e y dy y e y y y y ∫ ∫ ∫ ∫ ∞ ∞ − − ∞ ∞ − − ∞ ∞ − − ∞ ∞ − − 2 2 2 2 4 3 2 Inha University Department of Physics The first and third integrals are seen to be zero (see Example 5.7). The other two integrals may be found from tables, from symbolic-manipulation programs, or by any of the methods outlined at the end of this chapter or in Special Integrals for Harmonic Oscillators, preceding the solutions for Section 5.8 problems in this manual. The integrals are . , π π 4 3 2 1 2 2 4 2 · · ∫ ∫ ∞ ∞ − − ∞ ∞ − − dy e y dy e y y y An immediate result is that <x> = 0 for the first two states of any harmonic oscillator, and in fact <x> = 0 for any state of a harmonic oscillator (if x = 0 is the minimum of potential energy). A generalization of the above to any case where the potential energy is a symmetric function of x, which gives rise to wave functions that are either symmetric or antisymmetric, leads to <x> = 0. To find <x 2 > for the first two states, the necessary integrals are ; ) / ( * / / / k E m h m dy e y m m dx x y o 0 2 2 2 3 2 2 3 2 1 0 2 4 2 1 2 2 2 2 2 · · · , _ ¸ ¸ , _ ¸ ¸ · ∫ ∫ ∞ ∞ − − ∞ ∞ − ν π ν π ν π ν π ν ψ ψ h h h . ) / ( * / / / k E m h m dy e y m m dx x y 1 2 2 2 3 4 2 3 2 1 1 1 2 4 2 3 2 3 2 2 2 2 2 2 · · · , _ ¸ ¸ , _ ¸ ¸ · ∫ ∫ ∞ ∞ − − ∞ ∞ − ν π ν π ν π ν π ν ψ ψ h h h Inha University Department of Physics 33. A pendulum with a 1.00-g bob has a massless string 250 mm long. The period of the pendulum is 1.00 s. (a) What is its zero-point energy? Would you expect the zero-point oscillations to be detectable? (b) The pendulum swings with a very small amplitude such that its bob rises a maximum of 1.00 mm above its equilibrium position. What is the corresponding quantum number? ¡ ¼Sol¡ ½ (a) The zero-point energy would be In both of the above integrals, dy m dy dy dx dx ν π 2 h · · has been used, as well as Table 5.2 and Equation (5.64). , . ) . ( . eV 10 07 2 s 00 1 2 s eV 10 14 4 2 2 1 15 15 0 − − × · ⋅ × · · · T h h E ν which is not detectable. (b) The total energy is E = mgH (here, H is the maximum pendulum height, given as an uppercase letter to distinguish from Planck's constant), and solving Equation (5.70) for n, ( ) . . . . ) . )( . ( / 28 34 2 3 10 48 1 2 1 s J 10 63 6 s 00 1 m/s 80 9 kg 10 00 1 2 1 × · − ⋅ × × · · − · − − T h mgH h E n ν Inha University Department of Physics 37. Consider a beam of particles of kinetic energy E incident on a potential step at x = 0 that is U high, where E > U (Fig. 5.19). (a) Explain why the solution De -ik’ x (in the notation of appendix) has no physical meaning in this situation, so that D = 0. (b) Show that the transmission probability here is T = CC*v‘ /AA*v 1 = 4k 1 2 /(k 1 + k’ ) 2 . (c) A 1.00-mA beam of electrons moving at 2.00x10 6 m/s enters a region with a sharply defined boundary in which the electron speeds are reduced to 1.00x10 6 m/s by a difference in potential. Find the transmitted and reflected currents. ¡ ¼Sol¡ ½ (a) In the notation of the Appendix, the wave function in the two regions has the form , , x k i x k i II x ik x ik I De Ce Be Ae ′ − ′ − + · + · ψ ψ 1 1 where . ) ( , h h U E m k mE k − · ′ · 2 2 1 The terms corresponding to exp(ik 1 x) and exp(ik’ x) represent particles traveling to the left; this is possible in region I, due to reflection at the step at x = 0, but not in region II (the reasoning is the same as that which lead to setting G = 0 in Equation (5.82)). Therefore, the exp(-ik’ x) term is not physically meaningful, and D = 0. Inha University Department of Physics (b) The boundary condition at x= 0 are then . , C k k B A C k i B ik A ik C B A 1 1 1 or ′ · − ′ · − · + Adding to eliminate B, so 1 2 1 , C k k A , _ ¸ ¸ ′ + · . ) ( * * , 2 1 2 1 1 1 4 and 2 k k k AA CC k k k A C ′ + · ′ + · (c) The particle speeds are different in the two regions, so Equation (5.83) becomes . ) ) / (( ) / ( ) ( * * 2 1 1 2 1 1 1 1 2 2 1 4 4 + ′ ′ · ′ + ′ · ′ · ′ · k k k k k k k k k k AA CC v v T I II ψ ψ For the given situation, k 1 /k’ = v 1 /v’ = 2.00, so T = (4x2)/(2+1)2 = 8/9. The transmitted current is (T)(1.00 mA) = 0.889 mA, and the reflected current is 0.111mA. As a check on the last result, note that the ratio of the reflected current to the incident current is, in the notation of the Appendix, * * AA BB v v R I I · · + − 1 2 1 2 ψ ψ Eliminating C from the equations obtained in part (b) from the continuity condition as x = 0, T k k k k R k k B k k A − · · , _ ¸ ¸ + ′ − ′ · , _ ¸ ¸ ′ + · , _ ¸ ¸ ′ − 1 9 1 1 1 so 1 1 1 1 1 1 ) / ( ) / ( , Inha University Department of Physics ¡ ¼sol¡ ½ Whether in Cartesian (x, y , z ) or spherical coordinates, three quantities are needed to describe the variation of the wave function throughout space. The three quantum numbers needed to describe an atomic electron correspond to the variation in the radial direction, the variation in the azimuthal direction (the variation along the circumference of the classical orbit), and the variation with the polar direction (variation along the direction from the classical axis of rotation). Chapter 6 Problem Solutions 1. Why is it natural that three quantum numbers are needed to describe an atomic electron (apart from electron spin)? 3. Show that ¡ ¼sol¡ ½ For the given function, is a solution of Eq. (6.14) and that it is normalized. o a r e a r R 2 2 3 0 10 2 / / ) ( − · and 2 2 5 0 10 , / / o a r e a R dr d − − · Inha University Department of Physics 10 2 2 2 2 5 10 2 2 2 1 2 1 2 1 R a r a e a r r r a dr dR r dr d r o o a r o o o , _ ¸ ¸ − · , _ ¸ ¸ − − · , _ ¸ ¸ − / / This is the solution to Equation (6.14) if l=0 ( as indicated by the index of R 10 ), 2 2 2 2 2 4 or 4 2 2 me h a h me a o o o o ε π πε · · , which is the case, and 1 2 2 2 8 or 1 2 E a e E a E m o o o · − · − · πε , h again as indicated by the index of R 10 . To show normalization, , / ∫ ∫ ∫ ∞ − ∞ − ∞ · · 0 2 0 2 2 3 0 2 2 10 2 1 4 du e u dr e r a dr r R u a r o o where the substitution u=2r/a o has been made. The improper definite integral in u is known to have the value 2 and so the given function is normalized. Inha University Department of Physics 5. In Exercise 12 of Chap. 5 it was stated that an important property of the eigenfunctions of a system is that they are orthogonal to one another, which means that ¡ ¼sol¡ ½ From Equation (6.15) the integral, apart from the normalization constants, is m n dV m n ≠ · ∫ ∞ ∞ − 0 ψ ψ * l l m m m m d l l ′ ≠ Φ Φ ∫ ′ for 2 0 φ π * Verify that this is true for the azimuthal wave functions of the hydrogen atom by calculating l m Φ , * φ φ π φ φ π d e e d l l l l m i im m m ∫ ∫ ′ − ′ · Φ Φ 2 0 2 0 It is possible to express the integral in terms of real and imaginary parts, but it turns out to be more convenient to do the integral directly in terms of complex exponentials: [ ] 0 1 2 0 2 0 2 0 · − ′ · · ′ − ′ ′ − ′ ′ − ∫ ∫ π φ π φ π φ φ φ φ ) ( ) ( ) ( l l l l l l m m i l l m m i m i im e m m i d e d e e The above form for the integral is valid only for m l ≠ m l ’ , which is given for this case. In evaluating the integral at the limits, the fact that e i2πn = 1 for any integer n ( in this case (m l ’ – m l )) has been used. Inha University Department of Physics 7. Compare the angular momentum of a ground-state electron in the Bohr model of the hydrogen atom with its value in the quantum theory. 9. Under what circumstances, if any, is L z equal to L? ¡ ¼sol¡ ½ In the Bohr model, for the ground-state orbit of an electron in a hydrogen atom, λ = h/ mv = 2πr, and so L = pr = . In the quantum theory, zero-angular-momentum states (ψ spherically symmetric) are allowed, and L = 0 for a ground-state hydrogen atom. ¡ ¼sol¡ ½ From Equation (6.22), L z must be an integer multiple of ; for L to be equal to L z , the product l(1+1), from Equation(6.21), must be the square of some integer less than or equal to l. But, or any nonnegative l, with equality holding in the first relation only if l = 0. Therefore, l(l + 1) is the square of an integer only if l = 0, in which case L z = 0 and L = L z = 0. 2 2 1 1 ) ( ) ( + < + ≤ l l l l Inha University Department of Physics 11. What are the possible values of the magnetic quantum number m l of an atomic electron whose orbital quantum number is l = 4? 13. Find the percentage difference between L and the maximum value of L z for an atomic electron in p, d , and f states. ¡ ¼sol¡ ½ From Equation (6.22), the possible values for the magnetic quantum number m l are m l = 0, ±1, ±2, ±3, ±4, a total of nine possible values. ¡ ¼sol¡ ½ The fractional difference between L and the largest value of L z , is, for a given l, . ) ( ) ( max , 1 1 1 1 + − · + − + · − l l l l l l l L L L z % . % . % . 13 13 0 - 1 and 3 state, a For 18 18 0 - 1 and 2 state, a For 29 29 0 - 1 and 1 state, a For 4 3 3 2 2 1 · · · · · · · · · l f l d l p Inha University Department of Physics 15. In Sec. 6.7 it is stated that the most probable value of r for a 1s electron in a hydrogen atom is the Bohr radius a o . Verify this. ¡ ¼sol¡ ½ Using R 10 (r) from Table 6.1 in Equation (6.25), The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, . ) ( / o a r o e a r r P 2 3 2 4 − · o o a r o o a r a r r e a r r a r P dr d o , , ) ( / 0 and or 0 2 2 4 2 2 2 3 · · · , _ ¸ ¸ − · − for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dp/ dr → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = a o , which is the radius of the first Bohr orbit. Inha University Department of Physics 17. Find the most probable value of r for a 3d electron in a hydrogen atom. ¡ ¼sol¡ ½ Using R 20 (r) from Table 6.1 in Equation (6.25), and ignoring the leading constants (which would not affect the position of extremes), o a r e r r P 3 2 6 / ) ( − · The most probable value of r is that for which P(r) is a maximum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero, o o a r o a r a r r e a r r r P dr d o 9 0 and 3 2 6 or 0 3 2 6 6 5 3 2 6 5 , , ) ( / · · · , _ ¸ ¸ − · − for an extreme. At r = 0, P(r) = 0, and because P(r) is never negative, this must be a minimum. dP/d r → 0 as r → ∞, and this also corresponds to a minimum. The only maximum of P(r) is at r = 9a o , which is the radius of the third Bohr orbit. Inha University Department of Physics 19. How much more likely is the electron in a ground-state hydrogen atom to be at the distance a o from the nucleus than at the distance 2a o ? ¡ ¼sol¡ ½ For the ground state, n = 1, the wave function is independent of angle, as seen from the functions Φ(φ) and Θ(θ) in Table 6.1, where for n = 1, l = m l , = 0 (see Problem 6-14). The ratio of the probabilities is then the ratio of the product r 2 (R 10 (r)) 2 evaluated at the different distances. Specially, 47 1 4 4 1 2 2 1 2 2 2 2 2 2 . ) / ( ) / ( ) / ( ) ( / ) / ( / · · · · − − − − e e e e a e a dr a P dr a P o o o o a a o a a o o o 85 1 4 4 2 2 2 4 2 2 2 2 2 2 . ) ( ) ( ) ( ) ( / ) ( / · · · · − − − − e e e e a e a dr a P dr a P o o o o a a o a a o o o Similarly, Inha University Department of Physics 21. The probability of finding an atomic electron whose radial wave function is R(r) outside a sphere of radius r o centered on the nucleus is (a) Calculate the probability of finding a 1s electron in a hydrogen atom at a distance greater than a o from the nucleus. (b) When a 1s electron in a hydrogen atom is 2a o from the nucleus, all its energy is potential energy. According to classical physics, the electron therefore cannot ever exceed the distance 2a o from the nucleus. Find the probability r > 2a o for a 1s electron in a hydrogen atom. ¡ ¼sol¡ ½ (a) Using R 10 (r) for the 1s radial function from Table 6.1, ∫ ∞ o r dr r r R 2 2 ) ( , ) ( / du e u dr e r a dr r r R u a a r o a o o o ∫ ∫ ∫ ∞ − ∞ − ∞ · · 2 2 2 2 3 2 2 2 1 4 where the substitution u = 2r/a 0 has been made. Using the method outlined at the end of this chapter to find the improper definite integral leads to ( ) [ ] [ ] %, . 68 68 0 10 2 1 2 2 2 1 2 1 2 2 2 2 2 · · · + + − · − ∞ − ∞ − ∫ e u u e du e u u u Inha University Department of Physics (b) Repeating the above calculation with 2 a 0 as the lower limit of the integral, ( ) [ ] [ ] % , . 24 24 0 26 2 1 2 2 2 1 2 1 4 4 2 4 2 · · · + + − · − ∞ − ∞ − ∫ e u u e du e u u u 23. Unsold's theorem states that for any value of the orbital quantum number l, the probability densities summed over all possible states from m l = -1 to m l = +1 yield a constant independent of angles θ or φ that is, This theorem means that every closed subshell atom or ion (Sec. 7.6) has a spherically symmetric distribution of electric charge. Verify Unsold's theorem for l = 0, l = 1, and l = 2 with the help of Table 6. 1. ¡ ¼sol¡ ½For l = 0, only m l = 0 is allowed, Φ(φ) and Θ(θ) are both constants (from Table 6.1)), and the theorem is verified. For l = 1, the sum is constant 2 2 · Φ Θ ∑ + − · l l m l , s in cos s in π θ π θ π θ π 4 3 4 3 2 1 2 3 2 1 4 3 2 1 2 2 2 · + + Inha University Department of Physics . ) cos ( cos s in s in 2 2 2 2 4 1 3 16 10 2 1 4 15 2 1 2 16 15 2 1 2 − + + θ π θ θ π θ π The above may he simplified by extracting the commons constant factors, to ]. s in cos s in ) cos [( θ θ θ θ π 4 2 2 2 2 3 12 1 3 16 5 + + − Of the many ways of showing the term in brackets is indeed a constant, the one presented here, using a bit of hindsight, seems to be one of the more direct methods. Using the identity sin 2 θ = 1 - cos 2 θ to eliminate sin θ , , ) cos cos ( cos ) cos ( ) cos cos ( s in cos s in ) cos ( 1 2 1 3 1 12 1 6 9 3 12 1 3 4 2 2 2 2 4 4 2 2 2 2 · + − + − + + − · + + − θ θ θ θ θ θ θ θ θ θ and the theorem is verified. In the above, Φ*Φ= 1/2π, which holds for any l and m l , has been used. Note that one term appears twice, one for m l = -1 and once for m l = 1. For l = 2, combining the identical terms for m l = ±2 and m l = ±1, and again using Φ*Φ= 1/2π, the sum is Inha University Department of Physics 25. With the help of the wave functions listed in Table 6.1 verify that ∆l = ±1 for n = 2 àn = 1 transitions in the hydrogen atom. ¡ ¼sol¡ ½ In the integral of Equation (6.35), the radial integral will never. vanish, and only the angular functions Φ(φ) and Θ(θ) need to be considered. The ∆l = 0 transition is seen to be forbidden, in that the product π θ φ θ φ 4 1 00 0 00 0 · Θ Φ Θ Φ ∗ )) ( ) ( ( )) ( ) ( ( is spherically symmetric, and any integral of the form of Equation (6.35) must vanish, as the argument u = x, y or z will assume positive and negative values with equal probability amplitudes. If l = 1 in the initial state, the integral in Equation (6.35) will be seen to to vanish if u is chosen appropriately. If m l = 0 initially, and u = z = r cos θ is used, the integral (apart from constants) is 0 3 2 0 2 ≠ · ∫ θ θ θ π d s in cos If m l = ±1 initially, and u = x = r sin θ cos φ is used, the θ -integral is of the form 0 2 0 2 ≠ · ∫ π θ θ π d s in and the φ -integral is of the form 0 2 0 2 2 0 ≠ · · ∫ ∫ t π φ φ φ φ π π φ d d e i cos cos and the transition is allowed. Inha University Department of Physics 27. Verify that the n = 3 → n = 1 transition for the particle in a box of Sec. 5.8 is forbidden whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. ¡ ¼sol¡ ½ The relevant integrals are of the form . s in s in dx L x m L x n x L π π ∫ 0 The integrals may be found in a number of ways, including consulting tables or using symbolic- manipulation programs (see; for instance, the solution to Problem 5-15 for sample Maple commands that are easily adapted to this problem). One way to find a general form for the integral is to use the identity )] cos( ) [cos ( s in s in β α β α β α + − − · 2 1 and the indefinite integral (found from integration by parts) 2 1 k k x k k x x dx k x k k k x x dx k x x cos s in s in s in cos + · − · ∫ ∫ to find the above definite integral as , ) ( cos ) ( ) ( s in ) ( ) ( cos ) ( ) ( s in ) ( L L x m n m n L L x m n m n Lx L x m n m n L L x m n m n Lx 0 2 2 2 2 2 2 2 1 1 1 1 1 1 ] 1 ¸ + + + + + − − − + − − π π π π π π π Inha University Department of Physics where n ≠ m 2 is assumed. The terms involving sines vanish, with the result of . ) ( ) cos ( ) ( ) cos ( 1 ] 1 ¸ + − + − − − − 2 2 2 2 1 1 2 m n m n m n m n L π π π If n and m axe both odd or both even, n + m and n - m are even, the arguments of the cosine terms in the above expression are even-integral multiples of π, and the integral vanishes. Thus, the n = 3 → n = 1 transition is forbidden, while the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. To make use of symmetry arguments, consider that dx L x m L x n x dx L x m L x n L x L L ∫ ∫ · , _ ¸ ¸ − 0 0 2 π π π π s in s in s in s in for n ≠ m, because the integral of L times the product of the wave functions is zero; the wave functions were shown to be orthogonal in Chapter 5 (again, see Problem 5-15). Letting u=L/ 2 – x, , _ ¸ ¸ − · − · L u n n L u L n L x n π π π π 2 2 s in ) ) / (( s in s in This expression will be ± cos ( nπu/L ) for n odd and ±sin ( nπu/L ) for n even. The integrand is then an odd function of u when n and m are both even or both odd, and hence the integral is zero. If one of n or m is even and the other odd, the integrand is an even function of u and the integral is nonzero. Inha University Department of Physics 29. Show that the magnetic moment of an electron in a Bohr orbit of radius r n is proportional to 31. Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400-nm wavelength when a spectrometer whose resolution is 0.010 nm is used. ¡ ¼sol¡ ½ From Equation (6.39), the magnitude of the magnetic moment of an electron in a Bohr orbit is proportional to the magnitude of the angular momentum, and hence proportional to n. The orbital radius is proportional to n 2 (See Equation (4.13) or Problem 4-28), and so the magnetic moment is proportional to . ¡ ¼sol¡ ½ See Example 6.4; solving for B, n r n r T 34 1 C) 10 6 1 m/s) 10 0 3 kg 10 1 9 4 m) 10 (400 m 10 010 0 4 19 8 31 2 9 - 9 2 . . ( . )( . ( . · × × × × × · ∆ · − − − π π λ λ e mc B Inha University Department of Physics 1. A beam of electrons enters a uniform 1.20-T magnetic field. (a) Find the energy difference between electrons whose spins are parallel and antiparallel to the field. (b) Find the wavelength of the radiation that can cause the electrons whose spins are parallel to the field to flip so that their spins are antiparallel. (a) Using Equations (7.4) and (6.41), the energy difference is, Chapter 7 Problem Solutions eV 10 39 1 T 20 1 eV/T 10 79 5 2 2 2 4 5 − − × = × = = = ∆ . ) . )( . ( B B E B s z µ µ (b) The wavelength of the radiation that corresponds to this energy is Note that a more precise value of AB was needed in the intermediate calculation to avoid roundoff error. mm 93 8 eV 10 39 1 m eV 10 24 1 4 6 . . . = × ⋅ × = ∆ = − − E hc λ 3. Find the possible angles between the z axis and the direction of the spin angular-momentum vector S. For an electron, and so the possible angles axe given by , ) / ( , ) / ( h h 2 1 2 3 ± = = z s s o o 3 125 7 54 3 1 2 3 2 1 . , . a r ccos ) / ( ) / ( a r ccos =       =       ± h h ¡¼sol¡ ½ ¡ ¼sol¡ ½ Inha University Department of Physics 5. Protons and neutrons, like electrons, are spin- ½particles. The nuclei of ordinary helium atoms, , contain two protons and two neutrons each; the nuclei of another type of helium atom, , contain two protons and one neutron each. The properties of liquid and liquid are different because one type of helium atom obeys the exclusion principle but the other does not. Which is which, and why? He 4 2 He 4 2 He 3 2 He 3 2 atoms contain even numbers of spin-½ particles, which pair off to give zero or integral spins for the atoms. Such atoms do not obey the exclusion principle. atoms contain odd numbers of spin- ½particles, and so have net spins of and they obey the exclusion principle. He 4 2 He 3 2 , , 2 5 2 3 2 1 or 7. In what way does the electron structure of an alkali metal atom differ from that of a halogen atom? From that of an inert gas atom? ¡ ¼sol¡ ½ An alkali metal atom has one electron outside closed inner shells: A halogen atom lacks one electron of having a closed outer shell: An inert gas atom has a closed outer shell. ¡ ¼sol¡ ½ Inha University Department of Physics 9. How many electrons can occupy an f subshell? ¡ ¼sol¡ ½ For f subshell, with l = 3, the possible values of m l are ±3, ±2, ¡ ¾1 or 0, for a total of 2l +1=7 values of m l . Each state can have two electrons of opposite spins, for a total of 14 electrons. 11. If atoms could contain electrons with principal quantum numbers up to and including n = 6, how many elements would there be? ¡ ¼sol¡ ½ The number of elements would be the total number of electrons in all of the shells. Repeated use of Equation (7.14) gives 2n 2 + 2 (n - 1) 2 +... + 2 (1) 2 = 2 (36 + 25 + 16 + 9 + 4 + 1) = 182. In general, using the expression for the sum of the squares of the first n integers, the number of elements would be which gives a total of 182 elements when n = 6. ( ) )], )( ( [ ) )( ( 1 1 2 1 1 2 2 3 1 6 1 + + = + + n n n n n n Inha University Department of Physics 13. The ionization energies of Li, Na, K, Rb, and Cs are, respectively, 5.4, 5.1, 4.3, 4.2, and 3.9 eV. All are in group 1 of the periodic table. Account for the decrease in ionization energy with increasing atomic number. ¡ ¼sol¡ ½ All of the atoms are hydrogenlike, in that there is a completely filled subshell that screens the nuclear charge and causes the atom to "appear" to be a single charge. The outermost electron in each of these atoms is further from the nucleus for higher atomic number, and hence has a successively lower binding energy. 15. (a) Make a rough estimate of the effective nuclear charge that acts on each electron in the outer shell of the calcium (Z = 20) atom. Would you think that such an electron is relatively easy or relatively hard to detach from the atom? (b) Do the same for the sulfur (Z = 16) atom. ¡ ¼sol¡ ½ (a) See Table 7.4. The 3d subshell is empty, and so the effective nuclear charge is roughly +2e, and the outer electron is relatively easy to detach. (b) Again, see Table 7.4. The completely filled K and L shells shield +10e of the nuclear charge of = 16e; the filled 3s 2 subshell will partially shield the nuclear charge, but not to the same extent as the filled shells, so +6e is a rough estimate for the effective nuclear charge. This outer electron is then relatively hard to detach. Inha University Department of Physics ¡ ¼sol¡ ½ Cl - ions have closed shells, whereas a Cl atom is one electron short of having a closed shell and the relatively poorly shielded nuclear charge tends to attract an electron from another atom to fill the shell. Na + ions have closed shells, whereas an Na atom has a single outer electron that can be detached relatively easily in a chemical reaction with another atom. 17. Why are Cl atoms more chemically active than Cl - ions? Why are Na atoms more chemically active than Na + ions? ¡ ¼sol¡ ½ The Li atom (Z = 3) is larger because the effective nuclear charge acting on its outer electron is less than that acting on the outer electrons of the F atom (Z = 9). The Na atom (Z = 11) is larger because it has an additional electron shell (see Table 7.4). The Cl atom (Z = 17) atom is larger because has an additional electron shell. The Na atom is larger than the Si atom (Z = 14) for the same reason as given for the Li atom. 19. In each of the following pairs of atoms, which would you expect to be larger in size? Why? Li and F; Li and Na; F and Cl; Na and Si. Inha University Department of Physics ¡ ¼sol¡ ½The only way to produce a normal Zeeman effect is to have no net electron spin; because the electron spin is ±½, the total number of electrons must be even. If the total number of electrons were odd, the net spin would be nonzero, and the anomalous Zeeman effect would be observable. 21. Why is the normal Zeeman effect observed only in atoms with an even number of electrons? 23. The spin-orbit effect splits the 3P → 3S transition in sodium (which gives rise to the yellow light of sodium-vapor highway lamps) into two lines, 589.0 nm corresponding to 3P 3/2 →3S 1/2 and 589.6 nm corresponding to 3P 1/2 →3S 1/2 . Use these wavelengths to calculate the effective magnetic field experienced by the outer electron in the sodium atom as a result of its orbital motion. ¡ ¼sol¡ ½ See Example 7.6. Expressing the difference in energy levels as , for solving 1 1 2 2 1 B hc B E B ;         − = = ∆ λ λ µ T 5 18 m 10 6 589 1 m 10 0 589 1 eV/T 10 5.79 2 m eV 10 24 1 1 1 2 9 9 5 - 6 2 1 . . . . =       × − × × × ⋅ × =         − = − − − λ λ µ B hc B Inha University Department of Physics 25. If , what values of l are possible? 2 5 = j ¡¼sol¡ ½ The possible values of l are . 2 and 3 2 1 2 1 = − = + j j 27. What must be true of the subshells of an atom which has a 1 S 0 ground state? ¡ ¼sol¡ ½ For the ground state to be a singlet state with no net angular momentum, all of the subshells must be filled. ¡ ¼sol¡ ½ For this doublet state, L = 0, S = J = ½. There axe no other allowed states. This state has the lowest possible values of L and J , and is the only possible ground state. 29. The lithium atom has one 2s electron outside a filled inner shell. Its ground state is 2 S 1/2 . (a) What are the term symbols of the other allowed states, if any? (b) Why would you think the 2 S 1/2 state is the ground state? Inha University Department of Physics ¡ ¼sol¡ ½ The two 3s electrons have no orbital angular momentum, and their spins are aligned oppositely to give no net angular momentum. The 3p electron has l = 1, so L = 1, and in the ground state J = ½. The term symbol is 2 P 1/2 . 31. The aluminum atom has two 3s electrons and one 3p electron outside filled inner shells. Find the term symbol of its ground state. 33. Why is it impossible for a 2 2 D 3/2 state to exist? ¡ ¼sol¡ ½ A D state has L = 2; for a 2 2 D 3/2 state, n = 2 but L must always be strictly less than n , and so this state cannot exist. 35. Answer the questions of Exercise 34 for an f electron in an atom whose total angular momentum is provided by this electron. ¡ ¼sol¡ ½ (a) From Equation (7.17), . , 2 7 2 5 2 1 = ± = l j (b) Also from Equation (7.17), the corresponding angular momenta are h h 2 63 2 35 and Inha University Department of Physics (c) The values of L and S are . The law of cosines is h h 2 3 and 12 , cos LS S L J 2 2 2 2 − − = θ where θ is the angle between L and S; then the angles are, and o 132 3 2 2 3 12 2 4 3 12 4 35 =       − =       − − a r ccos ) / ( ) / ( ) / ( a r ccos o 0 60 2 1 2 3 12 2 4 3 12 4 63 . a r ccos ) / ( ) / ( ) / ( a r ccos =       =       − − (d) The multiplicity is 2(1/2) + 1 = 2, the state is an f state because the total angular momentum is provided by the f electron, and so the terms symbols are 2 F 5/2 and 2 F 7/2 . 37. The magnetic moment µ J of an atom in which LS coupling holds has the magnitude where µ B = e©¤/2m is the Bohr magneton and B J J µ µ g ) ( 1 + = ϑ ϑ 1) 2ϑ(ϑ 1) Σ(Σ 1) Λ(Λ 1) ϑ(ϑ + + + + − + + = 1 J g Inha University Department of Physics is the Landé g factor. (a) Derive this result with the help of the law of cosines starting from the fact that averaged over time, only the components of µ L and µ S parallel to J contribute to µ L . (b) Consider an atom that obeys LS coupling that is in a weak magnetic field B in which the coupling is preserved. How many substates are there for a given value of J ? What is the energy difference between different substates? In the above, the factor of 2 in 2µ B relating the electron spin magnetic moment to the Bohr magneton is from Equation (7.3). The middle term is obtained by using |S| cos α + |S| cos β = |J|. The above expression is equal to the product µ J because in this form, the magnitudes of the angular momenta include factors of h. From the law of cosines, ¡ ¼sol¡ ½ (a) In Figure 7.15, let the angle between J and S be α and the angle between J and L be β. Then, the product µ J has magnitude         + = + = + α µ α µ µ β µ α µ cos cos cos cos J S 1 J S J L S 2 B B B B B S J 2 S J L 2 2 2 − − − = α cos and so ) J(J ) S(S ) L(L ) J(J 1 2 1 1 1 J 2 S J L J S 2 2 2 2 + + + + − + = − − = α cos Inha University Department of Physics (b) There will be one substate for each value of M J , where M J = -J ... J , for a total of 2J + 1 substates. The difference in energy between the substates is and the expression for µ J in terms of the quantum numbers is re whe 1 ( or J J , ) , B J B J J g g µ µ µ µ + = = ϑ ϑ h ) ( ) ( ) ( ) ( 1 2 1 1 1 1 + + + + − + + = ϑ ϑ Σ Σ Λ Λ ϑ ϑ J g B M g E J B J µ = ∆ 39. Explain why the x-ray spectra of elements of nearby atomic numbers are qualitatively very similar, although the optical spectra of these elements may differ considerably. ¡ ¼sol¡ ½ The transitions that give rise to x-ray spectra are the same in all elements since the transitions involve only inner, closed-shell electrons. Optical spectra, however, depend upon the possible states of the outermost electrons, which, together with the transitions permitted for them, are different for atoms of different atomic number. Inha University Department of Physics ¡ ¼sol¡ ½ From either of Equations (7.21) or (7.22), E = (10.2 eV) (Z - 1) 2 = (10.2 eV) (144) = 1.47 keV. The wavelength is 41. Find the energy and the wavelength of the K α x-rays of aluminum. nm 0.844 m 10 44 8 eV 10 7 14 m eV 10 24 1 10 3 6 = × = × ⋅ × = = − − . . . E hc λ ϒ…sol¡ ½ In a singlet state, the spins of the outer electrons are antiparrallel. In a triplet state, they are parallel 43. Distinguish between singlet and triplet states in atoms with two outer electrons. Inha University Department of Physics 1. The energy needed to detach the electron from a hydrogen atom is 13.6 eV, but the energy needed to detach an electron from a hydrogen molecule is 15.7 eV. Why do you think the latter energy is greater? ¡ ¼sol¡ ½ The nuclear charge of +2e is concentrated at the nucleus, while the electron charges' densities are spread out in (presumably) the 1s subshell. This means that the additional attractive force of the two protons exceeds the mutual repulsion of the electrons to increase the binding energy. 3. At what temperature would the average kinetic energy of the molecules in a hydrogen sample be equal to their binding energy? ¡ ¼sol¡ ½ Using 4.5 eV for the binding energy of hydrogen, Chapter 8 Problem Solutions K 10 5 3 eV/K 10 8.62 eV 5 4 3 2 or eV 5 4 2 3 4 5 - × = × = = . . . T k T Inha University Department of Physics 5. When a molecule rotates, inertia causes its bonds to stretch. (This is why the earth bulges at the equator.) What effects does this stretching have on the rotational spectrum of the molecule? ¡ ¼sol¡ ½ The increase in bond lengths in the molecule increases its moment of inertia and accordingly decreases the frequencies in its rotational spectrum (see Equation (8.9)). In addition, the higher the quantum number J (and hence the greater the angular momentum), the faster the rotation and the greater the distortion, so the spectral lines are no longer evenly spaced. Quantitatively, the parameter I (the moment of inertia of the molecule) is a function of J, with I larger for higher J. Thus, all of the levels as given by Equation (8.11) are different, so that the spectral lines are not evenly spaced. (It should be noted that if I depends on J, the algebraic steps that lead to Equation (8.11) will not be valid.) 7. The J=0àJ=1 rotational absorption line occurs at 1.153x10 11 Hz in 12 C 16 O and at 1.102x10 11 Hz in ? C 16 O. Find the mass number of the unknown carbon isotope. ¡ ¼sol¡ ½ From Equation (8.11), the ratios of the frequencies will be the ratio of the moments of inertia. For the different isotopes, the atomic separation, which depends on the charges of the atoms, will be essentially the same. The ratio of the moments of inertia will then be the ratio of the reduced masses. Denoting the unknown mass number by x and the ratio of the frequencies as r, r in terms of x is Inha University Department of Physics 16 12 16 12 16 16 + ⋅ + ⋅ = x x r Solving for x in terms of r, r r x 3 7 48 − = Using r = (1.153)/(1.102) in the above expression gives x = 13.007, or the integer 13 to three significant figures. 9. The rotational spectrum of HCI contains the following wavelengths: 12.03 x 10 -5 m, 9.60 x 10 -5 m, 8.04 x 10 -5 m, 6.89 x 10 -5 m, 6.04 x 10 -5 m If the isotopes involved are 1 H and 35 Cl, find the distance between the hydrogen and chlorine nuclei in an HCl molecule. Inha University Department of Physics The average spacing of these frequencies is ∆v = 0.616 x 10 12 Hz. (A least-squares fit from a spreadsheet program gives 0.6151 if c = 2.998 x 10 8 m/s is used.) From Equation (8.11), the spacing of the frequencies should be ∆v = /2πI ; Solving for I and using ∆v as found above, The reduced mass of the HCI molecule is (35/36)rn H , and so the distance between the nuclei is (keeping extra significant figures in the intermediate calculation gives a result that is rounded to 0.130 nm to three significant figures). 2 47 12 34 m kg 10 73 2 Hz 10 6151 0 2 s J 10 055 1 2 ⋅ × = × ⋅ × = ∆ = − − . ) . ( . π ν π h I nm 129 0 kg) 10 67 1 35 m kg 10 73 2 36 27 2 47 . . ( ) . ( = × × ⋅ × × = = − − µ I R ¡¼sol¡ ½ The corresponding frequencies are, from ν = c/λ , and keeping an extra significant figure, in multiplies of 10 12 Hz: 2.484, 3.113, 4.337, 4.947 Inha University Department of Physics 11. A 200 Hg 35 Cl Molecule emits a 4.4-cm photon when it undergoes a rotational transition from j = 1 to j = 0. Find the interatomic distance in this molecule. ¡ ¼sol¡ ½ Using ν 1→0 = c/λ and I = m’ R 2 in Equation (8.11) and solving for R, For this atom, m’ = m H (200x35)/(200 + 35), and c m R ′ = π λ 2 2 h or 0.22 nm to two significant figures. nm 223 0 m/s 10 0 3 kg 10 67 1 2 m 10 4 4 s J 10 055 1 8 27 2 34 . ) . )( . ( ) . )( . ( = × × × ⋅ × = − − − π R Inha University Department of Physics ¡ ¼sol¡ ½ Equation (8.11) may be re-expressed in terms of the frequency of the emitted photon when the molecule drops from the J rotational level to the J - 1 rotational level, For large J, the angular momentum of the molecule in its initial state is Thus, for large J, the classical expression. . I J J J π ν 2 1 h = − → J J J J J L h h h ≈ + = + = / ) ( 1 1 1 , , I L I L ω π ν = ≈ or 2 13. In Sec. 4.6 it was shown that, for large quantum numbers, the frequency of the radiation from a hydrogen atom that drops from an initial state of quantumnumber n to a final state of quantum number n - 1 is equal to the classical frequency of revolution of an electron in the n-th Bohr orbit. This is an example of Bohr's correspondence principle. Show that a similar correspondence holds for a diatomic molecule rotating about its center of mass. Inha University Department of Physics 15. The hydrogen isotope deuterium has an atomic mass approximately twice that of ordinary hydrogen. Does H 2 or HD have the greater zero-point energy? How does this affect the binding energies of the two molecules? ¡ ¼sol¡ ½ The shape of the curve in Figure 8.18 will be the same for either isotope; that is, the value of k in Equation (8.14) will be the same. HD has the greater reduced mass, and hence the smaller frequency of vibration v o and the smaller zero- point energy. HD is the more tightly bound, and has the greater binding energy since its zero-point energy contributes less energy to the splitting of the molecule. 17. The force constant of the 1 H 19 F molecule is approximately 966 N/m. (a) Find the frequency of vibration of the molecule. (b) The bond length in 1 H 19 F is approximately 0.92 nm. Plot the potential energy of this molecule versus internuclear distance in the vicinity of 0.92 nm and show the vibrational energy levels as in Fig. 8.20. ¡ ¼sol¡ ½ (a) Using m'= (19/20)m H in Equation (8.15), Hz 10 24 1 19 20 kg 10 1.67 N/m 966 2 1 14 27 - × = × = . π ν o Inha University Department of Physics (b) = 4.11 X 10 -20 J. The levels are shown below, where the vertical scale is in units of 10 -20 J and the horizontal scale is in units of 10 -11 m. m k E o ′ = h 2 1 19. The lowest vibrational states of the 23 Na 35 Cl molecule are 0.063 eV apart. Find the approximate force constant of this molecule. ¡ ¼sol¡ ½ From Equation (8.16), the lower energy levels are separated by ∆E = h v o , and v o = ∆E /h. Solving Equation (8.15) for k ,       ∆ ′ = ′ = h E m m k o 2 2 ) ( πν Inha University Department of Physics Using m’ = m H (23· 35)/(23 + 35), or 2.1 x10 2 N/m to the given two significant figures. N/m 213 s eV 10 4.14 J/eV) 10 eV)(1.60 (0.063 kg) 10 67 1 58 35 23 15 - 19 - 27 =         ⋅ × × × ⋅ = − . ( k 21. The bond between the hydrogen and chlorine atoms in a 1 H 35 Cl molecule has a force constant of 516 N/m. Is it likely that an HCl molecule will be vibrating in its first excited vibrational state at room temperature? Atomic masses are given in the Appendix. ¡ ¼sol¡ ½Using An individual atom is not likely to he vibrating in its first excited level, but in a large collection of atoms, it is likely that some of these atoms will be in the first excited state. It's important to note that in the above calculations, the symbol "k " has been used for both a spring constant and Boltzmann's constant, quantities that are not interchangeable. , 36 35 and H o m m m k h E = ′ ′ = = ∆ h ν At room temperature of about 300 K, k T = (8.617 x 10 -5 eV/K) (300 K) = 0.026 eV. eV 371 0 J 10 94 5 35 36 kg 10 67 1 N/m 516 s) J 10 055 1 20 27 34 . . . . ( = × = × ⋅ × = ∆ − − − E Inha University Department of Physics Chapter 9 Problem Solutions 1. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in the n=2 energy level? 2 8 1 2 · · ) ( , ) ( ε ε g g k T k T e e n n / / ) ( ) ( ) ( 1 1 2 3 1 2 4 4 1000 1 ε ε ε ε ε · · · − − K 10 43 1 4000 eV/K 10 62 8 eV 6 13 4 3 4000 4 3 1 4 5 1 × · × · − , _ ¸ ¸ · − . ) )(ln . ( ) . )( / ( ln ) )( / ( ε k T eV 6 13 and 4 1 1 2 . , / − · · ε ε ε Then, where 3. The 3 2 P l/2 first excited sate in sodium is 2.093 eV above the 3 2 S 1/2 ground state. Find the ratio between the numbers of atoms in each state in sodium vapor at l200 K. (see Example 7.6.) 9 5 10 86 4 K 1200 eV/K 10 62 8 eV 09 2 1 3 − − × · , _ ¸ ¸ × − , _ ¸ ¸ . ) )( . ( . exp ¡ ¼sol¡ ½ ¡ ¼sol¡ ½ multiplicity of P-level : 2L+1=3, multiplicity of S-level : 1 The ratio of the numbers of atoms in the states is then, Inha University Department of Physics 5. The moment of inertia of the H 2 molecule is 4.64¡ ¿10 -48 kg· m 2 . (a) Find the relative popula- tions of the J=0,1,2,3, and 4 rotational states at 300 K. (b) can the populations of the J=2 and J=3 states ever be equal? If so, at what temperature does this occur? I J J J J g J 2 1 1 2 2 h ) ( , ) ( + · + · ε 0 0 · · J ε ) ( ) ( ) ( ] . )[ ( ) )( . )( . ( ) . ( exp ) ( exp ) ( ) ( exp ) ( ) ( ) ( 1 1 23 2 48 2 34 1 2 2 749 0 1 2 K 300 J/K 10 38 1 m kg 10 64 4 2 s J 10 06 1 1 2 2 1 2 2 1 1 2 0 + + − − − + + · 1 1 ] 1 ¸ , _ ¸ ¸ × ⋅ × ⋅ × − + · 1 1 ] 1 ¸ , _ ¸ ¸ − + · , _ ¸ ¸ + − + · · J J J J J J J J Ik T J Ik T J J J J N J N h h Applying this expression to J=0, 1, 2, 3, and 4 gives, respectively, 1 exactly, 1.68, 0.880, 0.217, and 0.0275. (b) Introduce the dimensionless parameter . Then, for the populations of the J=2 and J=3 states to be equal, 7 5 6 and 7 5 7 5 6 12 6 ln ln , · · · x x x x Using , and solving for T, ) / ( - ) / ( Ik T x 5 7 7 5 and 2 2 ln ln / ln · − · h ¡¼sol¡ ½ (a) Inha University Department of Physics K 10 55 1 4 1 J/K 10 38 1 m kg 10 64 4 2 s J 10 05 1 6 5 7 2 6 3 23 2 48 2 34 2 × · × ⋅ × ⋅ × · · − − − . ) . ln ( ) . )( . ( ) . ( ) / ln ( Ik T h 7. Find and v rms for an assembly of two molecules, one with a speed of 1.00 m/s and the other with a speed of 3.00 m/s. v (m/s) 24 2 00 3 00 1 (m/s) 00 2 00 3 00 1 2 2 2 1 2 1 . ] . . [ . ) . . ( · + · · + · rms v v 9. At what temperature will the average molecular kinetic energy in gaseous hydrogen equal the binding energy of a hydrogen atom? k T KE 2 3 · solving for T with 1 E KE − · K 10 05 1 eV/K 10 62 8 eV 6 13 3 2 3 2 5 5 1 × · × · , _ ¸ ¸ − · − . ) . ( ) . )( / ( k E T ¡¼sol¡ ½ ¡ ¼sol¡ ½ For a monatomic hydrogen, the kinetic energy is all translational and Inha University Department of Physics 11. Find the width due to the Doppler effect of the 656.3-nm spectral line emitted by a gas of atomic hydrogen at 500 K. m k T v / 3 · pm 15.4 m 10 54 1 m/s 10 0 3 kg 10 67 1 K 500 J/K 10 38 1 3 m 10 3 656 2 3 2 11 8 27 23 9 · × · × × × × · · ∆ − − − − . . ) . / ( ) )( . ( ) . ( / c m k T λ λ 13. Verify that the average value of 1/v for an ideal-gas molecule is . / k T m π 2 )] / ( : [ a dv ve av 2 1 Note 0 2 · ∫ ∞ − > < · · , _ ¸ ¸ , _ ¸ ¸ · , _ ¸ ¸ · · ∫ ∫ ∞ − ∞ v k T m m k T k T m dv ve k T m N N dv v n v N v k T mv 1 2 2 2 4 2 4 1 1 1 1 2 3 0 2 2 3 0 2 π π π π π / / / ) ( ¡ ¼sol¡ ½ For nonrelativistic atoms, the shift in wavelength will be between +λ(v/ c) and -λ(v/ c) a n d t h e widt h of t h e Doppler-br oa den ed lin e will be 2λ(v/ c). Us in g t h e r ms s peed fr om KE=(3/ 2)k T = (1/ 2)mv 2 , , an d ¡ ¼sol¡ ½ The average value of 1/v is Inha University Department of Physics 17. How many independent standing waves with wavelengths between 95 and 10.5 mm can occur in a cubical cavity 1 m on a side? How many with wavelengths between 99.5 and 100.5 mm? (Hint: First show that g(λ)dλ = 8πL 3 dλ/λ 4 .) Similarly, the number of waves between99.5mm and 100.5mm is 2.5x10 2 , lower by a factor of 10 4 . ν ν π ν ν d c L d g 3 2 3 8 · ) ( λ λ π λ λ λ π ν ν λ λ d L d c c c L d g d g 4 3 2 2 3 3 8 8 · , _ ¸ ¸ · · ) ( ) ( 6 4 3 10 5 2 mm 0 1 mm 10 m 1 8 × · · . ) . ( ) ( ) ( ) ( π λ λ d g Therefore the number of standing waves between 9.5mm and 10.5mm is l9. A thermograph measures the rate at which each small portion of a persons skin emits infrared radiation. To verify that a small difference in skin temperature means a significant difference in radiation rate, find the percentage difference between the total radiation from skin at 34 o and at 35 o C. ¡ ¼sol¡ ½ The number of standing waves in the cavity is Inha University Department of Physics ¡ ¼sol¡ ½ By the Stefan-Boltzmann law, the total energy density is proportional to the fourth power of the absolute temperature of the cavity walls, as The percentage difference is 4 T R σ · % . . 3 1 013 0 K 308 K 307 1 1 4 1 2 4 1 4 2 4 1 4 1 4 2 4 1 · · , _ ¸ ¸ − · , _ ¸ ¸ − · − · − T T T T T T T T σ σ σ For temperature variations this small, the fractional variation may be approximated by 013 0 K 308 K 1 3 3 3 4 3 4 4 . ) ( · · ∆ · ∆ · ∆ · ∆ T T T T T T T R R 21. At what rate would solar energy arrive at the earth if the solar surface had a temperature 10 percent lower than it is? %) ( . ) . )( . ( 66 kW/m 92 0 90 0 kW/m 4 1 2 4 2 · · ϒ…sol¡ ½ Lowering the Kelvin temperature by a given fraction will lower the radiation by a factor equal to the fourth power of the ratio of the temperatures. Using 1.4 kW/m 2 as the rate at which the sun’ s energy arrives at the surface of the earth Inha University Department of Physics 23. An object is at a temperature of 400 o C. At what temperature would it radiate energy twice as fast? ) ( ] ) [( / C 527 K 800 K 2 673 K 273 400 2 4 1 4 4 o T T · × · · + 25. At what rate does radiation escape from a hole l0 cm 2 in area in the wall of a furnace whose interior is at 700 o C? W 51 m 10 10 K 973 K W/(m 10 67 5 2 4 4 4 2 8 4 · × ⋅ × · · − − ) ( ) ))( . ( ' A T P σ 27. Find the surface area of a blackbody that radiates 100 kW when its temperature is 500 o C. If the blackbody is a sphere, what is its radius? 4 T Ae P σ · 2 2 2 4 4 2 8 3 4 cm 494 m 10 94 4 K 273 500 K W/(m 10 67 5 1 W 10 100 · × · + ⋅ × × · · − − . ) ) ))(( . )( ( T e P A σ ϒ…sol¡ ½ To radiate at twice the radiate, the fourth power of the Kelvin temperature would need to double. Thus, ¡ ¼sol¡ ½ The power radiated per unit area with unit emissivity in the wall is P=σT 4 . Then the power radiated for the hole in the wall is ¡ ¼sol¡ ½ The radiated power of the blackbody (assuming unit emissivity) is Inha University Department of Physics The radius of a sphere with this surface area is, then, cm 27 6 4 4 2 . / · · · π π A r r A 31. The brightest part of the spectrum of the star Sirius is located at a wavelength of about 290 nm. What is the surface temperature of Sirius? K 10 0 1 m 10 290 K m 10 898 2 K m 10 898 2 4 9 3 3 × · × ⋅ × · ⋅ × · − − − . . . ma x λ T 33. A gas cloud in our galaxy emits radiation at a rate of 1.0x10 27 W. The radiation has its maximum intensity at a wavelength of 10 µm. If the cloud is spherical and radiates like a blackbody, find its surface temperature and its diameter. C 17 K 290 K 10 9 2 m 10 10 K m 10 898 2 o 2 6 - 3 · · × · × ⋅ × · − . . T Assuming unit emissivity, the radiation rate is 2 4 D P A P T R π σ · · · where D is the cloud’ s diameter. Solving for D, m 10 9 8 K 290 K W/m 10 (5.67 W 10 0 1 11 2 1 4 4 2 8 - 27 4 × · , _ ¸ ¸ ⋅ × × · · . ) )( . / π πσT P D ¡¼sol¡ ½ From the Wien’ s displacement law, the surface temperature of Sirius is ¡ ¼sol¡ ½ From the Wien’ s displacement law, the surface temperature of cloud is Inha University Department of Physics 35. Find the specific heat at constant volume of 1.00 cm 3 of radiation in thermal equilibrium at 1000 K. 4 4 4 T V VaT Vu U c σ · · · The specific heat at constant volume is then J/K 10 03 3 m 10 0 1 K 1000 m/s 10 998 2 K W/m 10 67 5 16 16 12 3 6 3 8 4 2 8 3 − − − × · × × ⋅ × · · ∂ ∂ · . ) . ( ) ( . ) . ( V T c T U c V σ 37. Show that the median energy in a free-electron gas at T=0 is equal to ε F /2 2/3 =0.630ε F . ε ε ε ε ε ε d d F M ∫ ∫ · 0 2 1 0 ε ϒ…sol¡ ½ The total energy(U) is related to the energy density by U=Vu , where V is the volume. In terms of temperature, ¡ ¼sol¡ ½ At T=0, all states with energy less than the Fermi energy ε F are occupied, and all states with energy above the Fermi energy are empty. For 0≤ε≤ε F , the electron energy distribution is proportional to . The median energy is that energy for which there are many occupied states below the median as there are above. The median energy ε M is then the energy such that Inha University Department of Physics Evaluating the integrals, F F F M ε ε ε ε ε 63 0 or 2 3 2 1 M 2 3 3 1 2 3 3 2 . ) ( , ) ( ) ( / / / · · · 39. The Fermi energy in silver is 5.51 eV. (a)What is the average energy of the free electrons in silver at O K? (b)What temperature is necessary for the average molecular energy in an ideal gas to have this value? (c)What is the speed of an electron with this energy? eV 31 3 5 3 0 . · · F ε ε (b) Setting (3/2)kT=(3/5)ε F and solving for T, K 10 56 2 eV/K 10 8.62 eV 51 5 5 2 5 2 4 5 - × · × · · . . k T F ε (c) The speed in terms of the kinetic energy is m/s 10 08 1 kg 10 11 9 5 J/eV 10 602 1 eV 51 5 6 5 6 2 6 31 19 × · × × · · · − − . ) . ( ) . )( . ( m m KE v F ε 43. Show that, if the average occupancy of a state of energy ε F +∆ε is f l at any temperature, then the average occupancy of a state of energy ε F -∆ε is f 2 =1-f 1 . (This is the reason for the symmetry of the curves in Fig.9.10 about ε F .) ¡ ¼sol¡ ½ (a) The average energy at T=0 K is Inha University Department of Physics ¡ ¼sol¡ ½ Using the Fermi-Dirac distribution function 1 1 1 + · ∆ + · ∆ k T F FD e f f / ) ( ε ε ε 1 1 2 + · ∆ − · ∆ − k T F FD e f f / ) ( ε ε ε 1 1 1 1 1 1 1 1 2 1 · + + + · + + + · + ∆ ∆ ∆ ∆ − ∆ k T k T k T k T k T e e e e e f f / / / / / ε ε ε ε ε 45. The density of zinc is 7.l3 g/cm 3 and its atomic mass is 65.4 u. The electronic structure of zinc is given in Table 7.4, and the effective mass of an electron in zinc is 0.85 m e . Calculate the Fermi energy in zinc. eV 11 J 10 78 1 kg/u 10 66 (65.4u)(1. 8 kg/m 10 13 7 2 3 kg 10 11 2(0.85)(9. s J 10 626 6 8 2 3 2 18 3 2 27 - 3 3 31 - 2 34 3 2 2 · × · , _ ¸ ¸ × × , _ ¸ ¸ × ⋅ × · , _ ¸ ¸ · − − . ) ) . )( ( ) ) . ( ) ( / / * π π ρ ε Zn Zn F m m h ¡¼sol¡ ½ Zinc in its ground state has two electrons in 4s subshell and completely filled K, L, and M shells. Thus, there are two free electrons per atom. The number of atoms per unit volume is the ratio of the mass density ρ Zn to the mass per atom m Zn . Then, Inha University Department of Physics 47. Find the number of electron states per electronvolt at ε=ε F /2 in a 1.00-g sample of copper at O K. Are we justified in considering the electron energy distribution as continuous in a metal? ε ε ε 2 3 2 3 / ) ( ) ( − · F N n At ε=ε F /2, The number of atoms is the mass divided by the mass per atom, ( ) F N n F ε ε 8 3 2 · 21 27 3 10 48 9 kg/u 10 66 1 u 55 63 kg) 10 00 1 × · × × · − − . ) . )( . ( . ( N states/eV 10 43 1 eV 04 7 10 48 9 8 3 2 21 21 × · × · , _ ¸ ¸ . . . F n ε with the atomic mass of copper from the front endpapers and ε F =7.04 eV. The number of states per electronvolt is and the distribution may certainly be considered to be continuous. ¡ ¼sol¡ ½ At T=0, the electron distribution n(ε) is Inha University Department of Physics 49. The Bose-Einstein and Fermi-Dirac distribution functions both reduce to the Maxwell- Boltzmann function when e α e ε/ k T >>1. For energies in the neighborhood of k T, this approximation holds if e α >>1. Helium atoms have spin 0 and so obey Bose-Einstein statistics verify that f(ε)=1/e α e ε/ k T ≈Ae -ε/ kT is valid for He at STP (20 o C and atmospheric pressure, when the volume of 1 kmol of any gas is=22.4 m 3 ) by showing that of A<< l under these circumstances. To do this, use Eq(9.55) for g(ε)dε with a coefficient of 4 instead of 8 since a He atom does not have the two spin states of an electron, and employing the approximation, find A from the norma1ization condition n(ε)dε=N, where N is the total number of atoms in the sample. (A kilomole of He contains Avogadro’ s number No atoms, the atomic mass of He is 4.00 u and ∫ ∞ − · 0 2a a dx e x x / / π α ϒ…sol¡ ½ Using the approximation f(ε)=Ae -ε/ k T , and a factor of 4 instead of 8 in Equation (9.55), Equation (9.57) becomes ε ε π ε ε ε ε ε ε d e h Vm A d f g d n k T / / ) ( ) ( ) ( − · · 3 2 3 2 4 Integrating over all energies, ε ε π ε ε ε d e h Vm A d n N k T ∫ ∫ ∞ − ∞ · · 0 3 2 3 0 2 4 / / ) ( Inha University Department of Physics . The integral is that given in the problem with x= ε and a=k T, 2 3 3 3 3 2 3 2 2 2 4 / / ) ( ) ( mkT h V A k T h Vm A N π π π · · 2 3 3 2 / ) ( − · mkT h V N A π 2 3 0 ) ( / k T d e k T π ε ε ε · ∫ ∞ − , so that Solving for A, Using the given numerical values, which is much less than one. , 10 56 . 3 )] K 293 )( J/K 10 1 kg/u)(1.38 10 66 . 1 )( u 00 . 4 ( 2 [ ) s J 10 626 . 6 ( kg/kmol 22.4 kmol 10 022 . 6 6 2 / 3 23 - 27 3 34 1 26 − − − − − × · × × × ⋅ × × · π A Inha University Department of Physics 51. The Fermi-Dirac distribution function for the free electrons in a metal cannot be approximated by the Maxwell-Boltzmann function at STP for energies in the neighborhood of k T. Verify this by using the method of Exercise 49 to show that A>1 in copper if f(ε)≈Aexp(ε/k T). As calculated in Sec. 9.9 N/ V=8.48x10 28 electrons/m 3 for copper. Note that Eq.(9.55) must be used unchanged here. ¡ ¼sol¡ ½ Here, the original factor of 8 must be retained, with the result that , 10 50 . 3 )] K 293 )( J/K 10 38 . 1 )( 10 11 . 9 ( 2 [ ) s J 10 63 . 6 )( m 10 48 . 8 ( ) 2 ( 2 1 3 2 / 3 23 31 3 34 3 26 2 1 2 / 3 3 × · × × × ⋅ × × · · − − − − − − π π k T m h V N A e Which is much greater than one, and so the Fermi-Dirac distribution cannot be approximated by a Maxwell-Boltzmann distribution. 5. Two observers, A on earth and B in a spacecraft whose speed is 2.00 x 108 m/s, both set their watches to the same time when the ship is abreast of the earth. (a) How much time must elapse by A's reckoning before the watches differ by 1.00 s? (b) To A, B's watch seems to run slow. To B, does A's watch seem to run fast, run slow, or keep the same time as his own watch? 【Sol】 Note that the nonrelativistic approximation is not valid, as v/c = 2/3. (a) See Example 1.1. In Equation (1.3), with t representing both the time measured by A and the time as measured in A's frame for the clock in B's frame to advance by to, we need 2 2    1 − 1 − v  = t 1 − 1 −  2   = t × 0.255 = 1.00 s t − t0 = t      c2   3     from which t = 3.93 s. (b) A moving clock always seems to run slower. In this problem, the time t is the time that observer A measures as the time that B's clock takes to record a time change of to. Inha University Department of Physics 7. How fast must a spacecraft travel relative to the earth for each day on the spacecraft to correspond to 2 d on the earth? 【Sol】 From Equation (1.3), for the time t on the earth to correspond to twice the time t0 elapsed on the ship’s clock, v2 1 3 1 − 2 = , so v = c = 2.60 × 108 m/s, 2 2 c relating three significant figures. 9. A certain particle has a lifetime of 1.00 x10-7 s when measured at rest. How far does it go before decaying if its speed is 0.99c when it is created? 【Sol】 The lifetime of the particle is t0, and the distance the particle will travel is, from Equation (1.3), vt = vt 0 1 − v /c 2 2 = ( 0.99)( 3.0 × 108 m/s)(1.00 × 10− 7 s) 1 − ( 0.99) 2 = 210 m to two significant figures. Inha University Department of Physics 8). ν νo ν 1 + v /c where the sign convention for v is that of Equation (1. c 3.0 × 108 m/s so that λ= λ = λo 1 − v /c 1 + 0. For this problem. If one of the characteristic wavelengths of the light the galaxy emits is 550 nm. v 1.000 km/s. what is the corresponding wavelength measured by astronomers on the earth? 【Sol】 See Example 1.050 Inha University Department of Physics .3. note that c c νo 1 − v /c = = λo . for the intermediate calculations. which v positive for an approaching source and v negative for a receding source. A galaxy in the constellation Ursa Major is receding from the earth at 15.050 = (550 nm) = 578 nm 1 + v /c 1 − 0.11.050.50 × 107 km/s =− = −0. 6). If the receiver on earth can measure frequencies to the nearest hertz. 1 + ( u /c ) νr = νo 1 − (u /c ) 1 − (u /c )2 = νo 1 + (u /c ) 1 + (u /c ) The last expression for vo. In Equation (1. moving away from the earth (V < 0). which essentially incorporates the classical result (counting the number of ticks). In Equation (1. and allows expression of the ratio νc 1 = . all of which need to use the fact that when the frequencies due to the classical and relativistic effects are found.8)). where u is the speed of the spacecraft. are ν0 νc = . A spacecraft receding from the earth emits radio waves at a constant frequency of 109 Hz. while differing by 1 Hz.6). we have v = 0 and V = -u.13. at what spacecraft speed can the difference between the relativistic and classical Doppler effects be detected? For the classical effect.4). assume the earth is stationary. we have v = u (or v = -u in Equation (1. The classical and relativistic frequencies. is motivated by the derivation of Equation (1. will both be sufficiently close to vo = 109 Hz so that vo could be used for an approximation to either. those frequencies. 2 νr 1 − (u /c ) Inha University Department of Physics . vc and vr respectively. 【Sol】 This problem may be done in several ways. 2 2c which is solved for u = 2 ×10−9c = 1. with the result 1 u2 = 10−9 . and even numerical solutions would require a higher precision than is commonly available.4km/s. we can use 1 − 1 − (u /c )2 ≈ (1/ 2)(u /c )2 . The denominator will be indistinguishable from 1 at low speed.34×104 m/s= 13.Use of the above forms for the frequencies allows the calculation of the ratio ∆ν νc − ν r 1 − 1 − (u /c )2 1 Hz = = = 9 = 10−9 νo νo 1 + (u /c ) 10 Hz Attempts to solve this equation exactly are not likely to be met with success. recognizing that the numerator 1 − 1 − (u /c )2 is of the form that can be approximated using the methods outlined at the beginning of this chapter. However. Inha University Department of Physics . 1 − v /c 1 − v /c which is Equation (1. For an approaching source. which is Equation (1. the angle θ = ±π/2 (or ±90o). and the given expression becomes 1 − v 2 /c 2 1 + v /c ν = νo = νo . θ = π (or 180o). 1 + v /c 1 + v /c Inha University Department of Physics .8). and cos θ = 1. the frequency v the observer finds is given by 1 − v 2 /c 2 ν = νo 1 − (v /c ) cos θ where v is the relative speed of the source. 1 − v 2 /c 2 1 − v /c = νo . For a receding source. 【Sol】 The transverse Doppler effect corresponds to a direction of motion of the light source that is perpendicular to the direction from it to the observer. and ν = νo 1 − v 2 /c 2 .5). If the angle between the direction of motion of a light source of frequency vo and the direction from it to an observer is 0. Show that this formula includes Eqs.15. so cos θ = 0. (1. θ = 0.5) to (1. The given expression becomes ν = νo which is Equation (1.8).7) as special cases. cos θ = 1. 90c relative to the earth. 【Sol】 The time will be the length as measured by the observer divided by the speed.6 ft 19.9). and this is what any observer in the spacecraft will measure.0 × 10 m/s) Inha University Department of Physics .32 × 10− 8 s 8 v v (0. What is his height as measured by an observer in the same spacecraft? By an observer on the earth? 【Sol】 The astronaut’s proper length (height) is 6 ft.90)2 = 2. From Equation (1.17. or L Lo 1 − v 2 /c 2 (1.100)(3. How much time does a meter stick moving at 0.00 m) 1 − (0. An astronaut whose height on the earth is exactly 6 ft is lying parallel to the axis of a spacecraft moving at 0.100)2 t= = = = 3.100c relative to an observer take to pass the observer? The meter stick is parallel to its direction of motion. an observer on the earth would measure L = Lo 1 − v 2 /c 2 = ( 6 ft ) 1 − (0. 9). The angle as seen from the earth will then be    tan(10o )  L ′ sin(10o ) arctan  = arctan  = 14o. an observer on the earth would measure an angle θ given by tan θ = tan θo 1 − v 2 /c 2 Inha University Department of Physics . while the length perpendicular to the spacecraft's motion will appear unchanged. the length in the direction of the spacecraft's axis will be contracted as described by Equation (1. and the projection onto an axis perpendicular to the spacecraft's axis will have a length L'sin(10o).70c. A spacecraft antenna is at an angle of 10o relative to the axis of the spacecraft. L ′ cos(10o ) 1 − v 2 /c 2   1 − (0.70)2      The generalization of the above is that if the angle is 00 as measured by an observer on the spacecraft. If the spacecraft moves away from the earth at a speed of 0. To an observer on the earth. the projection of the antenna onto the spacecraft will have a length L'cos(10o). what is the angle of the antenna as seen from the earth? 【Sol】 If the antenna has a length L' as measured by an observer on the spacecraft (L' is not either L or LO in Equation (1.21.9)). Dynamite liberates about 5. not mv 27.4 x 106 J/kg when it explodes. in that a force on an object changes the object's momentum. momentum should be conserved in any inertial frame. In the absence of forces. at a speed of 0. A woman leaves the earth in a spacecraft that makes a round trip to the nearest star. symbolically. M × (3.0 × 10−11. 4 lightyears distant. and the conserved quantity is p = -γmv.23.4 × 106 J/kg) = 6. v 0.9 2 = 5 yr. 【Sol】 The age difference will be the difference in the times that each measures the round trip to take. F = dp/dt should still be valid.0 × 108 m/s)2 Inha University Department of Physics .9c. All definitions are arbitrary. or ∆t = 2 Lo 4 yr 1 − 1 − v 2 /c 2 = 2 1 − 1 − 0. What fraction of its total energy content is this? 【Sol】 For a given mass M. the ratio of the mass liberated to the mass energy is M × (5.9 ( ) ( ) 25. What is the objection to defining linear momentum as p = mv instead of the more complicated p = γ mv? 【Sol】 It is convenient to maintain the relationship from Newtonian mechanics. but some are more useful than others. 【Sol】 Classically.200 MeV × 1.60 × 10−19 J/eV 9. then E = 2mc2 and Equation (1.23) for v as a function of K.    e   e  2 2 Inha University Department of Physics . Find its speed according to classical and relativistic mechanics.60 × 108 m/s 2 31.88 × 108 m/s.23) reduces to 1 =2 2 2 1 − v /c (γ = 2 in the notation of Section 1.100 MeV.  me c 2   me c 2    1  = c 1−   = c 1−   v = c 1−   E  m c 2 + K   1 + K /(m c 2 )  .11 × 10− 31 kg 2 = 1. solving Equation (1. v = 2K = me 2 × 0. Relativistically. Solving for v.7).29. v = 3 c = 2. An electron has a kinetic energy of 0. At what speed does the kinetic energy of a particle equal its rest energy? 【Sol】 If the kinetic energy K = Eo = mc2. 1  v = c 1 −   = 0.64 × 108 m/s.511)  The two speeds are comparable. 【Sol】 Using Equation (1. A particle has a kinetic energy 20 times its rest energy.511 MeV) = 0.  21  2 2 2 Inha University Department of Physics .511. for larger values of the ratio of the kinetic and rest energies.100/0. larger discrepancies would be found.100) /(0.100 MeV)/(0. E = Eo + 20Eo. but not the same.  1  v = 3.9989c .0 × 108 m/s × 1 −   = 1. 33.23) and solving for v/c.22) in Equation (1.With K/(mec2) = (0.  1 + ( 0. v E  = 1−  o  c E  With E = 21Eo. Find the speed of the particle in terms of c. that is. 4 X 108 m/s? 【Sol】 The difference in energies will be.294 MeV 2 2  1 − ( 2.4 / 3. 【Sol】 Using the expression in Equation (1. does not equal the kinetic energy of a particle moving at relativistic speeds. from Equation (1. Prove that ½γmv2.0)    37. 2 c 2  γ − 1  2 c 2 1 − 1 − v 2 /c 2    Inha University Department of Physics .2 / 3.0) 1 − (1. How much work (in MeV) must be done to increase the speed of an electron from 1.511 MeV ) −  = 0. the ratio of the two quantities is 1 γmv 2 2 K  1 v2  γ  1 v2  1 =  =  .   1 1 m ec  −  2 2 2 2  1 − v 2 /c 1 − v1 /c    2   1 1 = ( 0.20) for the kinetic energy.35.23).2 x 108 m/s to 2. where. If the CM of the box is to remain in its original place. and when it is emitted. mS]. the final position of the center of mass is M  L  M  L   − m  + S  −  + m  − S  = 0.39. the radiation has the momentum p = Eo/c. the mass M of the box is equally divided between its two ends. the result MS = mL is obtained. also given by Einstein. the radiation must have transferred mass from one end to the other. L L Mc c c 2 Inha University Department of Physics .  2  2   2  2  Expanding the products and canceling similar terms [(M/2)(L/2). A burst of electromagnetic radiation of energy Eo is emitted by one end of the box. the box recoils with the speed v ≈ E01Mc so that the total momentum of the system remains zero.27 shows a rigid box of length L that rests on a frictionless surface. After a time t ≈ L/c the radiation reaches the other end of the box and is absorbed there. as shown in the problem statement. According to classical physics. Then. Figure 1. The distance 5 is the product vt. MS M E L E m= = = . is based on the principle that the location of the center of mass (CM) of an isolated system cannot be changed by any process that occurs inside the system. v ≈ E/Mc (approximate in the nonrelativistic limit M >> Elc2) and t ≈ L/c. Show that this amount of mass is m = EO 1c2. 【Sol】 Measured from the original center of the box. An alternative derivation of the mass-energy formula EO = mc2. which brings the box to a stop after having moved the distance S. so that the original position of the center of mass is 0. 600c ) 1 − (0. In its own frame of reference. where t is the time in the proton's frame that it takes to cross the galaxy. which is about 105 light-years in diameter. Lo Lo 105 ly 9 E = Eoγ = Eo ≈ Eo ≈ (10 eV ) × (3 × 107 s/yr ) = 1019 eV L ct c ( 300 s ) 43. Combining. is the same as the ratio LO/L.600) 2 = 0. of course).16).383 MeV /c Inha University Department of Physics . from Equation (1. 【Sol】 Taking magnitudes in Equation (1.9). with v ≈ c (but v < c. (a) What is the approximate energy of the proton in electronvolts?. a proton takes 5 min to cross the Milky Way galaxy. where L is the diameter of the galaxy in the proton's frame of reference. where EO is the proton's rest energy and γ = 1/ 1 − v 2 /c 2.511 MeV /c 2 )( 0. γ.600c. p= mev 1 − v /c 2 2 = ( 0. The energy of the proton will be E = Eoγ. (b) About how long would the proton take to cross the galaxy as measured by an observer in the galaxy's reference frame? 【Sol】 To cross the galaxy in a matter of minutes. Find the momentum (in MeV/c) of an electron whose speed is 0.41. the proton must be highly relativistic. and for the highly-relativistic proton L ≈ ct. However. 37 GeV/c Inha University Department of Physics . and Equation (1.500) 2 = 0.) The value of the speed may be substituted into Equation (1.938 /3. the total energy is twice the rest energy.17) gives 3 The result of Problem 1-29 could be used directly. or p= 3(mec 2 ) /c = 3( 511 keV /c ) = 1.45. p = (E /c )2 − (Eo /c )2 = (3.24) may be solved for the magnitude of the momentum. γ = 2.938 GeV/c )2 = 3. v = ( p= mec.500 GeV 【Sol】 Solving Equation (1.963c numerically 2. 3 47. Find the speed and momentum (in GeV/c) of a proton whose total energy is 3.23) for the speed v in terms of the rest energy EO and the total energy E.24) becomes 4 4 4mec 4 = mec 4 + p 2c 2 . as above. and Equation (1. v = c 1 − ( Eo / E ) = c 1 − ( 0.94 GeV /c /2)c.16) (or the result of Problem 1-46). (The result of Problem 1-32 does not give an answer accurate to three significant figures. Find the momentum of an electron whose kinetic energy equals its rest energy of 511 keV 【Sol】 When the kinetic energy of an electron is equal to its rest energy. or Equation (1.500 GeV /c )2 − (0.888 x 108 m/s. giving v = c2 p pc 335 MeV =c =c = 0. cancelling the m2c4 term. Expanding the binomial. a very convenient result is that of Problem 1-46.24). (mc 2 + K )2 = m 2c 4 + p 2c 2 = ( 335 MeV )2 − ( 62 MeV )2 = 874 MeV /c 2 . E 874 MeV + 62 MeV mc 2 + K m= ( pc )2 − K 2 Inha University Department of Physics . A particle has a kinetic energy of 62 MeV and a momentum of 335 MeV/c. 2c 2 K 2c 2 ( 62 MeV ) The particle's speed may be found any number of ways. 【Sol】 From E = mc2 + K and Equation (1.49. and solving for m. Find its mass (in MeV/c2) and speed (as a fraction of c).36c . An observer detects two explosions. Inserting this into Equation (1. the time interval that the second observer measures should be that given by Equation (1.998 × 108 m/s) 2 (For this calculation. (2.44). Algebraically and numerically.97 ms. is to note that when the relative speed of the observers (5. and a good cheek.00 ms later 100 km away.00 x 107 m/s) has been determined. = ( 2. the different methods give the same result. one that occurs near her at a certain time and another that occurs 2.00 × 107 m/s t 2.3) (but be careful of which time it t. x2 x2 t− 2 1− 2 2 ( x /t )2 tc c t t′ = =t = t 1− 2 2 2 2 c2 1 − ( x /ct ) 1 − x /c t = 1. Another observer finds that the two explosions occur at the.) An equally valid method. What time interval separates the explosions to the second observer? 【Sol】 The given observation that the two explosions occur at the same place to the second observer means that x' = 0 in Equation (1.00 ms ) 1 − (5. the approximation is 1 − (x /ct)2 ≈ 1 − (x 2 / 2c 2t 2 )valid to three significant figures. same place. and so the second observer is moving at a speed x 1.41).00 × 105 m v= = = 5.51.00 × 10−3 s with respect to the first observer.00 × 107 m/s)2 Inha University Department of Physics . which is to). 53. From Equation (1. Inha University Department of Physics . In the reference frame of the fixed stars. 2 x′ = x − vt 2 = r cos θ − ( −r /c ) 1 − v /c 2 2 =r cosθ + (v /c ) 1 − v /c 2 2 . one of which may be the source). In the unprimed frame. and the minus sign merely indicates that the signal was sent before it was received. where r is the distance from the source to the place where the ship receives the signal. in the unprimed frame (given here as the frame of the fixed stars. the speed of the spacecraft is v and the signal arrives at an angle θ to the axis of the spacecraft.41). the signal arrives at an angle 0 with respect to the positive x-direction. (b) What would you conclude from this result about the view of the stars from a porthole on the side of the spacecraft? 【Sol】 (a) A convenient choice for the origins of both the unprimed and primed coordinate systems is the point. Then. (a) With the help of the Lorentz transformation find the angle θ ' at which the signal arrives in the reference frame of the spacecraft. Take the direction of the ship's motion (assumed parallel to its axis) to be the positive x-direction. where the ship receives the signal. 1 − v /c and y’ = y = r sin θ. the signal was sent at a time t = -r/c. Then. so that in the frame of the fixed stars (the unprimed frame). in both space and time. A spacecraft moving in the +x direction receives a light signal from a source in the xy plane. x = r cos θ and y = r sin θ . 800c.900) 1 + vV x Inha University Department of Physics . including the stars.800 + 0. it can be seen that the numerator of the term in square brackets is less than sinθ . Looking out of a porthole. will appear to be in the directions close to the direction of the ship’s motion than they would for a ship with v = 0. to an observer in A. and so tan θ and θ’ < θ when v ≠ 0. sinθ 1 − v 2 /c 2  and θ ′ = arctan . The relative velocities will have opposite directions. A and B.988c .800)( 0. y′ sinθ tanθ ′ = = . As v àc. the sources. is then ′ Vx + v 0. cosθ + (v /c )     (b) From the form of the result of part (a). x ′ (cos θ + (v /c ))/ 1 − v 2 /c 2 55. ′ /c 2 1 + ( 0.900c. and all stars appear to be almost on the ship’s axis(farther forward in the field of view). θ’ à0. coming toward him from opposite directions at the respective speeds of 0. and the denominator is greater than cosθ . The speed with which B is seen to approach A. A man on the moon sees two spacecraft. (a) What does a man on A measure for the speed with which he is approaching the moon? For the speed with which he is approaching B? (b) What does a man on B measure for the speed with which he is approaching the moon? For the speed with which he is approaching A ? 【Sol】 (a) If the man on the moon sees A approaching with speed v = 0. but the relative speeds will be the same. then the observer on A will see the man in the moon approaching with speed v = 0.800c and 0.800 c.900 Vx = = c = 0. 1 + (0.900)(0.900 c.) S’(moon) S B Vx’ O’ v A Inha University Department of Physics .900 + 0.(b) Similarly.800 c = 0. to an observer on B.800) (Note that Equation (1. will be 0.49) is unchanged if Vx’ and v are interchanged. the observer on B will see the man on the moon approaching with speed 0.988c. and the apparent speed of A. while all other physical quantities. Inha University Department of Physics . KEmax = hν − φ = h(ν − νo ). remained the same. So that while KEmax is a linear function of the frequency ν of the incident light. KEmax is not proportional to the frequency. If Planck’s constant had a smaller value. 3. such as the speed of light. quantum phenomena would be less conspicuous than they are now. That is.9). If Planck's constant were smaller than it is. what would a correct statement of the relationship between KEmax and ν be? 【Sol】 No: the relation is given in Equation (2. would quantum phenomena be more or less conspicuous than they are now? 【Sol】 Planck’s constant gives a measure of the energy at which quantum effects are observed. quantum effects would be seen for phenomena that occur at higher frequencies or shorter wavelengths.Chapter 2 Problem Solutions 1.8) and Equation (2. Is it correct to say that the maximum photoelectron energy KEmax is proportional to the frequency ν of the incident light? If not. or P P 1. -9 700 × 10 m Or. A 1.0 × 108 m/s) E = = 2. in terms of joules.77 eV.63 × 10 J ⋅ s)(880 × 10 Hz) Inha University Department of Physics . Find the energy of a 700-nm photon. How many photons per second does it emit? 【Sol】 The number of photons per unit time is the total energy per unit time(the power) divided by the energy per photon.00-kW radio transmitter operates at a frequency of 880 kHz.84 × 10−19 J −9 700 × 10 m 7.72 × 1030 photons/s .63 × 10−34 J ⋅ s)(3.5. ( 6.24 × 10− 6 eV ⋅ m E = = 1.00 × 103 J/s = = = 1.11). 【Sol】 From Equation (2. 1. −34 3 E hν (6. 9.2 × 1045 photons/s .63 × 10.” for “astronomical unit. commonly abbreviated as “1 AU. at the rate of 1. hν (6.63 × 10.0 × 1026 J/s (6. P/A). or P /A 1.4 × 103 W/m 2 = = 4. even if there is no surface to absorb the energy. 2 ( P / A )4πRE −S = (1.4 x 103 W/m2 of area perpendicular to the direction of the light. where RE-S is the mean Earth-Sun distance. The total power is then. and how many photons per second does it emit? (c) How many photons per cubic meter are there near the earth? 【Sol】 (a) The number of photons per unit time per unit are will be the energy per unit time per unit area (the power per unit area. Inha University Department of Physics . all points at the same distance from the sun should have the same flux of energy.5 x 1011 m away. Light from the sun arrives at the earth.0 × 1014 Hz ) = 1.” The number of photons emitted per second is this power divided by the energy per photon. or 4. divided by the energy per photon.2 × 1021 photons/(s ⋅ m2 ).34 J ⋅ s)( 5.0 × 1026 W.0 x 1014 Hz. Assume that sunlight is monochromatic with a frequency of 5.4 × 103 W/m 2 )4π (1.34 J ⋅ s)(5.0 × 1014 Hz) (b) With the reasonable assumption that the sun radiates uniformly in all directions. (a) How many photons fall per second on each square meter of the earth's surface directly facing the sun? (b) What is the power output of the sun. an average of 1.5 × 1011 m )2 = 4. 2 × 1021 photons/(s ⋅ m 2 ) = 1.5 eV to be ejected? 【Sol】 Expressing Equation (2.5 eV )( 230 × 10− 9 m )  = ( 230 nm) 1 +  1.4 × 1013 photons/m3. hc λ   K λ= = λ0 1 + max o  (hc / λo ) + K max hc    (1.0 × 10 m/s 11. and in the same direction (spreading is not significant on the scale of the earth). and performing the needed algebraic manipulations.24 × 10− 6 eV ⋅ m   −1 −1 = 180 nm. 8 3. and so the number of photons per unit time per unit area is the product of the number per unit volume and the speed. The maximum wavelength for photoelectric emission in tungsten is 230 nm. Inha University Department of Physics . What wavelength of light must be used in order for electrons with a maximum energy of 1. Using the result from part (a).(c) The photons are all moving at the same speed c.9) in terms of λ = c/ν and λ0 = c/ν0. 4. 5 × 10− 6 J/s )( 400 × 10−9 m ) I =e =e =e = (1e ) = 0.9 eV.8). and hc 1. find the current in the cell. From Equation (2.5 mW of 400-nm light is directed at a photoelectric cell.48 µA E hc / λ hc 1. the available power is (1. 【Sol】 Because only 0. the current will be the product of the number of photoelectrons per unit time and the electron charge.10 percent of the incident photons produce photoelectrons.3 eV where the value of φ for sodium is taken from Table 2.24 × 10− 6 eV ⋅ m K max = hν − φ = −φ = − 2.5x10-3W) = 1.6 eV ⋅ m Inha University Department of Physics . If 0.1.0x10-3)(1.13.3 eV = 3.24 × 10−6 eV ⋅ m λmax = = = 539 nm φ 2. or P P Pλ (1.24 × 10. hc 1. −9 λ 200 × 10 m 15. What is the maximum wavelength of light that will cause photoelectrons to be emitted from sodium? What will the maximum kinetic energy of the photoelectrons be if 200-nm light falls on a sodium surface? 【Sol】 The maximum wavelength would correspond to the least energy that would allow an electron to be emitted. so the incident energy would be equal to the work function. 1.5x10-6 W.10% of the light creates photoelectrons. 0 × 1014 Hz − 8.34 J ⋅ s The work function φ may be obtained by substituting the above result into either of the above expressions relating the frequencies and the energies.ν2.2 ν1 − K max.0 × 1014 Hz − 8. while possibly more cumbersome than direct substitution.14 × 10−15 eV ⋅ s = 6.1 19.97 eV From these data find Planck's constant and the work function of the surface .1 ν 2 = hν1ν 2 − φν 2 .7 eV)(8. K max.52 eV The same surface illuminated by 12.5 × 1014 Hz to the allowed two significant figures.2 ν1 = hν 2ν1 − φν1.1 ν 2 (19.17. yielding φ = 3. subtracting to eliminate the product hν1ν2 and dividing by ν1 .1 × 10−15 eV ⋅ s ν 2 − ν1 12. K max. 【Sol】 Denoting the two energies and frequencies with subscripts 1 and 2.7 eV − 0.ν2 to obtain K max. a case of "swatting a fly with a sledgehammer".0 eV to the same two significant figures.0 × 1014 Hz) φ= = = 3. using such a program for this problem is. Keeping an extra figure gives h = 4.0 x 1014 Hz light emits electrons whose maximum energy is 1.52 eV)(12.0 eV ν 2 − ν1 (12. reflects the result of solving the system of equations using a symbolic-manipulation program. or the equations may be solved by rewriting them as K max.5 x 1014 Hz light emits electrons whose maximum energy is 0.64 × 10.) Inha University Department of Physics . K max. Subtracting to eliminate the work function φ and dividing by ν1 . A metal surface illuminated by 8.2 = hν 2 − φ .52 eV h= = = 4.5 × 1014 Hz) − (0.5 × 1014 Hz) (This last calculation. of course. K max. 2 − K max.1 = h ν1 − φ . and hence would have had more energy than after the interaction. and the photon would have had positive energy. However. Show that it is impossible for a photon to give up all its energy and momentum to a free electron. the electron's final kinetic energy is 2 p 2c 2 + mec 4 − mec 2 ≠ pc for nonzero p. and hence could not have existed. Equating the two expressions for Ee2 or 2 Ee = ( pc )2 + mec 2 = pc + mec 2 0 = 2( pc ) me c 2 . and if the electron were to attain all of the photon's momentum and energy. In this frame.19. so energy could not be conserved. the final electron kinetic energy must be KE = Eo = pc. ( ) ( ) 2 This is only possible if p = 0. To see the same result without using as much algebra. but before the interaction. the electron would have been moving (to conserve momentum). mec2. This is the reason why the photoelectric effect can take place only when photons strike bound electrons. for any electron we must have Ee2 = (pc)2 + (mec2)2. the final momentum of the electron must be pe = po = p. ( ) ( ( ) 2 ) 2 = ( pc )2 + 2( pc ) mec 2 + mec 2 . The photon's initial momentum is po = Eo/c. and so the final electron energy is Ee = pc + mec2. Inha University Department of Physics . in which case the photon had no initial momentum and no initial energy. 【Sol】 Consider the proposed interaction in the frame of the electron initially at rest. the final energy is the rest energy of the electron. An easier alternative is to consider the interaction in the frame where the electron is at rest after absorbing the photon. Electrons are accelerated in television tubes through potential differences of about 10 kV.9  2d   2 × 0.14 × 10−15 eV ⋅ s which corresponds to x-rays.030 nm  o θ = arcsin   = arcsin   = 2.4 × 1018 Hz h h 4.030-nm x-rays. The distance between adjacent atomic planes in calcite (CaCO3) is 0. What kind of waves are these? 【Sol】 For the highest frequency. The frequency of this radiation will be E eV (1e )(10 × 103 V) ν = = = = 2. and this energy will appear as the electromagnetic radiation emitted when these electrons strike the screen. Find the highest frequency of the electromagnetic waves emitted when these electrons strike the screen of the tube.300 nm  Inha University Department of Physics .  λ   0. Find the smallest angle of Bragg scattering for 0. 【Sol】 Solving Equation (2.21.13) for θ with n = 1. 23.300 nm. the electrons will acquire all of their kinetic energy from the accelerating voltage. λC .8 x 10-8 nm. which is much less than o. Show that this assumption is reasonable by calculating the Compton wavelength of a Na atom and comparing it with the typical x-ray wavelength of 0.7 the x-rays scattered by a crystal were assumed to undergo no change in wavelength.22).1 nm.Na h 6.0 × 1018 Hz h 6.0 × 108 m/s)(1. 2.1 × 10.63 × 10−34 J ⋅ s = = = 5.82 × 10. In See.23 kg ⋅ m/s) ν = = = 5. but with a sodium atom instead of an electron.) Inha University Department of Physics .0 × 108 m/s)(3. (Here.1 nm.63 × 10− 34 J ⋅ s 27.82 x 10-26 kg was taken from Problem 2-24. cp ( 3.8 × 10−17 m.1 x 10-23 kg m/s? 【Sol】 From Equation (2. 【Sol】 Following the steps that led to Equation (2.25. What is the frequency of an x-ray photon whose momentum is 1.26 kg) or 5. the rest mass MNa =3. cM Na (3.15). 43 × 10−12 m   1 λC  19 ν′ =  + = +  = 2. What is the wavelength of the x-rays in the direct beam? 【Sol】 Solving Equation (2.4 × 10 Hz 19 8 c  ν  3. An x-ray photon of initial frequency 3.0 × 10 Hz The above method avoids the intermediate calculation of wavelengths. and with cos 90o = 0. 31. A beam of x-rays is scattered by a target.426 pm)(1 − cos 45o ) = 1. At 45o from the beam direction the scattered x-rays have a wavelength of 2. the wavelength of the x-rays in the direct beam.0 x 1019 Hz collides with an electron and is scattered through 90o.2 pm − (2. with λ = c/ν and λ’ = c/ν’ . 【Sol】 Rewriting Equation (2. −1 Inha University Department of Physics .0 × 10 m/s   3.29.2 pm. Find its new frequency. c c = + λC ν′ ν and solving for ν’ gives −1  1 2.5 pm to the given two significant figures.23) for λ.23) in terms of frequencies. λ = λ ′ − λC (1 − cos φ ) = 2. 432 cos φ = 1 + − = 1+  −    E λC λC E′   100 keV 90 keV    from which φ = 64o to two significant figures. 35. ν′ =ν = = = λ′ 1 + ( ∆λ / λ ) 1 + ( 2λC / λ ) 1 + ( 2νλC /c ) where ∆λ = 2λC for φ = 180o. 【Sol】 For the electron to have the maximum recoil energy.  mc 2 mc 2  λ λ′  = 1 +  511 keV − 511 keV  = 0. With this expression. A photon of frequency ν is scattered by an electron initially at rest. 1 + ( 2νλC /c ) mc 2 Using λC = h/(mc) (which is Equation (2. Inha University Department of Physics . At what scattering angle will incident 100-keV x-rays leave a target with an energy of 90 keV? 【Sol】 Solving Equation (2. and Equation (2.23) for cos φ.33. the scattering angle must be 1800. Verify that the maximum kinetic energy of the recoil electron is KEmax = (2h2 ν2/mc2)/(1 + 2hν/mc2).22)) gives the desired result. consider λ ν ν ν .20) becomes mc2 KEmax = 2 (hv) (hv'). where KEmax = (hv . 2(h ν )(hν ′) 2(hν )2 /(mc 2 ) KE max = = . To simplify the algebra somewhat.hv') has been used. 37. 2(1 − cos φ ) At this point. there are many ways to proceed. so the above expression reduces to sin φ tanθ = . not the recoil angle of the scattering electron. and hence no interaction). what is the energy of the scattered photon? 【Sol】 As presented in the text. For an analytic solution which avoids the intermediate calculation of the scattering angle φ. with E = mc2. a numerical solution with θ = 40 o gives φ = 61. λC  ( ∆λ / λ ) + (1 − cos φ ) ( λC /λ )(1 − cos φ ) + (1 − cos φ )  1 + (1 − cos φ ) λ   For the given problem.cos2 φ = (1 + cos φ)(1 – cos φ) to obtain 1 + cos φ 4 tan 2 θ = 1 − cos φ (the factor 1 . If the electron moves off at an angle of 40o with the original photon direction. This may be re-expressed as Inha University Department of Physics . A photon whose energy equals the rest energy of the electron undergoes a Compton collision with an electron.cos φ may be divided. one method is to square both sides of the above relation and use the trigonometric identity sin2 φ = 1 . represents an undeflected photon. as cos φ = 1. the energy of the scattered photon is known in terms of the scattered angle. Consider the expression for the recoil angle as given preceding the solution to Problem 2-25: sin φ sin φ sin φ tan θ = = = . φ = 0. λ = hc/E = h/(mc) = λC.6 0 to three significant figures. 6 o more readily. φ = 2 arctan   2 2  2 tan θ  yielding the result θ = 61. λ 1 + 4 tan 2 θ 1 + 4 tan 2 ( 40o ) ′=E E =E = ( 511 keV) = 335 eV λ′ 3 + 4 tan 2 θ 3 + 4 tan 2 ( 40o ) An equivalent but slightly more cumbersome method is to use the trigonometric identities φ φ φ sin φ = 2 sin cos . 2 − cos φ = . tan θ = Inha University Department of Physics . or 2 3 + 4 tan2 θ 1 − cos φ = . 1 + 4 tan2 θ 1 + 4 tan2 θ Then with λ’ = λ + λC(1 – cos φ) = λC(2 – cos φ). 1 − cos φ = 2 sin 2 2 2 2 in the expression for tan θ to obtain 1 φ  1  cot .(1 − cos φ )( 4 tan 2 θ ) = 1 + cos φ = 2 − (1 − cos φ ). regardless of its initial energy. At this point. from which cos φ = ½ and φ = 60o. The scattered wavelength (a maximum) corresponding to this minimum energy is λ’max = (h/Emin ). so λC = 2λ’max .21 × 10−13 m = 0. Each particle had a kinetic energy of 1.24 × 10−6 eV ⋅ m λ= = = 8.23) with λ ' = λ'max) Inha University Department of Physics .821 pm E 1. h 2hc 2hc λC = = = .39. As an alternative. solving Equation (2. Show that.cos φ = ½ for λ ' = λC /2. A positron collides head on with an electron and both are annihilated. the angle at which the scattered photons will have wavelength λ’max can m be found as a function of the incoming photon energy E. The wavelength of each photon will be hc 1. 【Sol】 The energy of each photon will he the sum of one particle's rest and kinetic energies. and so 1 . a photon cannot undergo Compton scattering through an angle of more than 60o and still be able to produce an electron-positron pair.511 MeV (keeping an extra significant figure). it is possible to say that for the most energetic incoming photons. (Hint: Start by expressing the Compton wavelength of the electron in terms of the maximum photon wavelength needed for pair production.51 × 106 eV 41.) 【Sol】 Following the hint. mc 2mc 2 E min where Emin = 2mc2 is the minimum photon energy needed for pair production.00 MeV Find the wavelength of the resulting photons. 1. λ ~ 0. 693 I = I oe − µx ⇒ x 1/2 = = . which we know to be the case. ln10 2. 43. of an absorber required to reduce the intensity of a beam of radiation by a factor of 2 is given by x1/2 = 0. E must be greater than 2mc2.693/µ. cos φ = ½ and so φ = 60o. λC λC λC 2 E This expression shows that for E >> mc2.′ ′ λmax − λ λmax hc / E 1 mc 2 cos φ = 1 − = 1− + = + . (a) Show that the thickness x1/2. with Io/I = 10. so that ln 2 0.26) with Io/I = 2. 【Sol】 (a) The most direct way to get this result is to use Equation (2. (b) Find the absorber thickness needed to produce an intensity reduction of a factor of 10. because cos φ must always be less than 1.30 x1/10 = = . µ µ Inha University Department of Physics . µ µ (b) Similarly. but it also shows that. for pair production at any angle. 9 m-1 in water and 52 in -1 in lead. µ Pb 52 m -1 −3 = (10 × 10 m ) = 0. 【Sol】 From either Equation (2.26) or Problem 2-43 above. The linear absorption coefficient for 1-MeV gamma rays in lead is 78 m-1.45.693 x 1/ 2 = = = 8. Equation (2. The linear absorption coefficients for 2. find the thickness of lead required to reduce by half the intensity of a beam of such gamma rays.26) indicates that the ratios will be the same when the distances in water and lead are related by µ H 2O x H2O = µ Pb x Pb . or x H2O = x Pb or 11 cm two significant figures.9 m -1 Inha University Department of Physics . ln 2 0.0-MeV gamma rays are 4. What thickness of water would give the same shielding for such gamma rays as 10 mm of lead? 【Sol】 Rather than calculating the actual intensity ratios.9 mm µ 78 m -1 47.106 m µ H2O 4. 06 pm.26) or use of the result of Problem 2-43 gives ln 2 x 1/ 2 = = 1.06 × 10-12 m = 1.6 ) = 1. Inha University Department of Physics .0 x 1030 kg and its radius is 7.12 × 10.0 × 10 m/s) (7. 51. 【Sol】 In Equation (2. and ∆λ = λ GM c 2R = (500 × 10− 9 m)(2. the ratio ∆λ/λ will be the same as ∆ν/ν. What thickness of copper is needed to reduce the intensity of the beam in Exercise 48 by half. 4 -1 4. The sun's mass is 2.29).12 × 10− 6 2 8 2 4 -1 c R (3.67 × 10−11 N ⋅ m2 / kg)(2.7 × 10 m which is 0.49.0 × 1030 kg) = = 2.0 x 108 m. 【Sol】 Either a direct application of Equation (2.47 × 10−5 m.015 mm to two significant figures.0 × 10 m ) (keeping an extra significant figure) is so small that for an “approximate” red shift. the ratio GM (6. Find the approximate gravitational red shift in light of wavelength 500 nm emitted by the sun. is to realize that the nucleus will be moving nonrelativistically after the emission of the photon. So. Then. 2. the recoil momentum of the nucleus 2 2 2 is E/c.4 keV to reach its ground state. Such a source was used in the experiment described in See.cited 2576Fe nucleus is part of a 1. 12.4-keV gamma-ray photon after it has fallen 20 m near the earth's surface? 【Sol】 (a) The most convenient way to do this problem. This phenomenon is known as the Mössbauer effect. As discussed in Chap. The mass of a 57 Fe atom is 9. What is the original frequency and the change in frequency of a 14. and that the energy of the photon will be very close to E∞ = 14. for computational purposes. here M is the rest mass of the nucleus.9. certain atomic nuclei emit photons in undergoing transitions from "excited" energy states to their “ground” or normal states.4 keV. (a) The 57 Co nucleus 27 57 decays by K capture to 26 Fe .53. if the photon has an energy E. By how much is the photon energy reduced in this situation if the ex. the energy that the photon would have if the nucleus had been infinitely massive. When a nucleus emits a photon. These photons constitute gamma rays. instead of the individual atom. and its kinetic energy is p / 2M = E /(2Mc ) . conservation of energy implies Inha University Department of Physics .0-g crystal? (c) The essentially recoil-free emission of gamma rays in situations like that of b means that it is possible to construct a source of virtually monoenergetic and hence monochromatic photons. which then emits a photon in losing 14. it recoils in the opposite direction. By how much is the photon energy reduced from 26 the full 14.5 x 10-26 kg.4 keV available as a result of having to share energy and momentum with the recoiling atom? (b) In certain crystals the atoms are so tightly bound that the entire crystal recoils when a gamma-ray photon is emitted. and solution might be attempted by standard methods.9 × 103 eV. Mc   However. the dimensionless quantity E∞/(Mc2) is so small that standard calculators are not able to determine the difference between E and E∞. using (1 + x)1/2 ≈ 1 + (x/2) . and recognizing that E will be very close to E∞. but to find the change in energy due to the finite mass of the nucleus.5 × 10 kg)(3.60 × 10−16 J/keV) = = 1.4 keV) 2 (1. 2Mc 2 This is a quadratic in E.+ E = E∞ . is not made. and two terms must be kept to find the difference between E and E∞. the resulting quadratic is E 2 + 2Mc 2E − 2Mc 2E ∞ = 0.(x2/8). The square root must be expanded. .9 × 106 keV = 1.26 8 2 2(9. which is solved for   E E = Mc 2  1 + 2 ∞2 − 1. This approximation gives the previous result. E2 Inha University Department of Physics . the above relation may be expressed as 2 E2 E∞ E∞ − E = ≈ 2Mc 2 2Mc 2 (14.0 × 10 m/s) If the approximation E ≈ E∞. It so happens that a relativistic treatment of the recoiling nucleus gives the same numerical result, but without intermediate approximations or solution of a quadratic equation. The relativistic form expressing conservation of energy is, with pc = E and before, E 2 + (Mc2 )2 + E = Mc 2 + E∞ , or E 2 + (Mc 2 )2 = Mc 2 + E∞ − E . Squaring both sides, canceling E2 and (Mc2)2, and then solving for E, 2  1 + ( E ∞ /(2Mc 2 ))  E ∞ + 2Mc 2E ∞  E = = E∞  2  1 + (E /(Mc 2 )) . 2(Mc + E ∞ )  ∞  From this form, 2  E∞  1   E∞ − E =  , 2Mc 2  1 + E ∞ /(Mc 2 )   giving the same result. (b) For this situation, the above result applies, but the nonrelativistic approximation is by far the easiest for calculation; 2 E∞ (14.4 × 103 eV)2 (1.6 × 10−19 J/eV) E∞ − E = = = 1.8 × 10− 25 eV. 2 -3 8 2 2Mc 2(1.0 × 10 kg)(3.0 × 10 m/s) E∞ 14.4 × 103 eV (c) The original frequency is ν = = = 3.48 × 1018 Hz. −15 h 4.14 × 10 eV ⋅ s From Equation (2.28), the change in frequency is (9.8 m/s 2 )( 20 m )  gH  ∆ ν = ν ′ − ν =  2 ν = ( 3.48 × 1018 Hz) = 7.6 Hz. 8 2 c  (3.0 × 10 m/s) Inha University Department of Physics 55. The gravitational potential energy U relative to infinity of a body of mass m at a distance R from the center of a body of mass M is U = -GmM/R. (a) If R is the radius of the body of mass M, find the escape speed v, of the body, which is the minimum speed needed to leave it permanently. (b) Obtain a formula for the Schwarzschild radius of the body by setting vc = c, the speed of light, and solving for R. (Of course, a relativistic calculation is correct here, but it is interesting to see what a classical calculation produces.) 【Sol】 (a) To leave the body of mass M permanently, the body of mass m must have enough kinetic energy so that there is no radius at which its energy is positive. That is, its total energy must be nonnegative. The escape velocity ve is the speed (for a given radius, and assuming M >> m) that the body of mass m would have for a total energy of zero; 1 2GM 2 GMm mv e − = 0, or v e = . 2 R R (b) Solving the above expression for R in terms of ve, R= 2GM 2 ve , and if ve = c, Equation (2.30) is obtained. Inha University Department of Physics Chapter 3. Problem Solutions 1. A photon and a particle have the same wavelength. Can anything be said about how their linear momenta compare? About how the photon's energy compares with the particle's total energy? About how the photon’s energy compares with the particle's kinetic energy? 【Sol】 From Equation (3.1), any particle’s wavelength is determined by its momentum, and hence particles with the same wavelength have the same momenta. With a common momentum p, the photon’s energy is pc, and the particle’s energy is ( pc )2 + (mc 2 )2 , which is necessarily greater than pc for a massive particle. The particle’s kinetic energy is K = E − mc 2 = (pc )2 + (mc 2 ) 2 − mc 2 For low values of p (p<<mc for a nonrelativistic massive particle), the kinetic energy is K ≈ p2/2m, which is necessarily less than pc. For a relativistic massive particle, K ≈ pc – mc2, and K is less than the photon energy. The kinetic energy of a massive particle will always be less than pc, as can be seen by using E = (pc)2 + (mc2)2 to obtain ( pc )2 − K 2 = 2Kmc 2. Inha University Department of Physics h 6. Find the de Broglie wavelength of a 1.0-mg grain of sand blown by the wind at a speed of 20 m/s.71 × 10− 22 kg ⋅ m /s.100 + (0. Inha University Department of Physics . −6 mv (1. Problem Solutions 3. (K + mc ) − (mc ) 2 2 2 2 = ( 0.79 × 10− 22 kg ⋅ m/s.511)2 MeV /c = 1.3 × 10− 29 m. the percentage error in the wavelength will be the same as the percentage error in the reciprocal of the momentum. 【Sol】 For this nonrelativistic case. The nonrelativistic momentum is pnr = 2mK = 2( 9. with the nonrelativistic calculation giving the higher wavelength due to a lower calculated momentum.63 × 10−34 J ⋅ s λ= = = 3. By what percentage will a nonrelativistle calculation of the de Broglie wavelength of a 100-keV electron be in error? 【Sol】 Because the de Broglie wavelength depends only on the electron's momentum.1 × 10−31 kg)(100 × 103 eV)(1.6 × 10-19 J/eV) = 1.0 × 10 kg)(20 m/s) quantum effects certainly would not be noticed for such an object. and the relativistic momentum is pr = 1 c 5.Chapter 3. ) This energy is much less than the neutron's rest energy.282 nm. Problem Solutions keeping extra figures in the intermediate calculations.03 × 10− 3 eV 2 2 2 6 -9 2 2m 2mc 2mc λ 2(939.282 × 10 m) (Note that in the above calculation.282 nm. Find the kinetic energy (in eV) of a neutron with a de Broglie wavelength of 0.71 7. multiplication of numerator and denominator by c2 and use of the product hc in terms of electronvolts avoided further unit conversion.79 − 1. The percentage error in the computed de Broglie wavelength is then (h / pnr ) − (h / pr ) pr − pnr 1.Chapter 3. The atomic spacing in rock salt. Inha University Department of Physics .6 × 10 eV)(0.8 % . 【Sol】 A nonrelativistic calculation gives p 2 (hc /λ )2 (hc )2 (1. and so the nonrelativistic calculation is completely valid.71 = = = 4. NaCl. is 0. Is a relativistic calculation needed? Such neutrons can be used to study crystal structure. h / pr pnr 1.24 × 10−6 eV ⋅ m) 2 K = = = = = 1. Chapter 3. Problem Solutions Green light has a wavelength of about 550 nm. and so the de Broglie wavelengths would be the same. 2m 2mc 2 2(mc 2 )λ2 2( 511 × 103 eV)(550 × 10. 9. so the momentum of such a particle would be nearly the same as a photon with the same energy. 【Sol】 If E2 = (pc)2 + (mc2)2 >> (mc2)2.0 µV. Inha University Department of Physics . E = pc.24 × 10− 6 eV ⋅ m) 2 K = = = = = 5. its de Broglie wavelength is nearly the same as the wavelength of a photon with the same total energy. so the nonrelativistic calculation is valid.) 11.9 m) 2 so the electron would have to be accelerated through a potential difference of 5. then pc >> mc2 and E ≈ pc.0 × 10−6 eV. For a photon with the same energy. Note that the kinetic energy is very small compared to the electron rest energy. Show that if the total energy of a moving particle greatly exceeds its rest energy.0 x 10-6 V = 5. Through what potential difference must an electron be accelerated to have this wavelength? 【Sol】 A nonrelativistic calculation gives p 2 (hc / λ )2 (hc ) 2 (1. multiplication of numerator and denominator by c2 and use of the product he in terms of electronvolts avoided further unit conversion. (In the above calculation. the phase velocity vp = (ω/k) = u. the electron will have the longer wavelength by a factor of (mp/me) = 1838. 【Sol】 Suppose that the phase velocity is independent of wavelength. An electron and a proton have the same velocity Compare the wavelengths and the phase and group velocities of their de Broglie waves. From Equation (3. relativistic or nonrelativistic.16) they have the same group velocity. from Equation (3.3). there is no dispersion). a constant. then. and hence independent of the wave number k.3) the particles have the same phase velocity and from Equation (3. Verify the statement in the text that. if the phase velocity is the same for all wavelengths of a certain wave phenomenon (that is. 15.Chapter 3. Problem Solutions 13. dk Inha University Department of Physics . the group and phase velocities are the same. the momentum will be proportional to the mass. It follows that because ω = uk. and so the de Broglie wavelength will be inversely proportional to the mass. dω vg = = u = v p. 【Sol】 For massive particles of the same speed. dω 1 1 1 g 1ω 1 vg = = g = = = v p. where g is the acceleration of gravity. dk g gk ω2 ω 1 vg = = = = = vp. Find the group velocity of ocean waves 【Sol】 The phase velocity may be expressed in terms of the wave number k = 2π/λ as ω g vp = = . For those more comfortable with calculus. from which 2 ω k 2k 2 Inha University Department of Physics .Chapter 3. dω 2ω = 2ωv g = g . so that 2ω 2ωk 2ωk 2k 2 the same result. or ω = gk or ω 2 = gk. dk 2 k 2 k 2k 2 Using implicit differentiation in the formula for ω2(k). k k Finding the group velocity by differentiating ω(k) with respect to k. Problem Solutions 17. and v g = = v p. dω dk 1ω 1 = . The phase velocity of ocean waves is gλ / 2π . the dispersion relation may be expressed as 2 ln(ω ) = ln(k ) + ln(g ). 21.16). Find the phase and group velocities of the de Broglie waves of an electron whose kinetic energy is 500 keV. 【Sol】 1 K + mc 2 500 + 511 = = = 1.863c .16). and from Equation (3.Chapter 3. in that vg = v. 【Sol】 (a) Two equivalent methods will be presented here. p mv g γ mv g c Inha University Department of Physics . 2 vg h h h λ= = = 1− 2. v = c 1 − (1/γ )2 = c 1 − (1/1.16 c. Both will assume the validity of Equation (3. First: Express the wavelength x in terms of vg.978. The phase velocity is then vp = c2 /vg = 1. γ = 2 2 2 511 mc 1 − v /c Solving for v.978)2 = 0. For a kinetic energy of 500 keV. Problem Solutions 19. (a) Show that the phase velocity of the de Broglie waves of a particle of mass m and de Broglie wavelength λ is given by 2  mc λ  vp = c 1 +    h  (b) Compare the phase and group velocities of an electron whose de Broglie wavelength is exactly 1 x 10-13 m.863c. vg = v = 0. It should be noted that in the first method presented above could be used to find λ in terms of vp directly. gives the desired result. Inha University Department of Physics . 2 2 2 2 2 2 vp 1 + h /(mcλ ) h (mcλ ) + 1 1 + (mcλ ) /h which is an equivalent statement of the desired result.Multiplying by mvg.3). or course. squaring and solving for vg2 gives   mλc  2  h 2 2 vg = = c 1 +    . and in the second method the energy could be found in terms of vg. so that v p 1 + h 2 /(mc λ )2 c2 1 h 2 (mcλ )2 1 −1 = = 2 = . vp = c2/vg. 2 −1 Second: Consider the particle energy in terms of vp = c2 lvg. E 2 = ( pc )2 + mc 2 γ mc 2 ( ) 2 2 = 1−c ( )2 (mc 2 )2 2 /v 2 p 2  hc  =   + mc 2 . The final result is. the same. 2 2 2 ( λm ) + (h /c )   h     Taking the square root and using Equation (3.  λ  2 ( ) Dividing by (mc2)2 leads to c2 1 1− 2 = . 24 × 10 eV ⋅ m     In both of the above answers.99915c.00085c . write the result of part (a) as  mc 2λ   (511 × 103 eV)(1. Inha University Department of Physics .63 × 10 J⋅s   and vg = c2/vp = 0. vp = c 1 +  −34   6.00085c .  (9.(b) Using the result of part (a).0 × 108 m/s)(1. What effect on the scattering angle in the Davisson-Germer experiment does increasing the electron energy have? 【Sol】 Increasing the electron energy increases the electron's momentum. For a calculational shortcut.00 × 10-13 m)     = 1.1 × 10− 31 kg)(3.13). and hence decreases the electron's de Broglie wavelength.0 × 10-13 m)   = 1. the statement that the de Broglie wavelength is “exactly” 10-13 m means that the answers can be given to any desired precision. From Equation (2. vp = c 1 +  −6  hc  = c 1 +   1. 2 2 2 23. a smaller de Broglie wavelength results in a smaller scattering angle. Inha University Department of Physics .36 × 10 m/s) or 0.1 × 10 kg outside the crystal. Problem Solutions 25. the de Brogile wavelengths are found from h h 6.36 m/s − 31 m 9.11 × 10 kg)(4. 2K 2( 54 eV)(1.63 × 10− 34 J ⋅ s λ= = = = 1.167 nm outside the crystal. 3. which reduces its de Broglie wavelength.5 it was mentioned that the energy of an electron entering a crystal increase. with K = 80 eV).30 x 106 m/s inside the crystal (keeping an extra significant figure in both calculations). 【Sol】 (a) For the given energies. Consider a beam of 54-eV electrons directed at a nickel target.137 nm inside the crystal.60 × 10-19 J/eV) v = = = 4.Chapter 3. with a similar calculation giving 0. v = 5. (b) With the speeds found in part (a). The potential energy of an electron that enters the target changes by 26 eV. (b) Compare the respective de Broglie wavelengths. (a) Compare the electron speeds outside and inside the target.67 × 10−10 m. a nonrelativistic calculation is sufficient. −31 6 p mv ( 9. and (from a similar calculation. In Sec. 60 × 10-19 J/eV) = 4. What is the neutron's minimum energy? (The diameter of an atomic nucleus is of this order of magnitude.3 fm. How wide is the box? 【Sol】 The first excited state corresponds to n = 2 in Equation (3.00 × 10-14 m) 2 The minimum energy.) 【Sol】 From Equation (3.18). is 20. A proton in a one-dimensional box has an energy of 400 keV in its first excited state.18). Obtain an expression for the energy levels (in MeV) of a neutron confined to a one-dimensional box 1.53 × 10−14 m = 45. Inha University Department of Physics .00 x 10 -14 m wide. corresponding to n = 1.67 × 10− 27 kg)(1.Chapter 3. Problem Solutions 27.63 × 10− 34 J ⋅ s) 2 L =n =2 8mE 2 8(1.67 × 10− 27 kg)(400 × 103 eV)(1. h2 ( 6. 29.5 MeV. 8mL 2 8(1.28 × 10−13 J = n 2 20.5 MeV En = n 2 h2 =n 2 (6.63 × 10−34 J ⋅ s) 2 = n 2 3. Solving for the width L. 00 × 103 eV) Note that in the above calculation. while no such restriction holds for the molecules in an ideal gas. what is the percentage of uncertainty in its momentum? 【Sol】 The percentage uncertainty in the electron's momentum will be at least ∆p h h hc = = = p 4πp∆x 4π∆x 2mK 4π∆x 2(mc )2 K 4π (1.100 nm.otherwise the assembly of atoms would not be a solid.24 × 10− 6 eV ⋅ m) = 3. Inha University Department of Physics .626 x 10-34 J·s will of course give the same percentage uncertainty. = (1. If its position is located to within 0.Chapter 3. Problem Solutions 31.1 × 10− 2 = 3. The position and momentum of a 1.1 %. conversion of the mass of the electron into its energy equivalent in electronvolts is purely optional. The uncertainty in position of each atom is therefore finite. 33. The position of an ideal-gas molecule is not restricted. 【Sol】 Each atom in a solid is limited to a certain definite region of space . The atoms in a solid possess a certain minimum zero-point energy even at 0 K. converting the kinetic energy into joules and using h = 6.00-keV electron are simultaneously determined. Use the uncertainty principle to explain these statements. and its momentum and hence energy cannot be zero.00 × 10−10 m) 2(511 × 103 eV)(1. so the uncertainty in its position is effectively infinite and its momentum and hence energy can be zero. = 1. m m which is consistent with the previous result. In such a situation.00 keV of kinetic energy? 【Sol】 The proton will need to move a minimum distance h v∆t ≥ v . ∆ p2 p ∆E = = 2 ∆p = 2v∆p.44 × 10−13 m = 0. Problem Solutions 35. this is not inconsistent with Equation (3. ∆E was taken to be the (maximum) kinetic energy of the proton.21).00 × 103 eV) (See note to the solution to Problem 3-33 above).Chapter 3. ∆x ∆p ≥ h/4π . 4π∆E where v can be taken to be 2K 2 ∆E v = = . ( ) Inha University Department of Physics .144 pm. In the current problem.24 × 10−6 eV ⋅ m = 1. How accurately can the position of a proton with v << c be determined without giving it more than 1. The result for the product v∆t may be recognized as v∆t ≥ h/2πp. so that m m 2K h h hc v ∆t = = = m 4π∆E 2π 2mK 2π 2(mc 2 )K 2π 2( 938 × 106 MeV)(1. (b) What is the approximate minimum bandwidth (that is. The time needed for the reflections of these groups to return indicates the distance to a target.0 × 108 m/s)(8.0800 µs in duration.0 × 10-5 s) = 24 m. (8. The number of waves in each group is the pulse duration divided by the wave period.37.0 × 10− 8 s)(4900 × 106 Hz) = 752 waves.5 MHz.0 × 10−8 s)-1 = 12. (8. (a) Find the length of each group and the number of waves it contains. (b) The bandwidth is the reciprocal of the pulse duration. which is the pulse duration multiplied by the frequency. spread of frequencies) the radar receiver must be able to process? 【Sol】 (a) The length of each group is c∆t = (3. A marine radar operating at a frequency of 9400 MHz emits groups of electromagnetic waves 0. Inha University Department of Physics .  8π m  h      Inha University Department of Physics .  8π m  x 2   d E = 0. make the identification of p with ∆p and x with ∆x. and  h 2  1 C  2 E = E(x ) =  2  2 +   x . In classical physics the minimum energy of the oscillator is Emin = 0. Problem Solutions 39. The frequency of oscillation of a harmonic oscillator of mass m and spring constant C is The energy of the oscillator is E = p2/2m + Cx2/2. 2π mC Substution of this value into E(x) gives  h 2  2π mC   C  h  h C hν  +   E min =  2    2  2π mC  = 2π m = 2 .  4π m  x   which is solved for h x2 = . Differentiating with respect to x and setting dx  h2  1 −  2  3 + Cx = 0.Chapter 3. momentum when its displacement from the equilibrium position is x. so that p = h/ (4πx). where p is its ν = C /m / 2π . Use the uncertainty principle to find an expression for E in terms of x only and show that the minimum energy is actually Emin = hν/2 by setting dE/dx = 0 and solving for Emin . 【Sol】 To use the uncertainty principle. with no angular momentum with respect to the nucleus (an "Impact parameter" of zero.60 × 10 = = (8. the incident proton must be directed "head-on" to the nucleus.00-MeV protons incident on gold nuclei. and so the potential energy at closest approach will be the initial kinetic energy. Equating these energies. 3.Chapter 4. and gases and metals are overall electrically neutral. most of the volume of an atom is empty space.  4πε  K  −13 1. In this case. 【Sol】 For a "closest approach". see the Appendix to Chapter 4). K initial rmin Ze 2 = . To what conclusion about atomic structure does this observation lead? 【Sol】 The fact that most particles pass through undetected means that there is not much to deflect these particles. Problem Solutions 1.14 × 10−13 m. 4πεormin or −19  1  Ze 2 C) 2 9 2 2 (79 )(1 . at the point of closest approach the proton will have no kinetic energy. Determine the distance of closest approach of 1.99 × 10 N ⋅ m / C ) = 1.60 × 10 J  o  initial Inha University Department of Physics . The great majority of alpha particles pass through gases and thin metal foils with no deflections. taking the potential energy to be zero in the limit of very large separation. must be negative. For a classical particle subject to an inverse-square attractive force (such as two oppositely charged particles or two uniform spheres subject to gravitational attraction in a circular orbit. the shortest wavelength (highest energy) corresponds to the largest value of n. the electron is in constant motion. In the Bohr model. For n →∞. How can such an electron have a negative amount of energy? 【Sol】 While the kinetic energy of any particle is positive.5. the potential energy of any pair of particles that are mutually attracted is negative. the total energy.46 µm 7 -1 R 1.9). λ→ 16 16 = = 1. the sum of the positive kinetic energy and the total negative potential energy. the potential energy is twice the negative of the kinetic energy.46 × 10− 6 m = 1. What is the shortest wavelength present in the Brackett series of spectral lines? 【Sol】 The wavelengths in the Brackett series are given in Equation (4. For the system to be bound. Inha University Department of Physics .097 × 10 m 7. 【Sol】 (a) The velocity v.00 × 108 m/s ) Inha University Department of Physics .30 × 10−3. (b) Show that the value of α is very close to 1/137 and is a pure number with no dimensions. e2 e2 e4 v1 e2 1 2 v1 = = = 2 . (a) Show that α = v1/c.60 × 10−19 C)2 α = = 7. (1. is given by Equation (4. but α has nevertheless turned out to be a useful quantity in atomic physics. where v. Sommerfeld's approach was on the wrong track. (c) Show that αao = λc/2π. with r = r1 = ao. 4πεoma o  h 2ε o  4ε o h 2 c 2ε oh c  4πεom   πme 2    (b) From the above. where ao is the radius of the ground-state Bohr orbit and λc is the Compton wavelength of the electron. Because the magnetic behavior of a moving charge depends on its velocity. Combining to find v12 . so = = α.The fine structure constant is defined as α = e2/2εohc. 9. 2(8. is the velocity of the electron in the ground state of the Bohr atom. This quantity got its name because it first appeared in a theory by the German physicist Arnold Sommerfeld that tried to explain the fine structure in spectral lines (multiple lines close together instead of single lines) by assuming that elliptical as well as circular orbits are possible in the Bohr model.63 × 10−34 J ⋅ s )(3.4).85 × 10−12 C 2 / N ⋅ m 2 )(6. the small value of α is representative of the relative magnitudes of the magnetic and electric aspects of electron behavior in an atom. (c) Using the above expression for α and Equation (4. and will have the same numerical v alue in any system of units. α is a dimensionless quantity.1 to four significant figures. α accurate to better than 4 parts per billion. 2ε ohc πme 2 2π mc 2π where the Compton wavelength λC is given by Equation (2. The most accurate (November. e 2 h 2ε o 1 h λ αao = = = C. A close cheek of the units is worthwhile.13) with n = 1 for ao. 2001) value of 1/α is 1 = 137. [J] Thus. treating the units as algebraic quantities the units as given in the above calculation are [C2 ] [C2 ] [m] [J][s] [s] [N][m 2 ] = [N ⋅ m] = 1. Inha University Department of Physics .03599976.22).so that 1/α = 137. orbital speed and orbital radius of the earth known. Find the quantum number that characterizes the earth's orbit around the sun. ∆p > /2ao .0 x 104 m/s. 【Sol】 The uncertainty in position of an electron confined to such a region is.0 x 1024 kg.0 × 104 m/s)(1. its orbital radius is 1. The earth's mass is 6.5 x 1011 m. from Equation (3. Inha University Department of Physics .22).6 × 1074. 【Sol】 With the mass.11. and its orbital speed is 3.5 × 1011 m) n= = = = 2.) 13. while the magnitude of the linear momentum of an electron in the first Bohr orbit is h h h p= = = . Compare the uncertainty in the momentum of an electron confined to a region of linear dimension ao with the momentum of an electron in a ground-state Bohr orbit.13) is half of this momentum. λ 2πao ao the value of ∆p found from Equation (3. h h 1.06 × 10−34 J ⋅ s (The number of significant figures not of concern.0 × 1024 kg)(3. the earth's orbital angular momentum is known. L mvR (6. and the quantum number that would characterize the earth's orbit about the sun would be this angular momentum divided by . What is its wavelength? 【Sol】 It must assumed that the initial electrostatic potential energy is negligible. Inha University Department of Physics .24 × 10−6 eV ⋅ m λ= = = 9.15. 17. and the wavelength is hc 1.6 eV in the ultraviolet part of the spectrum (see.6 eV. so that the final energy of the hydrogen atom is E1 = -13. both at rest initially.12 × 10−8 m = 91. What effect would you expect the rapid random motion of the atoms of an excited gas to have on the spectral lines they produce? 【Sol】 The Doppler effect shifts the frequencies of the emitted light to both higher and lower frequencies to produce wider lines than atoms at rest would give rise to.2 nm. The energy of the photon emitted is then -El. − E1 13. for instance. combine to form a hydrogen atom in the ground state. the back endpapers of the text). A proton and an electron. A single photon is emitted in this process. 1 nm.21 × 10−8 m = 92. 100 1 100 1 λ= = = 9. A potential difference of 12. 1 8  ∆ E = E 3 − E1 = −E1 1 −  = (13.1 eV 9 9  (this assumes that few or none of the hydrogen atoms had electrons in the n = 2 level).097 × 10 m which is in the ultraviolet part of the spectrum (see.1 eV is necessary to accelerate the electrons to this energy. 7 -1 99 R 99 1. for instance. 21.18) with nf = 1 and ni = 10.19. Find the wavelength of the spectral line that corresponds to a transition in hydrogen from the n = 10 state to the ground state. A beam of electrons bombards a sample of hydrogen.6 eV) = 12.7) with n = 10 or Equation (4. In what part of the spectrum is this? 【Sol】 From either Equation (4. Through what potential difference must the electrons have been accelerated if the first line of the Balmer series is to be emitted? 【Sol】 The electrons’ energy must be at least the difference between the n = 1 and n = 3 levels. Inha University Department of Physics . the back endpapers of the text). and so the wavelength of the photon needed to ionize hydrogen is −1 −1  1 1  1 1    = λ= + +  = 91. (b) Use this formula to find ni for a 102.  n  λ  i  Inha University Department of Physics .6 nm    2 →1  As a check. The longest wavelength in the Lyman series is 121. note that this wavelength is R-1.13 nm. these are the energies that correspond to the longest wavelength (least energetic photon) in the Lyman series and the shortest wavelength (most energetic photon) in the Balmer series.55-nm photon. 【Sol】  1 1  = R 1 − 2 . 【Sol】 The energy needed to ionize hydrogen will be the energy needed to raise the energy from the ground state to the first excited state plus the energy needed to ionize an atom in the second excited state. An excited hydrogen atom emits a photon of wavelength λ in returning to the ground state. which is solved for (a) From Equation (4.23. The energies are proportional to the reciprocals of the wavelengths.5 nm 364.5 nm and the shortest wavelength in the Balmer series is 364. λ λ∞→ 2  121.7) with n = ni .6 nm. (a) Derive a formula that gives the quantum number of the initial ex cited state in terms of λ and R. Use the figures to find the longest wavelength of light that could ionize hydrogen. 25. 097x107 m-1) = 1. with the product λR = (102. (a) Modify Eq.  λR  λR − 1  (b) Either of the above forms gives n very close (four place) to 3. for which Ef . As a result. In this situation.16) is then −1/ 2 Inha University Department of Physics . Is the effect a major one? A nonrelativistic calculation is sufficient here. Equation (4. (b) Find the ratio between the recoil energy and the photon energy for the n = 3 → n = 2 transition in hydrogen. 27. the kinetic energy of the recoiling atom is p 2 (h ν /c ) 2 K = = .55x10-9 m)(1. 【Sol】 (a) A relativistic calculation would necessarily involve the change in mass of the atom due to the change in energy of the system. 2M 2M where m is the ftequency of the emitted photon and p = h/λ = hν/c is the magnitude of the momentum of both the photon and the recoiling atom. (4. the change is measured indirectly by measuring the energies of the emitted photons) means that a nonrelativistic calculation should suffice.16) to include this effect.Ei = 1. the linear momentum of the photon must be balanced by the recoil momentum of the atom.9 eV. When an excited atom emits a photon. The fact that this mass change is too small to measure (that is. some of the excitation energy of the atom goes into the kinetic energy of its recoil.125 rounded to four places as 9/8.1  λR  ni = 1 − = . specifically. n = 3 exactly. the result would be −1   1 Mc 2    . 2 6 2M 2M ∆E 2Mc 2 939 × 10 eV -9 to two significant figures. As in Problem 2-53. see part (b). E f − Ei = hν  1 + 1 +   2 hν      a form not often useful. a relativistic calculation is manageable.E in that problem.0 x 10 the front endpapers. (b) As indicated above and in the problem statement. the rest energy of the hydrogen atom is from or 1. a nonrelativistle calculation is sufficient. where hν = E∞.01 × 10− 9 . and the term p2/(2M) corresponds to E∞ . In the above. (h ν )2 ( ) Inha University Department of Physics . As in part (a). Ei − E f = hν + K = hν + p2 (∆E /c )2 .9 eV K = = = 1. and K = ∆E = 1. 2 2  2Mc 2Mc  This result is equivalent to that of Problem 2-53.hν   = hν  1 + . Note that in this form.19). Similarly. but with n replaced by n + 1 and p = 1. ν is positive because El is negative. 2E 1 fn = − 1 3 .17) with ni = n + 1 and nf = n. 【Sol】 There are many equivalent algebraic methods that may be used to derive Equation (4.29.20). ∆E  1 1  2E1  n + 1  2 ν= = − 2 = −  2 . hn  n + 2n + 1 n + 2n + 1      as the term in brackets is less than 1. 2 h h n (n + 1)2  (n + 1) n  This can be seen to be equivalent to the expression for v in terms of n and p that was found in the derivation of Equation (4. From this expression 2  n2 + 1n  2E1  n + 1 n  2 2 ν =− 3 2  = fn  2  < fn . Inha University Department of Physics . h n The frequency v of the photon emitted in going from the level n + 1 to the level n is obtained from Equation (4. and that result will be cited here. Show that the frequency of the photon emitted by a hydrogen atom in going from the level n + 1 to the level n is always intermediate between the frequencies of revolution of the electron in the respective orbits. (n + 1 )(n + 1) 2E1 (n + 1 )(n + 1) 2 2 ν=−  = fn +1   > fn +1. m µm p m µ = 207me .7). R' = R (m'/me) = 186R. 4/3 4 /3 λ= = = 6. as in Example 4.653 nm.53 × 10−10 m = 0. A µ− muon is in the n = 2 state of a muonic atom whose nucleus is a proton.097 × 10 m ) in the x-ray range. the Rydberg constant is multiplied by the ratio of the reduced masses of the muoninc atom and the hydrogen atom. m p = 1836me m′ = = 186me mµ + m p Inha University Department of Physics . from Equation (4.7. 3 2 2 h(n + 1)  n n    as the term in brackets is greater than 1. Find the wavelength of the photon emitted when the muonic atom drops to its ground state. 7 -1 R′ 186(1. In what part of the spectrum is this wavelength? 【Sol】 For a muonic atom. 31. a hydrogen isotope whose nucleus is approximately 3 times more massive than ordinary hydrogen. the wavelength would be λT = (36/5) (1/RT). m (m − mH )  2me ∆λ = λ  e T ≈λ . is excited and its spectrum observed. me (me + m H )  3m H   where the approximations me + rnH ≈ mH and mT ≈ 3mH have been used.33. The difference between the wavelengths would then be  R   λ  ∆ λ = λ − λT = λ 1 − T  = λ 1 − . where RT is the Rydberg constant evaluated with the reduced mass of the tritium atom replacing the reduced mass of the hydrogen atom. Inserting numerical values. RT memT /(me + mT ) mT (me + mH ) Using this in the above expression for ∆λ.097 × 107 m-1 ) 3(1.11 × 10− 31 kg) ∆λ = = 2. corresponding to n = 3 in Equation (4. (36 / 5) 2(9. have wavelengths of λ = (36/5) (1/R). For a tritium atom.38 × 10−10 m = 0. A mixture of ordinary hydrogen and tritium. (1. λ  RT     The values of R and RT are proportional to the respective reduced masses.6). How far apart in wavelength will the Hα lines of the two kinds of hydrogen be? 【Sol】 The Hα lines. and their ratio is R m m /(me + mH ) m H (me + mT ) = e H = .67 × 10− 27 kg) Inha University Department of Physics .238 nm. giving m ′Z 2e 4 1 En = − . the n = 2 level is the same as the n = 1 level for Hydrogen. and of course there are many more levels corresponding to higher n. The scale is close. Find the wavelength of the photon emitted in this process if the electron is assumed to have had no kinetic energy when it combined with the nucleus. which is an ion such as He + or Li2+ whose nuclear charge is +Ze and which contains a single electron. with Z = 2. 2 8πεoh 2 n 2 where the reduced mass m' will depend on the mass of the nucleus. for He +.15) are repeated. Inha University Department of Physics .35. (b) Sketch the energy levels of the He' ion and compare them with the energy levels of the H atom. and the n = 4 level is the same as the n = 2 level for hydrogen. 【Sol】 (a) The steps leading to Equation (4. with Ze2 instead of e2 and Z2e4 instead of e4. In the approximation that the reduced masses are the same. (b) A plot of the energy levels is given below. (a) Derive a formula for the energy levels of a hydrogenic atom. but not exact. (c) An electron joins a bare helium nucleus to form a He + ion. 6 eV) = -54.The energy levels for H and He +: (c) When the electron joins the Helium nucleus. the electron-nucleus system loses energy.8 nm.4 eV. − ∆E 54. where the result of part (a) has been used.4 eV Inha University Department of Physics .28 × 10−8 m = 22. The emitted photon's wavelength is hc 1. the emitted photon will have lost energy ∆E = 4 (-13.24 × 10− 6 eV ⋅ m λ= = = 2. 0 × 10 m/s) 39. and the scattering analysis.00-J pulses of light whose wavelength is 694 nm. which is the total energy of the pulse divided by the energy of each photon. based on a pointlike positively charged nucleus. What is the minimum number of Cr3+ ions in the ruby? 【Sol】 The minimum number of Cr3+ ions will he the minimum number of photons. The Rutherford scattering formula fails to agree with the data at very small scattering angles. Can you think of a reason? 【Sol】 Small angles correspond to particles that are not scattered much at all.00 J)(694 × 10-9 m) = = = 3. and the structure of the atom does not affect these particles. Inha University Department of Physics . the nucleus is either partially or completely screened by the atom's electron cloud. − 34 8 hc / λ hc ( 6.37. A certain ruby laser emits 1. E Eλ (1. To these nonpenetrating particles.49 × 1018 ions. is not applicable.63 × 10 J ⋅ s)(3. What fraction of a beam of 7. Inha University Department of Physics .6 × 10−13 m) = 11. θ (5.60 × 10-13 J/MeV) cot = ( 2. A 5. with f given in Equation (4.60 × 10−19 C)2 keeping extra significant figures.7-MeV alpha particles incident upon a gold foil 3. 2 (8.43.29).6 x 10-13 m.31).43) = 2 tan −1   = 10 .99 × 109 N ⋅ m 2 /C2 )(79)(1. Through what angle will it be scattered? 【Sol】 From Equation (4.f.43  43.0-MeV alpha particle approaches a gold nucleus with an impact parameter of 2. The scattering angle is then  1  o θ = 2 cot −1(11.  11.0 eV)(1. using the value for 1/4πεo given in the front endpapers.41.0 x 10-7 m thick is scattered by less than 1 o? 【Sol】 The fraction scattered by less than 1 o is 1 . is from Example 4.0 × 109 N ⋅ m2 /C 2 )2   ( 79)(1.2  Ze 2   1   Ze 2  2θ  cot  cot 2 θ f = πnt  = πnt     4πε   K   4πε K  2 2  o  o    2 2 = π (5.90 × 10 28 m-3)(3.7MeV)(1. and f ( 60o ) − f ( 90o ) cot 2( 30o ) − cot 2 ( 45o ) 3 − 1 = = = 2.31).6 × 10−19 C) 2   cot2 (0. the number of gold atoms per unit volume.16.f (90o). 2 45. × -13   ( 7.0 × 10− 7 m)(9.6 × 10 J/MeV)  where n. The fraction scattered by less than 1 o is 1 . 【Sol】 Regarding f as a function of 0 in Equation (4.8.f = 0.5o ) = 0. Inha University Department of Physics . o 2 o 1 f ( 90 ) cot ( 45 ) so twice as many particles are scattered between 60o and 90o than are scattered through angles greater than 90o.84. Show that twice as many alpha particles are scattered by a foil through angles between 60o and 90o as are scattered through angles of 90o or more. and the number scattered through angles greater than 90o is just f (90o). the number of particles scattered between 60o and 90o is f (60o) . 43 × 10− 4 deg = 0. 2 θ =  GM 2 tan −1  2 sun c R  sun −11  N ⋅ m 2 / kg 2 )( 2.67 × 10    = 2 tan 8 8    (3. as they do for the Coulomb force. Adapt Eq.0 x 1030 kg and 7. (4.0 × 10 m/s)(7.28) would be GM sunm F = .10. 【Sol】 If gravity acted on photons as if they were massive objects with mass m = Ev/c2. 1. 2 r 2 would cancel.0 × 1030 kg )  −1 ( 6. Using b = Rsun. Inha University Department of Physics . In special relativity. a conclusion supported by observations made during solar clipses as mentioned in Sec. The mass and radius of the sun are respectively 2. This suggests that we can treat a photon that passes near the sun in the same way as Rutherford treated an alpha particle that passes near a nucleus. a photon can be thought of as having a “mass” of m = Eν/c2.0 x 10 8 m.0 × 10 m)    = 2.87′′. In fact. general relativity shows that this result is exactly half the actual deflection. the magnitude of the force F in Equation (4.29) to this situation and find the angle of deflection θ for a photon that passes b = Rsun from the center of the sun. with an attractive gravitational force replacing the repulsive electrical force.47. 2 2 2 GM sun a result that is independent of the photon’s energy. and the result is the factors of r θ θ θ c 2b 2mc b sin = 2GM sun m cos and cot = . and cannot be a solution of Schrödinger's equation for all values of x. Inha University Department of Physics . (d) ψ = A exp(-x2). and is not a function at all. Figure (c) has discontinuous derivative in the shown interval.Chapter 5 Problem Solutions 1. neither ψ = A sec x or ψ = A tan x could be a solution of Schrödinger's equation for all values of x. 5. (c) ψ = A exp(x2). (b) ψ = A tan x. 【Sol】 The functions (a) and (b) are both infinite when cos x = 0. … ±(2n+1)π/2 for any integer n. Which of the wave functions in Fig. The function (c) diverges as x → ±∞. Figure (d) is finite everywhere in the shown interval. ±3π/2. and cannot have physical significance. Which of the following wave functions cannot be solutions of Schrödinger's equation for all values of x? Why not? (a) ψ =A sec x. 3. at x = ±π/2. Figure (f) is discontinuous in the shown interval.15 cannot have physical significance in the interval shown? Why not? 【Sol】 Figure (b) is double valued. Of the many ways to find this integral. evaluated between different limits for the two parts. including consulting tables and using symbolicmanipulation programs. The wave function of a certain particle is ψ = A cos2x for -π/2 < x < π /2. (a) Find the value of A. (b) Find the probability that the particle be found between x = 0 and x = π/4. and the normalization condition reduces to or A= 8 . 8 +π / 2 and ∫ +π / 2 cos 4xdx −π / 2 = 1 sin 4x −π /2 4 +π / 2 are seen to be vanish. 4 1 2 8 2 8 where the identity cos2 θ = ½(1+cos 2θ) has been used twice. a direct algebraic reduction gives cos 4 x = (cos 2 x )2 = [12 (1 + cos 2x )]2 = 1 [1 + 2 cos 2x + cos 2( 2x )] 4 = 1 [ + 2 cos 2x + 1 (1 + cos 4x )] = 3 + 1 cos 2x + 1 cos 4x . (a) The needed normalization condition is +π /2 ∗ 2 +π /2 4 ∫−π /2 ψ ψdx = A ∫−π / 2 cos xdx = A2 The integrals [∫ 3 +π / 2dx 8 −π / 2 + 1 +π / 2cos 2 xdx 2 −π / 2 ∫ + 1 +π /2 cos 4 xdx 8 −π / 2 ∫ ]= 1 ∫ +π /2 cos 2x dx −π / 2 = 1 sin 2x −π /2 2  3 1 = A 2  π .5. 3π Inha University Department of Physics . 【Sol】 Both parts involve the integral ∫cos4 xdx. (b) Evaluating the same integral between the different limits.46 32 4 32 4 7. and is differentiable to all orders with respect to both t and x. but is not normalizable. and in addition must be normalizable. continuous.9) gives the wave function of a particle moving freely (that is. Ψ∗Ψ = A*A is constant in both space and time. there can be no limit to its range. Equation (5. 5. and single-valued. or 1 [83 x + 14 sin 2x + 32 sin 4x ]π/4 = 3π + 1 . in order to give physically meaningful results in calculations a wave function and its partial derivatives must be finite. Does this wave function meet all the above requirements? If not. with no forces acting on it) in the +x direction as Ψ = Ae −(i /h)(Et − pc ) where E is the particle's total energy and p is its momentum. 0 32 4 A2 (3π + 1 ) = 38π (3π + 1 ) = 0. specifically. could a linear superposition of such wave functions meet these requirements? What is the significance of such a superposition of wave functions? 【Sol】 The given wave function satisfies the continuity condition. Inha University Department of Physics .1. ∫0 π /4 cos4 xdx = The probability of the particle being found between x = 0 and x = π/4 is the product of this integral and A2. As mentioned in Sec. and if the particle is to move freely. and so the integral of Ψ∗Ψ over an infinite region cannot be finite if A ≠ 0. h ∂ h ∂  ( px )Ψ = p( xΨ ) = ˆˆ ˆ ˆ ( xΨ ) =  Ψ + x Ψ . and hence a finite ∆x. multiply by /i and multiply by x. i ∂x i  ∂x   where the product rule for partial differentiation has been used." whereas the operator xp corresponds ˆˆ to "differentiate with respect to x.zero ∆p. Show that the expectation values <px> and <xp>) are related by <px> .<xp> = /i This result is described by saying that p and x do not commute. h ∂  h ∂  ( xp )Ψ = x ( pΨ) = x  ˆˆ ˆ ˆ Ψ  = x Ψ. corresponds to "multiply by x.A linear superposition of such waves could give a normalizable wave function. differentiate with respect to x and multiply by /i. the wave function is composed of different momentum states. i ∂x  i  ∂x    Inha University Department of Physics ." Using these operators. Such a superposition would necessarily have a non. at the expense of normalizing the wave function. Also. and it is intimately related to the uncertainty principle. and is localized. t). corresponding to a real particle. 9. when operating on the wave function Ψ(x. 【Sol】 ˆˆ It's crucial to realize that the expectation value <px> is found from the combined operator px . which. ∂ψ x  2π  2π x   = − A sin  2πνt − 2π  − A sin  2πνt − 2π . 【Sol】 Using λν = vp in Equation (3.5). t) normalized. Obtain Schrödinger’s steady-state equation from Eq.   x    = A cos 2πνt − 2π x . and using ψ instead of y. = ∂x λ  λ  λ λ   2 2 ∂ 2ψ 2π x   2π   2π  x    2π  = A cos  2πνt − 2π   − =  A cos 2πνt − 2π  = −  ψ λ λ  λ   λ  λ    λ  ∂x 2 Inha University Department of Physics . ( px − xp )Ψ = ˆ ˆ ˆˆ 11.1).Thus h Ψ i ∞ h h ∞ h and < px − xp >= ∫−∞ Ψ * Ψdx = ∫−∞ Ψ* Ψdx = i i i for Ψ(x.(3.5) with the help of de Broglie’s relationship λ = h/mv by letting y = ψ and finding ∂2ψ/∂x2 . ψ = A cos 2π  t −     v p  λ     Differentiating twice with respect to x using the chain rule for partial differentiation (similar to Example 5. The kinetic energy of a nonrelativistic particle is (E − U ) λ2 h 2 ∂ 2ψ 1 and 2 ψ Substituting the above expression relating ∂x 2 λ 2 2 2 ∂ψ 8π m 2m  2π  = −  ψ = − ( E − U )ψ = − 2 ( E − U )ψ .32)  λ  ∂x 2 h2 h so that p2 h  1 KE = E − U = =  . which is Equation (5. 2m  λ  2m 2 1 = 2m Inha University Department of Physics . One of the possible wave functions of a particle in the potential well of Fig. The wave function vanishes again where the potential V →∞. 5. and so the wavelength increases. 【Sol】 The wave function must vanish at x = 0. this condition would determine the allowed energies. The amplitude increases as the wavelength increases because a larger wavelength means a smaller momentum (indicated as well by the lower kinetic energy).17 is sketched there. and the particle is more likely to be found where the momentum has a lower magnitude.13. where V →∞. the particle's kinetic energy must decrease. Explain why the wavelength and amplitude of &P vary as they do. Inha University Department of Physics . As the potential energy increases with x. To show orthogonality. the stipulation n ≠ m means that α ≠ β and α ≠ -β and the integrals are of the form Inha University Department of Physics . but as the eigenfunctions in this problem are real. An important property of the eigenfunctions of a system is that they are orthogonal to one another. (A more general orthogonality relation would involve the integral of ψn*ψm. 2 as may be verified by expanding the cosines of the sum and difference of α and β. 【Sol】 The necessary integrals are of the form +∞ 2 L nπx mπx ∫−∞ψ nψ mdx = L ∫0 sin L sin L dx for integers n. which means that ∫ +∞ ψ ψ dV −∞ n m =0 n ≠m Verify this relationship for the eigenfunctions of a particle in a one-dimensional box given by Eq. m. the distinction need not be made. (5. a convenient identity to use is sin α sin β = 1 [cos(α − β ) − cos(α + β )].46).) To do the integrals directly. with n ≠ m and n ≠ -m.15. 【Sol】 Using Equation (5. As shown in the text. which means that its average position is the middle of the box.∫−∞ψ nψ mdx +∞ = 1 L ∫0 L (n − m )πx (n + m )πx   cos − cos  dx L L   L  L (n − m )πx L (n + m )πx  = sin − sin  = 0. the indefinite integral is nπx  L   L  3 ∫ x sin L dx =  nπ  ∫ u sin udu =  n π      where the substitution u = (nπ/L)x has been made.  4 4 8 6  Inha University Department of Physics .m)π = sin 0 = 0 has been used. the expectation value <x> of a particle trapped in a box L wide is L/2. 17. From either a table or repeated integration by parts. the expectation value <x2> is 2 L  n πx  < x 2 >n = ∫0 x 2 sin 2  dx . (n − m )π L (n + m )π L  o where sin(n . L  L  See the end of this chapter for an alternate analytic technique for evaluating this integral using Leibniz’s Rule.m)π = sin(n .46). Find the expectation value <x2>. 2 2 3 3 3  u u2 u 1 − sin 2u − cos 2u + sin 2u . and so the result is 3 2  L  (nπ )3 nπ  1  2 2 1 < x >n =  − . as found in Example 5. when x = 0 u = 0.   =L  − L  nπ   6 4  3 2n 2π 2    As a check. Each of the terms in the integral vanish at u = 0. 【Sol】 This is a special case of the probability that such a particle is between x1 and x2. Find the probability that a particle in a box L wide can be found between x = 0 and x = L/n when it is in the nth state. note that L2 2 lim < x >n = .This form makes evaluation of the definite integral a bit simpler. and the terms with sin 2u vanish at u = nπ.4. L 0 n  L 2nπ  L Inha University Department of Physics . n →∞ 3 which is the expectation value of <x2 > in the classical limit. for which the probability distribution is independent of position in the box. and when x = L u = nπ. cos 2u = cos 2nπ = 1. 19. With x1 = 0 and x2 = L. P0L 1 2nπx  1 x = − sin = .   L L L     Each integral above is equal to L/2 (from calculations identical to Equation (5. 3. 2. 【Sol】 The normalization constant. 2. A particle is in a cubic box with infinitely hard walls whose edges are L long (Fig. The result is  A   =1 2 2 L 3 or 2 A=  L  3/ 2 Inha University Department of Physics . assuming A to be real. The wave functions of the particle are given by n x = 1. 18). 3.21.43)).K L L L n z = 1. K Find the value of the normalization constant A. 2. 5. 3.K n xπ x n z πy n z πz ψ = A sin sin sin n y = 1. is given by ∫ψ * ψ dV = 1 = ∫ψ * ψdxdydz n yπy  L n πx  L n πz   L = A 2  ∫0 sin 2 x dx  ∫0 sin 2 dy  ∫0 sin 2 z dz . as that would mean ψ = 0 identically. substitution into Equation (5.) (b) Compare the ground-state energy of a particle in a one-dimensional box of length L with that of a particle in the three-dimensional box. (Hint: inside the box U = 0. Before substitution into Equation (5. 【Sol】 (a) For the wave function of Problem 5-21. E min = 3π 2h 2 2 − π2 2 2 2 (n x + n y + n z ) + ψ 2m . ny. (a) Find the possible energies of the particle in the box of Exercise 21 by substituting its wave function ψ in Schrödinger's equation and solving for E. ∂x 2 L ∂y 2 L ∂z 2 L Then.23. it is convenient and useful to note that for this wave function 2 2 2 n yπ 2 ∂ 2ψ nx π 2 ∂ 2ψ ∂ 2ψ n zπ 2 = − 2 ψ. 2 2mL (b) The lowest energy occurs when nx = ny = nz = 1. Equation (5. The minimum energy is then 2mL which is three times the ground-state energy of a particle in a one-dimensional box of length L. None of the integers nx.33) must be used to find the energy. L2 h2 π 2h 2 2 2 2 and so the energies are En x .33) gives E ψ = 0. Inha University Department of Physics .33). = − 2 ψ. = − 2 ψ.n y .nz = (n x + n y + n z ). or nz can be zero. 15 × 1010 m−1))2 = 0.200 × 10− 9 m ) Equation (5.1 × 10−31 kg)(1. but a more direct expression is p2 ( hk 2 )2 E = U − KE = U − =U − 2m 2m ((1. 【Sol】 Solving equation (5.15 × 1010 m -1 2L T 2( 0. The uncertainty principle dictates that such a particle would have an infinite uncertainty in momentum.00 eV high and 0. A beam of electrons is incident on a barrier 6.60) to find the energy they should have if 1. What bearing would you think the uncertainty principle has on the existence of the zero-point energy of a harmonic oscillator? 【Sol】 If a particle in a harmonic-oscillator potential had zero energy.6 × 10−19 J/eV ) 27. 1 1 1 k2 = ln = ln(100) = 1. (5. may be solved for the energy E. Use Eq. Inha University Department of Physics . the particle would have to be at rest at the position of the potential minimum.00 percent of them are to get through the barrier. and hence an infinite uncertainty in energy.95 eV = 6. This contradiction implies that the zero-point energy of a harmonic oscillator cannot be zero.200 nm wide.25. from the appendix.60) for k2.05 × 10− 34 J ⋅ s)(1.00 eV − 2( 9.86). 67) gives 2πmν hν mν 2 y= = 2π = 1. Show that for the n = 0 state of a harmonic oscillator whose classical amplitude of motion is A. −∞ ∞ and ∞ 2 ∫−∞ x ψ * ψ dx It is far more convenient to use the dimensionless variable y as defined in Equation (5. so A = . −∞ ∫ 2 ∞ y 4e −y dy .67).29. −∞ Inha University Department of Physics . 31. (5. y = 1 at x = A. where y is the quantity defined by Eq. Find the expectation values <x> and <x2> for the first two states of a harmonic oscillator. m and k.64) has been used to relate ν.67). −∞ ∫ 2 ∞ y 3e − y dy . 【Sol】 When the classical amplitude of motion is A. h k k where Equation (5. 2 2 k Using this for x in Equation (5. The necessary integrals will be proportional to ∫ 2 ∞ y 2e −y dy . 【Sol】 The expectation values will be of the forms ∫−∞ xψ * ψ dx ∫ 2 ∞ ye − y dy . the energy of the oscillator is 1 1 hν kA 2 = hν . The other two integrals may be found from tables. ∫−∞ ∫−∞ 2 4 An immediate result is that <x> = 0 for the first two states of any harmonic oscillator. A generalization of the above to any case where the potential energy is a symmetric function of x. and in fact <x> = 0 for any state of a harmonic oscillator (if x = 0 is the minimum of potential energy). preceding the solutions for Section 5. To find <x2> for the first two states. from symbolic-manipulation programs. k 2π 3/ 2mν 2 4π 2mν 2 1/2 ∫ ∞ x 2ψ 1* ψ1dx −∞  2mν  =   h  = h  h     2πmν  3/ 2 ∫ 2 ∞ 2y 4e − y dy −∞ 2π 3 /2mν 2 3 π ( 3/ 2)hν E1 = = .8 problems in this manual. the necessary integrals are ∫ ∞ x 2ψ o *ψ 0dx −∞  2mν  =   h  = h 1/2  h     2πmν  3/ 2 ∫ ∞ 2 −y2 y e dy −∞ π (1/ 2)hν E = = 0.7).The first and third integrals are seen to be zero (see Example 5. leads to <x> = 0. 2 k 4π 2mν 2 Inha University Department of Physics . or by any of the methods outlined at the end of this chapter or in Special Integrals for Harmonic Oscillators. y 4e − y dy = π. which gives rise to wave functions that are either symmetric or antisymmetric. The integrals are 2 2 ∞ 1 ∞ 3 y 2e −y dy = π. 14 × 10−15 eV ⋅ s E 0 = hν = = = 2.2 and Equation (5. (a) What is its zero-point energy? Would you expect the zero-point oscillations to be detectable? (b) The pendulum swings with a very small amplitude such that its bob rises a maximum of 1.00 mm above its equilibrium position.64). H is the maximum pendulum height.48 × 1028. The period of the pendulum is 1. 2 2T 2(1.00 s. E 1 mgH (1.00-g bob has a massless string 250 mm long.In both of the above integrals. and solving Equation (5. What is the corresponding quantum number? 【Sol】 (a) The zero-point energy would be 1 h 4. (b) The total energy is E = mgH (here. −34 hν 2 h /T 2 6. given as an uppercase letter to distinguish from Planck's constant). A pendulum with a 1.07 × 10−15 eV.63 × 10 J ⋅ s Inha University Department of Physics . dx dx = dy = dy h dy 2πmν has been used.70) for n. as well as Table 5. 33.80 m/s 2 )(1.00 s ) 1 n= − = = − = 1.00 s) which is not detectable.00 × 10− 3 kg )( 9. 82)). so that D = 0.00-mA beam of electrons moving at 2. Consider a beam of particles of kinetic energy E incident on a potential step at x = 0 that is U high. 5.37.00x106 m/s enters a region with a sharply defined boundary in which the electron speeds are reduced to 1. (a) Explain why the solution De-ik’x (in the notation of appendix) has no physical meaning in this situation. the wave function in the two regions has the form ψ I = Aeik1x + Be −ik1x . (c) A 1. this is possible in region I. 2mE 2m ( E − U ) . but not in region II (the reasoning is the same as that which lead to setting G = 0 in Equation (5. Therefore. where E > U (Fig. 【Sol】 (a) In the notation of the Appendix. due to reflection at the step at x = 0.19). Find the transmitted and reflected currents. k′ = . the exp(-ik’x) term is not physically meaningful. h h The terms corresponding to exp(ik1x) and exp(ik’x) represent particles traveling to the left. and D = 0.00x106 m/s by a difference in potential. Inha University Department of Physics . where k1 = ψ II = Ce ik ′x + De −ik ′x . (b) Show that the transmission probability here is T = CC*v‘/AA*v1 = 4k12/(k1 + k’)2. and the reflected current is 0. note that the ratio of the reflected current to the incident current is.00 mA) = 0. The transmitted current is (T)(1. 2 ′ A k1 + k AA * (k1 + k ′) (c) The particle speeds are different in the two regions. so Equation (5. 2 AA * k1 (k1 + k ′ )2 ((k1 /k ′) + 1)2 ψ I v1 For the given situation. so T = (4x2)/(2+1)2 = 8/9. R= ψ I − v1 ψ I + v1 2 2 = BB * AA * Eliminating C from the equations obtained in part (b) from the continuity condition as x = 0. As a check on the last result.    (k /k ′) − 1  1 k′  k′  A 1 −  = B 1 + .(b) The boundary condition at x= 0 are then A + B = C. in the notation of the Appendix.111mA.889 mA. and = . k1  k′  Adding to eliminate B. so  k   1 2 C 2k1 CC * 4k1 = . 2 A = 1 + C .00. k1/k’ = v1/v’ = 2. k′ ik1A − ik1B = ik ′C or A − B = C . so R =  1   k   (k /k ′) + 1  = 9 = 1 − T  k1     1  1  Inha University Department of Physics .83) becomes 2 ψ II v ′ CC * k ′ 4k1k ′ 4(k1 /k ′) T = = = = . three quantities are needed to describe the variation of the wave function throughout space. 3.14) and that it is normalized. d 2 R10 = − 5 / 2 e −r /ao . (6. 【sol】 For the given function. and the variation with the polar direction (variation along the direction from the classical axis of rotation). dr a0 and R10 (r ) = e −r / 2ao 3 /2 2 Inha University Department of Physics . y. the variation in the azimuthal direction (the variation along the circumference of the classical orbit). z) or spherical coordinates. Why is it natural that three quantum numbers are needed to describe an atomic electron (apart from electron spin)? 【sol】 Whether in Cartesian (x.Chapter 6 Problem Solutions 1. Show that a0 is a solution of Eq. The three quantum numbers needed to describe an atomic electron correspond to the variation in the radial direction. 14) if l=0 ( as indicated by the index of R10). or h2 ao again as indicated by the index of R10. 2 2me 2 = . 2m 1 e2 E =− = E1 8πεoao or ao = 4π 2εoh 2 me 2 Inha University Department of Physics . ao h 2 4πεo which is the case. and E = − 2 . 4 ∞ 2 − 2r /ao 1 ∞ ∞ 2 2 dr = ∫0 u 2e −udu.1 d  2 dR10  2 1  r 2  −r /ao e r  = − 5 /2 2  2r − 2 dr   dr  ao  r  ao r    1 2  R10 = 2− a r ao   o  This is the solution to Equation (6. ∫0 R10 r dr = a 3 ∫0 r e 2 o where the substitution u=2r/ao has been made. To show normalization. The improper definite integral in u is known to have the value 2 and so the given function is normalized. apart from the normalization constants. 5 it was stated that an important property of the eigenfunctions of a system is that they are orthogonal to one another. which means that Verify that this is true for the azimuthal wave functions Φml of the hydrogen atom by calculating 2π * ∫ Φml Φml′ dφ for ml ≠ ml′ 0 ∞ * ∫−∞ψ nψm dV = 0 n ≠ m 【sol】 From Equation (6.5. It is possible to express the integral in terms of real and imaginary parts. Inha University Department of Physics . but it turns out to be more convenient to do the integral directly in terms of complex exponentials: ∫0 2π e −imlφe imlφdφ = = ′ ∫0 2π i (ml′ −ml′ )φ e dφ 1 ′ ′ 2π e i(ml −ml )φ 0 = 0 i (ml′ − ml ) [ ] The above form for the integral is valid only for ml ≠ ml’. In evaluating the integral at the limits.15) the integral. the fact that ei2πn = 1 for any integer n ( in this case (ml’ – ml)) has been used. which is given for this case. In Exercise 12 of Chap. is 2π * 2π −im φ im ′φ ∫0 Φml Φml′dφ = ∫0 e l e l dφ . is Lz equal to L? 【sol】 From Equation (6. In the quantum theory. for the ground-state orbit of an electron in a hydrogen atom. from Equation(6. But.7. Under what circumstances. l 2 ≤ l (l + 1) < (l + 1)2 or any nonnegative l. for L to be equal to Lz. 9. if any. Therefore. Compare the angular momentum of a ground-state electron in the Bohr model of the hydrogen atom with its value in the quantum theory. Inha University Department of Physics . l(l + 1) is the square of an integer only if l = 0. the product l(1+1). in which case Lz = 0 and L = Lz = 0.21). 【sol】 In the Bohr model. and L = 0 for a ground-state hydrogen atom.22). must be the square of some integer less than or equal to l. zero-angular-momentum states (ψ spherically symmetric) are allowed. with equality holding in the first relation only if l = 0. and so L = pr = . λ = h/mv = 2πr. Lz must be an integer multiple of . is. Find the percentage difference between L and the maximum value of Lz for an atomic electron in p. the possible values for the magnetic quantum number ml are ml = 0. l = 3 and 1 1 2 2 3 3 4 = l (l + 1) − l l =1− . 13. l = 1 and 1 For a d state.22).max L For a p state. ±2.11. ±3. ±1. 【sol】 The fractional difference between L and the largest value of Lz. What are the possible values of the magnetic quantum number ml of an atomic electron whose orbital quantum number is l = 4? 【sol】 From Equation (6. and f states. a total of nine possible values.13 = 13% Inha University Department of Physics . l = 2 and 1 For a f state. for a given l. L − L z . ±4. d.29 = 29% = 0.18 = 18% = 0. l (l + 1) l +1 = 0. ao or for an extreme.7 it is stated that the most probable value of r for a 1s electron in a hydrogen atom is the Bohr radius ao. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero. this must be a minimum. Verify this. dp/dr → 0 as r → ∞. 6.  dr ao  ao   r2 r = ao and r = 0.1 in Equation (6. In Sec. P(r) = 0.15. which is the radius of the first Bohr orbit. 4r 2 − 2r /ao P (r ) = 3 e . At r = 0. d 4  2r 2  − 2r /ao e P (r ) = 3  2r − = 0.25). The only maximum of P(r) is at r = ao. ao The most probable value of r is that for which P(r) is a maximum. 【sol】 Using R10(r) from Table 6. and because P(r) is never negative. and this also corresponds to a minimum. Inha University Department of Physics . this must be a minimum. Inha University Department of Physics . and ignoring the leading constants (which would not affect the position of extremes). At r = 0. and this also corresponds to a minimum. Differentiating the above expression for P(r) with respect to r and setting the derivative equal to zero. dP/dr → 0 as r → ∞. and because P(r) is never negative. d P (r ) = dr 2r 6 6r = 3ao 5  5 2r 6  − 2r /3ao  6r − e = 0. Find the most probable value of r for a 3d electron in a hydrogen atom. which is the radius of the third Bohr orbit. 【sol】 Using R20 (r) from Table 6.1 in Equation (6. The only maximum of P(r) is at r = 9ao.25).17.  3ao    and r = 0. P(r) = 0. P (r ) = r 6e − 2r / 3ao The most probable value of r is that for which P(r) is a maximum. 9a o or for an extreme. 19. n = 1. Specially.1.85 P ( 2ao )dr ( 2a o )2 e − 2( 2ao )/ao ( 4)e − 4 4 − 2a /a Inha University Department of Physics . as seen from the functions Φ(φ) and Θ(θ) in Table 6. How much more likely is the electron in a ground-state hydrogen atom to be at the distance ao from the nucleus than at the distance 2ao? 【sol】 For the ground state. l = ml.47 P (ao / 2)dr (ao / 2)2 e − 2(ao / 2) /ao (1/ 4)e −1 e − 2a /a Similarly. 2 P (ao )dr aoe o o e −2 e2 = = = = 1. 2 P (ao )dr a oe o o e −2 4 = = = = 1. = 0 (see Problem 6-14). The ratio of the probabilities is then the ratio of the product r2 (R10 (r))2 evaluated at the different distances. the wave function is independent of angle. where for n = 1. 21.68 = 68%. 【sol】 (a) Using R10(r) for the 1s radial function from Table 6.1. 2 [ ] Inha University Department of Physics . Find the probability r > 2ao for a 1s electron in a hydrogen atom. The probability of finding an atomic electron whose radial wave function is R(r) outside a sphere of radius ro centered on the nucleus is ∫ ∞ ro R (r ) r 2dr 2 (a) Calculate the probability of finding a 1s electron in a hydrogen atom at a distance greater than ao from the nucleus. According to classical physics. ∫ao R(r ) r dr = a 3 ∫ao r e 2 o where the substitution u = 2r/a0 has been made. all its energy is potential energy. (b) When a 1s electron in a hydrogen atom is 2ao from the nucleus. the electron therefore cannot ever exceed the distance 2ao from the nucleus. Using the method outlined at the end of this chapter to find the improper definite integral leads to 1 ∞ 2 −u 1 −u 2 ∫2 u e du = 2 − e u + 2u + 2 2 [ ( )] ∞ 2 = 1 −2 e 10 = 0. ∞ 4 ∞ 2 − 2r /ao 1 ∞ 2 2 dr = ∫2 u 2e −udu . 2 ∫4 2 2 [ ( )] [ ] 23.(b) Repeating the above calculation with 2 a0 as the lower limit of the integral.24 = 24% . 【sol】For l = 0. 1. 7. Verify Unsold's theorem for l = 0. only ml = 0 is allowed. Unsold's theorem states that for any value of the orbital quantum number l.1)). ml = −l ∑Θ +l 2 Φ = constant 2 This theorem means that every closed subshell atom or ion (Sec. Φ(φ) and Θ(θ) are both constants (from Table 6. and l = 2 with the help of Table 6. the sum is 1 3 1 3 1 3 3 sin 2 θ + cos 2 θ + sin 2 θ = . 2π 4 2π 2 2π 4 4π Inha University Department of Physics . the probability densities summed over all possible states from ml = -1 to ml = +1 yield a constant independent of angles θ or φ that is. and the theorem is verified. l = 1.6) has a spherically symmetric distribution of electric charge. For l = 1. ∞ 1 ∞ 2 −u 1 1 u e du = − e −u u 2 + 2u + 2 4 = e − 4 26 = 0. the sum is 2 1 15 1 15 1 10 sin4 θ + 2 sin2 θ cos2 θ + (3 cos2 θ − 1)2 . combining the identical terms for ml = ±2 and ml = ±1. and the theorem is verified. one for ml = -1 and once for ml = 1. For l = 2. 2π 16 2π 4 2π 16 The above may he simplified by extracting the commons constant factors.cos2 θ to eliminate sin θ . to 5 [(3 cos2 θ − 1)2 + 12 sin2 θ cos2 θ + 3 sin4 θ ]. and again using Φ*Φ= 1/2π. which holds for any l and ml.In the above. Using the identity sin2 θ = 1 . 16π Of the many ways of showing the term in brackets is indeed a constant. ( 3 cos 2 θ − 1)2 + 12 sin 2 θ cos2 θ + 3 sin 4 θ = ( 9 cos4 θ − 6 cos 2 θ + 1) + 12(1 − cos 2 θ ) cos 2 θ + 3(1 − 2 cos 2 θ + cos4 θ ) = 1. Inha University Department of Physics . Note that one term appears twice. the one presented here. seems to be one of the more direct methods. has been used. Φ*Φ= 1/2π. using a bit of hindsight. Inha University Department of Physics . The ∆l = 0 transition is seen to be forbidden. and u = x = r sin θ cos φ is used. vanish.35). With the help of the wave functions listed in Table 6. If ml = 0 initially. and u = z = r cos θ is used. the integral in Equation (6. and only the angular functions Φ(φ) and Θ(θ) need to be considered. y or z will assume positive and negative values with equal probability amplitudes.35) must vanish. the integral (apart from constants) is π 2 cos 2 θ sin θdθ = ≠ 0 ∫0 3 If ml = ±1 initially. 【sol】 In the integral of Equation (6. the θ -integral is of the form and the φ -integral is of the form π 2 ∫0 sin θdθ = 2 ≠ 0 π ∫ 2π ±i φ e cos φdφ 0 = ∫ 2π cos2 φdφ 0 =π ≠0 and the transition is allowed.1 verify that ∆l = ±1 for n = 2 à n = 1 transitions in the hydrogen atom.25. and any integral of the form of Equation (6. in that the product 1 ( Φ0 (φ )Θ00(θ ))∗ (Φ 0 (φ )Θ00 (θ )) = 4π is spherically symmetric. as the argument u = x.35) will be seen to to vanish if u is chosen appropriately. the radial integral will never. If l = 1 in the initial state. including consulting tables or using symbolicmanipulation programs (see. 5.27. ∫0 L L The integrals may be found in a number of ways. Verify that the n = 3 → n = 1 transition for the particle in a box of Sec. 2 Lx (n + m )πx L2 (n + m )πx   −  sin + cos 2 2 (n + m )π L L  (n + m ) π 0   Inha University Department of Physics . 【sol】 The relevant integrals are of the form nπx mπx L x sin sin dx. for instance.8 is forbidden whereas the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. One way to find a general form for the integral is to use the identity sin α sin β = 1 [cos(α − β ) − cos(α + β )] 2 and the indefinite integral (found from integration by parts) x sin kx 1 x sin kx cos kx ∫ x cos kxdx = k − k ∫ sin kxdx = k + k 2 to find the above definite integral as L  Lx  (n − m )πx L2 (n − m )πx sin + cos   (n − m ) L L (n − m )2 π 2 1  . the solution to Problem 5-15 for sample Maple commands that are easily adapted to this problem). the n = 3 → n = 1 transition is forbidden. 2π 2  (n − m )2 (n + m )2  If n and m axe both odd or both even.m are even. consider that L L nπx mπx L nπx mπx sin dx = ∫0 x sin sin dx  x −  sin ∫0  2 L L L L for n ≠ m. see Problem 5-15). n + m and n . the arguments of the cosine terms in the above expression are even-integral multiples of π. Letting u=L/2 – x. and hence the integral is zero. because the integral of L times the product of the wave functions is zero. The integrand is then an odd function of u when n and m are both even or both odd. the wave functions were shown to be orthogonal in Chapter 5 (again. with the result of L2  cos(n − m )π − 1 cos(n + m )π − 1 −  .where n ≠ m2 is assumed. To make use of symmetry arguments. sin nπx nπ (( L / 2) − u )  nπ nπu  = sin = sin  −  L L 2 L   This expression will be ± cos ( nπu/L ) for n odd and ±sin ( nπu/L ) for n even. while the n = 3 → n = 2 and n = 2 → n = 1 transitions are allowed. If one of n or m is even and the other odd. and the integral vanishes. the integrand is an even function of u and the integral is nonzero. The terms involving sines vanish. Thus. Inha University Department of Physics . and hence proportional to n. the magnitude of the magnetic moment of an electron in a Bohr orbit is proportional to the magnitude of the angular momentum.1 × 10−31 kg )( 3.010 nm is used. and so the magnetic moment is proportional to rn .39). 31.010 × 10−9 m 4π ( 9.4.13) or Problem 4-28). The orbital radius is proportional to n2 (See Equation (4.0 × 108 m/s) B = 2 = = 1.29. Show that the magnetic moment of an electron in a Bohr orbit of radius rn is proportional to rn 【sol】 From Equation (6. 【sol】 See Example 6. Find the minimum magnetic field needed for the Zeeman effect to be observed in a spectral line of 400-nm wavelength when a spectrometer whose resolution is 0.6 × 10 C) Inha University Department of Physics . ∆ λ 4πmc 0. solving for B.34 T -9 2 −19 e λ (400 × 10 m) (1. Chapter 7 1. Problem Solutions A beam of electrons enters a uniform 1.20-T magnetic field. (a) Find the energy difference between electrons whose spins are parallel and antiparallel to the field. (b) Find the wavelength of the radiation that can cause the electrons whose spins are parallel to the field to flip so that their spins are antiparallel. 【sol】 (a) Using Equations (7.4) and (6.41), the energy difference is, ∆E = 2µsz B = 2µB B = 2(5.79 × 10−5 eV/T)(1.20 T) = 1.39 × 10−4 eV (b) The wavelength of the radiation that corresponds to this energy is hc 1.24 × 10− 6 eV ⋅ m λ= = = 8.93 mm ∆E 1.39 × 10− 4 eV Note that a more precise value of AB was needed in the intermediate calculation to avoid roundoff error. 3. Find the possible angles between the z axis and the direction of the spin angular-momentum vector S. 【sol】 For an electron, s = ( 3 /2)h, sz = ±(1/ 2)h, and so the possible angles axe given by  ± (1/ 2)h   1  o o arccos   = arccos   = 54.7 , 125.3  3  ( 3 / 2)h  Inha University Department of Physics 5. Protons and neutrons, like electrons, are spin- ½ particles. The nuclei of ordinary helium atoms, 4 He , contain two protons and two neutrons each; the nuclei of another type of helium 2 3 , contain two protons and one neutron each. The properties of liquid 4 atom, 2 He 2 He and liquid 3 2 He are different because one type of helium atom obeys the exclusion principle but the other does not. Which is which, and why? 【sol】 4 2 He atoms contain even numbers of spin-½ particles, which pair off to give zero or integral 3 spins for the atoms. Such atoms do not obey the exclusion principle. 2 He atoms contain odd numbers of spin- ½ particles, and so have net spins of 1 , 3 or 5 , and they obey the exclusion 2 2 2 principle. 7. In what way does the electron structure of an alkali metal atom differ from that of a halogen atom? From that of an inert gas atom? 【sol】 An alkali metal atom has one electron outside closed inner shells: A halogen atom lacks one electron of having a closed outer shell: An inert gas atom has a closed outer shell. Inha University Department of Physics 9. How many electrons can occupy an f subshell? 【sol】 For f subshell, with l = 3, the possible values of ml are ±3, ±2, ±1 or 0, for a total of 2l +1=7 values of ml. Each state can have two electrons of opposite spins, for a total of 14 electrons. 11. If atoms could contain electrons with principal quantum numbers up to and including n = 6, how many elements would there be? 【sol】 The number of elements would be the total number of electrons in all of the shells. Repeated use of Equation (7.14) gives 2n2 + 2 (n - 1)2 +... + 2 (1)2 = 2 (36 + 25 + 16 + 9 + 4 + 1) = 182. In general, using the expression for the sum of the squares of the first n integers, the number of elements would be 2(1 n(2n + 1)(n + 1)) = 1 [n( 2n + 1)(n + 1)], 6 3 which gives a total of 182 elements when n = 6. Inha University Department of Physics 4. 【sol】 All of the atoms are hydrogenlike. This outer electron is then relatively hard to d etach.9 eV. Na.4. 4.1. The completely filled K and L shells shield +10e of the nuclear charge of = 16e. 5. in that there is a completely filled subshell that screens the nuclear charge and causes the atom to "appear" to be a single charge. Rb. and so the effective nuclear charge is roughly +2e. Inha University Department of Physics . (b) Again. respectively. The 3d subshell is empty. see Table 7. Account for the decrease in ionization energy with increasing atomic number.2. the filled 3s2 subshell will partially shield the nuclear charge. and Cs are. and the outer electron is relatively easy to detach. Would you think that such an electron is relatively easy or relatively hard to detach from the atom? (b) Do the same for the sulfur (Z = 16) atom. The ionization energies of Li. K. (a) Make a rough estimate of the effective nuclear charge that acts on each electron in the outer shell of the calcium (Z = 20) atom. but not to the same extent as the filled shells.3.4. and 3. 4. and hence has a successively lower binding energy.13. 5. The outermost electron in each of these atoms is further from the nucleus for higher atomic number. All are in group 1 of the periodic table. 【sol】 (a) See Table 7. so +6e is a rough estimate for the effective nuclear charge. 15. 4). The Na atom (Z = 11) is larger because it has an additional electron shell (see Table 7. Na and Si. In each of the following pairs of atoms. which would you expect to be larger in size? Why? Li and F. The Cl atom (Z = 17) atom is larger because has an additional electron shell. Li and Na.ions have closed shells. whereas an Na atom has a single outer electron that can be detached relatively easily in a chemical reaction with another atom.17. F and Cl.ions? Why are Na atoms more chemically active than Na+ ions? 【sol】 Cl. 19. The Na atom is larger than the Si atom (Z = 14) for the same reason as given for the Li atom. Inha University Department of Physics . Why are Cl atoms more chemically active than Cl. 【sol】 The Li atom (Z = 3) is larger because the effective nuclear charge acting on its outer electron is less than that acting on the outer electrons of the F atom (Z = 9). Na+ ions have closed shells. whereas a Cl atom is one electron short of having a closed shell and the relatively poorly shielded nuclear charge tends to attract an electron from another atom to fill the shell. solving for B . the total number of electrons must be even.24 × 10−6 eV ⋅ m  1 1  = −   = 18.6 nm corresponding to 3P1/2 → 3S1/2 . λ λ   1 2 hc  1 1 B =  −  2 µB  λ1 λ2    1.6 × 10−9 m  Inha University Department of Physics .5 T 2 × 5.0 × 10−9 m 589. Why is the normal Zeeman effect observed only in atoms with an even number of electrons? 【sol】 The only way to produce a normal Zeeman effect is to have no net electron spin. the net spin would be nonzero. 【sol】 See Example 7. because the electron spin is ± ½.6. Use these wavelengths to calculate the effective magnetic field experienced by the outer electron in the sodium atom as a result of its orbital motion. If the total number of electrons were odd.0 nm corresponding to 3P3/2 → 3S1/2 and 589. 23.5 eV/T  589. and the anomalous Zeeman effect would be observable.21. The spin-orbit effect splits the 3P → 3S transition in sodium (which gives rise to the yellow light of sodium-vapor highway lamps) into two lines. 589. Expressing the difference in energy levels as 1 1  ∆ E = 2µ B B = hc  − .79 × 10. 29. The lithium atom has one 2s electron outside a filled inner shell. all of the subshells must be filled. What must be true of the subshells of an atom which has a 1S0 ground state? 【sol】 For the ground state to be a singlet state with no net angular momentum. S = J = ½. This state has the lowest possible values of L and J. L = 0. Its ground state is 2S1/2 . and is the only possible ground state. (a) What are the term symbols of the other allowed states.25. 27. There axe no other allowed states. what values of l are possible? 1 2 【sol】 The possible values of l are j + = 3 and j − 1 2 = 2. If j= 5 2 . Inha University Department of Physics . if any? (b) Why would you think the 2S 1/2 state is the ground state? 【sol】 For this doublet state. the corresponding angular momenta are h and 63 2 h Inha University Department of Physics . 【sol】 The two 3s electrons have no orbital angular momentum. and so this state cannot exist. The 3p electron has l = 1. Why is it impossible for a 2 2D3/2 state to exist? 【sol】 A D state has L = 2. The term symbol is 2P1/2 . n = 2 but L must always be strictly less than n. so L = 1. 【sol】 (a) From Equation (7.17).17). The aluminum atom has two 3s electrons and one 3p electron outside filled inner shells. and in the ground state J = ½ . j =l ± 1 2 = 5. Find the term symbol of its ground state.31. 33. 35. 2 2 35 2 (b) Also from Equation (7. and their spins are aligned oppositely to give no net angular momentum. Answer the questions of Exercise 34 for an f electron in an atom whose total angular momentum is provided by this electron. 7. for a 2 2D3/2 state. 37.0o 2 12( 3 / 2)  2   (d) The multiplicity is 2(1/2) + 1 = 2. 2LS where θ is the angle between L and S.  ( 35 / 4) − 12 − ( 3/ 4)   2 arccos   = arccos  −  = 132o 2 12( 3 / 2)  3   and  (63/ 4) − 12 − ( 3/ 4)  1 arccos   = arccos   = 60. the state is an f state because the total angular momentum is provided by the f electron.(c) The values of L and S are 12h and 23 h. The law of cosines is J 2 − L2 − S 2 cos θ = . The magnetic moment µ J of an atom in which LS coupling holds has the magnitude µJ = J( J + 1)g J µ B where µ B = e ħ/2m is the Bohr magneton and gJ = 1 + J(J + 1) − L(L + 1) + S(S + 1) 2J(J + 1) Inha University Department of Physics . and so the terms symbols are 2F5/2 and 2F7/2 . then the angles are. The middle term is obtained by using |S| cos α + |S| cos β = |J|. the product µJ has magnitude   S 2 µB S cos α + µB L cos β = µB J + µB S cos α = µB J  1 + cos α    J   In the above. the magnitudes of the angular momenta include factors of h. The above expression is equal to the product µJ because in this form. let the angle between J and S be α and the angle between J and L be β. L − J −S cos α = − 2J S and so S L −J −S J(J + 1) − L(L + 1) + S(S + 1) cos α = = 2 J 2J(J + 1) 2J 2 2 2 2 2 2 Inha University Department of Physics . How many substates are there for a given value of J? What is the energy difference between different substates? 【sol】 (a) In Figure 7. (b) Consider an atom that obeys LS coupling that is in a weak magnetic field B in which the coupling is preserved. (a) Derive this result with the help of the law of cosines starting from the fact that averaged over time.15. the factor of 2 in 2µB relating the electron spin magnetic moment to the Bohr magneton is from Equation (7. only the components of µ L and µ S parallel to J contribute to µ L .3).is the Landé g factor. Then. From the law of cosines. . Optical spectra. although the optical spectra of these elements may differ considerably. J. depend upon the possible states of the outermost electrons. 【sol】 The transitions that give rise to x-ray spectra are the same in all elements since the transitions involve only inner.and the expression for µJ in terms of the quantum numbers is µJ h = J gJ µ B . closed-shell electrons. together with the transitions permitted for them. which. for a total of 2J + 1 substates. however. where MJ = -J . or µJ = J(J + 1)g J µB . Explain why the x-ray spectra of elements of nearby atomic numbers are qualitatively very similar. Inha University Department of Physics .. are different for atoms of different atomic number. The difference in energy between the substates is ∆ E = g J µB M J B gJ = 1 + 39. whe re J(J + 1) − L (L + 1) + S(S + 1) 2J(J + 1) (b) There will be one substate for each value of MJ. 2 eV) (144) = 1.21) or (7.22). In a triplet state. the spins of the outer electrons are antiparrallel. 【sol】 From either of Equations (7.844 nm 3 E 14. they are parallel Inha University Department of Physics . The wavelength is hc 1.1)2 = (10. 【sol】 In a singlet state.2 eV) (Z .7 × 10 eV 43. E = (10.24 × 10−6 eV ⋅ m λ= = = 8. Distinguish between singlet and triplet states in atoms with two outer electrons.41. Find the energy and the wavelength of the Kα x-rays of aluminum.44 × 10−10 m = 0.47 keV. Why do you think the latter energy is greater? 【sol】 The nuclear charge of +2e is concentrated at the nucleus.7 eV.5 eV or T = = 3. The energy needed to detach the electron from a hydrogen atom is 13. At what temperature would the average kinetic energy of the molecules in a hydrogen sample be equal to their binding energy? 【sol】 Using 4. 3. but the energy needed to detach an electron from a hydrogen molecule is 15.6 eV.5 eV kT = 4. 3 2 4. while the electron charges' densities are spread out in (presumably) the 1s subshell. This means that the additional attractive force of the two protons exceeds the mutual repulsion of the electrons to increase the binding energy.62 × 10 eV/K Inha University Department of Physics .5 × 104 K -5 2 3 8.5 eV for the binding energy of hydrogen.Chapter 8 Problem Solutions 1. ) 7. r in terms of x is Inha University Department of Physics .5. Quantitatively. the faster the rotation and the greater the distortion.9)). inertia causes its bonds to stretch. the atomic separation.11) are different. the algebraic steps that lead to Equation (8. For the different isotopes. Thus. The J=0àJ=1 rotational absorption line occurs at 1.11). the parameter I (the moment of inertia of the molecule) is a function of J. (This is why the earth bulges at the equator. (It should be noted that if I depends on J.11) will not be valid. 【sol】 From Equation (8. will be essentially the same.) What effects does this stretching have on the rotational spectrum of the molecule? 【sol】 The increase in bond lengths in the molecule increases its moment of inertia and accordingly decreases the frequencies in its rotational spectrum (see Equation (8. Denoting the unknown mass number by x and the ratio of the frequencies as r.102x10 11 Hz in ?C16O. all of the levels as given by Equation (8. Find the mass number of the unknown carbon isotope. In addition.153x1011 Hz in 12C16O and at 1. so that the spectral lines are not evenly spaced. The ratio of the moments of inertia will then be the ratio of the reduced masses. which depends on the charges of the atoms. the higher the quantum number J (and hence the greater the angular momentum). the ratios of the frequencies will be the ratio of the moments of inertia. When a molecule rotates. so the spectral lines are no longer evenly spaced. with I larger for higher J. 03 x 10-5 m.102) in the above expression gives x = 13. or the integer 13 to three significant figures. find the distance between the hydrogen and chlorine nuclei in an HCl molecule. The rotational spectrum of HCI contains the following wavelengths: 12.04 x 10-5 m. x= 9.60 x 10-5 m.x ⋅ 16 r = x + 16 12 ⋅ 16 12 + 16 Solving for x in terms of r. 48r 7 − 3r Using r = (1.153)/(1. 8. 6. 9. Inha University Department of Physics .04 x 10-5 m If the isotopes involved are 1H and 35Cl.007.89 x 10-5 m. 6. 6151 × 1012 Hz ) The reduced mass of the HCI molecule is (35/36)rnH. h 1.11). 4. Solving for I and using ∆v as found above.67 × 10 − 27 kg) = 0. in multiplies of 1012 Hz: 2. and keeping an extra significant figure.998 x 108 m/s is used.6151 if c = 2.【sol】 The corresponding frequencies are.616 x 1012 Hz. 4.113.129 nm (keeping extra significant figures in the intermediate calculation gives a result that is rounded to 0. Inha University Department of Physics . 3.484.947 The average spacing of these frequencies is ∆v = 0.337. from ν = c/λ . (A least-squares fit from a spreadsheet program gives 0.73 × 10− 47 kg ⋅ m 2 2π∆ν 2π ( 0.055 × 10−34 J ⋅ s I = = = 2.) From Equation (8. and so the distance between the nuclei is R= I = µ 36 × ( 2. the spacing of the frequencies should be ∆v = /2πI .73 × 10−47 kg ⋅ m 2 ) 35 × (1.130 nm to three significant figures). 67 × 10 kg)(3.4-cm photon when it undergoes a rotational transition from j = 1 to j = 0.22 nm to two significant figures.11. Find the interatomic distance in this molecule.0 × 10 m/s) or 0. 【sol】 Using ν1→0 = c/λ and I = m’ R2 in Equation (8.055 × 10− 34 J ⋅ s)(4. Inha University Department of Physics . A 200Hg35Cl Molecule emits a 4. and (1. m’ = mH(200x35)/(200 + 35).223 nm −27 8 2π (1. hλ R2 = 2πm ′c For this atom.11) and solving for R.4 × 10− 2 m) R= = 0. 13. In Sec. 4.6 it was shown that, for large quantum numbers, the frequency of the radiation from a hydrogen atom that drops from an initial state of quantum number n to a final state of quantum number n - 1 is equal to the classical frequency of revolution of an electron in the n-th Bohr orbit. This is an example of Bohr's correspondence principle. Show that a similar correspondence holds for a diatomic molecule rotating about its center of mass. 【sol】 Equation (8.11) may be re-expressed in terms of the frequency of the emitted photon when the molecule drops from the J rotational level to the J - 1 rotational level, hJ ν J →J −1 = . 2πI For large J, the angular momentum of the molecule in its initial state is L = h J ( J + 1) = hJ 1 + 1/ J ≈ hJ Thus, for large J, L ν ≈ , or L = ωI , 2πI the classical expression. Inha University Department of Physics 15. The hydrogen isotope deuterium has an atomic mass approximately twice that of ordinary hydrogen. Does H2 or HD have the greater zero-point energy? How does this affect the binding energies of the two molecules? 【sol】 The shape of the curve in Figure 8.18 will be the same for either isotope; that is, the value of k in Equation (8.14) will be the same. HD has the greater reduced mass, and hence the smaller frequency of vibration vo and the smaller zero- point energy. HD is the more tightly bound, and has the greater binding energy since its zero-point energy contributes less energy to the splitting of the molecule. 17. The force constant of the 1H19F molecule is approximately 966 N/m. (a) Find the frequency of vibration of the molecule. (b) The bond length in 1H19F is approximately 0.92 nm. Plot the potential energy of this molecule versus internuclear distance in the vicinity of 0.92 nm and show the vibrational energy levels as in Fig. 8.20. 【sol】 (a) Using m'= (19/20)mH in Equation (8.15), νo = 1 2π 20 = 1.24 × 1014 Hz - 27 1.67 × 10 kg 19 966 N/m Inha University Department of Physics k = 4.11 X 10-20 J. The levels are shown below, where the vertical m′ scale is in units of 10-20 J and the horizontal scale is in units of 10-11 m. (b) Eo = 1 h 2 19. The lowest vibrational states of the 23Na35Cl molecule are 0.063 eV apart. Find the approximate force constant of this molecule. 【sol】 From Equation (8.16), the lower energy levels are separated by ∆E = hvo, and vo = ∆E /h. Solving Equation (8.15) for k,  ∆E  k = m ′( 2πνo )2 = m ′   h  Inha University Department of Physics 055 × 10−34 J ⋅ s) k T = (8.617 x 10-5 eV/K) (300 K) = 0. quantities that are not interchangeable.67 × 10 kg) -15   58 4.026 eV. Is it likely that an HCl molecule will be vibrating in its first excited vibrational state at room temperature? Atomic masses are given in the Appendix.60 × 10-19 J/eV)  23 ⋅ 35 − 27  = 213 N/m k= (1. it is likely that some of these atoms will be in the first excited state. The bond between the hydrogen and chlorine atoms in a 1H35Cl molecule has a force constant of 516 N/m. but in a large collection of atoms.063 eV)(1.Using m’ = mH (23·35)/(23 + 35).67 × 10− 27 kg 35 At room temperature of about 300 K. 【sol】 Using ∆E = hνo = h k m′ and m ′ = mH 35 . 21. ∆ E = (1. 36 36 = 5. the symbol "k" has been used for both a spring constant and Boltzmann's constant. 516 N/m Inha University Department of Physics .94 × 10− 20 J = 0.  (0.371 eV 1. It's important to note that in the above calculations.14 × 10 eV ⋅ s   or 2.1 x102 N/m to the given two significant figures. An individual atom is not likely to he vibrating in its first ex cited level. and ε1 = − 13.6 eV The 3 2Pl/2 first excited sate in sodium is 2.Chapter 9 Problem Solutions 1. multiplicity of S-level : 1 The ratio of the numbers of atoms in the states is then.62 × 10 − 5 eV/K )(1200 K )   1   Inha University Department of Physics .09 eV  3  = 4. n(ε 2 ) 1 = = 4 e −(ε2 −ε1)/kT = 4 e 3ε1 /kT n(ε1 ) 1000 (3/ 4)(13. At what temperature would one in a thousand of the atoms in a gas of atomic hydrogen be in the n=2 energy level? 【sol】 g(ε 2 ) = 8.86 × 10 − 9 exp −    (8. Find the ratio between the numbers of atoms in each state in sodium vapor at l200 K. 6 eV )  1  (3/ 4)(−ε 1 ) T =  = = 1. g (ε1 ) = 2 Then.   2.62 × 10 eV/K )(ln 4000 ) where 3.093 eV above the 3 2S1/2 ground state. (see Example 7. ε 2 = ε1 / 4.) 【sol】 multiplicity of P-level : 2L+1=3.43 × 10 4 K −5  k  ln 4000 (8.6. 2. (b) Introduce the dimensionless parameter . respectively. at what temperature does this occur? 【sol】 g (J ) = 2J + 1.ln ( 7/5) and solving for T. 749]J ( J +1) J ( J +1) Applying this expression to J=0. 1.5.0275.68. 0. 3. 1 exactly. (a) Find the relative populations of the J=0. (a) J (J + 1)h 2 εJ = ε J =0 = 0 2I J ( J +1)   J (J + 1)h 2   N (J ) h 2   = ( 2J + 1) exp −  = (2J + 1) exp −   2IkT   N (J = 0 ) 2IkT            (1. Inha University Department of Physics .64 × 10 − 48 kg ⋅ m 2 )(1. 1. Then.880. and 0. (b) can the populations of the J=2 and J=3 states ever be equal? If so.38 × 10 − 23 J/K )(300 K )        = (2J + 1)[0. for the populations of the J=2 and J=3 states to be equal. and 4 gives.1. x6 = 5 7 and 6 ln x = ln 5 7 Using . ln x = − h 2 / 2IkT and ln ( 5/7 ) = .06 × 10 −34 J ⋅ s)2  = (2J + 1) exp −  2(4. 2.3.217. 0. and 4 rotational states at 300 K. The moment of inertia of the H2 molecule is 4.64×10-48 kg·m2. 5x 6 = 7x 12 . 00 m/s.55 × 10 3 K − 48 2 − 23 2( 4.05 × 105 K −  = 3  k  (8.00 2 2 + 3. At what temperature will the average molecular kinetic energy in gaseous hydrogen equal the binding energy of a hydrogen atom? 【sol】 For a monatomic hydrogen.00 2 ] = 2 .00 + 3.24 (m/s) 9.4) 7.6h2 T = 2Ik ln(7 / 5) 6(1.05 × 10 − 34 J ⋅ s)2 = = 1.00) = 2. one with a speed of 1.00 m/s and the other with a speed of 3.62 × 10− 5 eV/K) Inha University Department of Physics . 【sol】 v = 1 (1.6 eV) = 1. Find v and vrms for an assembly of two molecules.38 × 10 J/K) ln(1.64 × 10 kg ⋅ m )(1.00 (m/s) 2 vrms = 1 [1. the kinetic energy is all translational and KE = 3 kT 2 solving for T with KE = − E1 T = 2  E1  (2 / 3)(13. Verify that the average value of 1/v for an ideal-gas molecule is 2m /πkT .11. Using the rms speed from KE=(3/2)kT = (1/2)mv2. 2 ∞ [Note : ∫0 ve −av dv = 1/( 2a )] 【sol】 1 1 ∞1 = n (v )dv The average value of 1/v is v N ∫0 v 1  m  = 4πN   N  2πkT  m  = 4π     2πkT  3/ 2 3/ 2 ∫ 2 ∞ ve − mv / 2kT dv 0  kT  =   m  2m 1 =2 πkT <v > Inha University Department of Physics .3 × 10 m) 3.3-nm spectral line emitted by a gas of atomic hydrogen at 500 K.54 × 10−11 m = 15. the shift in wavelength will be between +λ(v/c) and -λ(v/c) and the width of the Doppler-broadened line will be 2λ(v/c). 【sol】 For nonrelativistic atoms. Find the width due to the Doppler effect of the 656. v = 3kT /m . and ∆λ = 2λ 3kT /m c −9 3(1.0 × 108 m/s = 1.67 × 10−27 kg) = 2(656.38 × 10−23 J/K)(500 K)/(1.4 pm 13. ) 【sol】 The number of standing waves in the cavity is 8πL3ν 2 g(ν )dν = dν c3 2 g ( λ)dλ = g (ν)dν = 8πL3  c  c 8πL3 dλ = 4 dλ   c 3  λ  λ2 λ Therefore the number of standing waves between 9.5 and 100.0 mm ) = 2. Inha University Department of Physics .17.5 mm? (Hint: First show that g(λ)dλ = 8πL3 dλ/λ4.5mm and 10.5 mm can occur in a cubical cavity 1 m on a side? How many with wavelengths between 99. find the percentage difference between the total radiation from skin at 34o and at 35oC. A thermograph measures the rate at which each small portion of a persons skin emits infrared radiation.5 × 10 6 4 (10 mm ) Similarly. l9.5mm and 100. How many independent standing waves with wavelengths between 95 and 10. lower by a factor of 104. To verify that a small difference in skin temperature means a significant difference in radiation rate.5x102.5mm is 8π (1 m)3 g( λ )dλ = (1. the number of waves between99.5mm is 2. 92 kW/m 2 (= 66%) Inha University Department of Physics . the fractional variation may be approximated by ∆ R ∆(T 4 ) 3T 3∆T ∆T 1K = = =3 =3 = 0. 3% T  308 K    1 4 For temperature variations this small.【sol】 By the Stefan-Boltzmann law.4 kW/m2 as the rate at which the sun’s energy arrives at the surface of the earth (1.013 = 1. the total energy density is proportional to the fourth power of the absolute temperature of the cavity walls. At what rate would solar energy arrive at the earth if the solar surface had a temperature 10 percent lower than it is? 【sol】 Lowering the Kelvin temperature by a given fraction will lower the radiation by a factor equal to the fourth power of the ratio of the temperatures. Using 1.013 R T 308 K T4 T4 21.90)4 = 0.4 kW/m 2 )(0. as R = σT 4 The percentage difference is σT14 − σT 24 σT14 = T14 − T 24 T14 T   307 K  = 1−  2  = 1−   = 0. An object is at a temperature of 400oC. the fourth power of the Kelvin temperature would need to double. At what temperature would it radiate energy twice as fast? 【sol】 To radiate at twice the radiate. Then the power radiated for the hole in the wall is P ' = σT 4 A = (5. Find the surface area of a blackbody that radiates 100 kW when its temperature is 500oC. If the blackbody is a sphere. Thus. At what rate does radiation escape from a hole l0 cm2 in area in the wall of a furnace whose interior is at 700oC? 【sol】 The power radiated per unit area with unit emissivity in the wall is P=σT4.67 × 10−8 W/(m 2 ⋅ K 4 ))((500 + 273)K )4 = 4. 94 × 10− 2 m 2 = 494 cm2 Inha University Department of Physics .23. 2[( 400 + 273) K]4 = T 4 T = 673 × 21/ 4 K = 800 K(527 o C) 25. what is its radius? 【sol】 The radiated power of the blackbody (assuming unit emissivity) is P = Ae σT 4 P 100 × 103 W A= = eσT 4 (1)(5.67 × 10−8 W/(m 2 ⋅ K 4 ))( 973 K )4 (10 × 10− 4 m 2 ) = 51 W 27. A gas cloud in our galaxy emits radiation at a rate of 1.0 × 10 27 W  =  π (5. 【sol】 From the Wien’s displacement law. What is the surface temperature of Sirius? 【sol】 From the Wien’s displacement law.9 × 10 2 K = 290 K = 17 o C 10 × 10 . the surface temperature of cloud is 2. If the cloud is spherical and radiates like a blackbody.27 cm 31. Solving for D.9 × 1011 m Inha University Department of Physics .0x1027 W. The radiation has its maximum intensity at a wavelength of 10 µm.898 × 10 − 3 m ⋅ K T = = 2.6 m P P Assuming unit emissivity. the surface temperature of Sirius is 2.0 × 104 K −9 λmax 290 × 10 m 33. A = 4πr 2 r = A / 4π = 6.The radius of a sphere with this surface area is. then. find its surface temperature and its diameter. The brightest part of the spectrum of the star Sirius is located at a wavelength of about 290 nm. the radiation rate is R = σT 4 = = A πD 2 where D is the cloud’s diameter. D= P πσT 4   1.67 × 10 -8 W/m 2 ⋅ K 4 )(290 K )4    1 /2 = 8.898 × 10− 3 m ⋅ K T = = = 1.898 × 10− 3 m ⋅ K 2. where V is the volume.0 × 10 − 6 m 3 ) 2 . 【sol】 The total energy(U) is related to the energy density by U=Vu.03 × 10 −12 J/K 37.998 × 108 m/s = 3 .630εF.00 cm3 of radiation in thermal equilibrium at 1000 K. Find the specific heat at constant volume of 1. For 0≤ε≤εF. 【sol】 At T=0. Show that the median energy in a free-electron gas at T=0 is equal to ε F/22/3 =0. all states with energy less than the Fermi energy εF are occupied.67 × 10 − 8 W/m 2 ⋅ K 4 ) = (1000 K )3(1. U = Vu = VaT 4 = V The specific heat at constant volume is then ∂U 16σ 3 cV = = T V ∂T c 4σ T 4 c 16( 5. the electron energy distribution is proportional to ε . and all states with energy above the Fermi energy are empty. The median energy εM is then the energy such that ∫0 εM ε dε = 1 εF 2 0 ∫ ε dε Inha University Department of Physics . The median energy is that energy for which there are many occupied states below the median as there are above. In terms of temperature.35. then the average occupancy of a state of energy εF-∆ε is f2=1-f1. if the average occupancy of a state of energy εF+∆ε is fl at any temperature. The Fermi energy in silver is 5.Evaluating the integrals.51 eV.51 eV T = = = 2. (a)What is the average energy of the free electrons in silver at O K? (b)What temperature is necessary for the average molecular energy in an ideal gas to have this value? (c)What is the speed of an electron with this energy? 【sol】 (a) The average energy at T=0 K is ε 0 = 3 ε F = 3.08 × 10 6 m/s 43.11 × 10 − 31 kg) = 1.602 × 10−19 J/eV ) 5(9.10 about εF. 31 eV 5 (b) Setting (3/2)kT=(3/5)εF and solving for T. 2 εF 2 5. Show that.9.63 ε F 2 39.56 × 104 K -5 5 k 5 8.) Inha University Department of Physics .62 × 10 eV/K (c) The speed in terms of the kinetic energy is v = 2KE = m 6ε F = 5m 6(5.51 eV )(1. (This is the reason for the symmetry of the curves in Fig. 3 or ε M = ( 1 )3/ 2 ε F = 0. 2 (ε )3/ 2 3 M = 1 (ε F )3/ 2 . 11 × 10. The electronic structure of zinc is given in Table 7. L.31 kg)  8π (65. and the effective mass of an electron in zinc is 0. The density of zinc is 7. Calculate the Fermi energy in zinc.626 × 10 −34 J ⋅ s)2  3(2)(7.78 × 10−18 J = 11 eV 2 /3 Inha University Department of Physics .【sol】 Using the Fermi-Dirac distribution function f 1 = f FD (ε F + ∆ε ) = 1 e ∆ε /kT + 1 e − ∆ε /kT + 1 1 1 1 e ∆ε /kT f1 + f 2 = ∆ε /kT + − ∆ε /kT = ∆ε /kT + ∆ε /kT =1 e +1 e +1 e +1 e +1 f 2 = f FD (ε F − ∆ε ) = 1 45. there are two free electrons per atom. Then. Thus.4u)(1.l3 g/cm3 and its atomic mass is 65. The number of atoms per unit volume is the ratio of the mass density ρZn to the mass per atom mZn .4. 【sol】 Zinc in its ground state has two electrons in 4 s subshell and completely filled K. εF h2 = 2m *  3( 2) ρ Zn   8π m  Zn     2 /3  (6.13 × 103 kg/m 3 )    =  2(0.4 u.85 me.85)(9. and M shells.27 kg/u )     = 1.66 × 10. The number of states per electronvolt is 3 9.00 × 10− 3 kg) N = = 9.43 × 1021 states/eV 8 7. Are we justified in considering the electron energy distribution as continuous in a metal? 【sol】 3N (ε F )− 3 /2 ε At T=0. (1.55 u)(1. Inha University Department of Physics . n ε2 = 8 εF ( ) The number of atoms is the mass divided by the mass per atom.00-g sample of copper at O K. the electron distribution n(ε) is n (ε ) = 2 3 N F At ε=εF/2. Find the number of electron states per electronvolt at ε=εF/2 in a 1.04 eV  2  and the distribution may certainly be considered to be continuous.48 × 1021  εF  n  = = 1.04 eV.66 × 10− 27 kg/u ) with the atomic mass of copper from the front endpapers and εF =7.48 × 1021 ( 63.47. 4 m3) by showing that of A<< l under these circumstances.49. the atomic mass of He is 4. where N is the total number of atoms in the sample.57) becomes Vm 3/ 2 n(ε )dε = g (ε ) f (ε )dε = A 4 2π ε e − ε /kT dε h3 Integrating over all energies. Helium atoms have spin 0 and so obey Bose-Einstein statistics verify that f(ε)=1/eαeε/kT≈Ae-ε/kT is valid for He at STP (20oC and atmospheric pressure. and employing the approximation. For energies in the neighborhood of kT. Equation (9. The Bose-Einstein and Fermi-Dirac distribution functions both reduce to the MaxwellBoltzmann function when eαeε/kT>>1.55) for g(ε)dε with a coefficient of 4 instead of 8 since a He atom does not have the two spin states of an electron. this approximation holds if eα>>1.55). N = ∫ ∞ n (ε )dε 0 Vm 3 / 2 = A 4 2π h3 ∫ ∞ 0 ε e − ε /kT d ε Inha University Department of Physics . use Eq(9. To do this. when the volume of 1 kmol of any gas is=22. (A kilomole of He contains Avogadro’s number No atoms.00 u and ∫0 ∞ x e −αx dx = π / a / 2a 【sol】 Using the approximation f(ε)=Ae-ε/kT. and a factor of 4 instead of 8 in Equation (9. find A from the norma1ization condition n(ε)dε=N. 56 ×10 −6 .The integral is that given in the problem with x= ε and a=kT. so that Vm 3/2 π(kT )3 V N = A4 2π = A 3 ( 2πmkT )3/ 2 2 h3 h Solving for A.66 ×10 − 27 kg/u)(1. which is much less than one.. A= N 3 h ( 2πmkT )− 3/ 2 V Using the given numerical values.022 × 1026 kmol −1 (6.626 ×10− 34 J ⋅ s ) 3 × [ 2π( 4.38 1× 10-23 J/K )( 293K )]− 3 / 2 22.00 u)(1. Inha University Department of Physics .4 kg/kmol = 3. ∫ ∞ 0 εe −ε /kT dε = π (kT )3 2 . A= 6. 51. Inha University Department of Physics . Verify this by using the method of Exercise 49 to show that A>1 in copper if f(ε)≈Aexp(ε/kT).48x1028 electrons/m3 for copper.(9. 【sol】 Here. The Fermi-Dirac distribution function for the free electrons in a metal cannot be approximated by the Maxwell-Boltzmann function at STP for energies in the neighborhood of kT. As calculated in Sec. the original factor of 8 must be retained. with the result that A= 1N 3 h ( 2πm ekT ) − 3 / 2 2V = 1 (8.50 ×103 .63 ×10 −34 J ⋅ s ) 3 × [ 2π(9. 9.48 × 1026 m −3 )( 6.11 ×10 −31 )(1.38 ×10 −23 J/K )( 293 K)]− 3 / 2 2 = 3. and so the Fermi-Dirac distribution cannot be approximated by a Maxwell-Boltzmann distribution. Which is much greater than one.55) must be used unchanged here.9 N/V=8. Note that Eq.