AISM-09/C/ATSaskIITians Powered By IITians BRINGiiT on – Study Pack By ASKIITIANS.COM – powered by IITians SUBJECT – CHEMISTRY TOPIC – ATOMIC STRUCTURE COURSE CODE – AISM-09/C/ATS Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 1 AISM-09/C/ATS askIITians Powered By IITians Contents:- Atomic Structure Fundamental Particles..................................3 Discovery of Electron...................................4 Discovery of Proton.....................................8 Discovery of Neutron..................................12 Atomic Terms............................................13 Atomic Models...........................................17 Electronic Magnetic Radiation.......................24 Atomic Spectrum........................................28 Planck‟s Quantum Theory.............................29 Bohr‟s Atomic Model....................................32 Quantum Numbers......................................50 Pauli‟s Exclusion Principle..............................54 Shapes Ans Sizes of Orbitals..........................54 Hund‟s rule of maximum multiplicity................59 Electronic configuration of elements................60 Dual Character............................................68 Derivation of De-Broglie Equation...................69 Heisenberg‟s uncertainty principle……………………..76 Quantum mechanical model of model……………….80 Photoelectric Effect………………………………………………83 Answer to excercises…………………………………………..88 Miscellaneous excercises…………………………………….89 Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 2 AISM-09/C/ATS askIITians Powered By IITians FUNDAMENTAL PARTICLES Atoms are made up-essentially, of three fundamental particles, which differ in mass and electric charge as follows: Electron Symbol e or ep 1 Proton n 1 Neutron Approximate relative 1/1836 mass Approximate relative charge Mass in kg Mass in amu Actual charge/C 9.109534 5.4858026 1.6021892 10 31 4 19 1 +1 1.6726485 10 27 0 1.6749543 10 27 10 10 1.007276471 1.6021892 10 19 1.008665012 0 The atomic mass unit (amu) is 1/12th of the mass of an individual atom of 6C 12 , i.e. 1.660565 10 27 kg. The neutron and proton have approximately equal masses of 1 amu and the electron is about 1836 times lighter; its mass can sometimes be neglected as an approximation. The electron and proton have equal, but opposite, electric charges; the neutron is not charged. The existence of electrons in atoms was first suggested, by J.J. Thomson, as a result of experimental work on the conduction of electricity through gases at low pressures, which produces cathode rays and x-rays, and a study of radioactivity by Becquerel, the Curies and Rutherford. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 3 they do so under very low pressure and then patches of light are seen. When a current of high voltage (10. and the supposition that they existed within atoms came about as a result of Rutherford's experiments in which he bombarded elements with the . Thomson.rays and -rays were given off by radioactive elements. +91-120-4224248 Page 4 .03 mm) blue rays are seen emerging from the case.askiitians. and if it contains negatively charged electrons it must also contain some positively charged particles.J.AISM-09/C/ATS askIITians Powered By IITians An atom is electrically neutral. The passage of electricity through gases as studied by a number of physicists. DISCOVERY OF ELECTRON: CATHODE RAYS During the latter half of the nineteenth century.01 – 0. it was found that while normally dry gases do not conduct an electric current. Crookes and J. info@askiitians. particularly by Faraday. Davy. The neutron was discovered in 1932 by James Chadwick by bombarding beryllium with rays.com Tel: +91-120-4224242.com Email. These rays are called “Cathode Rays”. Ltd Website:www.000 volts) is passed through a gas of air kept at a very low pressure (0. Trans Web Educational Services Pvt. If the metal sheet is too thick to be penetrated the rays cast a shadow. Thomson and others are given below: Cathode rays come out at right angles to the surface of the cathode and move in straight lines.J. Their path is independent on the position of the anode.askiitians. Ltd Website:www.com Tel: +91-120-4224242. +91-120-4224248 Page 5 .com Email. The rays pass through thin sheet of metals.AISM-09/C/ATS askIITians Powered By IITians Cathode Rays Cathode Fluorescence Glass Tube To Vacuum Pump + (Anode) CATHODE RAY EXPERIMENT (DISCHARGE TUBE) Some of the important properties of the cathode rays studied by Sir J. Trans Web Educational Services Pvt. They blacken photographic plates. They produce phosphorescence on certain salts like ZnS and fluorescence on glass. They produce X ray when they strike a metal. info@askiitians. This shows that cathode rays contain material particles having both mass and velocity. They heat a substance on which they fall.com Email.askiitians. The mass of a particle present in cathode rays is found to be 1/1837 of H atom.AISM-09/C/ATS askIITians Powered By IITians The rays ionize a gas through which they pass. Ltd Website:www. No cathode ray was produced when the tube was completely evacuated. This indicates that the particles present in cathode rays are fundamental constituent of all matter.com Tel: +91-120-4224242. Cathode rays are deflected by a magnetic or an electric field showing the particle to be electrically charged. Different gases produce same cathode rays as they have the same e/m (charge/mass) ratio. info@askiitians. the direction of deflection shows that they are negatively charged. Cathode rays contain the smallest unit of negative charge. They rotate a light wheel placed in their paths. Trans Web Educational Services Pvt. This shows that the particle is of sub atomic nature. +91-120-4224248 Page 6 . where we can observe them with a microscope. Trans Web Educational Services Pvt.
[email protected] e/m = 96500/1.com Tel: +91-120-4224242.J. Millikan’s Oil Drop Method: Determination of Charge on an Electron: In 1909. Ltd Website:www. some of which pass through an opening into a viewing chamber. Often these droplets have an electric charge.askiitians.com Email. 108 C/g. In this method a spray of oil droplets is produced by an atomizer. Determination of Velocity and Charge/mass (e/m) ratio of Electrons: Sir J. which is picked up from the friction forming the oil droplets. Thomson (1897) extended the cathode ray experiment for the determination of velocity of electrons and their charge/mass ratio. A droplet may have one or more additional electrons in it. The value of e/m for an electron = 1.008 C/g. giving it a negative charge.J. Thomson named these negatively charged sub atomic particles as electron. For the H+ ion (proton). +91-120-4224248 Page 7 . Millikan measured the charge on an electron by his oil drop method. “A sub atomic particle which is a fundamental constituent of all matter having a mass 1/1837th of a H atom and which carries the smallest unit of negative charge is called an electron”.AISM-09/C/ATS askIITians Powered By IITians Sir J. where the value of e is 1.to charge ratio for the droplet. info@askiitians. Because we already know the mass of the droplet we can find the charge on it. +91-120-4224248 10–19C. The droplet can be suspended between them. Millikan's found that the charge on all droplets could be expressed as whole number multiples of e.602 DISCOVERY OF PROTON: POSITIVE RAYS OR CANAL RAYS Atoms are electrically neutral.AISM-09/C/ATS askIITians Powered By IITians As the droplet falls to the bottom of the chamber. attempts were made to discover the positively charged counterpart of electrons. Hence after the discovery of the negatively charged constituent (electron) of an atom.76 108 = 9.askiitians. We then use the voltage needed to establish this balance to calculate the mass . By using a discharge tube Trans Web Educational Services Pvt. ratio and 'e' we calculate mass of the electron Me e e/m 1.6022 10 19 1.com Tel: +91-120-4224242.104 10-31 kg This very small value shows that the electron is a subatomic particle.com Email. Ltd Website:www.602 10-19 C. we adjust the voltage in the plates so that the electrical attraction upward just balances the force of gravity downward. Thus charge on an electron = 1. Page 8 . it passes between two electrically charged plates. By combining e/m. Cathode Rays Positive Rays Anode Cathode To Vacuum Pump POSITIVE RAYS These are called the positive rays or canal rays. J. When hydrogen gas was filled in the discharge tube the positive charge on the positive rays was equal to the negative charge on an electron. +91-120-4224248 Page 9 .com Email.J. Thomson (1910) measured their charge by mass ratio from which he was able to deduce that these contain positive ions. The positive charge is either equal to or whole number multiple of the charge on an electron. Goldstein (1886) found that some rays passed through these holes in a direction opposite to that of the cathode rays. Trans Web Educational Services Pvt.askiitians.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians containing a perforated cathode. Ltd Website:www. info@askiitians. and the mass was less than the hydrogen atom. Their properties are: They are positively charged. com Tel: +91-120-4224242.com Email.AISM-09/C/ATS askIITians Powered By IITians Unlike cathode rays the properties of positive rays are characteristics of the gas in the tube. Ltd Website:www. info@askiitians. The charge/mass ratio also varies because the change in positive charge on the rays. The deflection of positive rays under the influence of an electric or magnetic field is smaller than that of the cathode rays for the same strength of field. Positive Rays H O H+ O+ Cathode Rays e– e– Trans Web Educational Services Pvt. +91-120-4224248 Page 10 . The lightest of all particles identified in positive rays from different elements was one with a mass very slightly less than that of hydrogen atom (or nearly equal to H atom). The mass of the positive rays depends on the atomic weights or molecular weights of the gases in the discharge tube. It may be either equal to or integral multiple of the charge on an electron.askiitians. Positive rays are atomic or molecular resides from which some electrons have been removed. This shows that the positive rays have a greater mass than that of electrons. The lightest positively charged particle is called a proton (P or P+). The removed electrons constitute the cathode rays and the positive residues form the positive or canal rays. +3… Variable.com Tel: +91-120-4224242. which is a fundamental constituent of all matter having a mass slightly less than that of H atom and which carries a positive charge equal in magnitude to the charge on an electron.askiitians. +91-120-4224248 Page 11 . Comparison of Positive (Canal) Rays and Cathode Rays: Properties Sign of Charge Mag. Ltd Website:www. of Charge Mass e/m Cathode Rays Negative Always –1 Definite value Definite value Canal Rays Positive Mostly +1. but also +2. A proton is denoted by p or p+ of +1p. info@askiitians. depends on ions Variable. “A sub atomic particle. depends on ions DISCOVERY OF NEUTRON After the discovery of electrons and protons.AISM-09/C/ATS askIITians Powered By IITians O2 O2 O2+ O22+ e– 2e– The mass of a proton is very slightly less than that of a H atom. Protons are fundamental constituent of matter because positive rays are produced by all substances. This shows that protons are sub atomic particles. In 1932. is called a proton”.com Email. Chadwick Trans Web Educational Services Pvt. Rutherford (1920) had predicted the existence of a neutral fundamental particle. It was found to have approximately the same mass as the protons.6753 10–27 kg) Mass of a neutron is 1. info@askiitians. The radiation was highly penetrating. is called a neutron”. It is denoted by n or 1 n . +91-120-4224248 Page 12 . Bombardment of beryllium by o carbon and neutrons are emitted.com Email.008930 amu (1. Trans Web Educational Services Pvt.6753 Neutron “A sub atomic particle. which is a fundamental constituent of matter having mass approximately equal to the hydrogen atom and which is electrically neutral. Ltd Website:www.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians bombarded the element Beryllium with particles and noticed the emission of a radiation having the following characteristics. proton and neutron is also a fundamental constituent of atoms (a single exception is 1H atom which does not contain any neutron) 1 10–24g or 1. The name „neutron‟ was given to this sub atomic particle.askiitians. Be particles results in the formation of C 1 0 n At present there are a number of evidences which confirm that like electron. The radiation was unaffected by magnetic and electric fields which show that it is electrically neutral. AISM-09/C/ATS askIITians Powered By IITians Illustration 1: The neutron is attracted towards (A) (B) (C) (D) Solution: Neutron is an uncharged particle.com Tel: +91-120-4224242.askiitians.e. protons and neutrons. info@askiitians. Hence (C) is correct. Isotopes: Trans Web Educational Services Pvt.. Ltd Website:www.com Email. ATOMIC TERM Nuclide: Various species of atoms in general. Nucleons: Sub-atomic particles in the nucleus of an atom. +91-120-4224248 Positive charged particles Negative charged particles Not attracted by any charge None of these Page 13 . i.
[email protected]/C/ATS askIITians Powered By IITians Atoms of an element with the same atomic number but different mass number. having the same mass numbers but different atomic numbers. 15 7 N. when subtracted from A. This. i. Atomic number (Z): The number of protons in the nucleus of an atom. e. 14 6 32 15P and 32 16S .g.. e. 16 8 O.com Tel: +91-120-4224242. Isobars: Atoms. the total number of nucleons. Isotones: Atoms having the same number of neutrons but different number of protons or mass number. gives the number of neutrons. Mass number (A): Sum of the number of protons and neutrons.askiitians. Ltd Website:www.g..e. C. +91-120-4224248 Page 14 ..com Email. Isoelectronic species: Trans Web Educational Services Pvt. g. CN-. Atomic mass unit: Exactly equal to 1/12th of the mass of 6C12 atom..m.u. info@askiitians. uranium X (half life 1. = 1. Hence (A) is correct CNO2(B) (D) O2+ N2+ 10–24 g 931.com Tel: +91-120-4224242.4 min) and uranium Z (half life 6.u. e.5 MeV Trans Web Educational Services Pvt. N2. Ltd Website:www.g.7 hours).m. (a.AISM-09/C/ATS askIITians Powered By IITians Atoms molecules or ions having the same number of electrons. +91-120-4224248 Page 15 .askiitians.66 Illustration 2: The ion that is isoelectronic with CO is (A) (C) Solution: Both CO and CN have 14 electrons.): 1 a. CO. e.com Email.. Nuclear isomers: Atoms with the same atomic and mass numbers but different radioactive properties. 006 1022 6.18 g 30 15 P = 0.
[email protected] P = 15 0. electrons and neutrons are present in 0. +91-120-4224248 Page 16 . Solution: Number of protons = number of electrons = 11 Number of neutrons = atomic mass – number of protons =23–11 = 12 Illustration 4: 30 How many protons.com Email. number of P atoms in 0.418 1022 and number of neutrons = 5. Ltd Website:www. Find the number of neutrons in a neutral atom having atomic mass 23 and number of electrons 11.006 mole = 0.askiitians. of neutrons in one atom = (30 – 15) = 15 0.02 1023 = 5.02 1023 Number of electrons in 0.006 mole of mole 30 15 P =Number of protons in 0.AISM-09/C/ATS askIITians Powered By IITians Illustration 3.418 Exercise 1: Trans Web Educational Services Pvt.006 30 15 6.com Tel: +91-120-4224242.18 g 15 P ? Solution: No.18 = 0.006 mole 30 30 15 Now. J.askiitians. Find the total number and total mass of protons present in 34 mg of NH3.AISM-09/C/ATS askIITians Powered By IITians (i) (ii) Find the mass number of a uninegative ion having 10 neutrons and 10 electrons in one mole of the ions.com Tel: +91-120-4224242. ATOMIC MODELS We know the fundamental particles of the atom. +91-120-4224248 Page 17 . electron Positive sphere This model could not satisfactorily explain the results of scattering experiment carried out by Rutherford who worked with Thomson. info@askiitians. This model is called the plum – pudding model after a type of Victorian dessert in which bits of plums were surrounded by matrix of pudding. Ltd Website:www. Thomson.com Email. in 1904. Now let us see. how these particles are arranged in an atom to suggest a model of the atom. proposed that there was an equal and opposite positive charge enveloping the electrons in a matrix. Thomson’s Model: J. Trans Web Educational Services Pvt. The zinc sulphide screen gives off a visible flash of light when struck by an particle. [For this experiment. +91-120-4224248 Page 18 . Rutherford specifically used particles because they are relatively heavy resulting in high momentum].com Email. Observation: Majority of the –particles pass straight through the gold strip with little or no deflection.0004 cm and determined the subsequent path of these particles with the help of a zinc sulphide fluorescent screen.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians Rutherford’s Model: – particles emitted by radioactive substance were shown to be dipositive Helium ions (He++) having a mass of 4 units and 2 units of positive charge.askiitians. info@askiitians. Rutherford allowed a narrow beam of –particles to fall on a very thin gold foil of thickness of the order of 0. Some –particles are deflected from their path and diverge. as ZnS has the remarkable property of converting kinetic energy of particle into visible light. Trans Web Educational Services Pvt. Ltd Website:www.
[email protected] are deflected at large angles indicates . There is far less difference between air and bullet than there is between gold atoms and distance of nucleus from where the -particle assuming of course that density of a gold atom is evenly distributed.particles must have come closer to or collided the presence of a heavy positively charge body i.particles passed straight through the metal foil indicates the most part of the atom is empty. Ltd Website:www. Trans Web Educational Services Pvt. +91-120-4224248 Page 19 . The fact that few deflections to occur . for such large with a massive positively charged body. The .e..particle returns back through 180 is called distance of closet approach and is given by r0 q1q2 1 4 0 mv 2 2 Conclusions: The fact that most of the . –particles are deflected backwards through angles greater Some were even scattered in the opposite direction at an angle of 180 [Rutherford was very much surprised by it and remarked that “It was as incredible as if you fired a 15–inch shell at a piece of tissue paper and it came back and hit you”].AISM-09/C/ATS askIITians Powered By IITians Very few than 90 .com Tel: +91-120-4224242.askiitians.com Email. 000 have deflected at 180° backwards indicates that volume occupied by this heavy positively charged body is very small in comparison to total volume of the atom.
[email protected]. [Given 1 amu = 1. +91-120-4224248 Page 20 .com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians The fact that one in 20. atomic mass of He = 4 and gold = 79 and 1 4 0 =9 109 Nm2C 2] Solution: We known that ro q1q2 1 4 0 mv 2 2 v2 2q1q2 4 0mr0 Now one He atom has charge (q1) = 2e One gold atom has charge (q2) = 79e Putting these values we get v 2 2 2 1.66 kg.6 10 19 79 1. Illustration 5: An 10 10 –particle is traveling towards gold nuclei returns back through m from it.66 10 10 10 v = 3. What is the velocity of the 10 27 .311 105 m/s Trans Web Educational Services Pvt.particle.6 10 27 19 9 109 4 1. Ltd Website:www.com Email. AISM-09/C/ATS askIITians Powered By IITians Conclusions of -Scattering Experiment: On the basis of the above observation, and having realized that the rebounding follows. All the +ve charge and nearly the total mass of an atom is present in a very small region at the centre of the atom. The atom‟s central core is called nucleus. The size of the nucleus is very small in comparison to the size of the atom. Diameter of the nucleus is about 10 –13cm while the atom has a diameter of the order of 10–8 cm. So, the size of atom is 105 times more than that of nucleus. Most of the space outside the nucleus is empty. The electrons, equal in number to the net nuclear positive charge, revolve around the nucleus with fast speed just like planets around the sun. The centrifugal force arising due to the fast speed of an electron balances the coulombic force of attraction of the nucleus and the Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 -particles had met something even more massive than themselves inside the gold atom, Rutherford proposed an atomic model as Page 21 AISM-09/C/ATS askIITians Powered By IITians electron remains stable in its path. Thus according to him atom consists of two parts (a) nucleus and (b) extra nuclear part. Defects in Rutherford’s Atomic Model: Position of electrons: The exact positions of the electrons from the nucleus are not mentioned. Stability of the atom: Bohr pointed out that Rutherford‟s atom should be highly unstable. According to the law of electro–dynamics, when a charged body moves under the influence of an attractive force, it loses energy continuously in the form of electromagnetic therefore, radiation. The emit + electron should continuously radiation and lose energy. As a result of this a moving electron will come closer and closer to the nucleus and after passing through a spiral path, it should ultimately fall into the nucleus. It was calculated that the electron should fall into the nucleus in less than 10–8 sec. But it is known that electrons keep moving outside the nucleus. To solve this problem Neils Bohr proposed an improved form of Rutherford‟s atomic model. Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 22 AISM-09/C/ATS askIITians Powered By IITians Before going into the details of Neils Bohr model we would like to introduce you some important atomic terms. Illustration 6: Prove that density of the nucleus is constant. Solution: Radius of the nucleus = 1.33 number = 1.33 10–11 A1/3 m Mass Volume kg A 1/ 3 3 10–13 A1/3 cm, where A is the mass Density of nucleus = = A 1.66 10 4 3 (1.33 10 27 11 = ) 1.66 10 27 3 4 3.14 (1.33 10 11 3 ) kg/m3 = constant Thus density of nucleus is constant, independent of the element under consideration. SOME IMPORTANT CHARACTERISTICS OF A WAVE A wave is a sort of disturbance which originates from some vibrating source and travels outward as a continuous sequence of alternating crests and Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 23 info@askiitians. etc.com Email. X–rays. +91-120-4224248 Page 24 . nanometers (1 nm =10–9 m) or Angstrom (1 Å=10–10 m). Frequency ( ): The frequency of a wave is the number of times a wave passes through a given point in a medium in one second. These radiations have wave characteristics and do not require any medium for their propagation. It is denoted by and is expressed in cm. frequency ( ). It is denoted by (nu) and is expressed in cycles per second (cps) or hertz (Hz) 1Hz = 1cps. wavelength ( ). Crest a Crest Trough Trough Electronic Magnetic Radiation: Ordinary light rays.AISM-09/C/ATS askIITians Powered By IITians troughs. Ltd Website:www. Wavelength ( ): The distance between two neighbouring troughs or crests is known as wavelength.askiitians. –rays. m. Every wave has five important characteristics. namely. The frequency of a wave is inversely proportional to its wave length ( ) Trans Web Educational Services Pvt.com Tel: +91-120-4224242. wave number and amplitude (a). velocity (c). are called electromagnetic radiations because similar waves can be produced by moving a charged body in a magnetic field or a magnet in an electric field. It and is expressed in cm–1.com Tel: +91-120-4224242.com Email.askiitians. = 1 or = c Amplitude: It is the height of the crest or depth of the trough of a wave and is denoted by a. c= or = c Wave number is denoted by : It is defined as number of wavelengths per cm. Wavelengths of Electromagnetic Radiations: Trans Web Educational Services Pvt. +91-120-4224248 Page 25 . info@askiitians. Ltd Website:www. It is denoted by c and is expressed in cm sec –1. It determines the intensity or brightness of the beam of light.AISM-09/C/ATS askIITians Powered By IITians 1 or = c Velocity: The distance travelled by the wave in one second is called its velocity. Electromagnetic Spectrum: The arrangement of the various types of electromagnetic radiation in order of increasing or decreasing wavelengths or frequencies is known as electromagnetic spectrum. 1 to 0.AISM-09/C/ATS askIITians Powered By IITians Electromagnetic radiations Radio waves Micro waves Infrared (IR) 1. +91-120-4224248 Page 26 . info@askiitians. transition occurs Trans Web Educational Services Pvt.com Tel: +91-120-4224242.01 0.01 to zero Illustration 7: Find out the longest wavelength of absorption line for hydrogen gas containing atoms in ground state.1.e.1 Visible Ultra violet (UV) X–rays Gamma rays Cosmic rays Wave length (Å) 3 1014 to 3 107 3 109 to 3 106 6 106 to 7600 7600 to 3800 3800 to 150 150 to 0. i. Solution: 1 RZ 2 1 2 n1 1 2 n2 For longest wavelength from n = 1 to n = 2 E should be smallest. Ltd Website:www.com Email.1 0.askiitians. 2157 10–5 cm = 121.02 1023 1 104 J or 96.com Email.6 10–19 = 9. +91-120-4224248 potential difference in volt Page 27 .AISM-09/C/ATS askIITians Powered By IITians n=7 n=6 n=5 n=4 n=3 n=2 n=1 i.e.632 6.askiitians.
[email protected] Tel: +91-120-4224242. 1 = 109673 cm–1 12 1 12 1 22 1 = 109673 3 4 cm–1 = 4 3 109673cm 1 = 1.32 kJ Trans Web Educational Services Pvt. Solution: Energy in joules = charge in coulombs = 1. Ltd Website:www.6 nm Illustration 8: Calculate the energy in kJ per mole of electronic charge accelerated by a potential of 1 volt. Types of Emission Spectra: Continuous spectra: When white light from any source such as sun or bulb is analyzed by passing through a prism. This line constitutes the emission spectrum.askiitians. This dark line constitutes the absorption spectrum.com Email. a dark line will appear in the spectrum. the electron passes from higher to a lower energy level.com Tel: +91-120-4224242. it splits up into seven different wide bands of colour from violet to red (like rainbow). energy is absorbed that means a specific wave length is absorbed. These Trans Web Educational Services Pvt. Consequently. energy is released and a spectral line of specific wavelength is emitted.AISM-09/C/ATS askIITians Powered By IITians Exercise 2: (i) What is the ratio between energies of two radiations one with a wavelength of 6000 Å and other with 2000 Å. (ii) Find the frequency and wave number of a radiation having wavelength of 1000 Å. +91-120-4224248 Page 28 . info@askiitians. If the atom loses energy. Ltd Website:www. ATOMIC SPECTRUM If the atom gains energy the electron passes from a lower energy level to a higher energy level. Ltd Website:www. it emits thermal radiations of different wavelengths or frequency. Each line in the spectrum corresponds to a particular wavelength. PLANCK’S QUANTUM THEORY When a black body is heated. +91-120-4224248 Page 29 .com Tel: +91-120-4224242. The smallest packet of energy is called quantum. To explain these radiations. The main points of quantum theory are Substances radiate or absorb energy discontinuously in the form of small packets or bundles of energy. Line spectra: When an electric discharge is passed through a gas at low pressure light is emitted. Each element gives its own characteristic spectrum.askiitians. info@askiitians. Trans Web Educational Services Pvt. Hence the spectrum is called as continuous spectrum. If this light is resolved by a spectroscope. Max Planck put forward a theory known as Planck‟s quantum theory. It is found that some isolated coloured lines are obtained on a photographic plate separated from each other by dark spaces.com Email. In case of light the quantum is known as photon.AISM-09/C/ATS askIITians Powered By IITians colour are so continuous that each of them merges into the next. This spectrum is called line spectrum. 626 108 cm/s 400 cm–1 Trans Web Educational Services Pvt. Find out its (a) (c) (e) Solution: Wavelength J per photon kJ per mol of photons (b) Frequency (d) kcal per mol of photons (a) (b) (c) 1 or c = 1 = 1 400 cm 1 = 2.3h ………. 2h .com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians The energy of a quantum is directly proportional to the frequency of the radiation. info@askiitians.. Neils Bohr used this theory to explain the structure of atom. (or) E = h where is the frequency of radiation 10–27 erg – sec or and h is Planck‟s constant having the value 6. where „n' is the positive integer.626 A body can radiate or absorb energy in whole number multiples of a quantum h .com Email.5 10–3 cm 400 cm–1 = 1. Ltd Website:www. E 6. +91-120-4224248 Page 30 . Illustration 9.2 10 7 s 1 = = c =3 hc 10–10 cm/s = h c 10–34 Js 3 Ephoton = h = = 6.askiitians. The wave number of a radiation is 400 cm 1.626 10–34 J–sec.nh . 14 kcal mol –1 103 J mol–1) (1 kJ/1000J) = 4. +91-120-4224248 Page 31 .95 = (4.7875 = 1.AISM-09/C/ATS askIITians Powered By IITians = 7.askiitians.95 10–21J 10–21J 10–21J 6. What would be the wavelength of the second photon? Solution: E= hc Wavelength of the 1st photon = E1 = hc 1 1 And that of second photon = E2 = hc 2 2 ETotal = E1 + E2= emitted energy hc hc 1 hc 2 Trans Web Educational Services Pvt. Ltd Website:www.com Email.
[email protected] Tel: +91-120-4224242.022 1023 mol–1 103 J mol–1) (1 kcal/4184 J) For 1 mol of photon energy = 7.7875 kJ mol –1 Illustration 10: A near ultraviolet photon of 300 nm is absorbed by a gas and then re emitted as two photons.95 (d) Ephoton = 7. One photon is red with wavelength 760 nm.7875 (e) E = (4. Ltd Website:www. Important Postulates: An atom consists of a dense nucleus situated at the centre with the electron revolving around it in circular orbits without emitting any energy.com Tel: +91-120-4224242. The force of attraction between the nucleus and an electron is equal to the centrifugal force of the moving electron. an electron can revolve only in those orbits whose angular momentum (mvr) is an integral multiple of factor h 2 mvr = nh 2 where. He applied quantum theory in considering the energy of an electron bound to the nucleus. info@askiitians. Of the finite number of circular orbits around the nucleus.com Email. m = mass of the electron Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians 1 300 1 2 1 760 1 300 1 2 1 760 1 2 760 300 760 300 2 = wavelength of the second photon = 496 nm BOHR’S ATOMIC MODEL Bohr developed a model for hydrogen atom and hydrogen like one–electron species (hydrogenic species).askiitians. +91-120-4224248 Page 32 . This new state of electron is called as excited state. Trans Web Educational Services Pvt.com Tel: +91-120-4224242. Such a stable state of the atom is called as ground state or normal state. atom will lose it‟s energy and come back to the ground state. etc.com Email. (from nucleus onwards) or K.) by absorbing one or more quanta of energy. The greater the distance of the energy level from the nucleus.4. Since the excited state is less stable. Each stationary state is associated with a definite amount of energy and it is also known as energy levels. M. Ordinarily an electron continues to move in a particular stationary state without losing energy. Ltd Website:www. it may jump (excite) instantaneously from lower energy (say 1) to higher energy level (say 2. Hence these orbits are called stationary states. If energy is supplied to an electron. the more is the energy associated with it. info@askiitians. +91-120-4224248 Page 33 . 4. The different energy levels are numbered as 1.3. 3.2. N etc. L.AISM-09/C/ATS askIITians Powered By IITians v = velocity of the electron n = orbit number in which electron is present r = radius of the orbit As long as an electron is revolving in an orbit it neither loses nor gains energy.askiitians. The quantum of energy absorbed is equal to the difference in energies of the two concerned levels. 6 eV then all photons are absorbed and excess energy appears as kinetic energy of emitted photo electron. is the frequency of radiation absorbed or emitted. r is the radius of the orbit in which electron is revolving.askiitians.AISM-09/C/ATS askIITians Powered By IITians Energy absorbed or released in an electron jump.com Tel: +91-120-4224242. Trans Web Educational Services Pvt. and Note: If the energy supplied to hydrogen atom is less than 13..6 eV. ( E) is given by E = E 2 – E1 = h where E1 and E2 are the energies of the electron in the first and second energy levels. But if energy supplied to hydrogen atom is more than 13.com Email. +91-120-4224248 Page 34 . Radius and Energy Levels of Hydrogen Atom: Consider an electron of mass „m‟ and charge „e‟ revolving around a nucleus of charge Ze (where. Z = atomic number and e is the charge of the proton) with a tangential velocity v. it will accept or absorb only those quanta which can take it to a certain higher energy level i. Ltd Website:www.e. info@askiitians. all those photons having energy less than or more than a particular energy level will not be absorbed by hydrogen atom. Ltd Website:www.G. info@askiitians. value of K = 1 dyne cm2 (esu)–2 The centrifugal force acting on the electron is mv 2 r Since the electrostatic force balances the centrifugal force.askiitians. we have mvr = nh 2 Trans Web Educational Services Pvt. units.AISM-09/C/ATS askIITians Powered By IITians By Coulomb‟s Law. mv 2 KZe2 = 2 r r … (1) or v2 = KZe2 mr … (2) According to Bohr‟s postulate of angular momentum quantization. the electrostatic force of attraction between the moving electron and nucleus is Coulombic force = KZe2 r2 K= 1 4 o (where o is permittivity of free space) K = 9 109 Nm2 C–2 In C.com Email. +91-120-4224248 Page 35 . for the stable electron orbit.com Tel: +91-120-4224242.S. 6 10 19 2 9 10 9 =5.529 Å Radius of nth orbit for an atom with atomic number Z is simply written as Trans Web Educational Services Pvt.com Tel: +91-120-4224242. Ltd Website:www. 3 .. The greater the value of n.14 2 6.Hence only certain orbits whose radii are given by the above equation are available for the electron..askiitians.com Email.29 10–11 m=0.e. 2. +91-120-4224248 Page 36 .AISM-09/C/ATS askIITians Powered By IITians v= nh 2 mr 2 2 v2 = n2 h 2 4 m r2 … (3) Equating (2) and (3) KZe2 mr n2h2 4 2m2r 2 Solving for r we get r = n2h2 4π mΚΖe2 2 Where n = 1. i... info@askiitians. ro n2h2 = 2 2 = 4 me K 4 12 3. farther the energy level from the nucleus the greater is the radius.626 10 31 34 2 9 10 1. The radius of the smallest orbit (n=1) for hydrogen atom (Z=1) is ro. AISM-09/C/ATS askIITians Powered By IITians rn = 0.529 n2 Å Z Calculation of Energy of an Electron: The total energy, E of the electron is the sum of kinetic energy and potential energy. Kinetic energy of the electron = ½ mv 2 Potential energy = KZe2 r Total energy = 1/2 mv2 – KZe2 r … (4) From equation (1) we know that mv 2 KZe2 = r r2 ½ mv2 = KZe2 2r Substituting this in equation (4) Total energy (E) = KZe2 KZe2 – = 2r r KZe2 2r Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 37 AISM-09/C/ATS askIITians Powered By IITians Substituting for r, gives us E= 2π2mZ2e4K 2 n2h2 where n = 1, 2, 3………. This expression shows that only certain energies are allowed to the electron. Since this energy expression consist of so many fundamental constant, we are giving you the following simplified expressions. E = –21.8 10–12 Z2 erg per atom n2 = –21.8 10–19 Z2 J per atom n2 = –13.6 Z2 eV per atom n2 (1eV = 3.83 10–23 kcal 1eV = 1.602 10–12 erg 1eV = 1.602 10–19J) E = –313.6 Z2 kcal / mole (1 cal = 4.18 J) n2 Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 38 AISM-09/C/ATS askIITians Powered By IITians The energies are negative since the energy of the electron in the atom is less than the energy of a free electron (i.e., the electron is at infinite distance from the nucleus) which is taken as zero. The lowest energy level of the atom corresponds to n=1, and as the quantum number increases, E becomes less negative. When n = , E = 0, which corresponds to an ionized atom i.e., the electron and nucleus are infinitely separated. H H++ e– (ionization). Calculation of Velocity: We know that mvr = nh nh ;v = 2 mr 2 By substituting for r we get v= 2 KZe2 nh Where except n and Z all are constants Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 39 53 10–10 m) Solution: Trans Web Educational Services Pvt. info@askiitians. Ltd Website:www. In each case of this kind. Illustration 12: Calculate the velocity of an electron in Bohr‟s first orbit of hydrogen atom (Given r = 0.18 106 m / s (B) (D) 1.36 106 m / s None of these vn v0 Z n . Hence (A) is correct.com Tel: +91-120-4224242. to other one electron species (Hydrogenic ion) such as He+ and Li2+.AISM-09/C/ATS askIITians Powered By IITians v = 2.askiitians. +91-120-4224248 Page 40 . Further application of Bohr‟s work was made. Illustration 11: The velocity of electron in the second orbit of He will be (A) (C) Solution: 2. Bohr‟s prediction of the spectrum was correct.18 108 Z n cm/sec.com Email.09 106 m / s 4. visible and IR regions.53 10–10m 10–34kg m2 s–1.62 10 kgm2 s 1 ) 10 = 2. +91-120-4224248 Page 41 .
[email protected]/C/ATS askIITians Powered By IITians According to Bohr‟s theory mvr = nh 2 or v nh 2 mr r = 0.askiitians.45 3 106 m/s HYDROGEN ATOM If an electric discharge is passed through hydrogen gas taken in a discharge tube under low pressure. Solution: 106 m/s. m = 9.1 34 n = 1. h = 6.1 10 7 31 10–31 kg 106 m/s 1 (6.com Tel: +91-120-4224242. it is found to consist of a series of sharp lines in the UV. This series of lines is known as line or atomic Trans Web Educational Services Pvt.19 10 2 = 1.62 v= 22 2 (9.18 m) kg) (0. and the emitted radiation is analyzed with the help of spectrograph.com Email. Ltd Website:www.17 velocity in 3rd orbit of He2+ ion. Calculate the vn = v 0 Z n 6 Hence v = 2.53 10 Illustration 13: The velocity of e in first Bohr‟s orbit is 2. etc Main spectral lines Ultra – violet Visible Infra – red Infra – red Infra – red Trans Web Educational Services Pvt. 5. 6 etc 5. 4. 4. 7 etc 6. The lines in the visible region can be directly seen on the photographic film. 6.com Email. = wave number = wave length R = Rydberg constant (109678 cm–1) n1 and n2 have integral values as follows Series Lyman Balmer Paschen Brackett Pfund n1 1 2 3 4 5 n2 2. 5 etc 4. The wavelength of all these series can be expressed by a single formula. etc 3. each series.AISM-09/C/ATS askIITians Powered By IITians spectrum of hydrogen. Each line of the spectrum corresponds to a light of definite wavelength. Ltd Website:www. Lyman. Brackett. Paschen. 7. 1 =R 1 2 n1 1 n2 2 Where. known after their discoverer as the Balmer. 3. Pfund and Humphrey series. info@askiitians. The entire spectrum consists of six series of lines. +91-120-4224248 Page 42 .com Tel: +91-120-4224242.askiitians. com Tel: +91-120-4224242. info@askiitians. i. +91-120-4224248 Page 43 .AISM-09/C/ATS askIITians Powered By IITians Note: All lines in the visible region are of Balmer series but reverse is not true.askiitians. Illustration 14: Find the wavelength of a spectral line produced when an electron in the H-atoms jumps from 4th level to 2nd level..e. all Balmer lines will not fall in visible region] The pattern of lines in atomic spectrum is characteristic of hydrogen. Solution: 1 RH 1 2 n1 1 n2 2 1 109678 1 22 0 1 cm 42 1 16 cm 3 109678 = 4863 A Merits of Bohr’s Theory: The experimental value of radii and energies in hydrogen atom are in good agreement with that calculated on the basis of Bohr‟s theory. Ltd Website:www. Trans Web Educational Services Pvt.com Email. +91-120-4224248 Page 44 .
[email protected] Tel: +91-120-4224242. each spectral line further splits into a number of lines. According to this principle “It is impossible to determine simultaneously the exact position and momentum of a small Trans Web Educational Services Pvt. It has particle and wave character. It was observed that when the source of a spectrum is placed in a strong magnetic or electric field.askiitians. Limitations of Bohr’s Theory It does not explain the spectra of atoms having more than one electron. Ltd Website:www. Bohr treated the electron only as particle.com Email.AISM-09/C/ATS askIITians Powered By IITians Bohr‟s concept of stationary state of electron explains the emission and absorption spectra of hydrogen like atoms. De Broglie suggested that electrons like light have dual character. This observation could not be explained on the basis of Bohr‟s model. Another objection to Bohr‟s theory came from Heisenberg‟s Uncertainty Principle. Bohr‟s atomic model failed to account for the effect of magnetic field (Zeeman effect) or electric field (Stark effect) on the spectra of atoms or ions. The experimental values of the spectral lines of the hydrogen spectrum are in close agreement with that calculated by Bohr‟s theory. AISM-09/C/ATS askIITians Powered By IITians moving particle like an electron”.11 10–31 kg) (A) (C) Solution: Trans Web Educational Services Pvt.askiitians. Hence (B) is correct. that electrons revolve in well defined orbits around the nucleus with well defined velocities is thus not tenable. The postulate of Bohr.7 nm 9. What is the wavelength of the emitted electron? (me = 9.40 eV (B) (D) 2. Ltd Website:www.com Tel: +91-120-4224242. n2 Illustration 16: An electron in a hydrogen atom in its ground state absorbs 1. Illustration 15: The radii of two of the first four Bohr orbits of the hydrogen atom are in the ratio 1 : 4.20 eV rn r0n2 and En Z E0 Z2 .50 times as much energy as the minimum required for its escape from the atom. +91-120-4224248 4. The energy difference between them may be (A) (C) Solution: 0.7 Å 9.
[email protected] Email.55 eV 8.40 nm Page 45 .4 Å (B) (D) 4.85 eV 3. 6eV 20.
[email protected] 10 10 m 4.6 eV is need for ionization Total energy absorbed = 1.6 10 Hence (A) is correct.5 13.AISM-09/C/ATS askIITians Powered By IITians Since 13.4eV 6. have particle properties (B) are emitted by atoms (D) have wave properties are absorbed by ions Trans Web Educational Services Pvt.625 10 2 9. Ltd Website:www. Hence (B) is correct.com Email.8 1.com Tel: +91-120-4224242.11 10 31 34 19 = 4.E.8 eV is converted into K. h 2m KE 6.askiitians. Illustration 17: The observation that electrons can be diffracted is an evidence that electrons (A) (C) Solution: Diffraction prove wave character of electron.7 Å 6. +91-120-4224248 Page 46 . e.2 nm Illustration 19: Trans Web Educational Services Pvt.askiitians.
[email protected] Email. 434.AISM-09/C/ATS askIITians Powered By IITians Illustration 18: A series of lines in the spectrum of atomic hydrogen lies at wavelengths 656. 1 =R 1 22 1 72 = 109673 1 4 1 49 = 397.17. 410.com Tel: +91-120-4224242. Ltd Website:www.29 10–7 cm and n1 = 2 n2 may be calculated for the last line 1 nm.46. 482. What is the =R 1 2 n1 1 2 n2 1 410.29 10 7 = 109673 1 22 1 n2 2 n2 = 6 Thus next line will be obtained during the jump of electron from 7 th to 2nd shell i.7.2 10–7 cm = 397.29 wavelength of next line in this series? Solution: The given series of lines are in the visible region and thus appears to be Balmer series Therefore n1 = 2 and n2=? for next line If = 410. +91-120-4224248 Page 47 . 178 10–18 (Z2) 1 2 n1 1 J / atom n2 2 For visible photon.com Email.63 1012 ergs Which hydrogen like ionic species has wavelength difference between the first line of Balmer and first line of Lyman series equal to 59. Ltd Website:www. Solution: Trans Web Educational Services Pvt. n1 = 2 For lowest energy transition n2 = 3 E = 2. Solution: E = 2.
[email protected] Illustration 20: 105J = 3.askiitians.023 1023) 2.3 10–9 m? Neglect the reduced mass effect.com Tel: +91-120-4224242.178 For 2.0 gm atom of hydrogen undergo transition giving spectral line of lowest energy in visible region of its atomic spectrum.AISM-09/C/ATS askIITians Powered By IITians Find the energy released in (ergs) when 2. +91-120-4224248 Page 48 .178 10–18 5 36 10–18 12 1 22 1 32 J/atom joules = 3.0g atom E = (2 6. Trans Web Educational Services Pvt. What is the energy required for its excitation to Bohr‟s second orbit.42 10–12 and –2.com Email.askiitians.12 105 J mol–1.AISM-09/C/ATS askIITians Powered By IITians Wave number of first Balmer line of an species with atomic number Z is given by RZ2 5RZ2 36 1 22 1 32 .com Tel: +91-120-4224242. the energy of an electron in first Bohr‟s orbit is – 13. 36 5RZ2 3 4 1 and 4 3 1 = 4 = 1 2 36 2 RZ 5 3RZ = 88 15RZ2 Z2 = or 59.097 107 =9 Z=3 ionic species is Li 2+ Exercise 3: (i) Calculate of of the radiations when the electron jumps from III to II atoms are –5. (ii) In a hydrogen atom. +91-120-4224248 Page 49 . Similarly wave number of v of first Lyman line is given by 1 1 22 = RZ2 12 – = RZ 2 .3 10 9 88 15 1.41 10–12 erg orbit of hydrogen atom. The electronic energy in II and III Bohr orbit hydrogen respectively. Ltd Website:www. info@askiitians. 677 cm 1) QUANTUM NUMBERS An atom contains large number of shells and subshells.AISM-09/C/ATS askIITians Powered By IITians (iii) What is the wavelength of the light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =? What is the colour corresponding to this wavelength? (Rydberg constant = 109. the type of orbital occupied and orientation of that orbital. It also tells the maximum number of electrons that a shell can accommodate is 2n 2. It tells us the address of the electron i.askiitians. These are distinguished from one another on the basis of their size. +91-120-4224248 M 3 N 4 Page 50 . where n is the principal quantum number.
[email protected] Tel: +91-120-4224242. the approximate distance of the electron from the nucleus and energy of that particular electron.e. The parameters are expressed in terms of different numbers called quantum numbers. Ltd Website:www. energy. location. Principal quantum number (n): It tells the main shell in which the electron resides. shape and orientation (direction) in space. Shell K L Principal quantum number (n) 1 2 Trans Web Educational Services Pvt.. Quantum numbers may be defined as a set of four numbers with the help of which we can get complete information about all the electrons in an atom.com Email. 1. p.… or s.askiitians. 4. +91-120-4224248 Page 51 . Ltd Website:www. d. The values allowed depends on the value of l. the electrons of a subshell can orient themselves in certain preferred regions of space around the nucleus called orbitals. The orbital angular momentum of the electron is given as particular value of „n‟ where possible 1 h 2 or 1 for a h 2 . This electric field is expected to produce a magnetic field. 2.AISM-09/C/ATS askIITians Powered By IITians Maximum number of electrons 2 8 18 32 Azimuthal or angular momentum quantum number : This represents the number of subshells present in the main shell. info@askiitians. 3. These subsidiary orbits within a shell will be denoted as 0. For a given value of n values of vary from 0 to n – 1. The magnetic quantum number (m): An electron due to its angular motion around the nucleus generates an electric field. +1 for associated with a particular value of = 1. Thus m can be –1.com Tel: +91-120-4224242. the angular momentum quantum number. Under the influence of external magnetic field. The magnetic quantum number determines the number of preferred orientations of the electron present in a subshell.com Email. Total values of m is given by 2 + 1. The spin quantum number (s): Just like earth which not only revolves around the sun but also spins about its own axis. 0. an electron in an atom not only revolves around the nucleus but also spins about Trans Web Educational Services Pvt. f… This tells the shape of the subshells. m can assume all integral values between – to + including zero. . When an electron goes to a vacant orbital.. 32 i. i.com Tel: +91-120-4224242. This quantum number helps to explain the magnetic properties of the substances. Illustration 21: If the principal quantum number n has a value of 3. i.askiitians.e. for any particular value of magnetic quantum number.e.e. i. spin quantum number can have two values.e. Since an electron can spin either in clockwise direction or in anticlockwise direction. +1/2 and –1/2 or these are represented by two arrows pointing in the opposite directions. what are permitted values of the quantum numbers „‟ and „m‟? Solution: If principal quantum number (n) has the value of 3. and . therefore. +1/2 or –1/2.AISM-09/C/ATS askIITians Powered By IITians its own axis.e.. Ltd Website:www. subsidary quantum number () will also have three values i. 0. 9 values in all and they are designated as under: n=3 = 0 m = 0 (s) = 1 m = – 1 (p) m=0 m=+1 = 2 m = – 2 (d) Trans Web Educational Services Pvt.com Email. +91-120-4224248 Page 52 .e. info@askiitians. 1 and 2 and magnetic quantum number (m) will have n 2. it can have a clockwise or anti clockwise spin i. Hence electron will residue in 4s orbital. 1. info@askiitians. m 0 n = 4. 0 for s subshell.com Tel: +91-120-4224242. +91-120-4224248 Page 53 . Hence electron will residue in 3pz orbital.askiitians. m 0 Solution: (i) (ii) 1 for p subshell and m = 0 for pz orbital.com Email. 0. (i) (ii) n = 3. Ltd Website:www. Illustration 23: Write all the quantum numbers for the following orbitals. (i) 3 d z 2 (ii) 3s (iii) 4px Solution: Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians m=–1 m=0 m = +1 m = +2 Illustration 22: In which orbital the electron will reside if it has the following values of quantum numbers. Two electrons in an orbital can be represented by or SHAPES AND SIZE OF ORBITALS An orbital is the region of space around the nucleus within which the probability of finding an electron of given energy is maximum (90–95%). Ltd Website:www. an orbital can contain a maximum number of two electrons and these two electrons must be of opposite spin. m 0.com Email. m 1 m . It is basically determined by the azimuthal quantum number . Trans Web Educational Services Pvt. +91-120-4224248 Page 54 . 1/ 2 Pauli’s Exclusion Principle: According to this principle. while the orientation of orbital depends on the magnetic quantum number (m). s 0.askiitians.com Tel: +91-120-4224242. 4. 0. s 1 or 1/ 2 1/ 2 1s . The shape of this region (electron cloud) gives the shape of the orbital. Let us now see the shapes of orbitals in the various subshells.AISM-09/C/ATS askIITians Powered By IITians (i) (ii) (iii) n n n 3. 2. 3. info@askiitians. Hence.com Email. The size of the orbital depends upon the value of principal quantum number (n). p–orbitals ( =1): The probability of finding the p–electron is maximum in two lobes on the opposite sides of the nucleus. where the probability of finding electron is zero. 0. m = –1.com Tel: +91-120-4224242. info@askiitians. 2s–orbital is larger than 1s orbital but both of them are non-directional and spherically symmetrical in shape. py & pz depending upon whether the density of electron is maximum along the x y and z axis respectively.AISM-09/C/ATS askIITians Powered By IITians s–orbitals: These orbitals are spherical and symmetrical probability of about the nucleus. Therefore. Greater the value of n. These are designated as p x. Trans Web Educational Services Pvt. The two lobes of p–orbitals are separated by a nodal plane. p–orbital have three different orientations. larger is the size of the orbital.askiitians. they have directional character. Ltd Website:www. +1. Thus. As they are not spherically symmetrical. This gives rise to a dumb–bell shape for the p–orbital. But there is no radial node for 1s orbital since it is starting from the nucleus. electron The is y finding the maximum near the nucleus and keep on decreasing as the distance from the nucleus increases. There is vacant space between two successive s–orbitals known as radial node. +91-120-4224248 Page 55 . For p–orbital 1s nucleus x 2s Z radial node = 1. Out of the five orbitals.
[email protected]. l = 2.AISM-09/C/ATS askIITians Powered By IITians Y py Z X px pz The three p-orbitals belonging to a particular energy shell have equal energies and are called degenerate orbitals.dzx) project in between the axis and the other two dz dx2 y2 2 and lie along the axis. +91-120-4224248 Page 56 . Hence m = –2. +1.–1. 0. +2.com Tel: +91-120-4224242. They have relatively complex geometry. d–orbitals ( = 2): For d–orbitals.com Email. Z X Y dz 2 dx 2 y2 Dough–nut shape or Baby soother shape X X Clover leaf shape Y Y dxy dxz Z dyz Z Trans Web Educational Services Pvt. Thus there are 5d orbitals. the three (dxy. Ltd Website:www. dyz. +91-120-4224248 Page 57 .com Email.com Tel: +91-120-4224242.askiitians. dz 2 Hence it has electron density in all xy.AISM-09/C/ATS askIITians Powered By IITians Illustration 24: An electron has a spin quantum number number 1. Hence (D) is correct. xz and yz plane. Ltd Website:www. It cannot occupy a/an (A) (C) Solution: For s orbital m = 0. Illustration 25: d orbital p orbital (B) (D) f orbital s orbital 1 2 and a magnetic quantum Which d-orbital has electron density in all the planes? Solution: orbital has lobe along Z-axis and a belt of electrons in xy plane. Trans Web Educational Services Pvt. info@askiitians. RULES FOR FILLING OF ELECTRONS IN VARIOUS ORBITALS The atom is built up by filling electrons in various orbitals according to the following rules. 5f.6s.3s.2s. Trans Web Educational Services Pvt.4d.6d.4s.AISM-09/C/ATS askIITians Powered By IITians Aufbau Principle: This principle states that the electrons are added one by one to the various orbitals in order of their increasing energy starting with the orbital of lowest energy.com Tel: +91-120-4224242.
[email protected]. the direction of the arrow gives the order of filling of orbitals.3p.5s.4p. The increasing order of energy of various orbital is 1s. Starting from the top. Ltd Website:www.com Email.askiitians.7p…………………… How to remember such a big sequence? To make it simple we are giving you the method to write the increasing order of the orbitals. the order of increasing energies of the various orbitals can be calculated on the basis of (n + ) rule.4f. 1s 2s 3s 4s 5s 6s 7s 2p 3p 4p 5p 6p 3d 4d 5d 4f Alternatively.6p.2p.5p. +91-120-4224248 Page 58 .5d. Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians The energy of an orbital depends upon the sum of values of the principal quantum number (n) and the azimuthal quantum number ( ).com Email. Ltd Website:www.com Tel: +91-120-4224242. info@askiitians. “Electron pairing in p. According to this rule. However. “In neutral isolated atom. Illustration of (n + ) Rule: Type of Value Values Values of orbitals of n 1s 2s 2p 3s 1 2 2 3 of (n+ ) 1+0=1 2+0=2 2+1=3 3+1=4 Relative energy Lowest energy Higher energy than 1s orbital 2p orbital (n=2) have lower energy than 3s orbital (n=3) 0 0 1 0 Hund’s Rule of Maximum Multiplicity: This rule deals with the filling of electrons in the equal energy (degenerate) orbitals of the same sub shell (p.askiitians. +91-120-4224248 Page 59 . This is called (n + ) rule. if the two different types of orbitals have the same value of (n + ). According to this rule. the lower the value of (n + ) for an orbital.d and f orbitals cannot occur until each orbital of a given subshell contains one electron each or is singly occupied”. lower is its energy. the orbitals with lower value of n has lower energy‟‟. d and f). info@askiitians. electronic configuration of any orbital can be simply represented by the notation] Number of electrons in the subshell x n Principal quantum number Symbol of subshell or orbitals (s.com Email. All the electrons in a degenerate set of orbitals will have same spin. This repulsion can.AISM-09/C/ATS askIITians Powered By IITians This is due to the fact that electrons being identical in charge. subshells and orbitals of an atom.d.f) Alternatively: Orbital can be represented by a box and an electron with its direction of spin by arrow. Electronic Configuration of Elements: Electronic configuration is the distribution of electrons into different shells. +91-120-4224248 Page 60 .com Tel: +91-120-4224242. sub–shell or orbital. just we need to know (i) the atomic number (ii) the order in which orbitals are to be filled (iii) maximum number of electrons in a shell.askiitians. Each orbital can accommodate two electrons Trans Web Educational Services Pvt. To write the electronic configuration. be minimized if two electrons move as far apart as possible by occupying different degenerate orbitals. Keeping in view the above mentioned rules. however. repel each other when present in the same orbital.p. Ltd Website:www. AISM-09/C/ATS askIITians Powered By IITians The number of electrons to be accomodated in a subshell is 2 numbers of degenerate orbitals. +91-120-4224248 Page 61 . info@askiitians. L. Importance of Knowing the Electronic Configuration: The chemical properties of an element are dependent on the relative arrangement of its electrons. Subshell s p d f Maximum number of electrons 2 6 10 14 The maximum number of electrons in each shell (K.com Tel: +91-120-4224242.askiitians. The maximum number of orbitals in a shell is given by n 2 where n is the principal quantum number. Ltd Website:www.com Email. Illustration 26: Write the electronic configuration of nitrogen (atomic number = 7) Trans Web Educational Services Pvt. N…) is given by 2n2 where n is the principal quantum number. M. AISM-09/C/ATS askIITians Powered By IITians Solution: 1s 2 2s 2 2p 3 Illustration 27: Write the electronic configuration of following: (i) S2 z 16 (ii) Fe2 z 26 Solution: (i) (ii) 1s2 2s2 2p6 3s2 3p6 1s2 2s2 2p6 3s2 3p6 3d6 Exceptional Configurations: Stability of Half Filled and Completely Filled Orbitals: Cu has 29 electrons. Ltd Website:www. info@askiitians. +91-120-4224248 Page 62 . Its expected electronic configuration is 1s22s22p63s23p64s23d9 But a shift of one electron from lower energy 4s orbital to higher energy 3d Trans Web Educational Services Pvt.com Tel: +91-120-4224242.askiitians.com Email. Illustration 29: The total spin resulting from a d7 configuration is (A) (C) 1 5 2 Mn Cu (B) (D) Cr Zn (B) (D) 2 3 2 Trans Web Educational Services Pvt. Thus the electronic configuration of Cu is 1s2 2s22p63s23p64s13d10 Fully filled and half filled orbitals are more stable. info@askiitians. Illustration 28: Which of the following metals has the highest value of exchange energy required for exchange stabilization? (A) (C) Solution: Cu has full filled d-subshell. Ltd Website:www.com Tel: +91-120-4224242.askiitians. Hence (C) is correct.com Email.AISM-09/C/ATS askIITians Powered By IITians orbital will make the distribution of electron symmetrical and hence will impart more stability. +91-120-4224248 Page 63 . askiitians. Hence (B) is correct. info@askiitians. where n = number of unpaired electrons = 3. This represents its (A) (C) Solution: It is ground state electronic configuration of Cr.com Email.com Tel: +91-120-4224242. Illustration 30: The electronic configuration of an element is 1s 22s22p63s23p63d54s1.AISM-09/C/ATS askIITians Powered By IITians Solution: S = n/2. Illustration 31: Write the electronic configuration of the following ions: (A) (C) Solution: H+ O2 (B) (D) Na+ F excited state cationic form (B) (D) ground state anionic form Trans Web Educational Services Pvt. Hence (D) is correct. +91-120-4224248 Page 64 . Ltd Website:www. all the singly occupied orbitals should have parallel spins. info@askiitians. 2px 2py 2pz ½ ½ ½ = 1½ The maximum multiplicity means that the total spin of unpaired electrons is maximum.e in the same direction eitherclockwise or anticlockwise. +91-120-4224248 Page 65 .askiitians. Explain how to arrange three electrons in p orbitals.com Tel: +91-120-4224242. i. Ltd Website:www. Trans Web Educational Services Pvt.com Email.AISM-09/C/ATS askIITians Powered By IITians (A) (B) (C) (D) 1s0 1s22s22p6 1s22s22p6 1s22s22p6 Illustration 32: We know the Hund‟s rule. Solution: The important point to be remembered is. 2p6.3d5.askiitians.AISM-09/C/ATS askIITians Powered By IITians Illustration 33: We know that fully filled and half filled orbitals are more stable. +91-120-4224248 Page 66 . The energy of 4s(4 + 0 = 4) is lower than that of 3d (3 + 2 = 5).com Email. Trans Web Educational Services Pvt. Solution: Cr (Z = 24) 1s2.3s2. Can you write the electronic configuration of Cr(Z = 24)?. Illustration 34: The configuration of potassium (19) is 1s22s22p63s24s1 and not 1s22s22p63s23p63d1. Why? Solution: According to Aufbau Principle‟s electron first fill in a sub-level of lower energy level (lower n+).3p6. 2s2.com Tel: +91-120-4224242. info@askiitians. Ltd Website:www.4s1. so the electronic configuration of K is 1s22s22p63s23p64s1. Since half filled orbital is more stable one 4s electron is shifted to 3d orbital. Illustration 37: Calculate the total number of d-electrons in molybdenum (atomic number = 42).AISM-09/C/ATS askIITians Powered By IITians Illustration 35: How many maximum number of electrons can be present in 3rd shell of an atom? Solution: Total number of electrons in a shell = 2n 2 =2 Illustration 36: How many maximum number of electrons a d-orbital can have? Solution: Any orbital can have maximum of 2 electrons.com Tel: +91-120-4224242. +91-120-4224248 Page 67 .com Email.askiitians. Ltd Website:www. Solution: 32 = 18 Trans Web Educational Services Pvt. info@askiitians. askiitians.
[email protected]/C/ATS askIITians Powered By IITians Electronic configuration of Mo is 1s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d4 Hence. the certain other phenomenon such as black body radiation and photoelectric effect can be Trans Web Educational Services Pvt. = 2 n = 2. total number of d-electrons = 10 + 4 = 14 Exercise 4: (i) Write the electronic configuration of the following: (a) Cr3+ (b) Mn2+ (c) Cu (ii) Write the orbital rotation for the following quantum numbers. +91-120-4224248 Page 68 . = 1 n = 3.com Tel: +91-120-4224242.com Email. (a) (b) (c) n = 4. However. = 0 DUAL CHARACTER (PARTICLE AND WAVE CHARACTER OF MATTER AND RADIATION) In case of light some phenomenon like diffraction and interference can be explained on the basis of its wave character. Ltd Website:www. askiitians. +91-120-4224248 Page 69 . Derivation of de-Broglie Equation: The wavelength of the wave associated with any material particle was calculated by analogy with photon. in 1924 extended the idea of photons to material particles such as electron and he proposed that matter also has a dual character-as wave and as particle. In case of photon. Ltd Website:www. its energy is given by E=h …(i) (according to the Planck‟s quantum theory) where is the frequency of the wave and „h‟ is Planck‟s constant If the photon is supposed to have particle character. light is said to have a dual character. its energy is given by E = mc2 … (ii) Trans Web Educational Services Pvt.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians explained only on the basis of its particle nature. info@askiitians. if it is assumed to have wave character. Louis de Broglie. Thus.com Email. Such studies on light were made by Einstein in 1905. E (E)=1/2 mv2 Trans Web Educational Services Pvt. „c‟ is the velocity of light. Ltd Website:www.com Email. = h/mv or = h p where mv = p is the momentum of the particle.AISM-09/C/ATS askIITians Powered By IITians (according to Einstein‟s equation) where „m‟ is the mass of photon.com Tel: +91-120-4224242. +91-120-4224248 Page 70 . By equating (i) and (ii) h But = mc2 = c/ h c = mc2 (or) = h /mc The above equation is applicable to material particle if the mass and velocity of photon is replaced by the mass and velocity of material particle. info@askiitians. Relation Between Kinetic Energy and Wavelength: K. Thus for any material particle like electron.askiitians. the electron revolves around the nucleus in circular orbits. the circumference of the orbit must be equal to an integral multiple of wave length ( ) Therefore 2 r = n where „n‟ is an integer and „r‟ is the radius of the orbit But = h/mv or mvr = n h/2 2 r = nh /mv which is Bohr‟s postulate of angular momentum. the electron is not only a particle but has a wave character also. If the wave is completely in phase.com Tel: +91-120-4224242.
[email protected]/C/ATS askIITians Powered By IITians v= 2E m h mv h 2E. where „n‟ is the principal quantum number. +91-120-4224248 Page 71 .com Email.m Derivation of Angular Momentum from de Broglie Equation: According to Bohr‟s model. Ltd Website:www. Trans Web Educational Services Pvt. According to de Broglie concept.askiitians.
[email protected] Tel: +91-120-4224242. +91-120-4224248 Page 72 .AISM-09/C/ATS askIITians Powered By IITians “Thus. (i. The distance it has to travel for one revolution 2 r. the circumference of the circle).e. Thus.askiitians. This energy equals the product of voltage and charge. the number of waves an electron makes in a particular Bohr orbit in one complete revolution is equal to the principal quantum number of the orbit”.. respectively. Alternatively: Number of waves „n‟ = 2 r = 2 mvr 2 r = h h mv Where v and r are the velocity of electron and radius of that particular Bohr orbit in which number of waves are to be calculated. Ltd Website:www.com Email. The electron is revolving around the nucleus in a circular orbit. How many revolutions it can make in one second? Let the velocity of electron be v m/sec. Since in SI units Trans Web Educational Services Pvt. the number of revolutions per second is = v 2 r Common unit of energy is electron volt which is amount of energy given when an electron is accelerated by a potential of exactly 1 volt. 1 60 = 1.6 10 34 0.com Tel: +91-120-4224242. Hence (B) is correct. +91-120-4224248 Page 73 .com Email.1 kg moving with a speed of 60 ms–1.AISM-09/C/ATS askIITians Powered By IITians coulombs volts = joules. Illustration 39: Calculate the de Broglie wavelength of a ball of mass 0. 4 E and E respectively.E .E of 16 E. a proton and an alpha particle have K. Trans Web Educational Services Pvt. info@askiitians. 1 eV numerically equals the electronic charge except that joule replaces coulombs. Solution: h mv = 6. What‟s the qualitative order of their Broglie wavelengths? (A) (C) Solution: e P > < P e > < a a (B) (D) e a > < p e = = a P = h 2mK.askiitians. Ltd Website:www.1 10–34 m. Illustration 38: An e-. com Tel: +91-120-4224242.wave nature of a sub atomic particle]: Photon 1. Although the de Broglie equation is applicable to all material objects but it has significance only in case of microscopic particles. 2.AISM-09/C/ATS askIITians Powered By IITians This is apparent that this wavelength is too small for ordinary observation. +91-120-4224248 Page 74 . Since. [Distinction between the wave.askiitians. Energy = h Wavelength = c Sub Atomic Particle Energy = 1 mv2 2 h mv Wavelength = Note: We should never interchange any of the above Illustration 40: Calculate the number of waves made by electron in 4th Bohr‟s orbit. Ltd Website:www. de Broglie relationship has no significance in everyday life. we come across macroscopic objects in our everyday life.particle nature of a photon and the particle.com Email. Solution: Trans Web Educational Services Pvt. info@askiitians. m 6.askiitians. of 5 eV? Solution: K.1 10 31 = 5. +91-120-4224248 Page 75 . Ltd Website:www.486 Illustration 42: 10 10 m Through what potential difference an electron must be accelerated to have a de-Broglie wavelength of 1 A .E. info@askiitians. m h mv Now.E.E.AISM-09/C/ATS askIITians Powered By IITians Number of waves made by electron in n th orbit = n Hence the number of waves in 4rd orbit = 4 Illustration 41: What is the de-Broglie wavelength of electron having K.com Email. o Trans Web Educational Services Pvt. = = h 2K.E. = v 1 mv 2 2 2K.62 10 2 5 1.com Tel: +91-120-4224242.6 10 19 34 9. askiitians. Heisenberg.AISM-09/C/ATS askIITians Powered By IITians Solution: = 10 10 m 1 mv 2 2 eV We know K. Page 76 . move along definite paths..62 10 2 10 10 34 2 h 2 1.com Email. a ball thrown in the air etc.6 10 19 9. Ltd Website:www.5 106 volt HEISENBERG’S UNCERTAINTY PRINCIPLE All moving objects that we see around us e. a car. in 1927 gave a principle about the uncertainties in simultaneous measurement of position and momentum (mass This Principle States: Trans Web Educational Services Pvt. Is it possible for subatomic particle also? As a consequence of dual nature of matter. = v= Now Hence v 2eV m h mv h 2eVm h2 2 2 em 6.g.com Tel: +91-120-4224242.E. +91-120-4224248 velocity) of small particles.1 10 31 = 1. Hence their position and velocity can be measured accurately at any instant of time. info@askiitians.. Thus.com Email. The uncertainty in position is . v where m is the mass of the particle and uncertainty in velocity) is equal to or greater than h/4 Planck‟s constant. p h/4 v is the where h is the Explanation of Heisenberg’s uncertainty principle Suppose we attempt to measure both the position and momentum of an electron. the other becomes less accurate. Ltd Website:www. As a result of the hitting. The shorter the wavelength..com Tel: +91-120-4224242. to pinpoint the position of the electron we have to use light so that the photon of light strikes the electron and the reflected photon is seen in the microscope. if an attempt is made to measure any one of these two quantities with higher accuracy. The accuracy with which the position of the particle can be measured depends upon the wavelength of the light used.AISM-09/C/ATS askIITians Powered By IITians “It is impossible to measure simultaneously the position and momentum of a small microscopic moving particle with absolute accuracy or certainty” i.e.askiitians. the mathematical expression for the Heisenberg‟s uncertainty principle is simply written as x. info@askiitians. the position as well as the velocity of the electron are disturbed. the greater is the accuracy. The product of the uncertainty in position ( x) and the uncertainty in the momentum ( p = m. But shorter wavelength means higher frequency and hence Trans Web Educational Services Pvt. +91-120-4224248 Page 77 .
[email protected]/C/ATS askIITians Powered By IITians higher energy. Hence Heisenberg‟s uncertainty principle is not applicable to macroscopic particles. Illustration 43: Why electron cannot exist inside the nucleus according to Heisenberg‟s uncertainty principle? Solution: Diameter of the atomic nucleus is of the order of 10 –15m The maximum uncertainty in the position of electron is 10 –15 m. Ltd Website:www.1 10 31 v = 5.com Tel: +91-120-4224242. Mass of electron = 9. p = x v= h 4 (m.80 possible. This high energy photon on striking the electron changes its speed as well as direction. x. But this is not true for macroscopic moving particle.1 10–31 kg. v) = h/4 h 4 1 6.com Email.63 10 = 22 x.m 4 7 34 10 15 1 9.askiitians. +91-120-4224248 Page 78 . Illustration 44: 1010 ms–1 This value is much higher than the velocity of light and hence not Trans Web Educational Services Pvt. if uncertainity in its velocity is 0.1 10 31 0.0058 x = 0.askiitians.002 0.0058 m/s? Solution: x v h 4 m x 6. Ltd Website:www.01 m Illustration 45: What is the uncertainity in the position of a ball of mass 10 g and which is moving with a velocity of 100 m/s with 0.
[email protected] 10 34 4 3.1 10 31 0.14 9.002 m/ s Now x 6.02 10 34 4 3.002 % uncertainity? Solution: x v h 4 m v 100 0.com Tel: +91-120-4224242.002 100 0. +91-120-4224248 Page 79 .0289 m Exercise 5: Trans Web Educational Services Pvt.14 9.com Email.AISM-09/C/ATS askIITians Powered By IITians What is the uncertainity in the position of electron. +91-120-4224248 Page 80 . The differential equation is 2 2 2 2 2 x y z 2 8 2m (E V) h2 0 where x. (iii) Calculate the uncertainty in position of a particle when uncertainty in the momentum is (a) 1 10–2 gm cm sec–1 and (b) zero. This model describes the electron as a three dimensional wave in the electronic field of positively charged nucleus.
[email protected]/C/ATS askIITians Powered By IITians (i) Two particles A and B are in motion. If the wavelength associated with particle A is 5 10–8 m. QUANTUM MECHANICAL MODEL OF ATOM The atomic model which is based on the particle and wave nature of the electron is known as wave or quantum mechanical model of the atom. V = potential energy of the electron. calculate the wavelength associated with particle B if its momentum is half of A.com Email.com Tel: +91-120-4224242. Trans Web Educational Services Pvt. y. z are certain coordinates of the electron. m = mass of the electron E = total energy of the electron. This was developed by Erwin Schrodinger in 1926. (ii) Calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV. Ltd Website:www. h = Planck‟s constant and (psi) = wave function of the electron. Schrodinger derived an equation which describes wave motion of an electron.askiitians.
[email protected]. Nodal Points and Planes: The point where there is zero probability of finding the electron is called nodal point. it is thus. The wave function may have positive or negative values depending upon the value of coordinates.AISM-09/C/ATS askIITians Powered By IITians Significance of : The wave function may be regarded as the amplitude function expressed in terms of coordinates x. There are two types of nodes: Radial nodes and angular nodes. The main aim of Schrodinger equation is to give solution for probability approach. +91-120-4224248 Page 81 . proper to use in Significance of : 2 is a probability factor. When the equation is solved. It describes the probability of finding an electron within a small space. The space in which there is maximum probability of finding an electron is termed as orbital.com Email.com Tel: +91-120-4224242. But the probability must 2 be always positive and cannot be negative. Ltd Website:www. it is observed that for some regions of space the value of favour of . Trans Web Educational Services Pvt. The important point of the solution of the wave equation is that it provides a set of numbers called quantum numbers which describe energies of the electron in atoms. The former is concerned with distance from the nucleus while latter is concerned with direction. 2 is negative. information about the shapes and orientations of the most probable distribution of electrons around nucleus. y and z. AISM-09/C/ATS askIITians Powered By IITians No. of radial nodes = n – – 1 No. of angular nodes = Total number of nodes = n – 1 Nodal planes are the planes of zero probability of finding the electron. The number of such planes is also equal to . Illustration 46: Calculate radial nodes and angular nodes for the following type of orbitals. (a) (c) (e) Solution: (a) (c) (e) 0, 0 1, 1 3, 0 (b) (d) (f) 0, 1 0, 2 1, 2 1s 3p 4s (b) (d) (f) 2p 3d 4d Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 82 AISM-09/C/ATS askIITians Powered By IITians PHOTOELECTRIC EFFECT Sir J.J. Thomson, observed that when a light of certain frequency strikes the surface of a metal, electrons are ejected from the metal. This phenomenon is known as photoelectric effect and the ejected electrons are called photoelectrons. A few metals, which are having low ionization energy like Cesium, show this effect under the action of visible light but many more show it under the action of more energetic ultraviolet light. Light + electrons – Evacuated quartz tube V A An evacuated tube contains two electrodes connected to a source of variable voltage, with the metal plate whose surface is irradiated as the anode. Some of the photoelectrons that emerge from this surface have enough energy to reach the cathode despite its negative polarity, and they constitute the measured current. The slower photoelectrons are repelled before they get to Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 83 AISM-09/C/ATS askIITians Powered By IITians the cathode. When the voltage is increased to a certain value V 0, of the order of several volts, no more photoelectrons arrive, as indicated by the current dropping to zero. This extinction voltage (or also referred as stopping potential) corresponds to the maximum photoelectron kinetic energy i.e., eVo = ½ mv2 The experimental findings are summarised as below: Electrons come out as soon as the light (of sufficient energy) strikes the metal surface. The light of any frequency will not be able to cause ejection of electrons from a metal surface. There is a minimum frequency, called the threshold (or critical) frequency, which can just cause the ejection. This frequency varies with the nature of the metal. The higher the frequency of the light, the more energy the photoelectrons have. Blue light results in faster electrons than red light. Photoelectric current is increased with increase in intensity of light of same frequency, if emission is permitted i.e., a bright light yields more photoelectrons than a dim one of the same frequency, but the electron energies remain the same. Light must have stream of energy particles or quanta of energy (h ). Suppose, the threshold frequency of light required to eject electrons from a Trans Web Educational Services Pvt. Ltd Website:www.askiitians.com Email.
[email protected] Tel: +91-120-4224242, +91-120-4224248 Page 84 E + ½ mv2 0 0 ½ mv2 = h –h where. when a photon of light of this frequency strikes a metal it imparts its entire energy (h 0) to the electron. o = frequency of the incident light Trans Web Educational Services Pvt. +91-120-4224248 Page 85 .Emax = h – h 0 “This energy enables the electron to break away from the atom by overcoming the attractive influence of the nucleus”. h =h h =h + K.askiitians. The excess energy would give a certain velocity (i. E=h 0 E>h 0 K.e.AISM-09/C/ATS askIITians Powered By IITians metal is 0. kinetic energy) to the electron.
[email protected] Email.com Tel: +91-120-4224242. If the frequency of light is higher than 0 0 there is no (let it be ). Thus each photon can eject one electron. the photon of this light having higher energy (h ). will impart some energy to the electron that is needed to remove it from the atom. If the frequency of light is less than ejection of electron. Ltd Website:www. E = 0 Metal K.
[email protected] of Photoelectrons Illustration 47: Work function of sodium is 2. if the energy of the ejected electrons is plotted as a function of frequency. Predict whether the wavelength 6500 A is suitable for a photoelectron ejection or not. K. It is constant for particular metal and is also equal to the ionization potential of gaseous atoms. it result in a straight line whose slope is equal to Planck‟s constant „h‟ and whose intercept is h 0. Solution: o Trans Web Educational Services Pvt. The kinetic energy of the photoelectrons increases linearly with the frequency of incident light. +91-120-4224248 Page 86 .5 eV. Ltd Website:www.AISM-09/C/ATS askIITians Powered By IITians 0 = threshold frequency = h h 0 is the threshold energy (or) the work function denoted by 0 (minimum energy of the photon to liberate electron). Thus.com Email.com Tel: +91-120-4224242. askiitians. Exercise 6: (i) Find the threshold wavelengths for photoelectric effect from a copper surface. 2.7 nm Trans Web Educational Services Pvt.3 eV and 1.com Tel: +91-120-4224242.5 eV.24 eV.9 eV 10 19 J Which is lower than work function.62 10 34 5 108 6500 10 10 = 3. +91-120-4224248 Page 87 . Hence no ejection will take place. Ltd Website:www.055 = 1.
[email protected] Email. (ii) Energy required to stop the ejection of electrons from Cu plate is 0.9 eV respectively. Calculate the work function when radiation of strikes the plate? = 253. The work function of these metal are 4. a sodium surface and a caesium surface.AISM-09/C/ATS askIITians Powered By IITians Energy of incident light hc 6. com Email.84 105 J mol –1 486 nm.com Tel: +91-120-4224242.
[email protected] 1022 10–5 kg (i) (ii) E1 E2 v hc 1 2 2 1 hc 2000 6000 107 m 1 1 3 3 1015 Hz.askiitians. Blue Exercise 4: (i) (a) (b) (c) 1s2 2s2 2p6 3s2 3p6 3d3 1s2 2s2 2p6 3s2 3p6 3d5 1s2 2s2 2p6 3s2 3p6 3d10 4s1 Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians ANSWER TO EXERCISES Exercise 1: (i) (ii) 19 (a) (b) Exercise 2: 1. Exercise 3: (i) (ii) (iii) 6603 Å 9.2044 2. +91-120-4224248 Page 88 . Ltd Website:www. 87 (a) (b) Exercise 6: (i) (ii) 276 nm.com Tel: +91-120-4224242.askiitians.AISM-09/C/ATS askIITians Powered By IITians (ii) (a) (b) (c) 4d 2p 3s Exercise 5: (i) (ii) (iii) 10–7 m 3. +91-120-4224248 Page 89 .com Email. Ltd Website:www.
[email protected] eV 10–11 m 5. 540 nm. 654 nm 4.27 10–28m Trans Web Educational Services Pvt. com Tel: +91-120-4224242. Exercise 4: The wave number of the first line of Balmer series of hydrogen is 15. +91-120-4224248 Page 90 . What is the wave number of the first line of Balmer series for the Li 2+ ion? Exercise 5: Trans Web Educational Services Pvt. Exercise 2: What is the fraction of volume occupied by the nucleus with respect to the total volume of an atom? Exercise 3: Calculate the minimum and maximum values of wavelength in Balmer series of a H atom.
[email protected] cm 1.com Email. Ltd Website:www.AISM-09/C/ATS askIITians Powered By IITians MISCELLANEOUS EXERCISES Exercise 1: Calculate the ratio of specific charge (e/m) of a proton and that of an -particle.askiitians. (given Planck constant h = 6. +91-120-4224248 Page 91 .askiitians.
[email protected] Tel: +91-120-4224242.0 g if the uncertainty in position is 10 10 34 5 m. Exercise 8: Calculate the de Broglie wavelength of a tennis ball of mass 60.63 Exercise 9: 10 34 Js). Ltd Website:www.6 Js). h = 6.com Email.AISM-09/C/ATS askIITians Powered By IITians What possible can be the ratio of the de Broglie wavelengths for two electrons having the same initial energy and accelerated through 50 volts and 200 volts? Exercise 6: What is likely to be the principle quantum number for a circular orbit of diameter 20 nm of the hydrogen atom if we assume Bohr orbit to be the same as that represented by the principle quantum number? Exercise 7: Calculate the uncertainty in the velocity of a particle weighing 25.0 g moving with a velocity of 10 metres per second (Planck constant. Trans Web Educational Services Pvt. 1 eV. If hydrogen atom in ground state excited by monochromatic light of energy 12. +91-120-4224248 Page 92 . Ltd Website:www. How many photons of green light ( needed to generate this minimum amount of energy? Exercise 10: Ionization potential of hydrogen atom 13.com Email.com Tel: +91-120-4224242. then what will be the total spectral lines emitted according to Bohr‟s theory? = 550 nm) are Trans Web Educational Services Pvt.
[email protected]/C/ATS askIITians Powered By IITians Suppose 10–17 J of energy is needed by the interior of human eye to see an object.6 eV.askiitians. Ltd Website:www.36.
[email protected]/C/ATS askIITians Powered By IITians ANSWERS TO MISCELLANEOUS EXERCISES Exercise 1: 1:2 Exercise 2: 10 15 Exercise 3: 3647 nm and 6564 nm Exercise 4: 1.askiitians.com Email. +91-120-4224248 1 Page 93 .800 cm Exercise 5: 2:1 Exercise 6: 14 Trans Web Educational Services Pvt.com Tel: +91-120-4224242. com Tel: +91-120-4224242.
[email protected]/C/ATS askIITians Powered By IITians Exercise 7: 2.com Email. +91-120-4224248 Page 94 .1 10 28 Exercise 8: 10 33 m Exercise 9: 28 Exercise 10: 3 Trans Web Educational Services Pvt. Ltd Website:www.askiitians. Ltd Website:www.
[email protected] Tel: +91-120-4224242. Problem 3: Trans Web Educational Services Pvt.com Email.AISM-09/C/ATS askIITians Powered By IITians SOLVED PROBLEMS Subjective: Board Type Questions Problem 1: Why are Bohr‟s orbits are called stationary states? Solution: This is because the energies of orbits in which the electrons revolve are fixed. +91-120-4224248 Page 95 . Solution: In the 3d104s1 the d-subshell is completely filled which is more stable. Problem 2: Explain why the electronic configuration of Cu is 3d 104s1 and not 3d94s2.askiitians. +91-120-4224248 Page 96 . Problem 4: Calculate the accelerating potential that must be applied to a proton beam to give it an effective wavelength of 0.com Email. Why? Solution: In Fe3+ ion 3d subshell is half filled hence more stable configuration.AISM-09/C/ATS askIITians Powered By IITians Fe3+ ion is more stable than Fe2+ ion.askiitians.com Tel: +91-120-4224242. Solution: Trans Web Educational Services Pvt. info@askiitians. Solution: v eV h m 1 mv 2 2 Putting the values we get V = 32.005 nm. Ltd Website:www.85 volt Problem 5: Give one example of isodiapheres. AISM-09/C/ATS askIITians Powered By IITians Isodiapheres have same difference between the number of neutrons and protons.askiitians.com Email. info@askiitians. For example 39 19 n p 1 K & 31 15 n p 1 P Trans Web Educational Services Pvt. +91-120-4224248 Page 97 .com Tel: +91-120-4224242. Ltd Website:www. Ltd Website:www. Solution: He RZ2 1 42 1 62 = 4R H 36 16 36 16 = 1 5R 36 1 =R 12 22 n2 vH R n2 vHe 5R R = 4 36 On solving above equation n2 = 9 n=3 Or corresponding transition from 3 atom has same frequency as that of 6 Problem 7: Trans Web Educational Services Pvt.com Email.AISM-09/C/ATS askIITians Powered By IITians IIT Level Questions Problem 6: Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n = 6 to n = 4 transition in He +. [Neglect reduced mass effect]. info@askiitians. +91-120-4224248 2 in Balmer series of hydrogen 4 transition in He+.askiitians.com Tel: +91-120-4224242. Page 98 . AISM-09/C/ATS askIITians Powered By IITians Calculate ionization potential in volts of (a) He + and (b) Li2+ Solution: I. +91-120-4224248 Page 99 . = –2K.com Email.E.
[email protected] Z2 [Z =2 for He+] 12 = 13.4 eV 13. Ltd Website:www.E and P.6 Problem 8: 9 = 122.askiitians.6Z2 n2 13.E. = = 13.6 4 = 54.E 1 =P. = Ze2 2r Ze2 r P.6 3 2 12 Similarly for Li 2+ = = 13.E 2 Problem 9: Trans Web Educational Services Pvt. = P.E.E of an electron in an orbit? Solution: K.E.E K.4 eV Calculate the ratio of K.com Tel: +91-120-4224242.
[email protected] Tel: +91-120-4224242. Ltd Website:www.0 (h = 6.askiitians. +91-120-4224248 Page 100 .com Email.63 105 ms–1 and (b) relative de Broglie wavelength of an atom of hydrogen and atom of oxygen moving with the same velocity 10–34 kg m2 s–1) Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians How many spectral lines are emitted by atomic hydrogen excited to nth energy level? Solution: 1 1 2 3 2 3 1 4 5 6 3 n=6 n=5 n=4 n=3 n=2 n=1 1 1+2=3 1+2+3= 6 Thus the number of lines emitted from nth energy level = 1 + 2 + 3 +………… n – 1 = n= n n 1 2 (n – 1) (n – 1) = n 1 n 1 1 2 = n 1 n 2 Number of spectral lines that appear in hydrogen spectrum when an electron jumps from nth energy level = n n-1 2 Problem 10: Calculate (a) the de Broglie wavelength of an electron moving with a velocity of 5. 46 10–9m An atom of oxygen has approximately 16 times the mass of an atom of hydrogen.63 10 9. Solution: K.E. Ltd Website:www. also constant.askiitians.0 10 ms Wavelength (b) = 1. = P. +91-120-4224248 Page 101 . for for the hydrogen atom would be 16 times greater than oxygen atom.11 10 31 34 kgm2 s 2 5 1 kg 5. In the formula h mv . h is constant while the conditions of problem make v. This means that and m are variables and varies inversely with m. Problem 11: A 1 MeV proton is sent against a gold leaf (Z = 79).com Tel: +91-120-4224242. 1 mv 2 2 1 Ze2 4 0 d Ze2 1 4 0 mv 2 2 Hence d = Trans Web Educational Services Pvt. info@askiitians. Calculate the distance of closest approach for head on collision.com Email.E.AISM-09/C/ATS askIITians Powered By IITians Solution: (a) = h mv = 6. Therefore. Solution: Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians 9 109 79 1.
[email protected] 10 2 9. 6. respectively.6 10 13 J m Problem 12: What is the wavelength associated with 150 eV electron Solution: = = h 2 m K.E.42 10–12 erg and –2.6 10 13 19 2 1.1 10 31 34 Js 19 = J 6.com Email. +91-120-4224248 Page 102 . Ltd Website:www. Calculate the wavelengths of emitted radiation when the electron drops from third to second orbit.41 10–12erg.6 10 Problem 13: The energy of electron in the second and third Bohr orbit of the hydrogen atom is –5.6 10 1.com Tel: +91-120-4224242.626 10 4368 10 34 50 = 10–10 m = 1 Å kg 150 1.137 10 13 as 1 MeV 1. 414 10–19J Trans Web Educational Services Pvt.42 10–12) = 3 1010 27 3 1010 = 6.268 10–19 + 3. +91-120-4224248 Page 103 . What is the maximum wavelength effective for photochemical dissociation of O2? Solution: O2 O2 ON + Oexcited ON + ON E = 498 103 J / mole = 498 103 J 6.626 10 12 3.01 10 = 6. The dissociation of O2 into two normal atoms of oxygen atoms requires 498kJ mole–1.146 10–19J Total energy required for photochemical dissociation of O2 = 8.023 1023 per molecule = 8.604 10–5 108 = 6604 Å O2 undergoes photochemical dissociation into one normal oxygen and one excited oxygen atom.414 10–19 J hc = 11.268 10–19 J Energy required for excitation = 1. Ltd Website:www.967 eV more energetic than normal.
[email protected] Email.AISM-09/C/ATS askIITians Powered By IITians E3 – E2 = h = 6.626 10 27 hc – 2.604 Problem 14: 10–5 cm = 6.askiitians.146 10–19 = 11.com Tel: +91-120-4224242. 1.967 eV = 3.41 10–12 – (– 5. Ltd Website:www.AISM-09/C/ATS askIITians Powered By IITians 34 3 108 = 6.414 10 Problem 15: Compare the wavelengths for the first three lines in the Balmer series with those which arise from similar transition in Be 3+ ion.5 Å 19 11.626 10 = 1.askiitians.com Email. (Neglect reduced mass effect). info@askiitians. 16 Trans Web Educational Services Pvt. Solution: vH R 12 1 22 1 22 1 n2 1 n2 vBe R 42 vBe vH H Be = 16 So we can conclude that all transitions in Be 3+ will occur at wavelengths 1 times the hydrogen wavelengths.7415 10–7 m = 1741.com Tel: +91-120-4224242. +91-120-4224248 Page 104 . askiitians. info@askiitians. orbital angular moment is (A) (C) Solution: 2 6 (B) (D) 2 Orbital angular momentum L = L for p electron = 1(1 1) (A) Problem 2: 2 ( 1) where h 2 For which of the following species.AISM-09/C/ATS askIITians Powered By IITians Objective: Problem 1: For a p-electron.com Tel: +91-120-4224242. +91-120-4224248 H Li2+ (B) (D) He+ Na+ Page 105 . Bohr theory doesn‟t apply (A) (C) Solution: Bohr theory is not applicable to multi electron species (D) Trans Web Educational Services Pvt. Ltd Website:www.com Email. askiitians.com Email.com Tel: +91-120-4224242. +91-120-4224248 Page 106 . Ltd Website:www. info@askiitians. The radius of third Bohr orbit will be (A) (C) 4 r2 9 9 r2 4 (B) (D) 4r2 9r2 Solution: r n2h2 4 2mZe2 r2 r3 22 32 9 r3 = 4 r2 (C) Problem 4: Number of waves made by an electron in one complete revolution in 3rd Bohr orbit is (A) (C) 2 4 (B) (D) 3 1 Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians Problem 3: If the radius of 2nd Bohr orbit of hydrogen atom is r2. circumference of 3rd orbit is three times the wavelength of electron or number of (B) Problem 5: RH is 16 waves made by Bohr electron in one complete revolution in 3rd orbit is three.askiitians. info@askiitians. = h mv = 2 r3 3 2 r3 = 3 i. +91-120-4224248 Page 107 . The degeneracy of the level of hydrogen atom that has energy (A) (C) Solution: 16 2 (B) (D) 4 1 Trans Web Educational Services Pvt. Ltd Website:www.com Email.e.AISM-09/C/ATS askIITians Powered By IITians Solution: Circumference of 3rd orbit = 2 r3 According to Bohr‟s angular momentum of electron in 3 rd orbit is mvr3 = 3 h 2 or h mv 2 r3 3 By de-Broglie equation.com Tel: +91-120-4224242. AISM-09/C/ATS askIITians Powered By IITians En = RH n2 RH n2 RH 16 i.55 10 9. Ltd Website:www.
[email protected] 25 10–25 2 4.1 10 31 103 = 7. for 4th sub-shell 1 n= 4 l = 0 m=0 one s -1 0 three p +1 -2 –1 0 five d +1 +2 –3 –2 –1 0 Seven f +1 +2 +3 2 3 i. What will KE = v2 = 1 2 mv2 = 4.com Tel: +91-120-4224242.28 10–7 m 3 10–5 m 10–25 J. +91-120-4224248 Page 108 .askiitians.55 be de Broglie wavelength for this electron? (A) (C) Solution: 5.e.28 10–7 m Trans Web Educational Services Pvt.1 10 31 1 106 v = 103 m/s de Broglie wavelength = h mv = 6.626 10 34 9. 1 + 3 + 5 + 7 = 16 Degeneracy is 16 Problem 6: An electron is moving with a kinetic energy of 4.com Email.28 10–7 m 2 10–10 m (B) (D) 7. AISM-09/C/ATS askIITians Powered By IITians (B) Problem 7: Suppose 10–17J of energy is needed by the interior of human eye to see an object.6 = 28 photons (B) Problem 8: The two electrons present in an orbital are distinguished by (A) (C) Solution: Trans Web Educational Services Pvt.askiitians.com Email.626 10 34 3 108 = 27. How many photons of green light ( needed to generate this minimum amount of energy? (A) (C) Solution: Let the number of photons required = n n hc 10 17 = 550 nm) are 14 39 (B) (D) 28 42 17 n = 10 hc = 10 17 550 10 9 6. +91-120-4224248 principal quantum number magnetic quantum number (B) (D) azimuthal quantum number spin quantum number Page 109 . Ltd Website:www.com Tel: +91-120-4224242. info@askiitians. Its velocity in the second orbit would be 1.AISM-09/C/ATS askIITians Powered By IITians They are distinguished by their spin.com Email.18 (A) (C) Solution: We know that velocity of electron in nth Bohr‟s orbit is given by = 2. Z = 1 v1 = v2 = (A) Problem 10: 2.09 2 106 m/s Trans Web Educational Services Pvt. Ltd Website:www.09 106 ms–1 5. +91-120-4224248 Page 110 .76 106 ms–1- m/s For H.com Tel: +91-120-4224242.18 106 m/s 1 2.38 106 ms–1 8.5 105 ms–1 (B) (D) 4.askiitians. info@askiitians. (D) Problem 9: The velocity of electron in the ground state hydrogen atom is 2.18 106 m/s = 1.18 106 Z n 106 ms–1. 18 10–18J) 2.askiitians.18 10 n2 18 J ( ionization energy of H = 2. +91-120-4224248 Page 111 .com Tel: +91-120-4224242. Different electromagnetic radiation have different wavelengths. The energy of an electron in its second orbit would be –1.AISM-09/C/ATS askIITians Powered By IITians The 2.09 10–18 J –4.18 10 22 18 J = –5. Ltd Website:www.18 (A) (C) Solution: ionization energy of the (B) (D) ground state hydrogen atom is 10–18J. hence different energy Trans Web Educational Services Pvt.45 10–19J (D) True and False Problem 11: All electromagnetic radiations have same energy.com Email.36 10–18J –2.18 10–18J –5.
[email protected] 10–19J Energy of electron in first Bohr‟s orbit of H–atom E= E2 = 2. Solution: False. info@askiitians. Problem 14: Isoelectronic species have same electronic configuration.askiitians. Solution: False. Li2+. Cathode rays contain electrons which remains same with every gas. He +. Ltd Website:www. Bohr‟s model is applicable for species having one electron like H.AISM-09/C/ATS askIITians Powered By IITians Problem 12: Bohr‟s model is applicable for H-atom only. Solution: Trans Web Educational Services Pvt.com Email. Problem 13: Specific charge of cathode rays remains same irrespective of the gas used in the discharge tube. +91-120-4224248 Page 112 .com Tel: +91-120-4224242. Solution: True. Isotopes have same atomic number but different atomic mass. Ltd Website:www.askiitians.com Email. Problem 15: Atoms with same atomic number but different number of neutrons are called isotopes. Trans Web Educational Services Pvt. Solution: True.
[email protected]/C/ATS askIITians Powered By IITians True. Isoelectronic species have same number of electrons.com Tel: +91-120-4224242. +91-120-4224248 Page 113 . Solution: Trans Web Educational Services Pvt.com Tel: +91-120-4224242. Ltd Website:www. Solution: Decreases Problem 18: If the magnetic moment of an ion An+ is 5.9.com Email.askiitians. the energy difference between adjacent energy levels in H-atom _________. +91-120-4224248 Page 114 .AISM-09/C/ATS askIITians Powered By IITians Fill in the Blanks Problem 16: The mass of positron is ___________ electron. info@askiitians. then the number of unpaired electrons present in An+ are ____________. Solution: Equal to Problem 17: With increasing principle quantum number. The energy required to abstract s-electrons is ____________ than for p electrons. Ltd Website:www.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians Five. info@askiitians. n1 3 Trans Web Educational Services Pvt. Applying = n n 2 . Solution: 16410 Å For Paschen series n2 4. +91-120-4224248 Page 115 . where n 5 Problem 19: s-electrons are more penetrating than p-electrons.com Email. Solution: Greater Problem 20: The wavelength of the first spectral line in Paschen series for H-atom is____________.askiitians. What are drawbacks of Rutherford‟s model atom? Explain why uncertainty principle is significant only for the motion of subatomic particles but not for macroscopic objects. Write down the electronic configuration of Mn 4+ and Cr3+. Write the number of protons. Ltd Website:www.askiitians. info@askiitians. 0 .com Email.com Tel: +91-120-4224242. 7. How many electrons in an atom have the following quantum numbers? (i) n = 4. m = 0 (ii) n = 2. Distinguish between absorption and emission spectra. Write correct orbital notation for each of the following sets of quantum numbers. (i) n = 1. How many unpaired electrons are present in them? 5. +91-120-4224248 Page 116 . 0 2. 4. 3.AISM-09/C/ATS askIITians Powered By IITians ASSIGNMENT PROBLEMS Subjective: Level – O 1. 1 . Trans Web Educational Services Pvt. electrons and neutrons in nitride ion. m = 1 6. ms = 1/2 (ii) n = 3. AISM-09/C/ATS askIITians Powered By IITians 8. Explain why half filled and full filled configurations are more stable. What is the maximum number of lines obtained when the excited electrons of a hydrogen atom in n = 6 drops to the ground state. 9. 13. Calculate the mass and charge of one mole of electrons.askiitians. +91-120-4224248 Page 117 . The wavelength of a moving body of mass 0. Calculate its kinetic energy h 6. Why we do not see a car moving in the form of a wave on a road? Why e/m ratio in case of anode rays depends upon the type of gas? Why splitting of spectral lines take place when the source giving the spectrum is placed in a magnetic field? 11. info@askiitians. Calculate the radius of the third orbit of a hydrogen atom. Trans Web Educational Services Pvt. Ltd Website:www. o 16.31 10 29 m.com Email.1 mg is 3. Will the negative and positive charged particles in cathode rays and anode rays be same? 17. 10. 12. In one discharge tube H2 gas is taken and in other O2 gas is taken.com Tel: +91-120-4224242. Calculate (a) wave number and (b) frequency of yellow radiations having wavelength of 5800 A . 15.625 10 34 J sec 14. com Tel: +91-120-4224242. info@askiitians. Some energy is absorbed by hydrogen atom due to which an electron in it jumped from ground state to the state having principle quantum number 5 and again jumped back to the original level. +91-120-4224248 Page 118 . Ltd Website:www.AISM-09/C/ATS askIITians Powered By IITians 18. What is the name of series? Trans Web Educational Services Pvt.askiitians.com Email. What type of spectrum is observed and in which region. 11 10 31kg. How the electron in the ground state of hydrogen atom is excited by means of monochromatic radiation of wavelength 970.c 3 10 10cmsec o 1 2. 4. Calculate the velocity of an electron revolving in the second orbit of a hydrogen atom 6. How many different lines are possible in the resulting emission spectrum? Find the longest wavelength among these? Trans Web Educational Services Pvt.com Email. +91-120-4224248 o Page 119 .AISM-09/C/ATS askIITians Powered By IITians Level – I 1.63 10 34 Js) 3.8 pm (Mass of electron 9. Calculate the energy of an electron in the second Bohr orbit of an excited hydrogen atom 5. Ltd Website:www. info@askiitians. Calculate the kinetic energy of a moving electron which has a wavelength of 4.h 6. Calculate de – Broglie wavelength of an electron moving with 1% of the speed of light. How many photons of light having a wavelength of 4000 A are necessary to provide 1 J of energy? (h = 6.63 10 27 ergsec.askiitians. What is the mass of a photon of sodium light with a wavelength of 5890 A ? h 6.6 A . c = 3 108 m/s) o 7.com Tel: +91-120-4224242.63 10-34Js. II and III permitted electron Bohr orbits in a hydrogen atom.6 10– kg m2sec–1) (c) Calculate the de-Broglie wavelength of electron accelerated through 100 volt. (a) (b) What is wavelength of a particle of mass 1 g moving with a velocity of 200 m/s? A moving electron has 4.com Tel: +91-120-4224242.com Email. A single electron system has ionization energy 11180 kJ / mol. Find the number of protons in the nucleus of the system. info@askiitians. Calculate the wavelength of the second line and the limiting line in Balmer series. 9. 11. +91-120-4224248 Page 120 . The wavelength of the first line in the Balmer series is 656 nm. Ltd Website:www.AISM-09/C/ATS askIITians Powered By IITians 8. Calculate the radii of the I. Also calculate the number of revolutions per second that this electron makes around the nucleus.1 10–31 kg and h = 6.55 34 10-25 joules of kinetic energy. 10. What are the corresponding values in the case of a singly ionised helium atom? Trans Web Educational Services Pvt. Calculate the velocity of an electron placed in the third orbit of hydrogen atom. Calculate its wavelength (mass = 9.askiitians. 12. com Tel: +91-120-4224242.02%. 1022 15.com Email. Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians 13. info@askiitians. A monochromatic source of light operating at 600 watt emits 2 photons per second. Ltd Website:www.askiitians. Calculate energy in kcal/mole necessary to remove an electron in a hydrogen atom in fourth principal quantum number to infinity. +91-120-4224248 Page 121 . 14. Find the wavelength of the light. Calculate the uncertainty in the position of an electron if it has a speed of 500 m/s with an uncertainity of 0. (a) The electron energy in hydrogen atom is given by E = – 21. A doubly ionised lithium atom is hydrogen like with an atomic no.com Tel: +91-120-4224242. Calculate de-Broglie wavelength of an electron moving with a speed of nearly 1 th that of light (3 20 108 ms–1) 4.askiitians.6 eV.09 Å. (a) (b) What Amount of accelerating potential is needed to produce an electron beam with an effective wavelength of 0. (a) Find the wavelength of the radiation required to excite the electron in Li from the first to the third Bohr orbit (ionisation energy of the hydrogen atom is equal to 13. The bulb is rated as 150 watt and 8% of the energy is emitted as light. Calculate the ionization energy for Li 2+ and Be3+ in the first excited state. Calculate the energy required to remove an electron completely from n = 2 orbit.com Email.AISM-09/C/ATS askIITians Powered By IITians Level – II 1. Ltd Website:www.7 10–12/n2 erg. +91-120-4224248 Page 122 . What is the longest wavelength (in cm) of light that can be used to cause this transition? (b) Ionization energy of hydrogen atom is 13. A bulb emits light of wave length 4500 Å. How many photons are emitted by bulb per second? 3.6 eV) Trans Web Educational Services Pvt. 3. 2. info@askiitians. of element Trans Web Educational Services Pvt. It is exposed to electromagnetic waves of 1028 Å and given out induced radiations. atom has nuclear charge +Ze where Z is atomic number and e is electronic charge. The study of spectra indicates that 27% of the atoms are in 3 rd energy level and 15% of atoms in 2nd energy level and the rest in ground state.com Email.AISM-09/C/ATS askIITians Powered By IITians (b) How many spectral lines are observed in the emission spectrum of the above excited system? 5. 7. info@askiitians. +91-120-4224248 Page 123 .41 10–12erg respectively.8 g hydrogen atoms are excited to radiations. Find (a) The atomic no.42 10–12 and –2.askiitians. It requires 16. Calculate of the radiations when the electron jumps from III to II orbit of hydrogen atom. Calculate (i) No. 1. 8. IP of H is 13. Find the wavelength of these induced radiations. The IP of H is 13. 6.6 eV.com Tel: +91-120-4224242. A single electron beam.6 eV. What is uncertainty in velocity of an electron if uncertainty in its position is 1Å? 9. The electronic energy in II and III Bohr orbit of hydrogen atoms are – 5. of atoms present in III and II energy level (ii) Total energy evolved when all the atoms return to ground state. Ltd Website:www.52 eV to excite the electron from the second Bohr orbit to third Bohr orbit. info@askiitians. Trans Web Educational Services Pvt. The energy required for transition of electron from first to third orbit.AISM-09/C/ATS askIITians Powered By IITians (b) (c) (d) 10.com Email. +91-120-4224248 Page 124 . Find out (a) (b) (c) The principal quantum numbers of initially excited electrons. Ltd Website:www. When the sample was exposed with radiation of energy 2. A sample of hydrogen gas containing some atoms in one excited state emitted three different types of photons.com Tel: +91-120-4224242. The kinetic energy of electron in first Bohr orbit.88 eV it emitted 10 different types of photons.05 eV.askiitians. The maximum and minimum energies of initially emitted photons. all having energy equal or less than 13. The principal quantum numbers of electrons in final excited state. Wavelength required to remove electron from first Bohr orbit to infinity. +91-120-4224248 Page 125 . 1:4 1:8 (B) (D) 1:1 1 : 16 Bohr model can explain spectrum of (A) (B) (C) (D) the hydrogen atom only all elements any atomic or ionic species having one electron only the hydrogen molecule 3. Which is the correct order of probability of being found close to the nucleus is (A) (C) s p p d d f f s (B) (D) f d d f p p s s 4. info@askiitians. Ltd Website:www. Ratio between longest wavelengths of H atom in Lyman series to the shortest wavelength in Balmer series of He + is (A) (C) 4 3 1 4 (B) (D) 36 5 5 9 Trans Web Educational Services Pvt.com Email.askiitians.AISM-09/C/ATS askIITians Powered By IITians Objective: Level – I 1. The ratio of energy of the electron in ground state of the hydrogen to electron in first excited state of He+ is (A) (C) 2.com Tel: +91-120-4224242. then de-Broglie wavelength of electron in 3rd orbit is nearly. info@askiitians. +91-120-4224248 Page 126 . The number of orbitals in a sub-shell are given by (A) (C) 2 2 + 1 (B) (D) n2 2n2 9.AISM-09/C/ATS askIITians Powered By IITians 5. the uncertainty in velocity would be (A) (C) 1 h 2m h (B) (D) h 2 h m 7.askiitians. (A) (C) 2 a1 9 a1 (B) (D) 6 a1 16 a1 6. If the radius of first Bohr orbit is a. the orbital angular momentum is (A) (C) 6 h/ 2 (B) (D) 2 h/ 2 h/2 2h/ 2 Trans Web Educational Services Pvt. If uncertainty in position and momentum are equal. Which of the following is a coloured ion? (A) (C) Cu+(aq) Cu2+ (aq) (B) (D) Na+(aq) K+ (aq) 8. Ltd Website:www. For a „d‟ electron.com Email.com Tel: +91-120-4224242. 11. Hydrogen atom consists of a single electron but so many lines appear in the spectrum of atomic hydrogen because (A) (B) (C) (D) Sample contains some impurity Experiment is done on collection of atoms. In what region of the electromagnetic spectrum does this line occur? (A) (C) far UV near IR (B) (D) near UV far IR 12. +91-120-4224248 Page 127 . The energy involved when 1. If a shell is having g sub-shell. Ltd Website:www. One molecule of a substance absorbs one quantum of energy. which is correct statement about principal quantum number n of this shell.AISM-09/C/ATS askIITians Powered By IITians 10.
[email protected] 105J (B) (D) 3.5 mole of the substance absorbs red light of frequency 7. Hydrogen atom splits to form more than one different species.028 10-2 nm-1 is emitted in the spectrum of atomic hydrogen.5 1014 sec–1 will be (A) (C) 2. A line with wave number 1.99 105 J 4. Some different isotope of hydrogen atom may be present.com Email. (A) (C) n 5 (B) (D) n 5 n=5 Cannot be determined Trans Web Educational Services Pvt.com Tel: +91-120-4224242.23 105J 2.askiitians.99 106 J 13. 87 Å (B) (D) 1397.93 10–23 kJ none of these 15. The total energy of electron in an atom Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians 14.5 Å none of these 16. The spectrum of He is expected to be similar to that of (A) (C) H Be2+ (B) (D) Li+ H+ 19. The wave number of a spectral line is 5 corresponding to this line will be (A) (C) 3. +91-120-4224248 Page 128 .askiitians.com Tel: +91-120-4224242.
[email protected] 10–23 kJ 3. Wave length of the radiation when electron jumps from second shell to 1st shell of H atom (RH = 109679 cm–1) (A) (C) 1215. The energy 9.com Email. The orbital which have lobe along the axis is/are (A) (C) dz 2 dx 2 y2 (B) (D) dxz dxy 17.6 Å 2395.45 10–24 J (B) (D) 105 m–1. Anode rays (A) (B) (C) (D) contain positively charged particles have constant e/m ratio irrespective of the type of gas used may contain He+ ions are always beams of protons 18. Ltd Website:www. +91-120-4224248 Page 129 .askiitians.
[email protected]/C/ATS askIITians Powered By IITians (A) (B) (C) (D) 20. is less than zero is the sum of kinetic energy and potential energy is equal to kinetic energy in magnitude is equal to potential energy in magnitude Which of the following orbitals have no nodal plane? (A) (C) px dz 2 (B) (D) dxy 4s Trans Web Educational Services Pvt.com Tel: +91-120-4224242. Ltd Website:www.com Email. If both Assertion and Reason are true and the Reason is a correct explanation of the Assertion. info@askiitians. (A) (B) (C) (D) (E) 21. 2s subshell has higher energy than 1s. Trans Web Educational Services Pvt.AISM-09/C/ATS askIITians Powered By IITians Assertion–Reason type questions The following questions consist of two statements each printed as Assertion and Reason. A: R: 24. A: R: 23. If the Assertion is false but the Reason is true. The energy of an electron is largely determined by its principle quantum number. Ltd Website:www. A: Cs is used in photoelectric cells. Cs is an alkali metal. While answering these questions. 22. A: R: A spectral line will be observed for 2px-2py transition. If both Assertion and Reason are false.com Email. If both Assertion and Reason are true but Reason is not a correct explanation of the Assertion. If Assertion is true but the Reason is false. The energy will be released in the form of electromagnetic radiations.com Tel: +91-120-4224242.askiitians. you are required to choose any one of the following five responses. Fe3+(g) ion is more stable than Fe2+(g). +91-120-4224248 Page 130 . askiitians.AISM-09/C/ATS askIITians Powered By IITians R: 25. A: R: Fe3+(g) ion has higher magnetic moment than Fe 2+(g) ion. de-Broglie‟s equation and uncertainity principle are applicable to moving bodies. These have significance to microscopic as well as macroscopic particles.com Email. +91-120-4224248 Page 131 . Trans Web Educational Services Pvt. info@askiitians. Ltd Website:www.com Tel: +91-120-4224242. II 1. Ltd Website:www. Orbit is a two dimensional track on which electron moves Atom has definite boundary Energies and angular momentum of orbits are quantized.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians Level . info@askiitians. +91-120-4224248 Page 132 .com Email. Which statement is wrong about Bohr‟s theory? (A) (B) (C) (D) 2. The introduction of a neutron into the nuclear composition of an atom would lead to a change in (A) (B) (C) (D) The number of electrons also The chemical nature of the atom Its atomic number Its atomic weight 3. Which statement is true? (A) (B) (C) (D) Spacing between energy levels n = 1 and n = 2 in hydrogen atom is greater than that of n = 2 and n = 3 Spacing between energy levels n = 1 and n = 2 in hydrogen atom is equal to that n = 2 and n = 3 Spacing between energy levels n = 1 and n = 3 in hydrogen atom is less than that of n = 2 and n = 3 None of these Trans Web Educational Services Pvt. Orbit is a three dimensional area where probability of finding electron is maximum.askiitians.
[email protected] nm 86. The number of valence electrons in the element is (A) (C) 1 3 (B) (D) 2 4 8. Energy Trans Web Educational Services Pvt. s = +1/2 = 3. The first four ionization energies of an element are 191. The radiation is emitted when a hydrogen atom goes from a high energy state to a lower energy state.0 nm 5. m = 0.872 and 5962 kcal. s = –1/2 7. If the wavelength of first line of the Balmer series of hydrogen atom is 656. +91-120-4224248 Page 133 .AISM-09/C/ATS askIITians Powered By IITians 4. s = –1/2 = 2. The wavelength of one line in visible region of atomic spectrum of hydrogen is 6.askiitians. Ltd Website:www.0 10–17 J (B) 1. Which set of quantum numbers is not possible for electron in 3 rd shell? (A) (B) (C) (D) n = 3. the wavelength of second line of this series would be (A) (C) 218. = 2.0 nm 640.com Tel: +91-120-4224242.com Email.5 difference between the two states is (A) 3. n = 3. n = 3. n = 3.0 nm (B) (D) 328. In absence of external magnetic field f sub-shell is (A) (C) 5 fold degenerate 7 fold degenerate (B) (D) 3 fold degenerate non-degenerate 6. m = – 1. m = 0. m = – 1.1 nm.0 10–18 J 10–9 m. s = +1/2 = 2. 578. Ltd Website:www. = 2. m = 0 12. (A) (B) (C) (D) V Cr Mn all of them will have equal magnetic moment Trans Web Educational Services Pvt. 5. info@askiitians. Mn (Z = 25) which will have highest magnetic moment. Which of the following sets of quantum numbers is/are not allowable? (A) (B) (C) (D) n = 3.askiitians.AISM-09/C/ATS askIITians Powered By IITians (C) 9. m = 0 = 0. m = + 1 = 0.0 10–10 J (D) 6. Which one of the following species is isoelectronic with P 3–? (A) (C) Kr Na+ (B) (D) Ca2+ F– 11. n = 4.com Tel: +91-120-4224242. Cr (Z = 24). n = 1.5 10–7 J The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of Be 3+ is (A) (C) 1:4 1 : 16 (B) (D) 1:8 16 : 1 10.com Email. Among V (Z = 23). +91-120-4224248 Page 134 . n = 2. m = – 1 = 3. Calculate the wavelength of a track star running 150 metre dash in 12.18 1017 sec–1 1019 sec–1 1010 sec–1 Trans Web Educational Services Pvt.92 10–37 m none of these 16. (A) (C) 9.com Email.28 7.askiitians.11 1.AISM-09/C/ATS askIITians Powered By IITians 13.com Tel: +91-120-4224242. m = +1 2. m = 2 3.89 5 1015 sec–1 (B) (D) 7. = 17. = = 1. info@askiitians. m = +2 0. +91-120-4224248 Page 135 . If velocity of an electron in 1st Bohr orbit of hydrogen atom is x. will be (A) (C) 5. Which of the following sets of quantum numbers is not allowed (A) (C) n = 3. Which element has a hydrogen like spectrum whose lines have wavelengths one fourth of atomic hydrogen? (A) (C) He+ Be3+ (B) (D) Li 2+ B4+ 15.1 sec if its weight is 50 kg. n = 3.12 10–34 m 10–45 metre (B) (D) 8. its velocity in 3rd orbit will be (A) (C) x 3 (B) (D) 3x x 9 9x 14. n= = 1. Ltd Website:www. Assuming that a 25 watt bulb emits monochromatic yellow light of wavelength 0.57 . m = 0 (B) (D) n = 3. The rate of emission of quanta per sec. and I P of Cl = 17.61 eV.
[email protected] Cl Cl + e. The energy required to remove both the electrons from the atom will be (A) (C) 59 eV 79 eV (B) (D) 81 eV None of these 20.24 2.askiitians.422 eV 1023 atoms 1015 atoms none of these 19.com Email. The binding energy of an electron in the ground state of the He atom is equal to 24 eV. for 6 1023 atoms (B) (D) 9.com Tel: +91-120-4224242. Ltd Website:www. The wave number of the shortest wave length transition in Balmer series of atomic hydrogen will be (A) (C) 4215 Å 3942 Å (B) (D) 1437 Å 3647 Å Trans Web Educational Services Pvt. How many chlorine atoms can you ionize in the process Cl by the energy liberated from the following process? Cl + e– (A) (C) 1.AISM-09/C/ATS askIITians Powered By IITians 18. +91-120-4224248 Page 136 .82 1020 atoms Given electron affinity of Cl = 3. e = 10.askiitians.AISM-09/C/ATS askIITians Powered By IITians ANSWERS TO ASSIGNMENT PROBLEMS Subjective: Level – O 1. (i) 1s (ii) 2px or 2py Refer to theory. n = 3. Cr3+ (Z = 24) = 2s22s22p63s23p63d3. Trans Web Educational Services Pvt. Half of them Nitride ion is N3 . +91-120-4224248 Page 137 . n = 7 3. We know that = h/mv. Mn4+ (Z = 25) = 1s22s2sp63s23p63d3. It also has three electrons.e. info@askiitians. Total electrons in n = 4 are 2n 2 i. 42 = 32. P = 7. (i) (ii) 2. Refer to theory. Ltd Website:www.com Tel: +91-120-4224242. 2 have ms = 1/2. 6. 5. 8. since mass of the car is large hence wavelength is very small.com Email. It has three unpaired electrons. 4. 0 means 3s orbitals which can have 2 electrons. 11. 14.72 106 m 1 5.askiitians. 18.com Email. 15. Ltd Website:www. 10.65 104 C 1. 13.172 1014 Hz 16. Refer to theory. +91-120-4224248 Page 138 . Emission spectrum. Anode rays are beams of positive ions which are different for different gases. info@askiitians. but anode rays will be different. Lyman series. 4. Trans Web Educational Services Pvt. Number of lines n n 1 2 6 6 1 2 15 12.AISM-09/C/ATS askIITians Powered By IITians 9.77 2 10 10 3 8 Mass = 5.com Tel: +91-120-4224242. In the presence of magnetic field the orbitals present in a subshell take up different orientations.486 (a) (b) 107 kg and charge = 9. Cathode rays are same in both cases. 17. 5. 8.545 10-11 erg 1. 4.432 105 m/sec 1014 10–33m 10–7m Trans Web Educational Services Pvt.4 nm Z=3 V3 = 7. 9.275 rps = 2.313 7. info@askiitians. Ltd Website:www.68 A 364. 10.09 108 cm/sec 2. 1. 3. 7.com Email. +91-120-4224248 11.25 Page 139 . (a) (b) 3.4A 0. 3. 6.75 10 33 g 2.AISM-09/C/ATS askIITians Powered By IITians Level – I 1.askiitians.4 10 10 mk 2.com Tel: +91-120-4224242.01 1018 = 1215.047 10 14 J 2. 77 Å : r3 = 2. 19. 15.12 Å for hydrogen : r2 = 1.askiitians.5 kcal/mole 6630 nm 5.227Å r1= 0. Ltd Website:www. 14.06 Å for He+ r3 = 4. +91-120-4224248 Page 140 .com Email.265 Å r2 = 2.77 10 4 Trans Web Educational Services Pvt.53 Å : r1 = 0.
[email protected] Å 13. 1.com Tel: +91-120-4224242.AISM-09/C/ATS askIITians Powered By IITians (c) 12. (iii) 1.com Tel: +91-120-4224242. 9.4 eV Trans Web Educational Services Pvt. 105 1216 Å.67 10–5 cm 30. Ltd Website:www. (i) (ii) 5.AISM-09/C/ATS askIITians Powered By IITians Level – II 1.4 eV for Be3+ 1018 1. 6.7 Å 3 6603 Å (i) (ii) 628.86 4. (a) (b) 2.79 (i) 3. 27. 3. +91-120-4224248 Page 141 . 54. (ii) 108.askiitians. 3. 10. 2= 7.50 kJ.6 eV for Li 2+.85 104 eV 10–11m 113.2 (a) (b) 4.01 (a) 2 and 3 10–6cm (iv) 122. info@askiitians. 3 = 6558 Å 5. 8.com Email.87 eV. 1 = 1028 Å.72 1021 atoms 832.
[email protected]/C/ATS askIITians Powered By IITians (b) 5 (c) 12. +91-120-4224248 Page 142 .09 eV and 1. Ltd Website:www.com Tel: +91-120-4224242.9 eV Trans Web Educational Services Pvt.com Email. 4. 5. 5. D B 3. 18. 18. 15. C C. 13. 11. C B C 2.AISM-09/C/ATS askIITians Powered By IITians Objective: Level –I 1. C D B Trans Web Educational Services Pvt. 10. 4. 6. 25. 17. Level – II 1.askiitians. C B C A B A. A D A B B A A A C B B A. +91-120-4224248 Page 143 . 7. 11. 9. Ltd Website:www. 14. 12. 22. 8. 19. 9. 20. 6.com Email. 14. A A A C A B. B. 12. 16. 20. 17. 15. A C C B A A C 2. 10.com Tel: +91-120-4224242. 13. info@askiitians. 19. 8. 7. 23. 16. 24. C A. 21. D C A B A D D 3.