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CHAPTER 12:- ATOMS, Atomic models, Plum pudding model:- This model was proposed by J. J. Thomson in 1898. According to this model, the, positive charge of the atom is uniformly distributed throughout the volume of the atom and the negatively, charged electrons are embedded in it like seeds in a watermelon., This model could not able to explain the discrete spectral lines of a atom or molecule., Rutherford’s planetary model:- According to this model, the entire positive charge and most of the mass of, the atom is concentrated in a small volume called the nucleus with electrons revolving around the nucleus just, as planets revolve around the sun., Geiger- Marsden alpha particle experiment:Experimental arrangement is as shown in the figure., They directed a beam of 5.5 MeV α-particles emitted, from a, radioactive source at a thin metal foil, made of gold. Alpha-particles emitted were collimated, into a narrow beam by their passage through lead, bricks. The beam was allowed to fall on a thin foil of, gold of thickness 2.1 × 10–7 m. The scattered alphaparticles were observed through a rotatable detector, consisting of zinc sulphide screen and a microscope., The scattered alpha-particles on striking the screen produced brief light flashes or scintillations. These flashes, may be viewed through a microscope and the distribution of the number of scattered particles may be studied, as a function of angle of scattering., Observations:- (i) Many of the α-particles pass through the foil without any collisions., (ii) Only about 0.14% of the incident α-particles scatter by more than 10., (iii) About 1 in 8000 deflect by more than 900. Rutherford argued that, to deflect the α-particle backwards, it, must experience a large repulsive force., Conclusions:- (i) Entire positive charge and most of the mass of the atom is concentrated in a small volume, called the nucleus as many α-particles pass through the foil without any collisions. (ii) Electrons are revolving, around the nucleus at a distance of about 10,000 to 1,00,000 times the size of the nucleus., Alpha particle trajectory:Impact parameter:- The perpendicular distance of the, initial velocity vector of the α-particle from the centre of, the nucleus is called impact parameter (b). It is seen that, an α-particle close to the nucleus (small impact, parameter) suffers large scattering. In case of head-on, collision, the impact parameter is minimum and the αparticle rebounds back ( θ ≅ π). For a large impact, parameter, the α-particle goes nearly undeviated and has, a small deflection. Note:- The distance of closest approach d between an α-particle and a nucleus is given by, d=, , where K is the kinetic energy of incoming α-particle and Z is atomic number of the target nucleus., , SMGPUC SIRSI, , Page 2

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Atomic Spectra:- Line emission spectrum:- A spectrum consisting of bright lines on a dark background is called, line emission spectrum. This spectrum is obtained when an atomic gas or vapour is excited by passing an, electric current through it. Line absorption spectrum:- A spectrum consisting of dark lines on a white, background is called line absorption spectrum. It is obtained when white light is passed through a gas., Spectral Series:Balmer series:- The spectral series obtained when electrons make transition from higher energy states to first, excited state is called Balmer series. The empirical formula for the wavelength is, , (, , ) where λ is, , the wavelength, R is a constant called the Rydberg constant and n = 3, 4, 5, etc. The value of R = 1.097 × 107, m–1. For the first member of the Balmer series, n = 3 and λ = 656.3 nm. It is called Hα line. For n = 4 we get, second member of the series Hβ. For series limit ( last member of the series) n= , λ∞= 364.6 nm., The other spectral series are, Lyman series. It is obtained when electron makes, transition from higher energy states to the ground state., It is observed in ultraviolet region. The empirical, (, , formula for the wavelength is, , ), n=, , 2,3,4,5 etc., Paschen series:- It is obtained when electron makes, transition from higher energy states to the second, excited state. The empirical formula for the wavelength, (, , is, , ), n= 4,5,6 etc. It is observed in, , infrared region., Brackett Series:- It is obtained when electron makes, transition from higher energy states to the third excited, state. The empirical formula for the wavelength is, (, , ), n= 5,6,7 etc. It is observed in far, , infrared region., Pfund series:- It is obtained when electron makes, transition from higher energy states to the fourth, excited state. The empirical formula for the wavelength, is, , (, , Energy level diagram of spectral series., , ), n= 6,7,8 etc. It is observed in far infrared region. Balmar formula can be written in terms of, , frequency follows. c = νλ,, , ,, , (, , ), , Failure of Rutherford’s atomic model:- According to classical electromagnetic theory, (i) An atom cannot be stable as an accelerated electron loses energy by emitting radiations., (ii) The spectrum emitted would be continuous as electron loses energy continuously., Rutherford’s model fails to explain these points., Bohr’s Atomic Model:- It was Niels Bohr who modified Rutherford model on the basis of quantum mechanics., , SMGPUC SIRSI, , Page 3

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Bohr’s Postulates:I., An electron in an atom could revolve in certain stable orbits without the emission of radiant energy., II., An electron revolves around the nucleus only in those orbits for which the angular momentum is an, integral multiple of h/2π where h is the Planck’s constant (6.6 × 10–34 Js). Thus, the, angular, momentum (L) of the orbiting electron is quantised. That is L = nh/2π., III., When an electron makes transition from higher energy state to lower energy state, a photon is emitted, whose energy is equal to energy difference between the two states. hν = Ei –Ef, Radius of an electron revolving in the nth orbit of hydrogen atom:Consider the hydrogen nucleus (Z=1) around Now, which an electron is revolving in circular orbit of, radius r as shown in the figure. The necessary, centripetal force for circular motion of the, electron is provided by the electrostatic force of, attraction between the nucleus and the electron., Fc = F e, , gives, , ( ), , ,, --------(1), , According to Bohr’s second postulate the orbital For nth orbit, angular momentum is integral multiple of h/2π,, i.e, , ------- (2), , Note:- Radius of the first orbit of hydrogen atom is called Bohr radius. Therefore, a0 = 5.29 x 10-11 m., , Bohr radius is denoted by a0 and is given by, , Velocity of an electron revolving in the nth orbit of hydrogen atom:Consider the hydrogen nucleus (Z=1) around Now, which an electron is revolving in circular orbit of, radius r as shown in the figure. The necessary, centripetal force for circular motion of the, electron is provided by the electrostatic force of, attraction between the nucleus and the electron., Fc = F e, , gives, , ,, -------(1), , For nth orbit, , According to Bohr’s second postulate the orbital, angular momentum is integral multiple of h/2π,, i.e, , SMGPUC SIRSI, , ------- (2), , Page 4

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Energy of an electron revolving in the nth orbit of hydrogen atom:Consider the hydrogen nucleus (Z=1) around which an electron is revolving, in circular orbit of radius r as shown in the figure. The necessary centripetal force, for circular motion of the electron is provided by the electrostatic force of, attraction between the nucleus and the electron. Fc = Fe, ,, The kinetic energy of the revolving electron is, Now the potential of the system is, , =, , =, , Assuming the expression for radius, Substituting this in the above equation, we get, , =, , =, , Total energy E = K + U, For the nth orbit, , Note:- (i), , , Substituting the values of constants, we get, , J,, , eV The negative sign of the total, , energy of an electron moving in an orbit means that the electron is, bound with the nucleus., (ii) The expression for energy other than hydrogen atom is, , =, , eV. Here Z is atomic number., , Ionisation Energy:- The minimum energy required to free an electron, from the ground state is called ionisation energy. For hydrogen atom, it is 13.6 eV., Excitation Energy:- The energy which excites an electron from lower, energy state to higher states is called excitation energy. It is the, difference between the two energy states. For hydrogen atom, to, excite an electron from ground state to 1st excited state, excitation, energy is = E2-E1 = -3.40 eV + 13.6 eV = 10.2 eV. To excite an electron, from ground state to 2nd excited state, excitation energy is = E3 – E1 =, -1.51 +13.6 = 12.09 eV., Energy level diagram of hydrogen atom, , SMGPUC SIRSI, , Page 5

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Expression for Rydberg constant:According to Bohr’s postulate, when an electron, (, )., makes transition from higher energy state to lower, energy state, it emits a photon. The energy of the This formula is called Rydberg formula., photon is hνif = Ei –Ef . Substituting the expression for In terms of wavelength, (, ),, energy, we, get, (, ). Comparing this with, , =, , (, , ), , spectral series formula, we get Rydberg constant, , ., , De Broglie’s explanation of Bohr’s Second Postulate of Quantisation:According to Bohr’s second postulate, an electron revolves around the nucleus only in those orbits for which, the angular momentum is an integral multiple of h/2π. That is L = nh/2π, where n = 1, 2, 3….., According to de Broglie, electrons revolving around the nucleus have matter waves. However only those, wavelengths survive which have nodes at the ends and form the standing wave in the orbit. It means that in a, orbit, standing waves are formed when the total distance travelled by a wave down and back is one, wavelength, two wavelengths, or any integral number of wavelengths. Thus circumference 2 rn = nλ., de Broglie wavelength is λ =, , . Therefore 2 rn =, , ,or, , , or, , Thus angular momentum of, , orbital electron is quantised., Limitations of Bohr model:(i) Bohr model is applicable only for hydrogen like atoms having only one electron., (ii) It could not explain the relative intensities of the spectral lines., Annual Examination Questions:, 1. Draw the schematic diagram of Geiger-Marsden (Rutherford’s) experiment. *March 2018+, 2. What is impact parameter? When is it minimum? [March 2017], 3. State and explain Bohr’s postulates of hydrogen atom. *March 2014, June 2015, March 2017, March, 2018, March 2019], 4. Obtain the expression for radius of nth orbit of H-atom. [March 2015], 5. Using Bohr's postulates, derive the expression for the radius of nth stationary orbit of electron in, hydrogen atom. Hence write the expression for Bohr radius.[March 2020], 6. Derive an expression for radius of electron in the nth Bohr orbit of hydrogen atom.[Sep 2020}, 7. Obtain the expression for energy of the electron in the n th orbit of H-atom. [July 2014, July 2016, June, 2017, July 2018, June 2019], 8. Write the expression for energy of the electron in the nth orbit of H-atom. [March 2019], 9. Name the spectral series of hydrogen which lies in the ultraviolet region of electromagnetic spectrum., [March 2015, March 2018], 10. Name the spectral series of hydrogen which lies in the visible region of electromagnetic spectrum., [June 2015, June 2019], 11. Mention the limitations of Bohr model. [March 2014, March 2018], SMGPUC SIRSI, , Page 6

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12. Give de Broglie’s explanation of Bohr’s second postulate of quantisation. *March 2018], 13. Write the limitations of Bohr’s atomic model. *Sep 2020+, Problems:, 1. Calculate the shortest and longest wavelength of Ballmer series of hydrogen atom., Given R = 1.097 x 107 m-1. [March 2016], 2. The first member of the Balmer series of hydrogen atom has wavelength 6563Å. Calculate the, wavelength and frequency of the second member of the same series., Given c = 3 × 108 ms-1. [March 2017], 3. Calculate the wave number, wavelength and frequency of Hα line of hydrogen spectrum., Given R = 1.097 x 107 m-1, c = 3 × 108 ms-1. [March 2017], , SMGPUC SIRSI, , Page 7