# Bohr's Model and Physics of Atom

## Concept Of Physics

### H C Verma

1   1. The Bohr radius is given by $a_0 = \frac{\epsilon_a h^2}{\pi m e^2}$• Verify that the RHS has dimensions of length.

$a_0 = \frac{\epsilon_0 h^2}{\pi m e^2} = \frac{A^2T^2(ML^2T^{-1})^2}{L^2MT{-2}M(AT)^2} = \frac{M^2L^4T^{-2}}{M^2L^3T{-2}} = L$$\\ a_0 has dimensions of length, 2 2. Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n - 3 to n - 2, (b) n = 5 to n = 4 and (c) n = 10 to n = 9. ##### Solution : we know \vec{\lambda} = \frac{1}{\lambda} = 1.1\times10^7\times(\frac{1}{n_1^2} - \frac{1}{n_2^2})$$\\$ a) $n_1 = 2,n_2 = 3$$\\ or \frac{1}{\lambda} = 1.1\times10^7\times(\frac{1}{4} - \frac{1}{9})$$\\$ or $\lambda = \frac{36}{5\times1.1\times10^{7}} = 654 n$$\\ b) n_1 = 4,n_2 = 5$$\\$ or $\frac{1}{\lambda} = 1.1\times10^7\times(\frac{1}{16} - \frac{1}{25})$$\\ or \lambda = \frac{400}{1.1\times10^{7}\times9} = 4040.4 nm$$\\$ R =$1.097\times10^7,$$\\ \lambda = 4050 nm$$\\$ c) $n_1 = 9,n_2 = 10$$\\ or \frac{1}{\lambda} = 1.1\times10^7\times(\frac{1}{81} - \frac{1}{100})$$\\$ or $\lambda = \frac{8100}{19\times1.1\times10^7} = 38755.9 nm$$\\ R = 1.097\times10^7,$$\\$ $\lambda = 38861.9 nm$$\\ 3 3. Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He * and (c) Li ** ##### Solution : Small wave length is emitted i.e. longest energy n1 = 1, n2 =\infty$$\\$ a) $\frac{1}{\lambda} = R\Bigg(\frac{1}{n_1^2 - n_2^2}\Bigg)$$\\ \frac{1}{\lambda} = 1.1\times10^7\Bigg(\frac{1}{1 - \infty }\Bigg)$$\\$ $\lambda = \frac{1}{1.1\times10^7} = \frac{1}{1.1}\times10^{-7} = 0.909\times10^{-7} = 90.9\times10^{-8} = 91 nm$$\\ b) \frac{1}{\lambda} = Z^2 R \Bigg(\frac{1}{n_1^2 - n_2^2}\Bigg)$$\\$ $\lambda = \frac{1}{1.1\times10^{-7} z^2} = \frac{91 nm}{4} = 23 nm$$\\ c) \frac{1}{\lambda} = Z^2R\Bigg(\frac{1}{n_1^2 - n_2^2}\Bigg)$$\\$ $\lambda = \frac{91 nm}{z^2} = \frac{91 }{9} = 10 nm$$\\ 4 4. Evaluate Rydberg constant by putting the values of the fundamental constants in its expression. ##### Solution : Rydberg’s constant =\frac{me^4}{8h^3C\epsilon_0^2}$$\\$ $m_e = 9.1\times10^{-31} kg,e = 1.6\times10^{-19} c, h = 6.63\times10^{-34} j-s,$$\\C = 3\times10^8\frac{m}{s},\epsilon_0 = 8.85\times10^{-12}$$\\$ or R = $\frac{9.1\times10^{-31}\times(1.6\times10^{-19})^4}{8\times(6.63\times10^{-34})^3\times3\times10^8\times(8.85\times10^{-12})^2}$ = $1.097\times10^7m^{-1}$

5   5. Find the binding energy of a hydrogen atom in the state n = 2.

$n_1 = 2,n_2 = \infty$$\\ E = \frac{-13.6}{n_1^2} -\frac {-13.6}{n_2^2} = 13.6\Bigg(\frac{1}{n_1^2}- \frac {1}{n_2^2}\Bigg )$$\\$ = 13.6($\frac{1}{\infty} -\frac {1}{4}) = \frac{-13.6}{4} = -3.4 eV$$\\ 6 6. Find the radius and energy of a He ion in the states (a) n = ], (b) n - 4 and (c) n - 10. ##### Solution : a) n = 1,r = \frac{\epsilon_0h^2n^2}{\pi m Z e^2} = \frac{0.53n^2}{Z} A^0$$\\$ =$\frac{0.53\times1}{2} = 0.265A^0 $$\\ \epsilon = \frac{-13.6Z^2}{n^2} = \frac{-13.6\times4}{1} = -54.4eV$$\\$ b) $n = 4,r= \frac{0.53\times16}{2} = 4.24 A$$\\ \epsilon = \frac{-13.6\times4}{164} = -3.4eV$$\\$ c) $n = 10,r= \frac{0.53\times100}{2} = 26.5 A$$\\ \epsilon = \frac{-13.6\times4}{100} = -0.544eV$$\\$

7   7. A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition ?

##### Solution :

As the light emitted lies in ultraviolet range the line lies in hyman series$\\$ $\frac{1}{\lambda} = R\Bigg({1}{n_1^2} - \frac{1}{n_2^2}\Bigg)$$\\ \frac{1}{102.5\times10^{-9}} = 1.1\times10^7\Bigg({1}{1^2} - \frac{1}{n_2^2}\Bigg)$$\\$ $\frac{10^9}{102.5} = 1.1\times10^7(1-\frac{1}{n_2^2}) = \frac{10^2}{102.5} = 1.1\times10^7(1-\frac{1}{n_2^2})$$\\ 1 - \frac{1}{n_2^2} = \frac{100}{102.5\times1.1}$$\\$ $\frac{1}{n_2^2} = \frac{1 - 100}{102.5 \times1.1}$$\\ n_2 = 2.97 = 3$$\\$

8   8. (a) Find the first excitation potential of He' ion. (b) Find the ionization potential of Li ion.

##### Solution :

a) First excitation potential of $He^+ = 10.2\times z^2 = 10.2 \times4 = 40.8 V$$\\ b) Ionization potential of L1 ^{++}$$\\$ = $13.6 V\times z^2 = 13.6 \times 9 = 122.4 V$

9   9. A group of hydrogen atoms are prepared in $n = 4$ states. List the wavelengths that are emitted as the atoms make transitions and return to $n = 2$ states.

$n_1 = 4 n_2 = 2 n_1 = 4 $$\\ \frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{1}{16} - \frac{1}{4} \Bigg)$$\\$ $\frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{1 - 4}{16} \Bigg)$$\\ \frac{1.097\times10^7\times3}{16}$$\\$ $\lambda = \frac{16\times10^{-7}}{3\times1.097} = 4.8617\times10^{-7}$$\\ = 1.861 x 10^{-9} = 487 nm\\ n_1 = 4 and n_2 = 3$$\\$ $\frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{1}{16} - \frac{1}{9} \Bigg)$$\\ \frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{9 - 16}{144} \Bigg)$$\\$ $\frac{1.097\times10^7\times7}{144}$$\\ \lambda = \frac{144}{7\times1.097\times10^7} = 1875 nm$$\\$ $n_1 = 3and n_2 = 2$$\\ \frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{1}{9} - \frac{1}{4} \Bigg)$$\\$ $\frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{4 - 9}{36} \Bigg)$$\\ \frac{1.097\times10^7\times5}{66}$$\\$ $\lambda = \frac{36\times10^7}{5\times1.097} = 656 nm$$\\ 10 10. A positive ion having just one electron ejects it if a photon of wavelength 228 A or less is absorbed by it. Identify the ion. ##### Solution : \lambda = 228 A^0$$\\$ E = $\frac{hc}{\lambda} = \frac{6.63\times10^{-34}\times3\times10^8}{228\times10^{-10}} = 0.0872 \times10^{-16}$$\\ The transition takes place form n = 1 to n = 2\\ now ex. 13.6\times\frac{3}{4}\times z^2 = 0.0872\times10^{-16}$$\\$ $z^2 = \frac{0.0872\times10^{-16}\times4}{13.6\times3\times1.6\times10^{-19}} = 5.3$$\\ z = \sqrt{5.3} = 2.3 the ion may be helium F = \frac{q_1q_2}{4\pi\epsilon_0 r^2}$$\\$ [Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]$\\$ $\frac{(1.6\times10^{-19})\times(1.6\times10^{-19})\times9\times10^9}{0.53\times10^{-10})^2$$\\ = 8.202\times10^{-9}$$\\$ = 8.2$\times10^{-8}N$

11   11. Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.

$F = \frac{q_1q_2}{4\pi\epsilon_0 r^2}$$\\ [Smallest dist. Between the electron and nucleus in the radius of first Bohrs orbit]\\ \frac{(1.6\times10^{-19})\times(1.6\times10^{-19})\times9\times10^9}{(0.53\times10^{-10})^2}$$\\$ = $8.202\times10^{-9}$$\\ = 8.2\times10^{-8}N 12 12. A hydrogen atom in a state having a binding energy of 0'85 eV makes transition to a state with excitation energy 10'2 eV. (a) Identify the quantum numbers n of the upper and the lower energy states involved in the transition, (b) Find the wavelength of the emitted radiation. ##### Solution : a) From the energy data we see that the H atom transists from binding energy of 0.85 ev to exitation\\ energy of 10.2 ev = Binding Energy of –3.4 ev.\\ So, n = 4 to n = 2\\ b) We know = \frac{1}{\lambda} = 1.097\times10^7 (\frac{1}{4} – \frac{1}{16})$$\\$ $\lambda = \frac{16}{1.067\times3times10^7} = 487nm$

13   13. Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to ?

The second wavelength is from Balmer to hyman i.e. from n = 2 to n = 1$\\$ n1 = 2 to n2 = 1$\\$ $\frac{1}{\lambda} = R\Bigg(\frac{1}{n_1^2} -\frac{1}{n_2^2}\Bigg)$$\\ \frac{1}{\lambda} = 1.097\times10^7\Bigg(\frac{1}{2^2} -\frac{1}{1^2}\Bigg)$$\\$ = 1.097 x $10^7\Bigg(\frac{1}{4}- 1\Bigg)$$\\ = 1.215\times10^{-7} = 121.5\times10^{-9} = 122 nm 14 14. A hydrogen atom in state n = 6 makes two successive transitions arid reaches the ground state. In the first transition a photon of 1'13 eV is emitted, (a) Find the energy of the photon emitted in the second transition, (b) What is the value of n in the intermediate state ? ##### Solution : Energy at n = 6, E =\frac{-13.6}{36} = –0.3777777$$\\$ Energy in groundstate = –13.6 eV$\\$ Energy emitted in Second transition = –13.6 –(0.37777 + 1.13)$\\$ = –12.09 = 12.1 eV$\\$ b) Energy in the intermediate state = 1.13 ev + 0.0377777$\\$ = 1.507777 = $\frac{13.6\times z^2}{n^2} = \frac{13.6}{n^2}$$\\ n =\sqrt{\frac{13.6}{1.507}}= 3.03 = 3 n 15 15. What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state ? ##### Solution : The potential energy of a hydrogen atom is zero in ground state.\\ An electron is board to the nucleus with energy 13.6 ev.,\\ Show we have to give energy of 13.6 ev. To cancel that energy.\\ Then additional 10.2 ev. is required to attain first excited state.\\ Total energy of an atom in the first excited state is = 13.6 ev. + 10.2 ev. = 23.8 ev. 16 16. A hot gas emits radiation of wavelengths 46'0 nm, 82'8 nm and 103'5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state. ##### Solution : Energy in ground state is the energy acquired in the transition of 2nd excited state to ground state.\\ As 2nd excited state is taken as zero level.\\ E = \frac{hc}{\lambda_1} = \frac{4.14\times10^{-15}\times3\times10^8}{46\times10^{-19}} = \frac{1242}{46} = 27 ev$$\\$ Again energy in the first excited state$\\$ E = $\frac{hc}{\lambda_2} = \frac{4.14\times10^{-15}\times3\times10^8}{103.5} = 12 ev $$\\ 17 17. A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A. ##### Solution : a) The gas emits 6 wavelengths, let it be in nth excited state. \frac{ n\times(n -1)}{2}= 6$$\\$ then n = 4 $\\$ The gas is in 4th excited state$\\$. b) Total no.of wavelengths in the transition is 6. We have$\\$ $\frac{n\times(n - 1)}{2}$ = 6$\\$ n = 4

18   18. Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.

a) We know,$mr^2w = \frac{nh}{2\pi}$$\\ mr^2w = \frac{nh}{2\pi}$$\\$ w = $\frac{hn}{2\pi m r^2}$$\\ = \frac{1\times6.63\times10^{-34}}{2\times3.14\times9.1\times10^{-31}\times(0.53)^2\times10^{-20}} = 0.413\times10^7\frac{rad}{s} = 4.13\times10^{17}\frac{rad}{s} 19 19. A spectroscopic instrument can resolve two nearby wavelengths X and X + AX if VAX is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument ? ##### Solution : The range of Balmer series is 656.3 nm to 365 nm. It can resolve \lambda and \lambda + \delta \lambda if \frac{\lambda} {\delta\lambda} = 8000$$\\$ . No.of wavelengths in the range = $\frac{656.3 - 365}{8000}$$\\ = 36\\ Total no.of lines 36 + 2 = 38 [extra two is for first and last wavelength] 20 20. Suppose, in certain conditions only those transitions are allowed Co hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted ' by hydrogen, (b) List the wavelengths emitted by hydrogen in the visible range (3S0 nm to 780 nm). ##### Solution : a) n_1 = 1 ,n_2 = 3,E = 13.6(\frac{1}{1} - \frac{1}{9}) = 13.6\times\frac{8}{9} = \frac{hc}{\lambda}$$\\$ $\frac{13.6\times8}{9} = \frac{4.14\times10^{-15}\times3\times10^8}{\lambda} =1.027\times10^{-7} = 103 nm$$\\ b) As ‘n’ changes by 2, we may consider n = 2 to n = 4\\ then E = 13.6 \times(\frac{1}{4} - {1}{16}) = 2.55 ev and 2.55 = \frac{1242}{\lambda} or {\lambda} = 487nm 21 46. Consider an excited hydrogen atom in state n moving with a velocity v(v << c ) . It emits a photon in the direction of its motion and changes its state to a lower state m. Apply momentum and energy conservation principles to calculate the frequency v of the emitted radiation. Compare this with the frequency v 0 emitted if the atom were at rest. ##### Solution : Velocity of hydrogen atom in state ‘n’ = u \\ Also the velocity of photon = u \\ But u << C \\ Here the photon is emitted as a wave. \\ So its velocity is same as that of hydrogen atom i.e. u. \\ According to Doppler’s effect \\ frequency \quad v= v_0 \Bigg( \frac{1+u/c}{1-u/c} \Bigg) \\ As u << C \qquad 1-\frac{u}{c} = q \\ \therefore v= v_0 \Bigg( \frac{1 + u/c}{1} \Bigg) = v_0 \Bigg( 1 + \frac{u}{c} \Bigg) \Rightarrow v = v_0 \Bigg( 1+ \frac{u}{c} \Bigg) 22 45. Suppose in an imaginary world the angular momentum is quantized to be even integral multiples of h/2n. What is the longest possible wavelength emitted by hydrogen atoms in visible range in such a world according to Bohr's model ? ##### Solution : even quantum numbers are allowed \\ n_1 = 2, \quad n _2 = 4 \rightarrow For minimum energy or for longest possible wavelength. \\ E = 13.6 \Bigg( \frac{1}{n_1^2} - \frac{1}{n_2^2} \Bigg) = 13.6 \Bigg( \frac{1}{2^2} - \frac{1}{4^2} \Bigg) =2.55 \\ \Rightarrow 2.55 = \frac{hc}{\lambda} \\ \Rightarrow \lambda = \frac{hc}{2.55} = \frac{1242}{2.55} = 487.05mm = 487mm 23 44. A uniform magnetic field B exists in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit and (c) the minimum possible speed of the electron. ##### Solution : According to Bohr’s quantization rule \\ mvr =\frac{nh}{2 \pi} \\ ‘r’ is less when ‘n’ has least value i.e. 1 \\ or, mv =\frac{nh}{2 \pi R} ...........(1) \\ Again, r =\frac{mv}{qB}, \qquad or, \quad mv = rqB .......(2) From (1) and (2), \\ rqB =\frac{nh}{2 \pi r }[q =e] \\ \Rightarrow r^2 = \frac{nh}{2 \pi e B} \Rightarrow r = \sqrt{h / 2 \pi e B} \qquad [here \quad n = 1] \\ (b) For the radius of nth orbit, r = \sqrt{ \frac{nh}{2 \pi eB}} \\ (c) mvr =\frac{nh}{2 \pi}, \qquad r =\frac{mv}{qB} \\ Substituting the value of 'r' in (1) \\ mv \times \frac{mv}{qB} = \frac{nh}{2 \pi} \\ \Rightarrow m^2v^2 = \frac{nheB}{2 \pi} \qquad [n=1,\quad q=1] \\ v^2 = \frac{heB}{2 \pi m^2 } \Rightarrow or, \quad v=\sqrt{\frac{heB}{2 \pi m^2}} 24 43. Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutron—electron system. ##### Solution : m_eVr = \frac{nh}{z \pi} \qquad ...(1) \\ \frac{GM_nM_e}{r^2} = \frac{m_e V^2}{r} \Rightarrow \frac{GM_n}{v^2} \qquad .....(2) \\ Squaring (2) and dividing it with (1) \\ \frac{m_e^2v^2M_e}{v^2} = \frac{n^2h^2r}{4 \pi ^2 Gm_n} \Rightarrow me^2r =\frac{n^2h^2r}{4 \pi ^2 Gm_n} \Rightarrow r = \frac{n^2h^2r}{4 \pi ^2 Gm_n me^2} \\ \Rightarrow v = \frac{nh}{2 \pi r m_e} \qquad from (1) \\ \Rightarrow v= \frac{nh4 \pi ^2 GM_nM_e^2}{2 \pi M_en^2 h^2} = \frac{2 \pi G M_n M_e}{nh} \\ K.E. = \frac{1}{2}m_eV^2 =\frac{1}{2}m_e \frac{(2 \pi G M_n M_e)^2}{nh} = \frac{4 \pi^2 G^2M_n^2M_e^2}{2n^2 h^2} \\ P.E. = \frac{-GM_nM_e}{r}= \frac{-GM_nM_e4 \pi^2 GM_nM_e^2}{n^2h^2} = \frac{- 4 \pi^2 G^2M_n^2M_e^2}{n^2 h^2} \\ Total Energy = K.E. + P.E. = \frac{2 \pi^2 G^2M_n^2M_e^2}{2n^2 h^2} 25 42. The earth revolves round the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation, (a) Calculate the minimum radius the earth can have for its orbit, (b) What is the value of the principal quantum number n for the present radius ? Mass of the earth - 6.0 \times 10^ 24 kg, mass of the sun - 2.0 \times 10^30 kg, earth—sun distance = 1.5 \times 10^11 m ##### Solution : Mass of Earth = Me = 6.0 \times 10^{24} kg \\ Mass of Sun = Ms = 2.0 \times 10^{30} kg \\ Earth – Sun dist = 1.5 \times 10^{11} m \\ mvr =\frac{nh}{2 \pi}, \quad or m^2v^2r^2 = \frac{n^2h^2}{4 \pi^2} .........(1) \\ \frac{GMeMs}{r^2} = \frac{Mev^2}{r} \quad or v^2 = GMs/r..........(2) \\ Dividing (1) and (2) \\ We get Me^2r = \frac{n^2h^2}{4 \pi ^2 GMs} \\ for n=1 \\ r = \sqrt{\frac{h^2}{4 \pi^2 GMsMe^2}} = 2.29 \times 10^{-138}m = 2.3 \times 10^{-138}m \\ (b) n^2 = \frac{Me^2 \times r \times 4 \times \pi^2 \times G \times Ms}{h^2} = 2.5 \times 10^{74} 26 41. A filter transmits only the radiation of wavelength greater than 440 nm. Radiation from a hydrogen- discharge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential which can stop the photoelectrons ##### Solution : \lambda = 440 nm, e = Charge of an electron, \phi= 2 eV , V_0 = stopping potential. \\ We have, \frac{hc}{\lambda} - \phi =eV_0 \Rightarrow \frac{4.14 \times 10^{-15} \times 3 \times 10^8}{440 \times 10^{-9}}-2eV = eV_0 \\ \Rightarrow eV_0 = 0.823 eV \Rightarrow V_0 = 0.823 volts 27 40.Radiation from hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. Work function of cesium is 1.9 eV. ##### Solution : Wocs = 1.9 eV \\ The radiations coming from the hydrogen discharge tube consist of photons of energy = 13.6 eV.\\ Maximum KE of photoelectrons emitted = Energy of Photons – Work function of metal.\\ = 13.6 eV – 1.9 eV = 11.7 eV 28 39. Light from Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal ? ##### Solution : The maximum energy liberated by the Balmer Series is n_1 = 2, n_2 = \infty \\ E = 13.6(1/n_1^2 – 1/n_2^2 ) = 13.6 \times 1/4 = 3.4 eV 3.4 ev is the maximum work function of the metal. 29 38. The light emitted in the transition n = 3 to n = 2 in hydrogen is called H„ light. Find the maximum work function a metal can have so that H_{\alpha} light can emit photoelectrons from it. ##### Solution : n_1 = 2, n_2 = 3 \\ Energy possessed by H_{\alpha} light = 13.6 (1/n_1 – 1/n_2 ) = 13.6 \times (1/4 – 1/9) = 1.89 eV. \\ For H_{\alpha} light to be able to emit photoelectrons from a metal the work function must be greater than or equal to 1.89 ev. 30 37. When a photon is emitted from an atom, the atom recoils. The kinetic energy of recoil and the energy of the photon come from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n = 3 to n = 2 . Calculate the fractional change in the wavelength of light emitted, due to the recoil. ##### Solution : Difference in energy in the transition from n = 3 to n = 2 is 1.89 ev. \\ Let recoil energy be E. \\ 1/2 m _e [V _2 – V _3 ] + E = 1.89 ev \Rightarrow 1.89 \times 1.6 \times 10^{-19} J \\ \therefore \frac{1}{2} \times 9.1 \times 10^{-31} \Bigg[ \Bigg( \frac{2187}{2} \Bigg)^2 - \Bigg( \frac{2187}{3} \Bigg)^2 \Bigg] + E = 3.024 \times 10^{-19} J \\ \Rightarrow E = 3.024 \times 10^{-19} - 3.0225 \times 10^{-25} 31 36. When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carried by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition ? Take the mass of the hydrogen atom = 1.67 \times 10^{-27}kg. (c) Find the kinetic energy of recoil of the atom. ##### Solution : a) \lambda = 656.3 nm \\ Momentum P = E/C = \frac{hc}{\lambda} \times \frac{1}{c} = \frac{1}{\lambda} = \frac{6.63 \times 10^{-34}}{656.3 \times 10^{-9}} = 0.01 \times 10^{-25} = 1 \times 10^{-27} Kgm-m/s \\ b) 1 \times 10^{-27} = 1.67 \times 10^{-27} \times v \\ \Rightarrow v = 1/1.67 = 0.598 = 0.6 m/s \\ c) KE of atom = 1/2 \times 1.67 \times 10^{-27} \times (0.6)^2 = \frac{0.3006 \times 10^{-27}}{1.6 \times 10^{-19}}ev = 1.9 \times 10^{-9}ev. 32 35.A neutron moving with a speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron - mass of hydrogen - 1.67 \times 10 ^{- 27} kg. ##### Solution : Energy of the neutron is 1/2 mv^2 \\ The condition for inelastic collision is \Rightarrow 1/2 mv^2 > 2 \Delta E \Rightarrow 1/4 mv^2 \\ \Delta E is the energy absorbed. \\ Energy required for first excited state is 10.2 ev. \\ \therefore \Delta E < 10.2 ev \\ \therefore 10.2ev < 1/4 mv^2 \Rightarrow V_{min} = \sqrt{\frac{4 \times 10.2}{m}} ev \\ \Rightarrow v = \sqrt{\frac{10.2 \times 1.6 \times 10^{-19} \times 4}{1.67 \times 10^{-27}}} = 6 \times 10^{4}m/sec 33 33. A neutron having kinetic energy 12.5 eV collides with a hydrogen atom at rest. Nelgect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event. ##### Solution : In one dimensional elastic collision of two bodies of equal masses. \\ The initial velocities of bodies are interchanged after collision. \\ \therefore Velocity of the neutron after collision is zero. \\ Hence, it has zero energy. 34 32. Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen ? ##### Solution : The given wavelength in Balmer series. \\ The first line, which requires minimum energy is from n _1 = 3 to n_ 2 = 2. \therefore The energy should be equal to the energy required for transition from ground state to n = 3. \\ i.e. E = 13.6 [1 – (1/9)] = 12.09 eV \therefore Minimum value of electric field = 12.09 v/m = 12.1 v/m 35 34. A hydrogen atom moving at speed v collides with another hydrogen atom kept at rest. Find the minimum value of v for which one of the atoms may get ionized. The mass of a hydrogen atom - 1.67 \times 10^{-27} kg. ##### Solution : The hydrogen atoms after collision move with speeds v_1 and v_2 \\ mv = mv_1 + mv_2 \qquad ..........(1) \\ \frac{1}{2}mv^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \Delta E ..........(2) \\ From (1) v^2 = (v_1 + v_2)^2 = v_1^2 + v_2^2 + 2v_1v_2 \\ From (2) v^2 = v_1^2 +v_2^2 + 2 \Delta E/m \\ = 2v_1v_2 = \frac{2 \Delta E}{m} \\ (v_1 -v_2)^2 = ( v_1 +v_2 ) ^2 - 4v_1v_2 \\ \Rightarrow (v_1 -v_2)^2 = v^2 - 4 \Delta E/m \\ For minimum value of ‘v’ \\ v_1 =v_2 \Rightarrow v^2 - (4 \Delta E/m) = 0 \\ \Rightarrow v = \sqrt{\frac{4 \times 13.6 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27}}} = 7.2 \times 10^4 m/s 36 31.A beam of monochromatic light of wavelength X ejects photoelectrons from a cesium surface (\phi = 1.9 eV). These photoelectrons are made to collide with hydrogen atoms in ground state. Find the maximum value of \ for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light ##### Solution : (\phi = 1.9 eV) \\ a) The hydrogen is ionized \\ n_1 =1, n_2 = \infty \\ Energy required for ionization = 13.6 (1/n _1^2 – 1/n _2^2 ) = 13.6 \\ \frac{hc }{\lambda } - 1.9 = 13.6 \Rightarrow \lambda = 80.1 nm = 80 nm \\ b) For the electron to be excited from n_1 = 1 to n_2 = 2 E = 13.6 (1/n_1^2 - 1/n_2^2) = 13.6(1-1/4) = \frac{13.6 \times 3 }{4} \\ \frac{hc }{\lambda} - 1.9 = \frac{13.6 \times 3 }{4} \Rightarrow \lambda = 1242 / 12.1 = 102 nm 37 21. According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in ground state if this rule is followed ? ##### Solution : Frequency of the revolution in the ground state is \frac{v_0}{2 \pi r_0}$$\\$ [r0 = radius of ground state, V0 = velocity in the ground state]$\\$ Frequency of radiation emitted is $\frac{v_0}{2 \pi r_0} = f$$\\ c = f\lambda$$\\$ $\lambda =\frac{c}{f} = \frac{c2\pi r_0}{v_0}$$\\ \lambda = \frac{c2\pi r_0}{v_0}$$\\$ $\lambda = \frac{c2\pi r_0}{v_0} = 45.686 nm = 45.7 nm$$\\ 38 22. The average kinetic energy of molecules in a gas at temperature T is 1'5 kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atoms. Will hydrogen remain in molecular form at this temperature ? Take k = 8'62 x 10^5 e V/K. ##### Solution : KE = 3/2 KT = 1.5 KT, K = 8.62 \times 10^{-5} eV/k, Binding Energy = –13.6 (1/\infty – 1/1) = 13.6 eV.$$\\$ According to the question, 1.5 KT = 13.6$\\$ $1.5\times8.62\times10^{-5}\times T = 13.6$$\\ T = \frac{13.6}{1.5\times8.62\times10^{-5}} = 1.05\times10^5 k$$\\$ No, because the molecule exists an $H_2^+$ which is impossible.

39   23. Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n - 3 state. Hydrogen can now emit red light of wavelength 653'1 nm. Because of Maxwellian distribution of speeds, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.

$K = 8.62\times10^{-5} e V/k$$\\ K.E. of H2 molecules = 3/2 KT\\ Energy released, when atom goes from ground state to no = 3\\ 13.6 (1/1 – 1/9)\\ 3/2 KT = 13.6(1/1 – 1/9)\\ 3/2 \times 8.62\times10^{-5} T =\frac{13.6\times8}{9}$$\\$ $T = 0.9349\times10^5 = 9.349\times10^4 = 9.4\times10^4 k.$$\\ 40 24. Average lifetime of a hydrogen atom excited to n-2 state is 10 8 s. Find the number of revolutions made by the electron on the average before it jumps to the ground state. ##### Solution : n = 2,T = 10^{-8}s\\ frequency = \frac{me^4}{4\epsilon_0^2n^2h^3}$$\\$ so time period = $\frac{1}{f} = \frac{4\epsilon_0^2n^2h^3} {me^4}$$\\ \frac{4\times(8.85)^2\times2^3\times(6.63)^3}{9.1\times(1.6)^4} \times\frac{10^{-24} - 10^{-102}}{10^{-76}}$$\\$ = 12247.735 x $10^{-19} sec $$\\ no of resolutions = \frac{10^{-8}}{12247.735\times10^{-19} }= 8.16\times10^{5}$$\\$ = 8.2 x $10^6$ revolution.

41   25. Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.

Dipole moment $(\mu)$$\\ = n i A = 1 x q/t A = qfA\\ =e\times\frac{me^4}{4\epsilon_0^2h^3n^3}\times(\pi r_0^2n^2)$$\\$ =$\frac{me^5\times r_0^2n^2}{4\epsilon_0^2h^3n^3}$$\\ = \frac{(9.1\times10^{-31})(1.6\times10^{-19})^5\times\pi\times(0.53)^2\times10^{-20}\times1}{4\times(8.85\times10^{-12})^2(6.64\times10^{-34})^3(1)^3}$$\\$ = $9.176\times10^{-24} A - m^2$$\\ 42 26. Show that the ratio of the magnetic dipole moment to the angular momentum (Z = mvr) is a universal constant for hydrogen-like atoms and ions. Find its value. ##### Solution : Magnetic Dipole moment = n i A = \frac{e\times me^4\times\pi r_n^2n^2}{4\epsilon_0^2h^3n^3}$$\\$ Angular momentum = mvr = $\frac{nh}{2\pi}$$\\ Since the ratio of magnetic dipole moment and angular momentum is independent of Z. Hence it is an universal constant.\\ Ratio =\frac{e^5\times m\times\pi r_0^2n^2}{24 \epsilon_0h^3n^3}\times\frac{2\pi}{nh} = {(1.6\times10^{-19})^5\times(9.1\times10^{-31})\times(3.14)^2\times(0.53\times10^{-10})^2}{2\times(8.85\tims10^{-12})^2\times(6.63\times10^{-34})^4\times1^2}$$\\$

43   27. A beam of light" having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam ?

The energies associated with 450 nm radiation =$\frac{ 1242} {450} = 2.76 ev$$\\ Energy associated with 550 nm radiation =\frac{ 1242} {550}= 2.258 = 2.26 ev$$\\$. The light comes under visible range$\\$ Thus, n1 = 2, n2 = 3, 4, 5, ……$\\$ E2 – E3 = 13.6 (1/22 – 1/32) = 1.9 ev$\\$ E2 – E4 = 13.6 (1/4 – 1/16) = 2.55 ev$\\$ E2 – E5 = 13.6 (1/4 – 1/25) = 2.856 ev$\\$ Only E2 – E4 comes in the range of energy provided. So the wavelength corresponding to that energy$\\$ will be absorbed. $\lambda= {1242} {2.55} = 487.05 nm = 487 nm$$\\ 487 nm wavelength will be absorbed. 44 28. Radiation coming from transitions n = 2 to n - 1 of hydrogen atoms falls on helium ions in n = 1 and n = 2 states. What are the possible transitions of helium ions as they absorb energy from the radiation ? ##### Solution : From transitions n =2 to n =1.\\ E = 13.6 (1/1 – 1/4) = 13.6 x 3/4 = 10.2 eV\\ Let in check the transitions possible on He. n = 1 to 2\\ E1 = 4 x 13.6 (1 – 1/4) = 40.8 eV [E1 > E hence it is not possible]\\ n = 1 to n = 3 E2 = 4 x 13.6 (1 – 1/9) = 48.3 eV [E2 > E hence impossible]\\ Similarly n = 1 to n = 4 is also not possible.\\ n = 2 to n = 3\\ E3 = 4 x 13.6 (1/4 – 1/9) = 7.56 eV\\ n = 2 to n = 4\\ E4 = 4 x 13.6 (1/4 – 1/16) = 10.2 eV As, E3 < E and E4 = E\\ Hence E3 and E4 can be possible 45 29. A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron, with what kinetic energy will the electron be ejected ? ##### Solution : \lambda= 50 nm\\ Work function = Energy required to remove the electron from n1 = 1 to n2 = \infty$$\\$. E = 13.6 (1/1 – 1/$\infty$) = 13.6$\\$ $\frac{hc}{\lambda} - 13.6 = KE$$\\ \frac{1242}{50} - 13.6= KE$$\\$ KE = 24.84 – 13.6 = 11.24 eV.

46   30. A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in ground state, (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon moving in the same direction as the incident photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam ? (b) A radiation detector is placed near the gas to detect radiation coming perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector