NCERT Solutions Class 12 Physics Chapter 12

Q.1 Choose the correct alternative from the clues given at the end of the each statement:

(a) The size of the atom in Thomson’s model is …… the atomic size in Rutherford’s model.
(much greater than/no different from /much less than.)

(b) In the ground state of …….. electrons are in stable equilibrium, while in ……. electrons always experience a net force.
(Thomson’s model/Rutherford’s model.)

(c) A classical atom based on……. is doomed to collapse.
(Thomson’s model / Rutherford’s model.)

(d) An atom has a nearly continuous mass distribution in a……. but has a highly non-uniform mass distribution in ……………..
(Thomson’s model /Rutherford’s model.)

(e) The positively charged part of the atom possesses most of the mass in …………
(Rutherford’s model/both the models.)

Ans.

(a) The size of the atom in Thomson’s model is no different from the atomic size in Rutherford’s model.

(b) In the ground state of Thomson’s model electrons are in stable equilibrium, while in Rutherford’s model electrons always experience a net force.

(c) A classical atom based on Rutherford’s model is doomed to collapse.

(d) An atom has a nearly continuous mass distribution in a Thomson’s model but has a highly non-uniform mass distribution in Rutherford’s model.

(e) The positively charged part of the atom possesses most of the mass in both the models.

Q.2 Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K) What results do you expect?

Ans.

In the alpha- particle scattering experiment if we use a thin sheet of solid hydrogen in place of gold foil, then the scattering angle would be very small. The reason behind it is that the mass of the target nucleus (Hydrogen of mass 1.67 × 10^-27 kg) is very small as compared to the mass of scattering particles (Alpha–particle of mass 6.64 × 10^-27 kg) and due to this repulsive force between them is very small and could not deflect the alpha particle backwards.

Q.3 What is the shortest wavelength present in the Paschen series of spectral lines?

Ans.

$\begin{array}{l}\text{The wavelength of the spectral lines forming Paschen series is given by}\\ \frac{\text{1}}{\text{λ}}\text{= R}\left(\frac{\text{1}}{{\text{3}}^{\text{2}}}\text{–}\frac{\text{1}}{{{\text{n}}^{\text{2}}}_{\text{i}}}\right)\text{,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}n = 4,5,6…}\\ \text{Where, R = 1.097}\times {\text{10}}^{\text{7}}{\text{m}}^{\text{-1}}\text{is Rydberg constant for shortest wavelength,}\\ {\text{n}}_{\text{i}}\text{=}\mathrm{\infty }\\ \frac{\text{1}}{\text{λ}}\text{=}\frac{\text{R}}{\text{9}}\text{or λ =}\frac{\text{9}}{\text{R}}\\ \text{λ =}\frac{\text{9}}{{\text{1.097×10}}^{\text{7}}{\text{\hspace{0.17em}\hspace{0.17em}m}}^{\text{-1}}}{\text{= 8.204×10}}^{\text{-7}}\text{m = 8204 Å = \hspace{0.17em}820.4 nm}\end{array}$

Q.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper level to the lower level?

Ans.

$\begin{array}{l}\text{Energy gap between two levels is, E = 2.3 eV,}\\ {\text{Given, 1 eV = 1.6×10}}^{\text{-19}}\text{\hspace{0.17em}J,}\\ {\text{E = 2.3 ×1.6×10}}^{\text{-19}}{\text{\hspace{0.17em}J = 3.68 ×10}}^{\text{-19}}\text{\hspace{0.17em}J}\\ \text{Energy\hspace{0.17em}of\hspace{0.17em}a\hspace{0.17em}photon\hspace{0.17em}is\hspace{0.17em}given\hspace{0.17em}by,\hspace{0.17em}\hspace{0.17em}E = h}\mathrm{\nu }\\ \text{where,\hspace{0.17em}\hspace{0.17em}E\hspace{0.17em}is\hspace{0.17em}the\hspace{0.17em}energy\hspace{0.17em}gap\hspace{0.17em}between\hspace{0.17em}two\hspace{0.17em}levels,}\\ \text{h is the Planck’s\hspace{0.17em}constant\hspace{0.17em}and}\\ \mathrm{\nu }\text{is the frequency\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}radiation\hspace{0.17em}emitted\hspace{0.17em}when\hspace{0.17em}the\hspace{0.17em}atom\hspace{0.17em}transits}\\ \text{from\hspace{0.17em}the\hspace{0.17em}upper\hspace{0.17em}level to\hspace{0.17em} the\hspace{0.17em}lower\hspace{0.17em}level}\\ \therefore \mathrm{\nu }\text{=}\frac{\text{E}}{\text{h}}\\ \mathrm{\nu }\text{=}\frac{{\text{3.68×10}}^{\text{-19}}\text{\hspace{0.17em}J}}{{\text{6.62×10}}^{\text{-34}}\text{\hspace{0.17em}Js}}\approx {\text{5.6×10}}^{\text{14}}\text{Hz}\\ {\text{Hence, the frequency of radiation is 5.6×10}}^{\text{14}}\text{\hspace{0.17em}Hz\hspace{0.17em}.}\end{array}$

Q.5 The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?

Ans.

As the potential energy of an electron of charge, e revolving in an orbit of radius, r around the nucleus of an atom is given by

\begin{array}{l}{\text{E}}_P\text{=}\frac{{\text{– e}}^{\text{2}}}{\text{4}\pi {\epsilon }_{\text{0}}\text{r}}\mathrm{\dots }\text{(i)}\\ \text{and its kinetic energy is}\\ {\text{E}}_{\text{K}}\text{=}\frac{{\text{e}}^{\text{2}}}{\text{2}\left(\text{4}\pi {\epsilon }_{\text{0}}\right)\text{r}}\mathrm{\dots }\text{(ii)}\\ {\text{E}}_{\text{K}}\text{=}\frac{\text{–}{\text{E}}_{\text{P}}}{\text{2}}\\ {\text{2E}}_{\text{K}}\text{= –}{\text{E}}_{\text{P}}\mathrm{\dots }\text{(iii)}\\ \text{Total energy of electron in ground state is}\\ \text{E =}{\text{E}}_{\text{P}}\text{+}\text{E}{}_{\text{K}}\text{= -13}\text{.6 eV}\\ \text{E = -2}{\text{E}}_{\text{K}}\text{+}\text{E}{}_{\text{K}}\text{= -13}\text{.6 eV}\text{(using}\text{(iii))}\\ {\text{E}}_{\text{K}}\text{= 13}\text{.6 eV and 2}\text{K}\text{.E=-}\text{P}\text{.E}\text{(from}\text{(iii))}\\ {\text{E}}_{\text{P}}\text{= -2×13}\text{.6 eV = -27}\text{.2}\text{eV}\\ \text{Thus, Kinetic energy = 13}\text{.6}\text{eV and Potential}\text{energy = -27}\text{.2}\text{eV}\end{array}

Q.6 A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.

Ans.

\begin{array}{l}{\text{As energy of an electron in the n}}^{\text{th}}\text{orbit of hydrogen atom is given by}\\ {\text{E}}_{\text{n}}\text{=}\frac{\text{-13}\text{.6 eV}}{{\text{n}}^{\text{2}}}\\ \text{For ground level n=1}\\ {\text{hence, E}}_{\text{1}}\text{=}\frac{\text{-13}\text{.6 eV}}{{\text{1}}^{\text{2}}}\text{= -13}\text{.6 eV}\\ \text{simillary, for n=4,}\\ {\text{E}}_{\text{4}}\text{=}\frac{\text{-13}\text{.6 eV}}{\text{16}}\text{= -0}\text{.85 eV}\\ \text{Thus, energy difference between energy levels for n = 1}\text{}\text{}\text{and}\text{n = 4}\\ {\text{E = E}}_{\text{4}}{\text{– E}}_{\text{1}}\text{= -0}\text{.85 eV + 13}\text{.6 eV =}\text{12}\text{.75 eV}\end{array}

\begin{array}{l}\text{The hydrogen atom will go to n = 4 from n = 1}\\ \text{if the photon of energy E = 12}\text{.75 eV is incident}\text{.}\\ \text{Thus, energy of photon, E = 12}\text{.75×1}{\text{.6×10}}^{\text{-19}}\text{J = 20}{\text{.4×10}}^{\text{-19}}{\text{J}}^{}\\ \text{From E =}\text{h}\nu \\ \text{where,}\text{E = 20}{\text{.4×10}}^{\text{-19}}\text{J,}\\ \text{h = Planck’s constant = 6}{\text{.6×10}}^{\text{-34}}\text{Js}\\ \nu \text{=frequency}\text{of}\text{the}\text{photon}\\ \therefore \nu \text{=}\frac{\text{20}{\text{.4×10}}^{\text{-19}}\text{J}}{\text{6}{\text{.63×10}}^{\text{-34}}\text{Js}}\text{= 3}{\text{.1×10}}^{\text{15}}\text{Hz}\\ \text{As λ =}\frac{\text{c}}{\nu }\\ \text{λ =}\frac{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{3}{\text{.1×10}}^{\text{15}}{\text{s}}^{\text{-1}}}\text{= 9}{\text{.7×10}}^{\text{-8}}\text{m}\end{array}

Q.7 (a) Using the Bohr’s model, calculate the speed of the electron in a hydrogen atom in the n = 1, 2 and 3 levels.
(b) Calculate the orbital period in each of these levels.

Ans.

$\begin{array}{l}\text{(a) Speed of an electron revolving in nth orbit of hydrogen atom is given by}\\ {\text{v}}_{\text{n}}\text{=}\frac{\text{c}}{\text{137 n}}\\ \text{Thus, for n = 1 ,}\\ {\text{v}}_{\text{1}}\text{=}\frac{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}{\left(\text{137×1}\right)}{\text{= 2.18×10}}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{For\hspace{0.17em} n = 2}\\ {\text{then, v}}_{\text{2}}\text{=}\frac{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}{\left(\text{137×2}\right)}{\text{= 1.09×10}}^{\text{6}}{\text{ms}}^{\text{-1}}\\ \text{For n = 3}\\ {\text{then,\hspace{0.17em} v}}_{\text{3}}\text{=}\frac{\left({\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}{\left(\text{137×3}\right)}{\text{= 7.27×10}}^{\text{5}}{\text{ms}}^{\text{-1}}\end{array}$

$\begin{array}{l}\text{(b) Orbital period of electron in the nth orbit of hydrogen atom,}\\ {\text{T}}_{\text{n}}\text{=}\frac{\text{2\hspace{0.17em}}\mathrm{\pi }{\text{\hspace{0.17em}r}}_{\text{n}}}{{\text{v}}_{\text{n}}}\text{=}\frac{\left(\text{2}\mathrm{\pi }{\text{a}}_{\text{0}}{\text{n}}^{\text{2}}\right)}{\left(\frac{\text{c}}{\text{137n}}\right)}\text{=}\frac{\left(\text{137×2\hspace{0.17em}}\mathrm{\pi }{\text{\hspace{0.17em}a}}_{\text{0}}{\text{n}}^{\text{3}}\right)}{\text{c}}\\ \text{Where, radius of the innermost electron orbit of hydrogen atom}\\ {\text{a}}_{\text{0}}{\text{= 0.53×10}}^{\text{-10}}\text{m,}\\ \text{when\hspace{0.17em} n = 1,}\\ {\text{T}}_{\text{1}}\text{=}\frac{\left({\text{137×2×3.14×0.53×10}}^{\text{-10}}\text{m×1}\right)}{{\text{3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}}{\text{= 1.52×10}}^{\text{-16}}\text{s}\\ \text{when\hspace{0.17em} n = 2 then,}\\ {\text{T}}_{\text{2}}\text{=}{\left(\text{2}\right)}^{\text{3}}\times {\text{T}}_{\text{1}}{\text{= 8×1.52×10}}^{\text{-16}}{\text{s = 1.22×10}}^{\text{-15}}\text{\hspace{0.17em}s}\\ \text{when\hspace{0.17em} n = 3 then,}\\ {\text{T}}_{\text{3}}\text{=}{\left(\text{3}\right)}^{\text{3}}\times {\text{T}}_{\text{1}}{\text{= 27×1.52×10}}^{\text{-16}}{\text{s = 4.11 x 10}}^{\text{-15}}\text{\hspace{0.17em}s}\end{array}$

Q.8 The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10^-11 m. What are the radii of the n = 2 and n = 3 orbits?

Ans.

$\begin{array}{l}\text{As radius of orbit of hydrogen atom is given by}\\ {\text{r}}_{\text{n}}\propto {\text{\hspace{0.17em}n}}^{\text{2}}\\ {\text{r}}_{\text{n}}{\text{= kn}}^{\text{2}}\\ \text{where\hspace{0.17em} k =}\frac{\text{4}{\mathrm{\pi \epsilon }}_0{\text{h}}^{\text{2}}}{\text{4}{\mathrm{\pi }}^2{\text{me}}^{\text{2}}}\\ \text{hence, for n = 1}\\ {\text{r}}_{\text{1}}{\text{= k(1)}}^{\text{2}}{\text{or\hspace{0.17em} k = r}}_{\text{1}}{\text{= 5.3×10}}^{\text{-11}}\text{\hspace{0.17em}m}\\ {\text{where\hspace{0.17em}r}}_{\text{1}}\text{is the radius of innermost electron orbit.}\\ \text{similarly for n = 2}\\ {\text{r}}_{\text{2}}{\text{= k(2)}}^{\text{2}}\text{= 4\hspace{0.17em}k}\\ {\text{where, k = 5.3×10}}^{\text{-11}}\text{\hspace{0.17em}m}\\ \therefore {\text{r}}_{\text{2}}{\text{= 4×5.3×10}}^{\text{-11}}{\text{m = 2.12×10}}^{\text{-10}}\text{\hspace{0.17em}m}\\ \text{similarly for n = 3}\\ {\text{r}}_{\text{3}}{\text{= k(3)}}^{\text{2}}\text{= 9k}\\ {\text{r}}_{\text{3}}{\text{= 9×5.3×10}}^{\text{-11}}{\text{m = 4.77×10}}^{\text{-10}}\text{\hspace{0.17em}m}\end{array}$

Q.9 A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelength will be emitted?

Ans.

$\begin{array}{l}\text{The energy of the gaseous hydrogen in its ground state is -13.6\hspace{0.17em}eV.}\\ \text{Energy of the electron in nth orbit is given by}\\ {\text{E}}_{\text{n}}\text{=}\frac{\text{-13.6 eV}}{{\text{n}}^{\text{2}}}\\ \text{Now,\hspace{0.17em}energy\hspace{0.17em}required\hspace{0.17em}to\hspace{0.17em}excite\hspace{0.17em}an\hspace{0.17em}electron\hspace{0.17em}from\hspace{0.17em}level\hspace{0.17em}1\hspace{0.17em}to\hspace{0.17em}n is\hspace{0.17em}given\hspace{0.17em}by}\\ {\text{E}}_{\text{n}}{\text{– E}}_{\text{1}}\text{= \hspace{0.17em}}\frac{\text{-13.6 eV}}{{\text{n}}^{\text{2}}}\text{–}\left(\frac{\text{-13.6 \hspace{0.17em}eV}}{{\text{1}}^{\text{2}}}\right)\text{,}\\ \text{According to question,\hspace{0.17em}12.5\hspace{0.17em}eV energy is available for the excitation of\hspace{0.17em}electron to higher level.}\\ \therefore {\text{\hspace{0.17em}E}}_{\text{n}}{\text{– E}}_{\text{1}}\text{=}\frac{\text{-13.6 eV}}{{\text{n}}^{\text{2}}}\text{+}\frac{\text{13.6 eV}}{{\text{1}}^{\text{2}}}\le \text{\hspace{0.17em}12.5\hspace{0.17em}eV}\\ \text{Now, for n = 2,}\\ {\text{E}}_{\text{2}}{\text{– E}}_{\text{1}}\text{=}\frac{\text{-13.6 eV}}{{\text{(2)}}^{\text{2}}}\text{+}\frac{\text{13.6 eV}}{{\text{1}}^{\text{2}}}\le \text{12.5\hspace{0.17em} eV}\\ {\text{E}}_{\text{2}}{\text{– E}}_{\text{1}}\text{= 10.2 eV \hspace{0.17em}}\le \text{12.5 \hspace{0.17em}eV\hspace{0.17em}}\\ \text{This\hspace{0.17em}shows\hspace{0.17em}that\hspace{0.17em}an\hspace{0.17em}electron\hspace{0.17em}can\hspace{0.17em}excite\hspace{0.17em}from\hspace{0.17em}level n = 1 \hspace{0.17em}to \hspace{0.17em}n = 2.}\\ \text{Hence,\hspace{0.17em}on\hspace{0.17em}de-excitation\hspace{0.17em}the\hspace{0.17em}electron\hspace{0.17em}\hspace{0.17em}may\hspace{0.17em}jump from \hspace{0.17em}n = 2 \hspace{0.17em}to \hspace{0.17em}n = 1\hspace{0.17em}giving\hspace{0.17em}rise\hspace{0.17em}to\hspace{0.17em}}\\ \text{Lyman \hspace{0.17em}series.\hspace{0.17em}}\\ \text{Similarly, for\hspace{0.17em}n = 3,}\\ {\text{E}}_{\text{3}}{\text{– E}}_{\text{1}}\text{=}\frac{\text{-13.6 eV}}{{\text{(3)}}^{\text{2}}}\text{+}\frac{\text{13.6 eV}}{{\text{1}}^{\text{2}}}\le \text{\hspace{0.17em}12.5\hspace{0.17em}eV}\\ {\text{E}}_{\text{3}}{\text{– E}}_{\text{1}}\text{= 12.08 eV \hspace{0.17em}}\le \text{\hspace{0.17em}12.5 \hspace{0.17em}eV}\\ \text{This\hspace{0.17em}shows\hspace{0.17em}that\hspace{0.17em}an\hspace{0.17em}electron\hspace{0.17em}can\hspace{0.17em}also\hspace{0.17em}excite\hspace{0.17em}from level \hspace{0.17em}n = 1 \hspace{0.17em}to \hspace{0.17em}n = 3.}\\ \text{Hence,\hspace{0.17em}on\hspace{0.17em}de-excitation\hspace{0.17em}the\hspace{0.17em}electron\hspace{0.17em}\hspace{0.17em}may\hspace{0.17em}jump from\hspace{0.17em} n = 3\hspace{0.17em} to \hspace{0.17em}n = 1\hspace{0.17em}directly\hspace{0.17em}and\hspace{0.17em}also\hspace{0.17em}}\\ \text{from\hspace{0.17em} n = 3 \hspace{0.17em}to \hspace{0.17em}n = 2.}\\ \text{Thus,\hspace{0.17em}the\hspace{0.17em}transistion\hspace{0.17em}which\hspace{0.17em}takes\hspace{0.17em}place\hspace{0.17em}from\hspace{0.17em} n = 3\hspace{0.17em} to\hspace{0.17em} n = 1 give\hspace{0.17em}rise\hspace{0.17em}to\hspace{0.17em}Lyman\hspace{0.17em}series\hspace{0.17em}}\\ \text{and\hspace{0.17em} the \hspace{0.17em}transistion\hspace{0.17em}which\hspace{0.17em}takes place\hspace{0.17em}from\hspace{0.17em}n = 3\hspace{0.17em} to \hspace{0.17em}n = 2\hspace{0.17em}give\hspace{0.17em}rise\hspace{0.17em}to\hspace{0.17em}Balmer\hspace{0.17em}series.\hspace{0.17em}}\\ \text{Simillary, for n = 4,}\\ {\text{E}}_{\text{4}}{\text{– E}}_{\text{1}}\text{=}\frac{\text{-13.6 eV}}{{\text{(4)}}^{\text{2}}}\text{+}\frac{\text{13.6 eV}}{{\text{1}}^{\text{2}}}\le \text{12.5 eV}\\ {\text{E}}_{\text{4}}{\text{– E}}_{\text{1}}\text{= 12.75 eV \hspace{0.17em}> 12.5 eV}\\ \text{Here,\hspace{0.17em}energy\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}electron\hspace{0.17em}beam\hspace{0.17em}is\hspace{0.17em}not\hspace{0.17em}sufficient\hspace{0.17em}for the\hspace{0.17em}excitation\hspace{0.17em}of\hspace{0.17em}electron\hspace{0.17em}to\hspace{0.17em}level\hspace{0.17em}n=4.}\end{array}$

Q.10 In accordance with the Bohr’s model, find the quantum number that characterizes the earth’s revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ ms^-1. (Mass of earth 6.0 × 10²⁴ kg)

Ans.

$\begin{array}{l}\text{According to Bohr’s postulate of quantization of angular momentum,}\\ \text{mvr =}\frac{\text{nh}}{\text{2}\mathrm{\pi }}\\ \text{Thus, n =}\frac{\text{2}\mathrm{\pi }\text{mvr}}{\text{h}}\\ \text{n =}\frac{\left({\text{2×3.14×6×10}}^{\text{24}}{\text{kg×3×10}}^{\text{4}}{\text{ms}}^{\text{-1}}{\text{×1.5×10}}^{\text{11}}\text{m}\right)}{{\text{6.63×10}}^{\text{-34}}\text{Js}}{\text{= 2.56×10}}^{\text{74}}\end{array}$

Q.11 Answer the following questions, which help you understand the difference between Thomson’s model and Rutherford’s model better.
(a) Is the average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model much less, about the same, or much greater than that predicted by Rutherford’s model?
(b) Is the probability of backward scattering (i.e. scattering of α -particles at angles greater than 90°) predicted by Thomson’s model much less, about the same or much greater than that predicted by Rutherford’s model?
(c) Keeping other factors fixed, it is found experimentally that for small thickness t, the number of α-particles scattered at moderate angles is proportional to t. What clue does this linear dependence on t provide?
(d) In which model is it completely wrong to ignore multiple scattering for the calculation of average angle of scattering of α -particle by a thin foil?

Ans.

(a) The average angle of deflection of α-particles by a thin gold foil predicted by Thomson’s model is about the same as predicted by Rutherford’s model because we are considering average deflection angle.

(b) The probability of backward scattering (i.e. scattering of α-particles at angles greater than 90°) predicted by Thomson’s model is much less than the predicted value because the massive core is absent in Thomson’s model.

(c) It suggest that the scattering of α-particle is primarily because of the single collision. The chance of single collision increases with the number of target atoms hence, collision increases linearly with the thickness of the foil.

(d) In Thomson’s model of atom, as single collision causes very little deflection thus the observed average scattering can be explained only by considering multiple scattering. Multiple collisions can‘t be ignored in this model because positive charge is spread throughout in this model.

Q.12 The gravitational attraction between electron and proton in a hydrogen atom is weaker than the coulomb attraction by a factor of about 10^-40. An alternative way of looking at this fact is to estimate the radius of the first Bohr orbit of a hydrogen atom if the electron and proton were bound by gravitational attraction. You will find the answer interesting.

Ans.

$\begin{array}{l}\text{If electron and proton were bound by gravitational attraction, then centrifugal force in electron}\\ \text{due to its velocity is \hspace{0.17em}}\frac{{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}}{\text{r}}\text{\hspace{0.17em} and gravitational force of attraction between the nucleus, containing}\\ \text{only proton, and electron revolving around the nucleus will be \hspace{0.17em}}\frac{{\text{Gm}}_{\text{e}}{\text{m}}_{\text{p}}}{{\text{r}}^{\text{2}}}\text{.}\\ \text{Hence,}\\ \frac{{\text{m}}_{\text{e}}{\text{v}}^{\text{2}}}{\text{r}}\text{=}\frac{{\text{Gm}}_{\text{e}}{\text{m}}_{\text{p}}}{{\text{r}}^{\text{2}}}...\text{(1)}\\ \text{According to Bohr’s quantization condition,}\\ {\text{m}}_{\text{e}}\text{vr =}\frac{\text{nh}}{\text{2}\mathrm{\pi }}\\ \text{Thus squaring both sides, we get}\\ {{\text{m}}^{\text{2}}}_{\text{e}}{\text{v}}^{\text{2}}{\text{r}}^{\text{2}}\text{=}\frac{{\text{n}}^{\text{2}}{\text{h}}^{\text{2}}}{\text{4}{\mathrm{\pi }}^{\text{2}}}...\text{(2)}\\ \text{Dividing}\left(2\right)\text{by}\left(\text{1}\right)\text{,}\\ {\text{m}}_{\text{e}}\text{r =}\frac{{\text{n}}^{\text{2}}{\text{h}}^{\text{2}}}{\text{4}{\mathrm{\pi }}^{\text{2}}{\text{Gm}}_{\text{e}}{\text{m}}_{\text{p}}}\\ \text{r=}\frac{{\text{n}}^{\text{2}}{\text{h}}^{\text{2}}}{\text{4}{\mathrm{\pi }}^{\text{2}}{{\text{Gm}}^{\text{2}}}_{\text{e}}{\text{m}}_{\text{p}}}\\ {\text{For 1}}^{\text{st}}\text{orbit, n = 1, G = 6.67}\times {\text{10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}{\text{, m}}_{\text{e}}\text{= 9.1}\times {\text{10}}^{\text{-31}}\text{kg and}\\ {\text{m}}_{\text{p}}\text{= 1.67}\times {\text{10}}^{\text{-27}}\text{kg}\\ \text{r=}\frac{{\left(\text{1}\right)}^{\text{2}}\text{×}{\left({\text{6.626×10}}^{\text{-34}}\text{Js}\right)}^{\text{2}}}{{\text{(4×9.87×6.67×10}}^{\text{-11}}{\text{Nm}}^{\text{2}}{\text{kg}}^{\text{-2}}\text{×}{\left({\text{9.1×10}}^{\text{-31}}\text{kg}\right)}^{\text{2}}{\text{×1.67×10}}^{\text{-27}}\text{kg}}\\ {\text{r \hspace{0.17em}\hspace{0.17em}= 1.21×10}}^{\text{29}}\text{m}\\ \text{This value of r is much greater than the size of universe.}\end{array}$

Q.13 Obtain an expression for the frequency of radiation emitted when a hydrogen atom de-excites from level n to level (n-1) . For large n, show that this frequency equals the classical frequency of revolution of the electron in the orbit.

Ans.

\begin{array}{l}\text{The energy of an electron in the nth orbit of hydrogen atom is given by}\\ {\text{E}}_{\text{n}}\text{=}\frac{\text{-2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{n}}^{\text{2}}{\text{h}}^{\text{2}}}\mathrm{\dots \dots \dots \dots }\text{(1)}\\ {\text{Also, for the p}}^{\text{th}}\text{orbit,}\\ {\text{E}}_{\text{p}}\text{=}\frac{\text{-2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{p}}^{\text{2}}{\text{h}}^{\text{2}}}\mathrm{\dots \dots \dots \dots }\text{(2)}\\ \text{where,}\text{p = n – 1}\\ {\text{The frequency of radiation emitted for the transition of electron from n}}^{\text{th}}{\text{to p}}^{\text{th}}\text{orbit is}\\ {\text{E}}_{\text{n}}{\text{– E}}_{\text{p}}\text{= h}\nu \\ \nu \text{=}\frac{{\text{E}}_{\text{n}}{\text{– E}}_{\text{p}}}{\text{h}}\\ \text{which implies,}\\ \nu \text{= 2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}\frac{\left[\frac{\text{1}}{{\text{p}}^{\text{2}}}\text{–}\frac{\text{1}}{{\text{n}}^{\text{2}}}\right]}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}}\\ \text{For n = n and p = n – 1, we have}\\ \nu \text{=}\frac{{\text{E}}_{\text{n}}{\text{– E}}_{\text{n-1}}}{\text{h}}\\ \nu \text{=}\frac{\text{2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}\left[\frac{\text{1}}{{\left(\text{n-1}\right)}^{\text{2}}}\text{–}\frac{\text{1}}{{\text{n}}^{\text{2}}}\right]}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}}\text{,}\\ \nu \text{=}\frac{\text{2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}\left[\frac{{\text{n}}^{\text{2}}\text{–}{\left(\text{n-1}\right)}^{\text{2}}}{{\text{n}}^{\text{2}}{\left(\text{n-1}\right)}^{\text{2}}}\right]}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}}\text{=}\frac{\text{2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}\left[\frac{\left(\text{2n- 1}\right)}{{\text{n}}^{\text{2}}{\left(\text{n-1}\right)}^{\text{2}}}\right]}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}}\text{,}\\ \text{For large value of}\text{n, (2n-1}\sim \text{2n) and (n-1}\sim \text{n)}\\ \text{Thus,}\\ \nu \text{=}\frac{\text{2}{\pi }^{\text{2}}{\text{me}}^{\text{4}}\text{×2n}}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}{\text{n}}^{\text{4}}}\text{=}\frac{\text{4}{\pi }^{\text{2}}{\text{me}}^{\text{4}}}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}{\text{n}}^{\text{3}}}\mathrm{\dots .}\text{(3)}\\ \text{As orbital classical frequency of revolution of electron is given by}\\ {\nu }_{\text{c}}\text{=}\frac{\text{v}}{\text{2}\pi \text{r}}\mathrm{\dots .}\text{(4)}\end{array}

\begin{array}{l}\text{But according to Bohr’s condition of angular momentum quantization,}\\ \text{mvr =}\frac{\text{nh}}{\text{2π}}\text{,}\\ \text{v =}\frac{\text{nh}}{\text{2}\pi \text{mr}}\mathrm{\dots }\left(\text{5}\right)\\ \text{By using}\left(\text{5}\right)\text{in}\left(\text{4}\right)\text{, we get}\\ {\text{ν}}_{\text{c}}\text{=}\frac{\text{nh}}{\text{4}{\pi }^{\text{2}}{\text{mr}}^{\text{2}}}\mathrm{\dots }\left(\text{6}\right)\\ \text{But, r =}\frac{\left(\text{4}\pi {\epsilon }_{\text{0}}\right){\text{n}}^{\text{2}}{\text{h}}^{\text{2}}}{\text{4}{\pi }^{\text{2}}{\text{me}}^{\text{2}}}\\ \text{From}\left(\text{6}\right)\\ {\text{ν}}_{\text{c}}\text{=}\frac{\text{4}{\pi }^{\text{2}}{\text{me}}^{\text{4}}}{{\left(\text{4}\pi {\epsilon }_{\text{0}}\right)}^{\text{2}}{\text{h}}^{\text{3}}{\text{n}}^{\text{3}}}\mathrm{\dots \dots \dots .}\left(\text{7}\right)\\ \text{For a large n,}{\text{ν = ν}}_{\text{c}}\left(\text{using}\left(\text{3}\right)\text{and}\left(\text{7}\right)\right)\\ \text{i}\text{.e frequencies are equal}\text{.}\\ \text{This is called the Bohr’s correspondence principle}\text{.}\end{array}

Q.14 Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10^-10 m).

(a) Construct a quantity with the dimensions of length from the fundamental constants e, m_e and c. Determine its numerical value.

(b) You will find that the length obtained in (a) is many orders of magnitude smaller than the atomic dimensions. Further, it involves c. But energies of atoms are mostly in non-relativistic domain where c is not expected to play any role. This is what may have suggested Bohr to discard c and look for ‘something else’ to get the right atomic size. Now, the Planck’s constant h had already made its appearance elsewhere. Bohr’s great insight lay in recognising that h, m_e and e will yield the right atomic size. Construct a quantity with the dimension of length from h, m_e and e and confirm that its numerical value has indeed the
correct order of magnitude.

Ans.

$\begin{array}{l}\text{(a) Given, charge on electron, e = 1}{\text{.6×10}}^{\text{-19}}\text{C}\\ {\text{mass of an electron, m}}_{\text{e}}\text{= 9}{\text{.1×10}}^{\text{-31}}\text{kg}\\ {\text{Speed of light, c = 3×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\\ {\text{Permittivity of free space, ε}}_{\text{0}}\text{= 8}{\text{.85×10}}^{\text{-12}}{\text{Fm}}^{\text{-1}}\\ \text{Let’s take a quantity involving the given quantities as}\left(\frac{{\text{e}}^{\text{2}}}{\text{4}\pi {\text{ε}}_{\text{0}}{\text{m}}_{\text{e}}{\text{c}}^{\text{2}}}\right)\text{,}\\ \text{Value of this quantity =}\frac{{\left(\text{1}{\text{.6×10}}^{\text{-19}}\text{C}\right)}^{\text{2}}}{\left(\frac{\text{1}}{{\text{9×10}}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{\text{-2}}}\right)\text{×}\left(\text{9}{\text{.11×10}}^{\text{-31}}\text{kg}\right)\text{×}{\left({\text{3 ×10}}^{\text{8}}{\text{ms}}^{\text{-1}}\right)}^{\text{2}}}\text{= 2}{\text{.81×10}}^{\text{-15}}\text{m}\\ \text{The magnitude of this length is very small as compared to the size of the atom}\text{.}\end{array}$

\begin{array}{l}\text{(b) Charge of on electron, e = 1}{\text{.6×10}}^{\text{-19}}\text{C}\\ {\text{Mass of an electron, m}}_{\text{e}}\text{= 9}{\text{.1×10}}^{\text{-31}}\text{kg}\\ \text{Planck’s constant, h = 6}{\text{.626×10}}^{\text{-34}}\text{Js}\\ \text{Let’s take a quantity involving the given quantities as}\frac{\left[{\text{4πε}}_{\text{0}}{\left(\frac{\text{h}}{\text{2}\pi }\right)}^{\text{2}}\right]}{{\text{me}}^{\text{2}}}\\ \text{value}\text{of}\text{this}\text{quantity}\text{is =}\frac{\left[\left(\frac{\text{1}}{{\text{9×10}}^{\text{9}}{\text{Nm}}^{\text{2}}{\text{C}}^{\text{-2}}}\right)\text{×}{\left(\frac{\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}}{\text{2×3}\text{.14}}\right)}^{\text{2}}\right]}{\text{9}{\text{.1×10}}^{\text{-31}}\text{kg×}{\left(\text{1}{\text{.6 x 10}}^{\text{-19}}\text{C}\right)}^{\text{2}}}\text{= 0}\text{.53 Å}\\ \text{This is nearly equal to the size of the atom}\text{.}\end{array}

Q.15 The total energy of an electron in the first excited state of the hydrogen atom is about – 3.4 eV.

(a) What is the kinetic energy of the electron in this state?
(b) What is the potential energy of the electron in this state?
(c) Which of the answers above would change if the choice of the zero of potential energy is changed?

Ans.

$\begin{array}{l}\text{In Bohr’s model,}\\ \text{K\hspace{0.17em}.\hspace{0.17em}E =}\frac{{\text{mv}}^{\text{2}}}{\text{2}}\text{=}\frac{{\text{Ze}}^{\text{2}}}{\text{8}\mathrm{\pi }{\text{ε}}_{\text{0}}\text{r}}\\ \text{P\hspace{0.17em}.\hspace{0.17em}E = –}\left(\frac{{\text{Ze}}^{\text{2}}}{4\mathrm{\pi }{\text{ε}}_{\text{0}}\text{r}}\right)\\ \therefore \text{P\hspace{0.17em}.\hspace{0.17em}E = -2}\left(\text{K\hspace{0.17em}.\hspace{0.17em}E}\right)\text{\hspace{0.17em} and E = P\hspace{0.17em}.\hspace{0.17em}E + K\hspace{0.17em}.\hspace{0.17em}E = -\hspace{0.17em}K.E}\\ \end{array}$

(a) Using E = – K.E, the kinetic energy of the electron in this state is +3.4 eV.

(b) As P.E = – 2K.E, thus P.E of the electron is – 6.8 eV.

(c) Kinetic energy does not depend upon the choice of zero of potential energy. Therefore, its value remains unchanged; however, the potential energy gets changed with the change in the zero level of potential energy.

Q.16 If Bohr’s quantization postulate

\left(\text{angular momentum}=\frac{\text{nh}}{\text{2}\pi }\right)

is a basic law of nature, it should be equally valid for the case of planetary motion also. Why then do we never speak of quantization of orbits of planets around the sun ?

Ans.

\begin{array}{l}\text{Applying the Bohr’s quantization postulate,}\\ \text{mvr =}\frac{\text{nh}}{\text{2}\pi }\\ \text{we get,}\\ \text{n =}\frac{\text{mvr(2}\pi )}{\text{h}}\\ \text{For the motion of a planet}\left(\text{say earth}\right)\text{,}\\ {\text{m = 6×10}}^{\text{24}}\text{kg, v =}\text{30000}{\text{ms}}^{\text{-1}}\\ \text{r = 1}{\text{.49×10}}^{\text{11}}\text{m and h =}\text{6}{\text{.626×10}}^{\text{-34}}\text{J}\text{s}\\ \therefore \text{n =}\frac{{\text{6×10}}^{\text{24}}{\text{kg×30000 ms}}^{\text{-1}}\text{×1}{\text{.49×10}}^{\text{11}}\text{m×2×3}\text{.14}}{\text{6}{\text{.626×10}}^{\text{-34}}\text{Js}}\text{= 2}{\text{.49×10}}^{\text{74}}\\ \text{i}\text{.e}\text{., n is very large}\text{.}\\ \text{Since n is very large, the differences in the successive energies and angular momenta}\\ \text{of quantized levels of the Bohr’s model are very small}\text{.}\\ \text{Hence, levels may be considered continuous}\text{.}\end{array}

Q.17 Obtain the first Bohr’s radius and the ground state energy of a muonic hydrogen atom (i.e, an atom in which a negatively charged muon (μ^–) of mass about 207 m_e orbits around a proton).

Ans.

$\begin{array}{l}{\text{Here mass of the particle revolving around the proton is m = 207m}}_{\text{e}}{\text{= 207×9.1×10}}^{\text{-31}}\text{\hspace{0.17em}kg}\\ \text{Radius of ground state energy orbit of muonic hydrogen atom is given by}\\ {\text{r}}_{\text{m}}\text{=}\frac{\left(\text{4}{\mathrm{\pi \epsilon }}_{\text{0}}\right){\text{\hspace{0.17em}h}}^{\text{2}}}{\text{4}{\mathrm{\pi }}^{\text{2}}{\text{me}}^{\text{2}}}\\ {\text{r}}_{\text{m}}\text{=}\frac{\left(\frac{\text{1}}{{\text{9×10}}^{\text{9}}{\text{\hspace{0.17em}Nm}}^{\text{2}}{\text{C}}^{\text{-2}}}\right)\text{×}{\left({\text{6.626×10}}^{\text{-34}}\text{\hspace{0.17em}Js}\right)}^{\text{2}}}{{\text{4×9.87×207×9.1×10}}^{\text{-31}}\text{kg×}{\left({\text{1.6×10}}^{\text{-19}}\text{\hspace{0.17em}C}\right)}^{\text{2}}}{\text{= 2.56×10}}^{\text{-13}}\text{\hspace{0.17em}m}\\ \text{Also, energy of this level ,}\\ {\text{E}}_{\text{µ}}{\text{= E}}_{\text{e}}\frac{{\text{m}}_{\text{µ}}}{{\text{m}}_{\text{e}}}\\ {\text{E}}_{\text{µ}}\text{= –}\left(\text{13.6×207}\right)\text{\hspace{0.17em}eV}\\ {\text{E}}_{\text{µ}}\text{= -2.81\hspace{0.17em} keV}\end{array}$