NCERT Exemplar for Class 12 Physics Chapter-12 (Book Solutions)

Exercise

MULTIPLE CHOICE QUESTIONS

12.01 Taking the Bohr radius as ${a_0} = 53{\text{ pm}}$, the radius of \[L{i^{ +  + }}\] ion in its ground state, on the basis of Bohr's model, will be about

(a) $53{\text{ pm}}$

(b) $27{\text{ pm}}$

(c) $18{\text{ pm}}$

(d) $13{\text{ pm}}$

Ans: Option (c) 

The radius of any ion in its ground state is given as $r = \dfrac{{{a_0}}}{Z}$ where ${a_0}$ is the Bohr’s radius and Z is the atomic number. The Bohr’s radius is equal to $53\,pm$ and the atomic number of lithium is three. So, the radius of \[L{i^{ +  + }}\] ion in the ground state is $r = \dfrac{{{a_0}}}{Z} = \dfrac{{53}}{3}pm = 17.67\,pm \approx 18\,pm$. So, option (c) is the correct answer. 

12.2 The binding energy of an H-atom, considering an electron moving around a fixed nuclei (proton), is $B =  - \dfrac{{m{e^4}}}{{8{n^2}\varepsilon _0^2{h^2}}}$ (m=mass of electron) If one decides to work in a frame of reference where the electron is at rest, the proton would be moving around it. By similar arguments, the binding energy would be

$B =  - \dfrac{{m{e^4}}}{{8{n^2}\varepsilon _0^2{h^2}}}$ (M = proton mass)

This last expression is not correct because

(a) n would not be integral. 

(b) Bohr - quantization applies only to electron 

(c) The frame in which the electron is at rest is not inertial.

(d) The motion of the proton would not be in circular orbits, even approximately. 

Ans: Option (c)

Electrons keep on moving continuously in the orbitals. So, the frame in which an electron is at rest cannot be inertial. Also, the electrostatic force cannot provide the centripetal force ${F_p} = \dfrac{{{m_p}{v^2}}}{r}$ as the mass of electron is very less (negligible) as compared to proton. So, option (c) is the correct answer. 

12.3 The simple Bohr model cannot be directly applied to calculate the energy levels of an atom with many electrons. This is because 

(a) Of the electrons, not being subject to a central force.

(b) Of the electrons colliding with each other

(c) Of screening effects

(d) The force between the nucleus and an electron will no longer be given by Coulomb’s law.

Ans: Option (a)

The simple Bohr model cannot be directly applied to calculate the energy levels of a multi-electron system because the electronic repulsions between the electrons will interfere and the net force will not be subject to the central force. Hence, option (a) is the correct answer. 

12.4 For the ground state, the electron in the H-atom has an angular momentum = h, according to the simple Bohr model. Angular momentum is a vector and hence there will be infinitely many orbits with the vector pointing in all possible directions. In actuality, this is not true,

(a) Because Bohr’s model gives incorrect values of angular momentum.

(b) Because only one of these would have a minimum energy.

(c) Angular momentum must be in the direction of spin of electron. 

(d) Because electrons go around only in horizontal orbits. 

Ans: Option (a)

Angular momentum of a revolving electron is an integral multiple of $\dfrac{h}{{2\pi }}$ according to Bohr's second postulate. Hence, Bohr's model does not give the correct value of angular momentum. So, option (a) is the correct answer. 

12.5 $O_2$ molecule consists of two oxygen atoms. In the molecule, nuclear force between the nuclei of the two atoms

(a) is not important because nuclear forces are short-ranged.

(b) is as important as electrostatic force for binding the two atoms.

(c) Cancels the repulsive electrostatic force between the nuclei.

(d) is not important because oxygen nuclei have an equal number of neutrons and protons.

Ans: Option (a)

The nuclear forces are short ranged. Since the distance between the two oxygen atoms is larger as compared to the electrons of the two atoms. Hence, nuclear force between the nuclei of the two atoms is not important because nuclear forces are short-ranged. So, option (a) is the correct answer. 

12.6 Two H atoms in the ground state collide elastically. The maximum amount by which their combined kinetic energy is reduced is 

(a) 10.20 eV

(b) 20.40 eV

(c) 13.6 eV 

(d) 27.2 eV

Ans: Option (a)

First we calculate the total energy of the two H atoms in the ground state. 

So, total energy of two H-atom in ground state \[ = 2\left( { - 13.6} \right) =  - 27.2{\text{ }}eV\]

The maximum amount by which their combined kinetic energy is reduced when any one hydrogen atom goes into first excited state after the inelastic collision i.e., the total energy of two hydrogen atoms after inelastic collision is:

$E = \dfrac{{13.6}}{{{n^2}}} + 13.6$

Now, for n equals two. We have, 

$E = \dfrac{{13.6}}{{{2^2}}} + 13.6$

$ \Rightarrow E = 3.4 + 13.6 = 17eV$

So, Loss in kinetic energy due to inelastic collision $ = 27.2 - 17.0 = 10.2{\text{ eV}}$

12.7 A set of atoms in an excited state decays.

(a) in general to any of the states with lower energy.

(b) Into a lower state only when excited by an external electric field.

(c) All together simultaneously into a lower state.

(d) To emit photons only when they collide. 

Ans: Option (a) 

A set of atoms in an excited state decays in general to any of the states with lower energy.

MULTIPLE CHOICE QUESTIONS-II MORE THAN ONE OPTION

12.8 An ionized H-molecule consists of an electron and two protons. The protons are separated by a small distance of the order of angstrom. In the ground state,

(a) The electron would not move in circular orbits.

(b) The energy would be (2)4 times that of an H-atom.

(c) The electrons' orbit would go around the protons.

(d) The molecule will soon decay in a proton and an H-atom.

Ans: Options (a) and (c) 

The nucleus of the two hydrogen atoms are separated by a small distance which is of order of angstrom. Also, the nuclear forces are of the same order. So, one electron will revolve around the protons. Hence, options (a) and (c) are the correct answers. 

12.9 Consider aiming a beam of free electrons towards free protons. When they scatter, an electron and a proton cannot combine to produce an H-atom,

(a) Because of energy conservation.

(b) Without simultaneously releasing energy in the form of radiation.

(c) Because of momentum conservation.

(d) Because of angular momentum conservation.

Ans: Options (a) and (b) 

Larger amount of energy is required for an electron to reach near the proton as the binding energy of the hydrogen atom is larger as compared to the energy of a free electron. So, due to the energy conservation principle, it is not possible for electrons and protons to combine and form hydrogen atoms. Hence, options (a) and (b) are the correct answers.

12.10 The Bohr model for the spectra of an H-atom

(a) Will not be applicable to hydrogen in the molecular form.

(b) Will not be applicable as it is for a He-atom.

(c) is valid only at room temperature.

(d) Predicts continuous as well as discrete spectral lines.

Ans: Options (a) and (b)

The Bohr model for the spectra of a hydrogen atom is applicable for only one electron system. So, it cannot be applied to hydrogen in the molecular form and for helium atoms. 

12.11 The Balmer series for the H-atom can be observed

(a) If we measure the frequencies of light emitted when an excited atom falls to the ground state.

(b) If we measure the frequencies of light emitted due to transitions between excited states and the first excited state.

(c) In any transition in a H-atom.

(d) As a sequence of frequencies with the higher frequencies getting closely packed.

Ans: (b) and (d): The spectrum of light formed when the electron jumps from a higher energy level to the first energy level is called the Balmer series. The spectral lines start becoming close as the electron jumps from higher energy level to the ground level. Hence, options (b) and (d) are the correct answers. 

12.12 Let ${E_n} = \dfrac{{ - 1m{e^4}}}{{8\varepsilon _0^2{n^2}{h^2}}}$ be the energy of the nth level of H-atom. If all the H atoms are in the ground state and radiation of frequency $\dfrac{{\left( {{E_2} - {E_1}} \right)}}{h}$ falls on it,

(a) It will not be absorbed at all

(b) Some of the atoms will move to the first excited state.

(c) All atoms will be excited to the ${\text{n}} = 2$ state.

(d) No atoms will make a transition to the ${\text{n}} = 3$ state.

Ans: (b) and (d): When the energy of radiation of photons is$\dfrac{{\left( {{E_2} - {E_1}} \right)n}}{h}.$, then electron jumps in next energy level $({\text{n}} = 2)$ after receiving this energy equal to the \[{E_2} - {E_1}\] energy. The new state is its unstable state. Electrons jump from $E_2$ to $E_1$ by radiating the energy of the same frequency i.e., to ($E_2$ - $E_1$). Electrons can jump in the next orbit. So electrons from the ground state will jump at ${\text{n}} = 2$ not ${\text{n}} = 3$.

12.13 The simple Bohr model is not applicable to $H{e_4}$ atom because

(a) $H{e_4}$ is an inert gas.

(b) $H{e_4}$ has neutrons in the nucleus. 

(c) $H{e_4}$ has one more electron.

(d) Electrons are not subject to central forces.

Ans: (c) and (d): The Bohr’s atomic model is applicable only for one electron system and there are two electrons in $H{e_4}$ atom. In multi electron systems, electrons are not subject to central forces due to longer distance than nuclear size. Hence, options (c) and (d) are the correct answers.

VERY SHORT ANSWER TYPE QUESTIONS

12.14 The mass of a H-atom is less than the sum of the masses of a proton and electron. Why is this? 

Ans: Some energy is needed to bind the nucleons together in the nucleus in the process of formation of an atom with electrons and nucleons. This energy can be calculated from from Einstein mass energy relation

${\text{E}} = \Delta {\text{m}}{{\text{C}}^2}$

Where $\Delta {\text{m}} = \left[ {{\text{Zm}}{{\text{p}}^ + }({\text{A}} - {\text{Z}}){\text{mn}}} \right] - {\text{M}}$

So the mass of an H atom is equal to

${m_p} + {m_p} - \dfrac{{B \cdot E}}{{{C^2}}}$

Where B.E. represents the binding energy and is equal to $13.6{\text{ eV}}$. 

12.15 Imagine removing one electron from \[{}_2H{e^4}\] and \[{}_2H{e^3}\]. Their energy levels, as worked out on the basis of the Bohr model will be very close. Explain why.

Ans: If we remove one electron from the isotopes of \[{}_2H{e^4}\] and \[{}_2H{e^3}\], both the isotopes will become one electron systems but the mass of the nucleus of the isotopes will be heavier than the hydrogen atom. So the energy levels as worked out on the basis of the Bohr model will be very close.  

12.16 When an electron falls from a higher energy to a lower energy level, the difference in the energies appears in the form of electromagnetic radiation. Why can't it be emitted as other forms of energy?

Ans: When an electron falls from a higher energy level to a lower energy level, the charged particle experiences acceleration. The accelerated charged particle can produce electromagnetic waves only.

12.17 Would the Bohr formula for the H-atom remain unchanged if proton had a charge $\left( {\dfrac{{ + 4}}{3}} \right)e$ and electron a charge$\left( {\dfrac{{ - 3}}{4}} \right)e\quad $, where ${{e}} = {{1}}.{{6}} \times {{1}}{{{0}}^{ - {{19}}}}{{C}}$ Give reasons for your answer?

Ans: The new charges on the proton and electron are given as $\left( {\dfrac{{ + 4}}{3}} \right)e$ and $\left( {\dfrac{{ - 3}}{4}} \right)e\quad $ respectively. So, we find the product of the charges in the new case as the product of charges matters in the coulomb forces rather than the individual charges. 

$\dfrac{{ + 4}}{3}e \times \dfrac{{ - 3}}{4}e =  - {e^2}$

Originally, charge on electrons was $ - e$ and charge on protons was $e$. 

Product is $ - {e^2}$. 

So, the product of charges is the same as earlier. As there is no change in position of protons and electrons and the product of charges is also the same. Hence, the Bohr's formula for the new H-atom remains the same.

12.18 Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but the same orbital angular momentum according to the Bohr model?

Ans: In excited state of electrons of two H-atoms, electrons may be in orbit or energy level either \[n = 2,3{\text{ }} \ldots .\]and can have same energy but angular momentum by Bohr’s model is $L = \dfrac{{nh}}{{2\pi }}.$. As n for both may be different so both H-atoms will have different angular momentum. 

SHORT ANSWER TYPE QUESTIONS

12.19 Positronium is just like a H-atom with the proton replaced by the positively charged anti-particle of the electron (called the positron which is as massive as the electron). What would be the ground state energy of positronium?

Ans: In the new species, we have a positron in place of a proton in comparison to a hydrogen atom.

So, mass of positronium $ = m = {m_g}^ -  + {m_g}^ + $

Hence, ${m_g}^ -  = {m_g}^ +  = \dfrac{{{m_g}}}{2}$

As ${E_n} =  - 13.6$ and so energy of positron

\[E_n^\prime  = \dfrac{{ - m_e^ + {e^4}}}{{8{\varepsilon _0}{n^2}{h^2}}} = \dfrac{{ - \left( {\dfrac{{{m_e}}}{2}} \right){e^4}}}{{8{\varepsilon _0}{n^2}{h^2}}} = \dfrac{{ - 13.6}}{2}\]

Therefore, we get, \[E_n^\prime  = \dfrac{{13.6}}{2}\left( {\because {m_e} = \dfrac{m}{2}} \right)\]

\[E_n^\prime  =  - 6.8\,eV\]

12.20 Assume that there is no repulsive force between the electrons in an atom but the force between positive and negative charges is given by Coulomb’s law as usual. Under such circumstances, calculate the ground state energy of a He-atom.

Ans: For ${\text{H atom Z}} = 1$ and ${\text{n}} = 1$

$\therefore {E_n} = \dfrac{{ - {m_e}{e^4}}}{{8\varepsilon _0^2{1^2}{h^2}}} = 13.6{\text{ eV}}$

$For\,He - atom,$ Z=4 and $n = 1$

$\therefore {E_n} = \dfrac{{ - 4{m_e}{e^4}}}{{8\varepsilon _0^2{1^2}{h^2}}} =  - 4 \times 13.6 =  - 54.4{\text{ eV}} =  - 54.4{\text{ eV}}$

12.21 Using the Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Ans: Let the velocity of electrons of hydrogen atoms in the orbit be v meters per second. 

So, radius of orbit $ = {a_0}$, where ${a_0}$ is the Bohr’s radius.  

So, the number of revolutions per second $ = \dfrac{{2\pi {a_0}}}{v}$

Now, we know that ${\text{I}} = {}^ - {\text{ev}}$

Since we have, $v = \dfrac{{2\pi {a_0}}}{v}$

So, we get, ${\text{I}} = \dfrac{{{}^ - e2\pi {a_0}}}{v}$

(-) sign shows that the direction of current is opposite to the direction of motion of electrons.

12.22 Show that the first few frequencies of light that is emitted when electrons fall to the nth level from levels higher than ${\text{n}}$, are approximate harmonics (i.e. in the ratio ${1:2:3 \ldots {\text{ ) when n}} >  > 1.{\text{ }}} $

Ans: When an electron falls from ${\left( {n + n'} \right)^{th}}$ to \[{n^{th}}\] energy level, the frequency of radiations in spectrum Hydrogen atom like atoms is given by: 

$v = CR{Z^2}\left[ {\dfrac{1}{{\left( {n + {n^\prime }} \right)}} - \dfrac{1}{{{n^2}}}} \right]$

In this case, $n' >  > n$

So, $n' = 1,2,3,....$ and R is the Rydberg’s constant. So, we get,

$v = CR{Z^2}\left[ {\dfrac{1}{{{n^2}{{\left[ {1 + \dfrac{{{n^\prime }}}{{{n^2}}}} \right]}^2}}} - \dfrac{1}{{{n^2}}}} \right] = CR{Z^2}\left[ {\dfrac{1}{{{n^2}}}{{\left[ {1 + \dfrac{n}{n}} \right]}^{ - 2}} - \dfrac{1}{{{n^2}}}} \right]$

Neglecting the higher terms as $n >  > {n^2}$,

$ = CR{Z^2}\left[ {\dfrac{1}{{{n^2}}}\left( {1 - \dfrac{{2{n^\prime }}}{n}} \right) - \dfrac{1}{{{n^2}}}} \right] = CR{Z^2}\left[ {\dfrac{1}{{{n^2}}} - \dfrac{{2{n^\prime }}}{{{n^3}}} - \dfrac{1}{{{n^2}}}} \right]$

$ = \dfrac{{ - CR{Z^2}2{n^\prime }}}{{{n^3}}} = \left( {\dfrac{{2CR{Z^2}}}{{{n^3}}}} \right){n^\prime }$

So the first few frequencies of light that is emitted when electrons fall from $\left( {n + n'} \right)$ to \[{n^{th}}\] energy level are in the ratio of $n' = 1,2,3,....$ when $n >  > 1$. 

12.23 What is the minimum energy that must be given to a ${\text{H}}$ atom in ground state so that it can emit an ${{H}}\gamma $ line in the Balmer series? If the angular momentum of the system is conserved, what would be the angular momentum of such ${\text{H}}\chi $ photon?

Ans: We know that ${\text{H}}$ spectral line in the Balmer series formed when an electron falls from ${\text{n}} = 5$ to ${\text{n}} = 1$. Here the electron is in ground state i.e., ${\text{n}} = 1$ and must be taken to ${\text{n}} = 5$ for ${\text{H}}$ line. So the energy of

$H? = {{\text{E}}_5} - {\text{E}}{}_1 = \left( {\dfrac{{13.6}}{{{5^2}}}} \right) - ( - 13.6) =  - 0.54 + 13.6 = 13.06{\text{eV}}$

Since angular momentum is conserved, so the angular momentum of 

${{H}}\gamma =$ Change in angular momentum of electron 

$ = {{\text{L}}_5} - {{\text{L}}_2} = 5\;{\text{ h}} - 2\;{\text{ h}}$ 

$ = 3\;{\text{ h}} = 3 \times 6.63 \times 10 - 34 = 19.89 \times 10 - 34{\text{ kg}}{{\text{m}}^2}/{\text{s}}$

LONG ANSWER TYPE QUESTIONS

12.24 The first four spectral lines in the Lyman series of a H-atom are \[{{\lambda }}{\text{ }} = {\text{ }}{{1218}}{\text{ }}{{\mathop A\limits^ \circ}},{\text{ }}{{1028\mathop A\limits^ \circ}},{\text{ }}{{974}}.{{3}}{\text{ }}{{\mathop A\limits^ \circ}}{\text{ }}{{and}}{\text{ }}{{951}}.{{4\mathop A\limits^ \circ}}.\]If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Ans: Reduced mass of H atom (mass defect) ${\mu _H}$ then

$\dfrac{1}{{{\mu _H}}} = \dfrac{1}{{{m_\varepsilon }}} + \dfrac{1}{M}({\text{M}}$ is mass of ${\text{H}}$ atom )

${\dfrac{1}{{{\mu _H}}} = \dfrac{{M + {m_e}}}{{M \cdot {m_e}}} = \dfrac{{M\left[ {1 + \dfrac{{{m_e}}}{M}} \right]}}{{M \cdot {m_e}}}\left( {\because  \cdot M >  >  > {{\text{m}}_{\text{e}}}} \right)} \\ {\therefore {\mu _H} = {m_e}{{\left[ {1 + \dfrac{{{m_e}}}{M}} \right]}^{ - 1}} = {m_e}\left[ {1 - \dfrac{{{m_e}}}{M}} \right]} $

For Deuterium ${\text{M}} = 2{\text{M}}$

Reduced mass of Deuterium $ = {\mu _D}$

${\mu _D} = {m_\varepsilon }\left[ {1 - \dfrac{{{m_e}}}{M}} \right]\left[ {1 + \dfrac{{{m_e}}}{M}} \right]$

Total energy $E_n$ of the electron revolving in an nth stationary orbit,

${E_n} = \dfrac{{ - m{e^4}}}{{8{n^2}ge_0^2{h^2}}}$

Here, $m$ is the reduced mass of the electron and proton in the H atom.

So $hv = E{n_i} - E{n_f}$

$v = \dfrac{{m{e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{{n_i^2}} - \dfrac{1}{{n_f^2}}} \right] = \dfrac{c}{\lambda }$

$\nu  \propto m$ (Reduced mass)

$\dfrac{1}{\lambda } \propto \mu  \Rightarrow \lambda  \propto \dfrac{1}{\mu }$

For hydrogen and deuterium $ = \dfrac{{{\lambda _H}}}{{{\lambda _D}}}$

${\text{ So }}\dfrac{{{\lambda _H}}}{{{\lambda _D}}} = \dfrac{{{\mu _H}}}{{{\mu _D}}} = \dfrac{{{m_e}\left[ {1 - \dfrac{{{m_e}}}{M}} \right]}}{{{m_e}\left[ {1 - \dfrac{{{m_e}}}{M}} \right]\left[ {1 + \dfrac{{{m_e}}}{{2M}}} \right]}}$

${\lambda _D} = {\left[ {1 + \dfrac{{{m_e}}}{{2M}}} \right]^{ - 1}}{\lambda _H} = \left( {1 - \dfrac{{{m_e}}}{M}} \right){\lambda _H}$

${\lambda _{\text{D}}} = {\lambda _{\text{H}}}(0.99973)$

${\lambda _H} = 1218 \mathop A\limits^ \circ  , 1028 \mathop A\limits^ \circ  , 974.3 \mathop A\limits^ \circ  {\text{ and }} 951.4\mathop A\limits^ \circ  $ (Given)

${\lambda _{D1}} = \mathop A\limits^{\mathop A\limits^ \circ  }  = 1214$

${\lambda _{D2}} = 0.9973 \times 1028 = 1025\mathop A\limits^ \circ  $

${\lambda _{D2}} = 0.9973 \times 974 = 971.37\mathop A\limits^ \circ  $

${\lambda _{D2}} = 0.9973 \times 954 = 951.42\mathop A\limits^ \circ  $

12.25 Deuterium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in 1H1and 1H2. This is because the wavelength of transition depends to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common center of mass. Such a system is equivalent to a single particle with a reduced mass μ, revolving around the nucleus at a distance equal to the electron-nucleus separation. Here $\mu  = \dfrac{{{m_d}M}}{{(M + m)}}$ where M is the nuclear mass and me is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in \[{}_1H^1\] and \[{}_1H^2\]. (Mass of \[{}_1H^1\] nucleus is $1.6725 \times 10 - 27\;{\text{ kg}}$, Mass of \[{}_1H^2\] nucleus is $3.3374 \times $ 10-27 kg, Mass of electron $ = 9.109 \times 10 - 31\;{\text{kg}}.)$

Ans: Total energy of electron in nth stable orbit in H or like atom

${E_n} = \dfrac{{\mu {Z^2}{e^4}}}{{8\varepsilon _0^2{h^2}{n^2}}}$

$\mu  = $ Reduced mass of electron, proton and neutron (mass defect)

${E_H} = \dfrac{{{\mu _H}{{(1)}^2}{e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = \dfrac{{{\mu _H}{e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{1} - \dfrac{1}{{{2^2}}}} \right] = \dfrac{{{\mu _H}{e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{3}{4}} \right]$

$E = hv = \dfrac{h}{\lambda }{\text{ or }}{\lambda _H} = \dfrac{h}{{{E_H}}}$

$\therefore h{v_H} = \dfrac{{\mu _{{H^e}}^4\dfrac{3}{{8\varepsilon _0^2{h^2}}}\dfrac{1}{4}}}{2}$

${v_H} = \dfrac{{{\mu _H}{e^4}}}{{8\varepsilon _0^2{h^3}}} \cdot \dfrac{3}{4}$

$ = \dfrac{{\left( {{\lambda _D} - {\lambda _H}} \right)}}{{{\lambda _H}}} \times 100$Percent change in wavelength

% change $\left. {\Delta E = \left[ {\dfrac{{{\lambda _D}}}{{{\lambda _H}}} - 1} \right] \times 100\because \Delta {\text{E}} = {{\text{E}}_1} - {{\text{E}}_2}} \right) \ldots  \ldots $ (I)

${hv = \dfrac{{\mu {e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{{n_2^2}} - \dfrac{1}{{n_2^2}}} \right]} \\ {v = \dfrac{{\mu {e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]} \\  {\dfrac{c}{\lambda } = \dfrac{{\mu {e^4}}}{{8\varepsilon _0^2{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]} \\  {\dfrac{1}{\lambda } = \dfrac{{\mu {e^4}}}{{8\varepsilon _0^2c{h^2}}}\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right]} $

As $\mu  = $ mass defect, \[e,\varepsilon oc\], and $h$ are constants for an atom.

$\therefore \lambda  \propto \dfrac{1}{h}$

So eqn. Ist can be written as percentage change in the wavelength $ = \left[ {\dfrac{{{\mu _H}}}{{{\mu _D}}} - 1} \right] \times 100$

$\because \mu  = \dfrac{{{m_e}M}}{{\left( {M + {m_t}} \right)}}({\text{ Given }})$

$\therefore $ Percentage change in wavelength $ = \left[ {\dfrac{{\dfrac{{{m_e}{M_H}}}{{\left( {{M_H} + {m_e}} \right)}}}}{{\dfrac{{{m_e} + {M_D}}}{{\left( {{M_D} + {m_f}} \right)}}}}} \right]100$

$\dfrac{{\Delta \lambda }}{{{\lambda _H}}} \times 100 = \left[ {\dfrac{{{M_H}}}{{{M_D}}}\dfrac{{\left( {{M_D} + {m_\theta }} \right)}}{{\left( {{M_H} + {m_f}} \right)}} - 1} \right] \times 100$

${ = \left[ {\dfrac{{{M_H}}}{{{M_D}}}\dfrac{{{M_D}\left( {1 + \dfrac{{{m_e}}}{{{M_D}}}} \right)}}{{{M_H}\left( {1 + \dfrac{{{m_e}}}{{{M_H}}}} \right)}} - 1} \right] \times 100} \\ { = \left[ {\left( {1 + \dfrac{{{m_e}}}{{{M_D}}}} \right){{\left( {1 + \dfrac{{{m_e}}}{{{M_H}}}} \right)}^{ - 1}} - 1} \right] \times 100} \\  { = \left[ {\left( {1 + \dfrac{{{m_e}}}{{{M_D}}}} \right)\left( {1 - \dfrac{{{m_e}}}{{{M_H}}}} \right) - 1} \right] \times 100} $

${\text{Me}} \ll {{\text{M}}_{\text{D}}}$ So neglecting the higher degree term

$\dfrac{{\Delta \lambda }}{{{\lambda _H}}} \times 100 = \left[ {1 - \dfrac{{{m_e}}}{{{M_H}}} + \dfrac{{{m_e}}}{{{M_D}}} - \dfrac{{{m_e}{m_e}}}{{{M_D} - {M_H}}} - 1} \right] \times 100$

${ = {m_e}\left[ {\dfrac{1}{{{M_D}}} - \dfrac{1}{{{M_H}}}} \right] \times 100} \\ { = 9.1 \times {{10}^{ - 31}}\left[ {\dfrac{1}{{3.3374 \times {{10}^{ - 27}}}} - \dfrac{1}{{1.6725 \times {{10}^{ - 27}}}}} \right] \times 100} \\  { = \dfrac{{9.1 \times {{10}^{ - 31 + 2}}}}{{{{10}^{ - 27}}}}\left[ {\dfrac{{1.6725 - 3.3374}}{{3.3374 \times 1.6725}}} \right]} \\  {\dfrac{{\Delta \lambda  \times 100}}{{{\lambda _H}}} = \dfrac{{ - 9.1 \times {{10}^{ - 20 + 27}} \times 0.6649}}{{3.3374 \times 1.6725}} = \dfrac{{ - 6.05059 \times {{10}^{ - 2}}}}{{5.5180}}} $

$\dfrac{{\Delta \lambda  \times 100}}{{{\lambda _H}}} =  - 1.084 \times {10^{ - 2}}$ % Decrease in wavelength.

(-) sign shows that ${\lambda _{\text{D}}} < {\lambda _{\text{H}}}$

12.26 If a proton had a radius ${\text{R}}$ and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when 

(i) ${{R}} = $ $0.1\mathop A\limits^ \circ$

Ans: Consider the H atom nucleus as a point charge electron revolving around a nucleus with speed $v$ and radius $r_A$. The Colombian force provides centripetal force to revolve around the nucleus.

$\therefore \dfrac{{{m_e}{v^2}}}{{{r_A}}} = \dfrac{{ - K{e^2}}}{{r_A^2}} \ldots {\text{ (I) }}$

Here $K = \dfrac{1}{{4\pi {\varepsilon _0}}}$

(-) sign shows the force of attraction.

By Bohr’s postulate, angular momentum$ = \dfrac{{nh}}{{2\pi }}$

${mv{r_A} = \dfrac{{nh}}{{2\pi }}} \\  {v = \dfrac{{nh}}{{2\pi m{r_A}}}} \\ {\dfrac{{m{n^2}{h^2}}}{{4{\pi ^2}{m^2}r_A^2{r_A}}} = \dfrac{{K{e^2}}}{{r_A^2}}[{\text{From I}}]} \\  {{r_A} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mK{e^2}}} \cdots {\text{ (II) }}} $

For ground state $n = 1$

${r_A} = \dfrac{{{h^2}}}{{4{\pi ^2}mK{e^2}}}$

${ = \dfrac{{6.63 \times {{10}^{ - 34}}6.63 \times {{10}^{ - 34}}}}{{{{(2 \times 3.14)}^2}9.1 \times {{10}^{ - 31}} \times 9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}} \\ { = \dfrac{{6.63 \times 6.63 \times {{10}^{ - 68 + 38 + 31 - 9}}}}{{9.19 \times 1.6 \times 1.6 \times 4 \times 3.14 \times 4 \times 3.14}}} $

${ = {r_{\text{A}}} = 0.53\mathop A\limits^ \circ} \\ {P.E. = \dfrac{{ - K{e^2}}}{{{r_A}}} = \dfrac{{ - 9 \times {{10}^9} \times 1.6 \times {{10}^{ - 19}} \times 1.6 \times {{10}^{ - 19}}}}{{0.53 \times {{10}^{ - 10}}}}J} \\  { = \dfrac{{ - 9 \times 1.6 \times 1.6 \times {{10}^{ - 19}} \times {{10}^{ - 19}} \times {{10}^9}}}{{0.53 \times {{10}^{ - 10}} \times 1.6 \times {{10}^{ - 19}}}}J} \\ {P.E. = \dfrac{{14.4}}{{0.53}} = 27.17 = 27.2{\text{eV}}} $

${K.E = \dfrac{1}{2}m{v^2}} \\  { = \dfrac{1}{2}\dfrac{{m \cdot {n^2}{h^2}}}{{4{\pi ^2}{m^2}{r^2}}} = \dfrac{1}{2}\dfrac{{{h^2}}}{{4{\pi ^2}m{r^2}}}({\text{n}} = 1{\text{ for ground state }})} \\  { = \dfrac{1}{2}\dfrac{{6.62 \times {{10}^{ - 34}} \times 6.62 \times {{10}^{ - 34}}}}{{4 \times 3.14 \times 3.14 \times 9 \times {{10}^{ - 31}} \times 0.53 \times {{10}^{ - 10}} \times 0.53 \times {{10}^{ - 10}}}}J} $

${ = \dfrac{{6.62 \times 6.62 \times {{10}^{ - 68 + 31 + 10 + 10}}}}{{4 \times 3.14 \times 3.14 \times 18 \times 0.53 \times 0.53 \times 1.6 \times {{10}^{ - 19}}}}\,eV} \\ { = \dfrac{{6.62 \times 6.62 \times {{10}^{ - 68 + 51 + 19}}}}{{4 \times 3.14 \times 3.14 \times 18 \times 0.53 \times 0.53}}} \\ {{\text{K}}.{\text{E}}.{\text{ = 0}}.{\text{1373}} \times {\text{1}}{{\text{0}}^{\text{2}}}{\text{eV = 13}}.{\text{7 eV}}} \\ {{\text{ P}}{\text{.E}}{\text{. = 27}}{\text{.2 eV}}} $

(ii) $R = 10\mathop A\limits^ \circ$.

Ans: Now for spherical nucleus of radius, R, electron moves charge inside the nucleus

$R > > r_b$ then the electron moves inside the nucleus. Then ($r_b$ is radius of new Bohr’s orbit of revolving electron)

Charge

$e = \dfrac{{e \cdot \left( {\dfrac{4}{3}\pi r_b^3} \right)}}{{\dfrac{4}{3}\pi {R^3}}}$

${e^\prime } = {q_2} = \dfrac{{er_b^3}}{{{R^3}}}$

${q_1} = e$

$\dfrac{{m{v^2}}}{{{r_b}}} = \dfrac{{Ke{e^\prime }}}{{r_b^2}}($ By Coulomb's law)

$m{v_b} = \dfrac{{nh}}{{2\pi }} \Rightarrow {v_b} = \dfrac{{nh}}{{2\pi m{r_b}}}$ (By Bohr's postulate)

$\therefore \dfrac{m}{{{r_b}}}\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}r_b^2}} = \dfrac{{ke{e^1}}}{{r_b^2}}$

${r_b} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}mke{e^\prime }}}$

Now for ground state of ${\text{H}},{\text{n}} = 1$ and${e^\prime } = \dfrac{{er_A^3}}{{{R^3}}}$, then

$\therefore {r_b} = \dfrac{{{h^2}}}{{4{\pi ^2}mK \cdot e \cdot e \cdot \dfrac{{r_b^3}}{{{R^3}}}}} = \left( {\dfrac{{{h^2}}}{{4{\pi ^2}mK{e^2}}}} \right) \times \dfrac{{{R^3}}}{{r_b^3}} = {r_A} \times \dfrac{{{R^3}}}{{r_b^3}}$

${r_A} = \dfrac{{{h^2}}}{{4{\pi ^2}mK{e^2}}} = 0.53A^\circ $.........Calculated in part (i)

${{r_b} = {r_A}{{\left( {\dfrac{R}{{{r_b}}}} \right)}^3}} \\ {r_b^4 = {r_A}{R^3} = 0.53\mathop A\limits^ \circ \times {{(10\mathop A\limits^ \circ)}^3}} \\ {r_b^4 = 0.53 \times 1000{{(\mathop A\limits^ \circ)}^4}\left( {\because {{\text{r}}_{\text{A}}} = 0.53\;{{\mathop A\limits^ \circ)}}} \right.} $

${{{\text{r}}_{\text{b}}} = {{\left[ {530\;{{\text{A}}^4}} \right]}^{1/4}}{{{\mathop A\limits^ \circ}}}}\\ {K.E. = \dfrac{1}{2}m{v^2} = \dfrac{m}{2} \cdot \dfrac{{{h^2}}}{{4{\pi ^2}{m^2}r_b^2}} = \dfrac{{{h^2}}}{{8{\pi ^2}mr_b^2}}\left[ {\because v = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m^2}r_b^2}}} \right]} $

${ = \dfrac{{6.62 \times 6.62 \times {{10}^{ - 34}} \times {{10}^{ - 34}}}}{{8 \times 3.14 \times 3.14 \times 9.1 \times {{10}^{ - 31}} \times 4.8 \times 4.8 \times {{10}^{ - 20}} \times 1.6 \times {{10}^{ - 19}}}}{\text{eV}}} $

${ = \dfrac{{6.62 \times 6.62 \times {{10}^{ - 68 + 31 + 20 + 19}}}}{{8 \times 3.14 \times 3.14 \times 9.1 \times 4.8 \times 4.8 \times 1.6}}{\text{eV}}} \\ { = \dfrac{{43.8244}}{{26460.2}}{{10}^{ - 68 + 70}} = 0.001656 \times {{10}^2}{\text{eV}}}$

${{\text{K}}.{\text{E}} = 0.167{\text{eV}}} \\ {P.E. = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\dfrac{{\left( {r_b^2 - 3{R^2}} \right)}}{{{R^3}}}\left( {P.E = \dfrac{{{\text{K}}{{\text{q}}_1}{q_2}}}{r}} \right)} $

$P.E. = \left[ {\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r_A}}}} \right]\dfrac{{{r_A}\left( {r_b^2 - 3{R^2}} \right)}}{{{R^3}}}\left[ {{\text{ multiplying by }}\dfrac{{{r_A}}}{{{r_A}}}} \right]$

From part (i)

$P.E. = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}{r_A}}} = 27.2{\text{eV}}$

$P.E. = 272\left[ {\dfrac{{0.53\left( {\sqrt {530}  - 300} \right)}}{{1000}}} \right]$

Since we know that ${r_b} = {\left( {530} \right)^{\dfrac{1}{4}}}\mathop A\limits^ \circ  $ and $R = 10\mathop A\limits^ \circ  $

So, $P.E. = \dfrac{{272eV \times 0.53 \times \left( {23.02 - 300} \right)}}{{1000}} = \dfrac{{272 \times 0.53 \times \left( { - 276.9} \right)}}{{1000}}eV$

$P.E. =  - \dfrac{{3992.9}}{{1000}}eV =  - 3.99eV$

$K.E. = 0.167eV$

12.27 In the Auger process an atom makes a transition to a lower state without emitting a photon. The excess energy is transferred to an outer electron which may be ejected by the atom. (This is called an Auger electron). Assuming the nucleus to be massive, calculate the kinetic energy of an ${{n}} = {{4}}$ Auger electron emitted by Chromium by absorbing the energy from a ${\text{n}} = 2$ to ${{n}} = {{1}}$ transition.

Ans: The energy \[{E_n}\] of the nth state

${{E_n} =  + {Z^2}R\left[ {\dfrac{1}{{n_1^2}} - \dfrac{1}{{n_2^2}}} \right] = {Z^2}R\left( {\dfrac{1}{1} - \dfrac{1}{4}} \right)\left( {{\text{ for }}{{\text{n}}_1} = 1,{{\text{n}}_2} = 2} \right)} \\ {{\text{Z}} = 24} $

${{\text{R}} = {\text{ Rydberg constant}}{\text{. }}} \\ {\therefore {E_n} = \dfrac{3}{4}{Z^2}R} $

The energy required to eject an electron from ${\text{n}} = 4$ state is

${E_4} = {Z^2}R\dfrac{1}{{{4^2}}} = \dfrac{1}{{16}}{Z^2}R$

Energy given to electrons is converted into K.E. of ejected electrons.

Hence, the K.E. of Auger (ejected) electron= En-E4

${K.E. = {Z^2}R\dfrac{3}{4} - \dfrac{1}{{16}}{Z^2}R = \dfrac{{11}}{{16}}{Z^2}R = \dfrac{{11}}{{16}} \times 24 \times 24 \times 13.6{\text{ eV}}} \\ {{\text{ K}}{\text{.E}}{\text{. }} = 11 \times 36 \times 13.6 = 5385.6{\text{ eV}}} $

12.28 The inverse square law in electrostatics is $|F| = \dfrac{{{e^2}}}{{\left( {4\pi {\varepsilon _0}} \right){r^2}}}$ for the force between an electron and a proton. The $1/{{r}}$ dependence of $|{{F}}|$ can be understood in quantum theory as being due to the fact that the 'particle' of light (photon) is massless. If photons had a mass${m_p}$, force would be modified to $\left[ {F = \dfrac{{{e^2}}}{{\left( {4\pi {\varepsilon _0}} \right){r^2}}}\left[ {\dfrac{1}{{{r^2}}} + \dfrac{\lambda }{r}} \right]{e^{ - {\lambda _r}}}} \right]$ where $\lambda  = \dfrac{{{m_p}c}}{\hbar }$ and $\hbar  = \dfrac{h}{{2\pi }}$

Estimate the change in the ground state energy of a H-atom if mp were 10=6 times the mass of an electron.

Ans: Mass of photon$ = 9.1 \times {10^{ - 31}} \times {10^{ - 6}}\;{\text{ kg}} = 9.1 \times {10^{ - 37}}\;{\text{ kg}}$

Wavelength associated with a photon $ = \dfrac{h}{{{m_p}c}}$

${\lambda  = \dfrac{{6.62 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 37}} \times 3 \times {{10}^8}}}} = \dfrac{{6.62}}{{9.1 \times 3}} \times {{10}^{ - 34 + 37 - 8}} = 2.4 \times {{10}^{ - 7}} \gg {{\text{r}}_{\text{A}}}({\text{ see Q}}.26) \\ {\lambda  \ll \dfrac{1}{{{r_A}}} < e{{\hat A}_A} <  < 1} $

$U(r) = \dfrac{{ - {e^2}{e^{ - {\lambda _e}}}}}{{\pi {\varepsilon _0}r}}  \\ mvr = \dfrac{h}{{2\pi }} = {\hbar _{0r}}v = \dfrac{\hbar }{{mr}} \ldots ({\text{I}})  \\ $

${\dfrac{{m{v^2}}}{v}\dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{1}{{{r^2}}} + \dfrac{\lambda }{r}} \right]\left[ {\because F = \dfrac{{{e^2}}}{{\left( {4\pi {\varepsilon _0}} \right){r^2}}}\left[ {\dfrac{1}{{{r^2}}} + \dfrac{\lambda }{r}} \right]{e^{ - ir}}{\text{ given }}} \right]} $

${\dfrac{m}{r} \cdot \dfrac{{{\hbar ^2}}}{{{m^2}{r^2}}} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}\left[ {\dfrac{{1 + \lambda r}}{{{r^2}}}} \right]} \\ {\dfrac{{{\hbar ^2}}}{{mv}} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}(1 + \lambda r)} $

${\dfrac{{{\hbar ^2}4\pi {\varepsilon _0}}}{{m{e^2}}} = \left( {r + \lambda {r^2}} \right)} $

${{\text{ If }}\lambda  = 0,\dfrac{{{\hbar ^2}4\pi {\varepsilon _0}}}{{{m_0}}}[{\text{neglecting }}{{\text{r}}^{\text{2}}}]} \\ {\dfrac{\hbar }{m} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}}}{r_A}\left( {{r_A} = r + \lambda {r^2}} \right)} $

$\because {\lambda _A} \gg {{\text{r}}_{\text{B}}}$ And $\delta $

Taking $\left. {{{\text{r}}_{\text{A}}} + {\text{r}} + \lambda {{\text{r}}^2}[\because \delta )} \right]$

$\left. {{r_{\text{A}}} = \left( {{r_{\text{A}}}\delta } \right) + {\text{M}}\left( {{{\text{r}}_{\text{A}}}\delta } \right){\text{ (put }}{\mathcal{S}^2}} \right)$

$0 = \delta  + \lambda r_A^2 + 2{r_\lambda }\delta \lambda $ (Neglecting small term ${\delta ^2}$ )

${0 = \delta  + 2{r_A}\delta \lambda  + \lambda r_A^2} \\ { \Rightarrow \delta \left[ {1 + 2{r_A}\lambda } \right] =  - \lambda r_A^2} $

${\delta  = \dfrac{{ - \lambda r_A^2}}{{\left( {1 + 2{r_A}\lambda } \right)}} =  - \lambda r_A^2{{\left( {1 + 2{r_A}\lambda } \right)}^{ - 1}}} \\ {\delta  =  - \lambda r_A^2\left[ {1 - 2{r_A}\lambda } \right] =  - \lambda r_A^2 + 2r_A^3{\lambda ^2}} $

$\therefore \lambda $ And ${{\text{r}}_{\text{A}}} \ll 1$ so $r_A^3{\lambda ^2}$ is very small so by neglecting it we get,

$\delta  =  - \lambda r_A^2$

${V(r) = \dfrac{{ - {e^2}}}{{4\pi {\varepsilon _0}}} = \dfrac{{{e^{\left( { - 2\delta  - \lambda {r_A}} \right)}}}}{{\left( {{r_A} + \delta } \right)}} = \dfrac{{ - {e^2}}}{{4\pi {\varepsilon _0}}} \cdot \dfrac{{{e^{ - 2(\delta  + r)}}}}{{{r_A}\left( {1 + \dfrac{\delta }{{{r_A}}}} \right)}}} \\ { = \dfrac{{ - {e^2}}}{{4\pi {\varepsilon _0}{r_A}}}{e^{ - \lambda r}}{{\left( {1 + \dfrac{\delta }{{{r_A}}}} \right)}^{ - 1}} = \dfrac{{ - {e^2}{e^{ - \lambda r}}}}{{4\pi r{\varepsilon _0}{r_A}}}\left( {1 - \dfrac{\delta }{{{r_A}}}} \right)} $

${\text{V}}({\text{r}}) =  - 27.2{\text{eV}}$ Remains unchanged

${K.E. = \dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{{\left( {\dfrac{\hbar }{{mr}}} \right)}^2} = \dfrac{1}{2}\dfrac{{{\hbar ^2}}}{{m{r^2}}}\left( {{\text{ From }}I,v = \dfrac{\hbar }{{mr}}} \right)} \\ { = \dfrac{{ - {\hbar ^2}}}{{2m{{\left( {{r_A} + \delta } \right)}^2}}} = \dfrac{{ - {\hbar ^2}}}{{2mr_A^2{{\left( {1 + \dfrac{\delta }{{{r_A}}}} \right)}^2}}} = \dfrac{\hbar }{{2mr_A^2}}\left( {1 - \dfrac{{2\delta }}{{{r_A}}}} \right)} $

${ = \dfrac{{{\hbar ^2}}}{{2m{r_A}}}\left( {1 - \dfrac{{ - \lambda r_A^2}}{{{r_A}}}} \right) = \dfrac{{{\hbar ^2}}}{{2m{r_A}}}\left( {1 + 2\lambda {r_A}} \right)} \\ { = 13.6{\text{eV}}\left( {1 + 2\lambda {r_{\text{A}}}} \right)} $

Total energy $ = \dfrac{{ - {e^2}}}{{4\pi {\varepsilon _0}{r_A}}} + \dfrac{{{\hbar ^2}}}{{2mr_A^2}}\left( {1 + 2\lambda {r_A}} \right)$

$ = \left[ { - 27.2 + 13.6\left( {1 + 2\lambda {{\text{r}}_{\text{A}}}} \right)} \right]{\text{eV}}$

$ =  - 27.2 + 13.6 + 27.2\lambda {{\text{r}}_{\text{A}}}{\text{eV}}$

Total ${\text{E}} =  - 13.6 + 27.2\lambda {{\text{r}}_{\text{A}}}$

Change in energy $ =  - 13.6 + 27.2\lambda {{\text{r}}_{\text{A}}} - ( - 13.6) = 27.2\lambda {{\text{r}}_{\text{A}}}{\text{ eV}}$

12.29 The Bohr model for the H-atom relies on Coulomb's law of electrostatics. Coulomb's law has not directly been verified for very short distances of the order of angstroms. Supposing Coulomb's law between two opposite’s charges $ + {{q}}1, - {{q2}}$ is modified to

${|F| = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}\left( {r \geqslant {R_0}} \right)} \\ {|F| = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}R_0^2}}{{\left[ {\dfrac{{{R_0}}}{r}} \right]}^z}\left( {r \leqslant {R_0}} \right)} $

Calculate in such a case, the ground state energy of a ${{H}} - {{a}}$ tom, if $\varepsilon  = 0.1,{{R}}0 = 1$ \[\mathop A\limits^ \circ  \].

Ans: $Case\,I:$ when $r \leqslant {R_0} = 1\mathop A\limits^ \circ$

Let $\varepsilon  = 2 + \delta $

$F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}R_0^2}}\dfrac{{R_0^{r + \delta }}}{{{r^{2 + \delta }}}} = \left( {\dfrac{{e \cdot ( - e)}}{{4\pi {\varepsilon _0}}}} \right)\dfrac{{{R^\delta }}}{{{r^{2 + \delta }}}}\left[ {{{\text{q}}_1} = {\text{e}},{{\text{q}}_2} =  - {\text{e}}} \right]$

$|F| =  - {\left( {1.6 \times {{10}^{ - 18}}} \right)^2} \times 9 \times {10^9}\dfrac{{{R^\delta }}}{{{r^{2 + \bar b}}}}[( - )$ [(-)ve sign shows force of attraction ]

$|F| = 23.04 \times {10^{ - 29}}\dfrac{{{R^\delta }}}{{{r^{2 + \delta }}}}$

The electrostatic force of attraction between positively charged nucleus and negatively charged electron provides necessary centripetal force.

$\dfrac{{m{v^2}}}{r} = \left( {23.04 \times {{10}^{ - 29}}} \right)\dfrac{{R_0^\delta }}{{{r^{2 + \delta }}}}$

Let $23.04 \times {10^{ - 29}} = \Lambda $

$\therefore {v^2} = \Lambda \dfrac{{R_0^\delta }}{{{r^{2 + \delta }}}} \cdot \dfrac{r}{m} = \dfrac{{\Lambda R_0^\delta }}{{m{r^{1 + \delta }}}}$

By Bohr's IInd postulate, angular momentum

${L = \dfrac{{nh}}{{2\pi }}{\text{ and }}\hbar  = \dfrac{h}{{2\pi }}} \\ {\therefore {\text{L}} = {\text{nh}}} \\ {{\text{mvr}} = {\text{nh}}} $

$r = \dfrac{{n\hbar }}{{mv}} = \dfrac{{n\hbar }}{m}\sqrt {\dfrac{{m{r^{1 + \delta }}}}{{\Lambda {R_0}^\delta }}}  = \dfrac{{n\hbar }}{m}\sqrt {\dfrac{m}{{\Lambda {R_0}^\delta }}}  \times \left( {{r^{\dfrac{{1 + \delta }}{2}}}} \right)$

${r^{1 - \dfrac{1}{2} - \dfrac{\delta }{2}}} = \dfrac{{n\hbar }}{m}{\left[ {\dfrac{m}{{\Lambda R_0^\delta }}} \right]^{1/2}}$

${r^{\dfrac{1}{2}(1 - \tilde \theta )}} = \dfrac{{n\hbar }}{m}{\left[ {\dfrac{m}{{\Lambda R_0^\delta }}} \right]^{ - 1/2}}$

${\gamma _n} = {\left[ {\dfrac{{{n^2}{h^2}}}{{m\Lambda R_0^\delta }}} \right]^{\dfrac{1}{{1 - \delta }}}}$

${ = {{\left[ {\dfrac{{{\hbar ^2}}}{{m\Lambda R_0^\delta }}} \right]}^{\dfrac{1}{{1 - \delta }}}}{\text{ (n = 1 for ground state) }}} \\ {{r_1} = {{\left[ {\dfrac{{{{\left( {1.05 \times {{10}^{ - 34}}} \right)}^2}}}{{9.1 \times {{10}^{ - 31}} \times 23.04 \times {{10}^{ - 29}} \times {{10}^{19}}}}} \right]}^{\dfrac{1}{{29}}}}} \\ {\left[ {\because \hbar  = \dfrac{h}{{2\pi }} = \dfrac{{6.6 \times {{10}^{ - 34}}}}{{2 \times 3.14}} = \left( {1.05 \times {{10}^{ - 34}}} \right)} \right]} $

Where ${\text{h}} = 1.05 \times {10^{ - 34}}{\text{J}}{{\text{S}}^{ - 1}}$

${R_0} = {10^{19}}{\text{ and }}1 - \delta  = 2.9$

${{\text{R}}_1} = 8 \times {10^{ - 11}}\;{\text{m}} = 0.08\;{\text{nm}}$

This is a radius of orbit of electron in ground state of hydrogen atom

Velocity of electron in ground state By IInd postulate of Bohr’s Atomic model

$m{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}\left( {\hbar  = \dfrac{h}{{2\pi }}} \right)$

${v_n} = \dfrac{{n\hbar }}{{m{r_n}}} = \dfrac{{n\hbar }}{m}{\left[ {\dfrac{{m\Lambda R_0^b}}{{{n^2}{\hbar ^2}}}} \right]^{\dfrac{1}{{1 - \delta }}}}$

${{\text{ Case II: n}} = 1,{v_1} = \dfrac{\hbar }{{m{r_1}}} = \dfrac{{1.05 \times {{10}^{ - 34}}}}{{9.1 \times {{10}^{ - 31}} \times 0.08 \times {{10}^{ - 9}}}} = \dfrac{{1.05}}{{0.728}} \times {{10}^{ - 34 + 31 + 9}}} \\ { = 1.44 \times {{10}^{ - 34 + 40}} = 1.44 \times {{10}^6}\;{\text{m}}/{\text{s}}} \\ {{v_1} = 1.44 \times {{10}^6}\;{\text{m}}/{\text{s}}} $

${KE = \dfrac{1}{2}mv_1^2 - 9.43 \times {{10}^{ - 19}}} \\ { = \dfrac{1}{2} \times 9.1 \times {{10}^{ - 31}} \times {{\left( {1.44 \times {{10}^5}} \right)}^2} - 9.43 \times {{10}^{ - 19}}\;{\text{J}}} \\ { = 5.9{\text{eV}}} $

P.E. from ${{\text{R}}_0}$ to $\dfrac{{ - \Lambda }}{{{R_0}}}$

P.E from ${{\text{R}}_0}$ to $r =  + \Lambda R_0^\theta \int_{{R_0}}^r {\dfrac{{dr}}{{{r^{2 + \delta }}}}} $

This is the P.E. of electron in ground state ${{\text{R}}_0}$ to $r = \dfrac{{ - \Lambda }}{{{R_0}}}$

${P.E. =  - \dfrac{{\Lambda R_0^\delta }}{{(1 + \delta )}}\left[ {\dfrac{1}{{{r^{1 + \delta }}}} - \dfrac{1}{{R_0^{1 + \dot \delta }}}} \right]} \\ { = \dfrac{{ - \Lambda }}{{(1 + \delta )}}\left[ {\dfrac{{R_0^\delta }}{{{r^{1 + \dot \theta }}}} - \dfrac{1}{{{R_0}}}} \right] = \dfrac{{ - \Lambda }}{{(1 + \delta )}}\left[ {\dfrac{{R_0^\delta }}{{{r^{1 + \dot \theta }}}} - \dfrac{1}{{{R_0}}} + \dfrac{{1 + \delta }}{{{R_0}}}} \right]} $

Put $\delta  =  - 1.9$

${P.E. = \dfrac{{ - \Lambda }}{{(1 + ( - 1.9)}}\left[ {\dfrac{{R_0^{ - 1.9}}}{{{r^{ - 0.9}}}} - \dfrac{1}{{{R_0}}} - \dfrac{{1.9}}{{{R_0}}}} \right]} \\ { = \dfrac{{ - \Lambda }}{{ - 0.9}}\left[ {\dfrac{{R_0^{ - 19}}}{{{r^{ - 0.9}}}} - \dfrac{{1.9}}{{{R_0}}}} \right] = \dfrac{{23.04 \times {{10}^{ - 29}}}}{{0.9}}\left[ {{{(0.8)}^{0.9}} - 1.9} \right]J} \\ { = 17.3{\text{ eV}}} \\ {{\text{ Total E =  P}}{\text{.E}}{\text{.  +  K}}{\text{.E}}{\text{. }}} \\ { = ( - 17.3 + 5.9){\text{ eV}}} $

Total energy = -11.4 eV 

This is the required total energy of an electron in the ground state of H-atom.

Key Topics for NCERT Physics Exemplar Solutions Class 12 Chapter 12 Atoms (Book Solutions)

Let's look at some of the key topics and explanations related to this Chapter:

Atoms – The smallest unit of a chemical element is known as an Atom. The main components of an Atom consist of three tiny particles: protons, electrons, and neutrons, etc. Protons are positively particles, electrons are poorly charged and neutrons have no electrical charges at all.

Electron orbits - Electrons pick up shells around the nucleus of an Atom. He can move in any direction but a limited space or routes. Each orbit has a name. Each shell may contain a fixed number of electrons. The first shell contains 2 electrons. The second shell contains 8 electrons and more. The standard formula of the nth shell is 2 electrons (n2).

Rutherford's Atomic Model - Rutherford's Atomic Model is also known as the nuclear Model. Why a nuclear Model? Because this was the first Atomic Model in which the nucleus is considered to be its main component. This Model states that Atoms are mainly located in space, electrons orbiting the nucleus with positive orbits.

Bohr Model of a Hydrogen Atom - Bohr Model of Hydrogen Atom states that there is a positively charged nucleus comprising (protons, neutrons) and negatively charged electrons orbiting the nucleus at different energy levels. These levels are called shells. He assumes that electrons revolve around different circuits and that orbits with a specific radio are allowed. This produces an absorption spectrum.

Why use NCERT Physics Exemplar Solutions Class 12 Chapter 12: Atoms by Vedantu?

Solving NCERT Solutions Grade 12 Physics Model Chapter 12: Atoms via Vedantu can be a great resource for students. Experts create solutions in a way that makes you feel comfortable while solving these tricky NCERT Exemplar issues. You do not have to rush through your syllabus as these solutions are free. These Vedantu solutions also help you with time management skills. Regular practice of this NCERT Problem Exemplar Case created by Vedantu will boost your confidence and help you protect a good level from competing tests.

Important Questions

• When the alpha particle scattering test is re-used a thin sheet of solid Hydrogen instead of gold foil. (Hydrogen is stable at temperatures below 14K.) Get results?

• Name the shortest wavelength in the Paschen series of spectral lines?

• When an Atom makes a transition from a high to a low level when the 5 eV difference divides the two levels of Atomic energy, how much radiation is emitted?