NCERT Exercise Q & A

NCERT Exercise Problems and Solution

Q12.1: 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 \(\ldots\) the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) In the ground state of \(\ldots\) electrons are in stable equilibrium, while in \(\ldots \ldots \ldots\). electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on \(\ldots \ldots \ldots\) is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a \(\ldots\) 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 \(\ldots\) (Rutherford’s model/both the models.)

Answer: (a) The sizes of the atoms taken in Thomson’s model and Rutherford’s model have the same order of magnitude.
(b) In the ground state of Thomson’s model, the electrons are in stable equilibrium. However, in Rutherford’s model, the 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 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.

Q12.2: 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?

Answer: In the alpha-particle scattering experiment, if a thin sheet of solid hydrogen is used in place of a gold foil, then the scattering angle would not be large enough. This is because the mass of hydrogen \(\left(1.67 \times 10^{-27} \mathrm{~kg}\right)\) is less than the mass of incident \(\alpha\)-particles \(\left(6.64 \times 10^{-27} \mathrm{~kg}\right)\). Thus, the mass of the scattering particle is more than the target nucleus (hydrogen). As a result, the a-particles would not bounce back if solid hydrogen is used in the \(\alpha\)-particle scattering experiment.

Q12.3: A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Answer: Rydberg’s formula is given as:
\(
\frac{h c}{\lambda}=21.76 \times 10^{-19}\left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]
\)
Where, \(h=\) Planck’s constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
\(c=\) Speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
( \(n_1\) and \(n_2\) are integers)
The shortest wavelength present in the Paschen series of the spectral lines is given for values \(n_1=3\) and \(n_2=\infty\).
\(
\begin{aligned}
\frac{h c}{\lambda} & =21.76 \times 10^{-19}\left[\frac{1}{(3)^2}-\frac{1}{(\infty)^2}\right] \\
\lambda & =\frac{6.6 \times 10^{-34} \times 3 \times 10^8 \times 9}{21.76 \times 10^{-19}} \\
& =8.189 \times 10^{-7} \mathrm{~m} \\
& =818.9 \mathrm{~nm}
\end{aligned}
\)

Q12.4: The ground state energy of hydrogen atom is -13.6 eV. What are the kinetic and potential energies of the electron in this state?

Answer: Ground state energy of hydrogen atom, \(E=-13.6 \mathrm{eV}\)
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of the total energy.
Kinetic energy \(=-E=-(-13.6)=13.6 \mathrm{eV}\)
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy \(=-2 \times(13.6)=-27.2 \mathrm{eV}\)

Q12.5: 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.

Answer: For ground level, \(n_1=1\)
Let \(E_1\) be the energy of this level. It is known that \(E_1\) is related with \(n_1\) as:
\(
\begin{aligned}
E_1 & =\frac{-13.6}{n_1^2} \mathrm{eV} \\
& =\frac{-13.6}{1^2}=-13.6 \mathrm{eV}
\end{aligned}
\)
The atom is excited to a higher level, \(n_2=4\).
Let \(E_2\) be the energy of this level.
\(
\begin{aligned}
\therefore E_2 & =\frac{-13.6}{n_2^2} \mathrm{eV} \\
& =\frac{-13.6}{4^2}=-\frac{13.6}{16} \mathrm{eV}
\end{aligned}
\)
The amount of energy absorbed by the photon is given as:
\(
\begin{aligned}
E & =E_2-E_1 \\
& =\frac{-13.6}{16}-\left(-\frac{13.6}{1}\right) \\
& =\frac{13.6 \times 15}{16} \mathrm{eV} \\
& =\frac{13.6 \times 15}{16} \times 1.6 \times 10^{-19}=2.04 \times 10^{-18} \mathrm{~J}
\end{aligned}
\)
For a photon of wavelength \(\lambda\), the expression of energy is written as:
\(
E=\frac{h c}{\lambda}
\)
Where, \(h=\) Planck’s constant \(=6.6 \times 10^{-34} \mathrm{Js}\)
\(c=\) Speed of light \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
\(
\begin{aligned}
\therefore \lambda & =\frac{h c}{E} \\
& =\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{2.04 \times 10^{-18}} \\
& =9.7 \times 10^{-8} \mathrm{~m}=97 \mathrm{~nm}
\end{aligned}
\)
And, frequency of a photon is given by the relation,
\(
\begin{aligned}
\nu & =\frac{c}{\lambda} \\
& =\frac{3 \times 10^8}{9.7 \times 10^{-8}} \approx 3.1 \times 10^{15} \mathrm{~Hz}
\end{aligned}
\)
Hence, the wavelength of the photon is 97 nm while the frequency is \(3.1 \times 10^{15} \mathrm{~Hz}\).

Q12.6: (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.

Answer: (a) Let \(v_1\) be the orbital speed of the electron in a hydrogen atom in the ground state level, \(n_1=1\). For charge (e) of an electron, \(v_1\) is given by the relation,
\(
v_1=\frac{e^2}{n_1 4 \pi \epsilon_0(h / 2 \pi)}=\frac{e^2}{2 \epsilon_0 h}
\)
Where,
\(
e=1.6 \times 10^{-19} \mathrm{C}
\)
\(
\epsilon_0=\text { Permittivity of free space }=8.85 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2}
\)
\(
\begin{aligned}
& h=\text { Planck’s constant }=6.62 \times 10^{-34} \mathrm{Js} \\
& \begin{aligned}
\therefore v_1 & =\frac{\left(1.6 \times 10^{-19}\right)^2}{2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}} \\
& =0.0218 \times 10^8=2.18 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)
For level \(n_2=2\), we can write the relation for the corresponding orbital speed as:
\(
\begin{aligned}
v_2 & =\frac{e^2}{n_2 2 \epsilon_0 h} \\
& =\frac{\left(1.6 \times 10^{-19}\right)^2}{2 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}} \\
& =1.09 \times 10^6 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
And, for \(n_3=3\), we can write the relation for the corresponding orbital speed as:
\(
\begin{aligned}
v_3 & =\frac{e^2}{n_3 2 \epsilon_0 h} \\
& =\frac{\left(1.6 \times 10^{-19}\right)^2}{3 \times 2 \times 8.85 \times 10^{-12} \times 6.62 \times 10^{-34}} \\
& =7.27 \times 10^5 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Hence, the speed of the electron in a hydrogen atom in \(n=1, n=2\), and \(n=3\) is \(2.18 \times\) \(10^6 \mathrm{~m} / \mathrm{s}, 1.09 \times 10^6 \mathrm{~m} / \mathrm{s}, 7.27 \times 10^5 \mathrm{~m} / \mathrm{s}\) respectively.
(b) Let \(T_1\) be the orbital period of the electron when it is in level \(n_1=1\).
The orbital period is related to orbital speed as:
\(
T_1=\frac{2 \pi r_1}{v_1}
\)
Where,
\(r_1=\) Radius of the orbit
\(
=\frac{n_1^2 h^2 \epsilon_0}{\pi m e^2}
\)
\(h=\) Planck’s constant \(=6.62 \times 10^{-34} \mathrm{Js}\)
\(e=\) Charge on an electron \(=1.6 \times 10^{-19} \mathrm{C}\)
\(\epsilon_0=\) Permittivity of free space \(=8.85 \times 10^{-12} \mathrm{~N}^{-1} \mathrm{C}^2 \mathrm{~m}^{-2}\)
\(
\begin{aligned}
&\begin{aligned}
m & =\text { Mass of an electron }=9.1 \times 10^{-31} \mathrm{~kg} \\
\therefore T_1 & =\frac{2 \pi r_{\mathrm{i}}}{v_1} \\
& =\frac{2 \pi \times(1)^2 \times\left(6.62 \times 10^{-34}\right)^2 \times 8.85 \times 10^{-12}}{2.18 \times 10^6 \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^2} \\
& =15.27 \times 10^{-17}=1.527 \times 10^{-16} \mathrm{~s}
\end{aligned}\\
&\text { For level } n_2=2 \text {, we can write the period as: }
\end{aligned}
\)
\(
T_2=\frac{2 \pi r_2}{v_2}
\)
Where,
\(r_2=\) Radius of the electron in \(n_2=2\)
\(
=\frac{\left(n_2\right)^2 h^2 \epsilon_0}{\pi m e^2}
\)
\(
\therefore T_2=\frac{2 \pi r_2}{v_2}
\)
\(
=\frac{2 \pi \times(2)^2 \times\left(6.62 \times 10^{-34}\right)^2 \times 8.85 \times 10^{-12}}{1.09 \times 10^6 \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^2}
\)
\(
=1.22 \times 10^{-15} \mathrm{~s}
\)
And, for level \(n_3=3\), we can write the period as:
\(
T_3=\frac{2 \pi r_3}{v_3}
\)
Where, \(r_3=\) Radius of the electron in \(n_3=3\)
\(
=\frac{\left(n_3\right)^2 h^2 \epsilon_0}{\pi m e^2}
\)
\(
\begin{aligned}
\therefore T_3 & =\frac{2 \pi r_3}{v_3} \\
& =\frac{2 \pi \times(3)^2 \times\left(6.62 \times 10^{-34}\right)^2 \times 8.85 \times 10^{-12}}{7.27 \times 10^5 \times \pi \times 9.1 \times 10^{-31} \times\left(1.6 \times 10^{-19}\right)^2} \\
& =4.12 \times 10^{-15} \mathrm{~s}
\end{aligned}
\)
Hence, the orbital period in each of these levels is \(1.52 \times 10^{-16} \mathrm{~s}, 1.22 \times 10^{-15} \mathrm{~s}\), and \(4.12 \times 10^{-15} \mathrm{~s}\) respectively.

Q12.7: The radius of the innermost electron orbit of a hydrogen atom is \(5.3 \times 10^{-11} \mathrm{~m}\). What are the radii of the \(n=2\) and \(n=3\) orbits?

Answer: The radius of the innermost orbit of a hydrogen atom, \(r_1=5.3 \times 10^{-11} \mathrm{~m}\).
Let \(r_2\) be the radius of the orbit at \(n=2\). It is related to the radius of the innermost orbit as:
\(
\begin{aligned}
r_2 & =(n)^2 r_1 \\
& =4 \times 5.3 \times 10^{-11}=2.12 \times 10^{-10} \mathrm{~m}
\end{aligned}
\)
For \(n=3\), we can write the corresponding electron radius as:
\(
\begin{aligned}
r_3 & =(n)^2 r_1 \\
& =9 \times 5.3 \times 10^{-11}=4.77 \times 10^{-10} \mathrm{~m}
\end{aligned}
\)
Hence, the radii of an electron for \(n=2\) and \(n=3\) orbits are \(2.12 \times 10^{-10} \mathrm{~m}\) and \(4.77 \times\) \(10^{-10} \mathrm{~m}\) respectively.

Q12.8: A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer: It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV. Also, the energy of the gaseous hydrogen in its ground state at room temperature is -13.6 eV.
When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes \(-13.6+12.5 \mathrm{eV}\) i.e., -1.1 eV.
Orbital energy is related to orbit level ( \(n\) ) as:
\(
E=\frac{-13.6}{(n)^2} \mathrm{eV}
\)
For \(n=3\),
\(
E=\frac{-13.6}{9}=-1.5 \mathrm{eV}
\)
This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from \(n=1\) to \(n=3\) level.
During its de-excitation, the electrons can jump from \(n=3\) to \(n=1\) directly, which forms a line of the Lyman series of the hydrogen spectrum.
We have the relation for wave number for the Lyman series as:
\(
\frac{1}{\lambda}=R_y\left(\frac{1}{1^2}-\frac{1}{n^2}\right)
\)
Where,
\(
R_y=\text { Rydberg constant }=1.097 \times 10^7 \mathrm{~m}^{-1}
\)
\(\lambda=\) Wavelength of radiation emitted by the transition of the electron
For \(n=3\), we can obtain \(\lambda\) as:
\(
\begin{aligned}
\frac{1}{\lambda} & =1.097 \times 10^7\left(\frac{1}{1^2}-\frac{1}{3^2}\right) \\
& =1.097 \times 10^7\left(1-\frac{1}{9}\right)=1.097 \times 10^7 \times \frac{8}{9} \\
\lambda & =\frac{9}{8 \times 1.097 \times 10^7}=102.55 \mathrm{~nm}
\end{aligned}
\)
If the electron jumps from \(n=2\) to \(n=1\), then the wavelength of the radiation is given as:
\(
\begin{aligned}
\frac{1}{\lambda} & =1.097 \times 10^7\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \\
& =1.097 \times 10^7\left(1-\frac{1}{4}\right)=1.097 \times 10^7 \times \frac{3}{4} \\
\lambda & =\frac{4}{1.097 \times 10^7 \times 3}=121.54 \mathrm{~nm}
\end{aligned}
\)
If the transition takes place from \(n=3\) to \(n=2\), then the wavelength of the radiation is given as:
\(
\begin{aligned}
\frac{1}{\lambda} & =1.097 \times 10^7\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\
& =1.097 \times 10^7\left(\frac{1}{4}-\frac{1}{9}\right)=1.097 \times 10^7 \times \frac{5}{36} \\
\lambda & =\frac{36}{5 \times 1.097 \times 10^7}=656.33 \mathrm{~nm}
\end{aligned}
\)
This radiation corresponds to the Balmer series of the hydrogen spectrum.
Hence, in the Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

Q12.9: In accordance with the Bohr’s model, find the quantum number that characterises the earth’s revolution around the sun in an orbit of radius \(1.5 \times 10^{11} \mathrm{~m}\) with orbital speed \(3 \times 10^4 \mathrm{~m} / \mathrm{s}\). (Mass of earth \(=6.0 \times 10^{24} \mathrm{~kg}\).)

Answer: Radius of the orbit of the Earth around the Sun, \(r=1.5 \times 10^{11} \mathrm{~m}\)
Orbital speed of the Earth, \(v=3 \times 10^4 \mathrm{~m} / \mathrm{s}\)
Mass of the Earth, \(m=6.0 \times 10^{24} \mathrm{~kg}\)
According to Bohr’s model, angular momentum is quantized and given as:
\(
m v r=\frac{n h}{2 \pi}
\)
Where,
\(
\begin{aligned}
& h=\text { Planck’s constant }=6.62 \times 10^{-34} \mathrm{Js} \\
& n=\text { Quantum number } \\
& \begin{aligned}
\therefore n & =\frac{m v r 2 \pi}{h} \\
& =\frac{2 \pi \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.62 \times 10^{-34}} \\
& =25.61 \times 10^{73}=2.6 \times 10^{74}
\end{aligned}
\end{aligned}
\)
Hence, the quanta number that characterizes the Earth’s revolution is \(2.6 \times 10^{74}\).

Exemplar Section

VSA

Q12.14: The mass of a H -atom is less than the sum of the masses of a proton and electron. Why is this?

Answer: Since, the difference in mass of a nucleus and its constituents, \(\Delta M\), is called the mass defect and is given by
\(
\Delta M=\left[Z m_p+(A-Z) m_n\right]-M
\)
Also, the binding energy is given by \(B=\) mass defect \((\Delta M) \times c^2\)
Thus, the mass of an H -atom is \(m_p+m_e-\frac{B}{c^2}\), where \(B \approx 13.6 \mathrm{eV}\) is the binding energy.

Q12.15: Imagine removing one electron from \(\mathrm{He}^4\) and \(\mathrm{He}^3\). Their energy levels, as worked out on the basis of Bohr model will be very close. Explain why.

Answer: On removing one electron from \(H e^4\) and \(H e^3\), the energy levels, as worked out on the basis of Bohr model will be very close as both the nuclei are very heavy as compared to electron mass. Also after removing one electron from \(\mathrm{He}^4\) and \(\mathrm{He}^3\) atoms contain one electron and are hydrogen like atoms.

Q12.16: 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 cannot it be emitted as other forms of energy?

Answer: The electrons are charged particles. When an electron falls from a higher energy to a lower energy level, it accelerates. We know accelerating charged particle radiates energy in the form of electromagnetic radiation.

Q12.17: Would the Bohr formula for the H -atom remain unchanged if proton had a charge \((+4 / 3) e\) and electron a charge \((-3 / 4) e\), where \(e=1.6 \times 10^{-19} \mathrm{C}\). Give reasons for your answer.

Answer: According to Bohr, for an electron around a stationary nucleus the electrostatics force of attraction provides the necessary centripetal force.
i.e. \(\frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}=\frac{m v^2}{r} \dots(i)\)
Hence the magnitude of electrostatic force \(F \propto q_1 \times q_2\)
If proton had a charge \((+4 / 3)e\) and electron a charge \((-3 / 4)e\), then the Bohr formula for the H -atom remains same since the Bohr formula involves only the product of the charges which remain constant for given values of charges.

Q12.18: 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?

Answer: The energy of an electron in a hydrogen atom in the \(n\)th energy level is given by the formula:
\(
E_n=-\frac{13.6 \mathrm{eV}}{n^2}
\)
where \(n\) is the principal quantum number.
Different Energies:
If we have two hydrogen atoms with their electrons in excited states, we denote the principal quantum numbers for these states as \(n_1\) and \(n_2\). Since the energies are different, we can say:
\(
E_1 \neq E_2 \Longrightarrow n_1 \neq n_2
\)
This means that the two electrons must be in different energy levels.
Orbital Angular Momentum:
According to the Bohr model, the orbital angular momentum \(L_n\) of an electron in the \(n\)th orbit is given by:
\(
L_n=n \frac{h}{2 \pi} \quad L_n=m v_n r_n=n\left(\frac{h}{2 \pi}\right)
\)
where \(h\) is Planck’s constant.
Calculating Angular Momentum:
For the two hydrogen atoms, the angular momentum values are:
\(
\begin{aligned}
& L_1=n_1 \frac{h}{2 \pi} \\
& L_2=n_2 \frac{h}{2 \pi}
\end{aligned}
\)
Comparing Angular Momentum:
Since we established that \(n_1 \neq n_2\) (because their energies are different), it follows that:
\(
L_1 \neq L_2
\)
Therefore, the angular momentum values for the two electrons cannot be the same.
Conclusion: Based on the above analysis, it is not possible for the electrons in the two different hydrogen atoms to have different energies while having the same orbital angular momentum.

SA

Q12.19: Positronium is just like an 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?

Answer: Step 1: Understand the System
Positronium is similar to a hydrogen atom, but instead of a proton, it has a positron (the antiparticle of the electron). Both the electron and positron have the same mass, denoted as \(m_e\).
Step 2: Use the Formula for Energy Levels
The energy levels of a hydrogen-like atom can be expressed as:
\(
E_n=-\frac{m r^4}{8 \epsilon_0^2 n^2 h^2}
\)
where:
\(m\) is the reduced mass of the system,
\(\epsilon_0\) is the permittivity of free space,
\(-n\) is the principal quantum number,
\(\boldsymbol{h}\) is Planck’s constant.
Step 3: Calculate the Reduced Mass
The reduced mass \(\mu\) for two particles with equal mass \(m_e\) is given by:
\(
\mu=\frac{m_e \cdot m_e}{m_e+m_e}=\frac{m_e^2}{2 m_e}=\frac{m_e}{2}
\)
Step 4: Substitute the Reduced Mass into the Energy Formula
Now, substituting the reduced mass into the energy formula for the ground state (where \(n=1\) ):
\(
E_1=-\frac{\mu e^4}{8 \epsilon_0^2 h^2}
\)
Substituting \(\mu=\frac{m_e}{2}\) :
\(
E_1=-\frac{\frac{m_e}{2} e^4}{8 \epsilon_0^2 h^2}
\)
This simplifies to:
\(
E_1=-\frac{m_e e^4}{16 \epsilon_0^2 h^2}
\)
Step 5: Relate to the Hydrogen Atom Energy
For hydrogen, the ground state energy is:
\(
E_H=-13.6 \mathrm{eV}
\)
Since the only difference is the reduced mass, we can relate the energies:
\(
E_{P_s}=\frac{1}{2} E_H=-\frac{13.6 \mathrm{eV}}{2}=-6.8 \mathrm{eV}
\)
Thus, the ground state energy of positronium is:
\(
E_{P s}=-6.8 \mathrm{eV}
\)

Q12.20: 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.

Answer: Total energy (for Hydrogen and \(\mathbf{H}_2\) like Atoms): The total energy of the electron in the mth stationary states of the hydrogen. An atom of the hydrogenlike atom of atomic number \(Z\) is given by
\(
\begin{aligned}
& E=-\left(\frac{m e^4}{8 \varepsilon_0^2 h^2}\right) \frac{Z^2}{n^2}=-\left(\frac{m e^4}{8 \varepsilon_0^2 c h^3}\right) \operatorname{ch} \frac{Z^2}{n^2} \\
& =-R \operatorname{ch} \frac{Z^2}{n^2} \\
& =-13.6 \frac{Z^2}{n^2} \mathrm{eV}
\end{aligned}
\)
For a He-nucleus \({Z}=2\), and for ground state \({n}=1\).
Thus, the ground state energy of a He -atom.
\(
\begin{aligned}
& E_n=-13.6 \frac{Z^2}{n^2} \mathrm{eV} \\
& =-13.6 \frac{2^2}{1^2} \mathrm{eV} \\
& =-54.4 \mathrm{eV}
\end{aligned}
\)
Thus, the ground state will have two electrons each of energy E and the total ground state energy would be \(-(4 \times\) 13.6) \(\mathrm{eV}=-54.4 \mathrm{eV}\)

Q12.21: Using Bohr model, calculate the electric current created by the electron when the H-atom is in the ground state.

Answer: Key concept: The equivalent electric current due to rotation of charge is given by \(i=\frac{Q}{T}=Q\left(\frac{1}{T}\right)=Q \times f\), where \(f\) is the frequency.

In Hydrogen atom an electron in the ground state revolves on a circular orbit whose radius is equal to the Bohr radius \(\left(a_0\right)\). Let the velocity of the electron is \(v\).
\(\therefore\) Number of revolutions per unit time \(f=\frac{2 \pi a_0}{v}\)
The electric current is given by \(i=Q \cdot f\), if \(Q\) charge flows in time \(T\). Here, \(Q=e\).
The electric current is given by \(i=e\left(\frac{2 \pi a_0}{v}\right)=\frac{2 \pi a_0}{v} e\).

Q12.22: Show that the first few frequencies of light that is emitted when electrons fall to the \(n^{\text {th }}\) level from levels higher than \(n\) are approximate harmonics (i.e. in the ratio \(1: 2: 3 \ldots\) ) when \(n \gg 1\).

Answer: According to Bohr’s formula, the frequency of electromagnetic radiation emitted when an electron falls from \(p^{\text {th }}\) level to the \(n^{\text {th }}\) level is given by
\(
v_{\operatorname{mn}}=c R Z^2\left[\frac{1}{(n+p)^2}-\frac{1}{n^2}\right],
\)
where \(m=n+p,(p=1,2,3, \ldots)\) and \(R\) is Rydberg constant.
For \(p \ll n\).
\(
\begin{aligned}
&v_{m n}=c R Z^2\left[\frac{1}{n^2}\left(1+\frac{p}{n}\right)^{-2}-\frac{1}{n^2}\right]\\
&v_{m n}=c R Z^2\left[\frac{1}{n^2}-\frac{2 p}{n^3}-\frac{1}{n^2}\right]\\
&v_{\pi n}=c R Z^2 \frac{2 p}{n^3} \simeq\left(\frac{2 c R Z^2}{n^3}\right) p
\end{aligned}
\)
Thus, \(v_{\mathrm{mn}}\) are approximately in the order \(1,2,3 \ldots \ldots \ldots\).

Q12.23: What is the minimum energy that must be given to an H atom in ground state so that it can emit an \(H_\gamma\) line in Balmer series. If the angular momentum of the system is conserved, what would be the angular momentum of such \(H_\gamma\) photon?

Answer: \(H_\gamma\) in Balmer series corresponds to transition \(n=5\) to \(n=2\). So the electron in ground state \(n=1\) must first be put in state \(n=5\). Energy required \(=E_1-E_5=13.6-0.54=13.06 \mathrm{eV}\).
If angular momentum is conserved, the angular momentum of photon \(=\) change in angular momentum of the electron
\(
\begin{aligned}
& =L_5-L_2=5 h-2 h=3 h=3 \times 1.06 \times 10^{-34} \\
& =3.18 \times 10^{-34} \mathrm{~kg} \mathrm{~m}^2 / \mathrm{s}
\end{aligned}
\)

LA

Q12.24: The first four spectral lines in the Lyman serics of an H -atom are \(\lambda=1218 Å, 1028 Å, 974.3 Å\) and \(951.4 Å\). If instead of Hydrogen, we consider Deuterium, calculate the shift in the wavelength of these lines.

Answer: Let \(\mu_H\) and \(\mu_{\mathrm{D}}\) are the reduced masses of electron for hydrogen and deuterium respectively.
We know that \(\frac{1}{\lambda}=R\left[\frac{1}{n_f^2}-\frac{1}{n_i^2}\right]\)
As \(n_i\) and \(n_f\) are fixed for by mass series for hydrogen and deuterium.
\(
\lambda \propto \frac{1}{R} \text { or } \frac{\lambda_D}{\lambda_H}=\frac{R_H}{R_D} \dots(i)
\)
\(
\begin{aligned}
& R_R=\frac{m_e e^4}{8 \varepsilon_0 c h^3}=\frac{\mu_H e^4}{8 \varepsilon_0 c h^3} \\
& R_D=\frac{m_e e^4}{8 \varepsilon_0 c h^3}=\frac{\mu_D e^4}{8 \varepsilon_0 c h^3} \\
& \frac{R_H}{R_D}=\frac{\mu_H}{\mu_D} \dots(ii)
\end{aligned}
\)
From equation (i) and (ii):
\(
\frac{\lambda_D}{\lambda_H}=\frac{\mu_H}{\mu_D} \dots(iii)
\)
Reduced mass for hydrogen,
\(
\mu_H=\frac{m_e}{1+m_e / M} \simeq m_e\left(1-\frac{m_e}{M}\right)
\)
Reduced mass for deuterium,
\(
\mu_D=\frac{2 M \cdot m_e}{2 M\left(1+\frac{m_e}{2 M}\right)} \simeq m_e\left(1-\frac{m_e}{2 M}\right)
\)
\(
\begin{aligned}
&\text { where } M \text { is mass of proton }\\
&\begin{aligned}
\frac{\mu_H}{\mu_D} & =\frac{m_e\left(1-\frac{m_e}{2 M}\right)}{m_e\left(1-\frac{m_e}{2 M}\right)}=\left(1-\frac{m_e}{M}\right)\left(1-\frac{m_e}{2 M}\right)^{-1} \\
& =\left(1-\frac{m_e}{M}\right)\left(1+\frac{m_e}{2 M}\right) \\
\Rightarrow \quad \frac{\mu_H}{\mu_D} & =\left(1-\frac{m_e}{2 M}\right)
\end{aligned}
\end{aligned}
\)
\(
\text { or } \quad \frac{\mu_H}{\mu_D}=\left(1-\frac{1}{2 \times 1840}\right)=0.99973 \dots(iv)
\)
\(
\begin{aligned}
&\left(\because \quad M=1840 m_e\right)\\
&\text { From (iii) and (iv) }\\
&\frac{\lambda_D}{\lambda_H}=0.99973, \quad \lambda_D=0.99973 \lambda_H
\end{aligned}
\)
Using \(\lambda_H=1218 Å, 1028 Å, 974.3 Å\) and \(951.4 Å\), we get
\(
\lambda_D=1217.7 Å, 1027.7 Å, 974.04 Å, 951.1 Å
\)
Shift in wavelength \(\left(\lambda_H-\lambda_D\right) \approx 0.3 Å\).

Q12.25: Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in \({ }^1 \mathrm{H}\) and \({ }^2 \mathrm{H}\). This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account then the electrons and nucleus revolve around their common centre of mass. Such a system is equivalent to a single particle with a reduced mass \(\mu\), revolving around the nucleus at a distance equal to the electron-nucleus separation. Here \(\mu=m_e M /\left(m_e+M\right)\) where \(M\) is the nuclear mass and \(m_e\) is the electronic mass. Estimate the percentage difference in wavelength for the 1st line of the Lyman series in \({ }^1 \mathrm{H}\) and \({ }^2 \mathrm{H}\). (Mass of \({ }^1 \mathrm{H}\) nucleus is \(1.6725 \times 10^{-27} \mathrm{~kg}\), Mass of \({ }^2 \mathrm{H}\) nucleus is \(3.3374 \times 10^{-27} \mathrm{~kg}\), Mass of electron \(=9.109 \times 10^{-31} \mathrm{~kg}\).)

Answer: Taking into account the nuclear motion, the stationary state energies shall be, \(E_n=-\frac{\mu Z^2 e^4}{8 \varepsilon_0^2 h^2}\left(\frac{1}{n^2}\right)\). Let \(\mu_H\) be the reduced mass of Hydrogen and \(\mu_D\) that of Deutrium. Then the frequency of the \(1^{\text {st }}\) Lyman line in Hydrogen is \(h \nu_H=\frac{\mu_H e^4}{8 \varepsilon_0^2 h^2}\left(1-\frac{1}{4}\right)=\frac{3}{4} \frac{\mu_H e^4}{8 \varepsilon_0^2 h^2}\). Thus the wavelength of the transition is \(\lambda_H=\frac{3}{4} \frac{\mu_H e^4}{8 \varepsilon_0^2 h^3 c}\). The wavelength of the transition for the same line in Deutrium is \(\lambda_D=\frac{3}{4} \frac{\mu_D e^4}{8 \varepsilon_0^2 h^3 c}\).
\(
\begin{aligned}
&\therefore \Delta \lambda=\lambda_D-\lambda_H\\
&\text { Hence the percentage difference is }\\
&100 \times \frac{\Delta \lambda}{\lambda_H}=\frac{\lambda_D-\lambda_H}{\lambda_H} \times 100=\frac{\mu_D-\mu_H}{\mu_H} \times 100
\end{aligned}
\)
\(
=\frac{\frac{m_e M_D}{\left(m_e+M_D\right)}-\frac{m_e M_H}{\left(m_e+M_H\right)}}{m_e M_H /\left(m_e+M_H\right)} \times 100
\)
\(
\begin{aligned}
&=\left[\left(\frac{m_e+M_H}{m_e+M_D}\right) \frac{M_D}{M_H}-1\right] \times 100\\
&\text { Since } m_e \ll M_H<M_D
\end{aligned}
\)
\(
\begin{aligned}
\frac{\Delta \lambda}{\lambda_H} \times 100=\left[\frac{M_H}{M_D}\right. & \left.\times \frac{M_D}{M_H}\left(\frac{1+m_e / M_H}{1+m_e / M_D}\right)-1\right] \times 100 \\
& =\left[\left(1+m_e / M_H\right)\left(1+m_e / M_D\right)^{-1}-1\right] \times 100 \\
& \simeq\left[\left(1+\frac{m_e}{M_H}-\frac{m_e}{M_D}-1\right] \times 100\right. \\
& \approx m_e\left[\frac{1}{M_H}-\frac{1}{M_D}\right] \times 100 \\
& =9.1 \times 10^{-31}\left[\frac{1}{\left.1.6725 \times 10^{-27}-\frac{1}{3.3374 \times 10^{-27}}\right] \times 100}\right. \\
& =9.1 \times 10^{-4}[0.5979-0.2996] \times 100 \\
& =2.714 \times 10^{-2} \%
\end{aligned}
\)

Q12.26: If a proton had a radius \(R\) and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of an H -atom when (i) \(R=0.1 Å\), and (ii) \(R=10 Å\).

Answer: In an H -atom in the ground state, the electron revolves around the point-size proton in a circular orbit of radius \(\mathrm{r}_{{B}}\) (Bohr’s radius).
As \(m v r_B=h\) and \(\frac{m v^2}{r_B}=\frac{-1 \times e \times e}{4 \pi \varepsilon_0 r_B^2}\)
\(
\begin{aligned}
& \frac{m}{r_B}\left(\frac{-h^2}{m^2 r_B^2}\right)=\frac{e^2}{4 \pi \varepsilon_0 r_B^2} \\
& r_B=\frac{4 \pi \varepsilon_0 h^2}{e^2} \frac{h^2}{m}=0.53 Å
\end{aligned}
\)
\(
\text { K.E. }=\frac{1}{2} m v^2=\left(\frac{m}{2}\right)\left(\frac{h}{m r_B}\right)^2
\)
\(
=\frac{h^2}{2 m r_B^2}=13.6 \mathrm{eV}
\)
PE of the electron and proton,
\(
\mathrm{U}=\frac{1}{4 \pi \varepsilon_0} \frac{e(-e)}{r_B}=-\frac{e^2}{4 \pi \varepsilon_0 r_B}=-27.2 \mathrm{eV}
\)
The total energy of the electron, i.e.,
\(
\mathrm{E}=\mathrm{K}+\mathrm{U}=+13.6 \mathrm{eV}-27.2 \mathrm{eV}=-13.6 \mathrm{eV}
\)
(i) When \(\mathrm{R}=0.1 Å: \mathrm{R}<\mathrm{r}_{\mathrm{B}}\left(\right.\) as \(\left.\mathrm{r}_{\mathrm{B}}=0.51 Å\right)\) and the ground state energy is the same as obtained earlier for pointsize proto 13.6 eV
(ii) When \(R=10 Å: \mathrm{R} \gg \mathrm{r}_{\mathrm{B}}\), the electron moves inside the proton (assumed to be a sphere of radius R ) with new
Bohr’s radius \(r_B^{\prime}\)
\(\text { Clearly, } r_B^{\prime}=\frac{4 \pi \varepsilon_0 h^2}{m(e)\left(e^{\prime}\right)} \ldots . .\left[\text { Replacing } \mathrm{e}^2 \text { by (e) } (\mathrm{e}^{\prime}) \text { where } {e}^{\prime} \text { is the charge on the sphere of radius } r_B^{\prime}\right. \text { ] }\)
\(
\begin{aligned}
& \text { Since } e^{\prime}=\left[\frac{e}{\left(\frac{4 \pi}{3}\right) R^3}\right]\left[\left(\frac{4 \pi}{3}\right) r_B^{\prime 3}\right]=\frac{r_B^{\prime 3}}{R^3} \\
& r_B^{\prime}=\frac{4 \pi \varepsilon_0 h^2}{m_e\left(e r_B^{\prime 3} R^3\right)}=\left(\frac{4 \pi \varepsilon_0 h^2}{m e^2}\right)\left(\frac{R^3}{r_B^{\prime 3}}\right)
\end{aligned}
\)
or \(r_B^{\prime 4}=\left(\frac{4 \pi \varepsilon_0 h^2}{m e^2}\right) R^3\)
\(
=(0.51 Å)(10 Å)^3=\left(510 Å^4\right)
\)
\(\therefore r_B^{\prime}=4.8 Å\), which is less than \({R}(=10 Å)\)
KE of the electron,
\(
\begin{aligned}
& K^{\prime}=\frac{1}{2} m v^{\prime 2}=\left(\frac{m}{2}\right)\left(\frac{h^2}{m^2 r_B^{\prime 2}}\right)=\frac{h}{2 m r_B^{\prime 2}} \\
& =\left(\frac{h^2}{2 m r_B^2}\right)\left(\frac{r_B}{r_B^{\prime}}\right)=13.6 \mathrm{eV}\left(\frac{0.51 Å}{4.8 Å}\right)^2=0.16 \mathrm{eV}
\end{aligned}
\)
Potential at a point inside the charged proton i.e.,
\(
V=\frac{k_e e}{R}\left(3-\frac{r_B^{\prime 2}}{R^2}\right)=k_e e\left(\frac{3 R^2-r_B^{\prime 2}}{R^3}\right)
\)
The potential energy of electron and proton, \(\mu=-\mathrm{eV}\)
\(
\begin{aligned}
&\begin{aligned}
& =-e\left(k_e e\right)\left[\frac{3 R^2-r_B^{\prime 2}}{R^3}\right] \\
& =-\left(\frac{e^2}{4 \pi \varepsilon_0 r^B}\right)\left[\frac{r_B\left(3 R^2-r_B^{\prime 2}\right)}{R^3}\right] \\
& =-(27.2 \mathrm{eV})\left[\frac{(0.51 Å)(300 Å-23.03 Å)}{(1000 Å)}\right] \\
& =-3.83 \mathrm{eV}
\end{aligned}\\
&\text { Total energy of the electron, } \mathrm{E}=\mathrm{K}+\mathrm{U}=0.16 \mathrm{eV}-3.83 \mathrm{eV}=-3.67 \mathrm{eV}
\end{aligned}
\)

Q12.27: 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 \(n=2\) to \(n=1\) transition.

Answer: As the nucleus is massive, recoil momentum of the atom may be neglected and the entire energy of the transition may be considered transferred to the Auger electron. As there is a single valence electron in Cr, the energy states may be thought of as given by the Bohr model.
The energy of the \(n\)th state \(E_n=-Z^2 R \frac{1}{n^2}\) where \(R\) is the Rydberg constant and \(Z=24\).
The energy released in a transition from 2 to 1 is \(\Delta E=Z^2 R\left(1-\frac{1}{4}\right)=\frac{3}{4} Z^2 R\). The energy required to eject a \(n=4\) electron is \(E_4=Z^2 R \frac{1}{16}\).
Thus the kinetic energy of the Auger electron is
\(
\begin{aligned}
& K . E=Z^2 R\left(\frac{3}{4}-\frac{1}{16}\right)=\frac{1}{16} Z^2 R \\
& =\frac{11}{16} \times 24 \times 24 \times 13.6 \mathrm{eV} \\
& =5385.6 \mathrm{eV}
\end{aligned}
\)

Q12.28: The inverse square law in electrostatics is \(|\mathbf{F}|=\frac{e^2}{\left(4 \pi \varepsilon_0\right) \cdot r^2}\) for the force between an electron and a proton. The \(\left(\frac{1}{r}\right)\) dependence of | \(\mathbf{F}\) | 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 \(|\mathbf{F}|=\frac{e^2}{\left(4 \pi \varepsilon_0\right) r^2}\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \cdot \exp (-\lambda r)\) where \(\lambda=m_p c / \hbar\) and \(\hbar=\frac{h}{2 \pi}\). Estimate the change in the ground state energy of an H-atom if \(m_p\) were \(10^{-6}\) times the mass of an electron.

Answer: 

\(
\begin{aligned}
m_{\mathrm{p}} c^2 & =10^{-6} \times \text { electron mass } \times c^2 \\
& \approx 10^{-6} \times 0.5 \mathrm{MeV} \\
& \approx 10^{-6} \times 0.5 \times 1.6 \times 10^{-13} \\
& \approx 0.8 \times 10^{-19} \mathrm{~J} \\
\frac{\hbar}{m_p c} & =\frac{\hbar c}{m_p c^2}=\frac{10^{-34} \times 3 \times 10^8}{0.8 \times 10^{-19}} \approx 4 \times 10^{-7} \mathrm{~m} \gg \text { Bohr radius. }
\end{aligned}
\)
\(
|\mathbf{F}|=\frac{e^2}{4 \pi \varepsilon_0}\left[\frac{1}{r^2}+\frac{\lambda}{r}\right] \exp (-\lambda r)
\)
where \(\lambda^{-1}=\frac{\hbar}{m_p c} \approx 4 \times 10^{-7} \mathrm{~m} \gg r_B\)
\(
\begin{aligned}
& \therefore \lambda \ll \frac{1}{r_B} \text { i.e } \lambda r_B \ll 1 \\
& U(r)=-\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{\exp (-\lambda r)}{r} \\
& m v r=\hbar \therefore v=\frac{\hbar}{m r}
\end{aligned}
\)
Also : \(\frac{m v^2}{r}=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{r^2}+\frac{\lambda}{r}\right]\)
\(
\therefore \frac{\hbar^2}{m r^3}=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[\frac{1}{r^2}+\frac{\lambda}{r}\right]
\)
\(
\therefore \frac{\hbar^2}{m}=\left(\frac{e^2}{4 \pi \varepsilon_0}\right)\left[r+\lambda r^2\right]
\)
\(
\begin{aligned}
& \text { If } \lambda=0 ; r=r_B=\frac{\hbar}{m} \cdot \frac{4 \pi \varepsilon_0}{e^2} \\
& \frac{\hbar^2}{m}=\frac{e^2}{4 \pi \varepsilon_0} \cdot r_B \\
& \text { Since } \lambda^{-1} \gg r_B, \text { put } r=r_B+\delta \\
& \therefore r_B=r_B+\delta+\lambda\left(r_B^2+\delta^2+2 \delta r_B\right) ; \text { negect } \delta^2 \\
& \text { or } 0=\lambda r_B^2+\delta\left(1+2 \lambda r_B\right) \\
& \delta=\frac{-\lambda r_B^2}{1+2 \lambda r_B} \approx \lambda r_B^2\left(1-2 \lambda r_B\right)=-\lambda r_B^2 \text { since } \lambda r_B \ll 1
\end{aligned}
\)
\(
\therefore V(r)=-\frac{e^2}{4 \pi \varepsilon_0} \cdot \frac{\exp \left(-\lambda \delta-\lambda r_B\right)}{r_B+\delta}
\)
\(
\therefore V(r)=-\frac{e^2}{4 \pi \varepsilon_0} \frac{1}{r_B}\left[\left(1-\frac{\delta}{r_B}\right) \cdot\left(1-\lambda r_B\right)\right]
\)
\(\equiv(-27.2 \mathrm{eV})\) remains unchanged.
\(
\begin{aligned}
& K . E=-\frac{1}{2} m v^2=\frac{1}{2} m \cdot \frac{\hbar^2}{m r^2}=\frac{\hbar^2}{2\left(r_B+\delta\right)^2}=\frac{\hbar^2}{2 r_B^2}\left(1-\frac{2 \delta}{r_B}\right) \\
& =(13.6 \mathrm{eV})\left[1+2 \lambda r_B\right]
\end{aligned}
\)
\(
\begin{gathered}
\text { Total energy }=\quad-\frac{e^2}{4 \pi \varepsilon_0 r_B}+\frac{\hbar^2}{2 r_B^2}\left[1+2 \lambda r_B\right] \\
=-27.2+13.6\left[1+2 \lambda r_B\right] \mathrm{eV}
\end{gathered}
\)
\(
\text { Change in energy }=13.6 \times 2 \lambda r_B \mathrm{eV}=27.2 \lambda r_B \mathrm{eV}
\)

Q12.29: The Bohr model for the H-atom relies on the 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 charge \(+q_1,-q_2\) is modified to
\(
\begin{aligned}
& |\mathbf{F}|=\frac{q_1 q_2}{\left(4 \pi \varepsilon_0\right)} \frac{1}{r^2}, \quad r \geq R_0 \\
& =\frac{q_1 q_2}{4 \pi \varepsilon_0} \frac{1}{R_0^2}\left(\frac{R_0}{r}\right)^{\varepsilon}, r \leq R_0
\end{aligned}
\)
Calculate in such a case, the ground state energy of a H -atom, if \(\varepsilon=0.1, R_o=1 Å\).

Answer: Let us consider the case, when \(r \leq R_0=1 Å\)
Let \(\varepsilon=2+\boldsymbol{\delta}\)
\(
F=\frac{q_1 q_2}{4 \pi \varepsilon_0} \cdot \frac{R_0^\delta}{r^{2+\delta}}=\frac{x R_0^\delta}{r^{2+\delta}}
\)
where, \(\quad \frac{q_1 q_2}{4 \pi_0 \varepsilon_0}=x=\left(1.6 \times 10^{-19}\right)^2 \times 9 \times 10^9\)
\(
=23.04 \times 10^{-29} \mathrm{~N} \mathrm{~m}^2
\)
The electrostatic force of attraction between the positively charged nucleus and negatively charged electrons (Coulombian force) provides necessary centripetal force.
\(
\frac{m v^2}{r}=\frac{x R_0^\delta}{r^{2+\delta}} \text { or } v^2=\frac{x R_0^\delta}{m r^{1+\delta}} \dots(i)
\)
\(
m v r=n \hbar \Rightarrow r=\frac{n \hbar}{m v}=\frac{n h}{m}\left[\frac{m}{x R_0^\delta}\right]^{1 / 2} r^{(1+\delta) / 2} \text { [Applying Bohr’s second postulates] }
\)
\(
\text { Solving this for } r \text {, we get } r_n=\left[\frac{n^2 \hbar^2}{m x R_0^\delta}\right]^{\frac{1}{1-\delta}}
\)
where, \(r_n\) is the radius of \(n\)th orbit of electron.
For \(n=1\) and substituting the values of constant, we get
\(
\begin{aligned}
r_1 & =\left[\frac{\hbar^2}{m x R_0^\delta}\right]^{\frac{1}{1-\delta}} \\
\Rightarrow \quad r_1 & =\left[\frac{1.05^2 \times 10^{-68}}{9.1 \times 10^{-31} \times 2.3 \times 10^{-28} \times 10^{+19}}\right]^{\frac{1}{2.9}} \\
& =8 \times 10^{-11}=0.08 \mathrm{~nm}(<0.1 \mathrm{~nm})
\end{aligned}
\)
This is the radius of orbit of electron in ground state of hydrogen atom. Again using Bohr’s second postulate, the speed of the electron
\(
v_n=\frac{n \hbar}{m r_n}=n \hbar\left(\frac{m x R_0^\delta}{n^2 \hbar^2}\right)^{\frac{1}{1-\delta}}
\)
For \(n=1\), the speed of electron in ground state \(v_1=\frac{\hbar}{m r_1}=1.44 \times 10^6 \mathrm{~m} / \mathrm{s}\)
The kinetic energy of electron in the ground state
\(
\mathrm{KE}=\frac{1}{2} m v_1^2-9.43 \times 10^{-19} \mathrm{~J}=5.9 \mathrm{eV}
\)
The potential energy of electron in the ground state till \(R_0\)
\(
U=\int_0^{R_0} F d r=\int_0^{R_0} \frac{x}{r^2} d r=-\frac{x}{R_0}
\)
Potential energy from \(R_0\) to \(r, U=\int_{R_0}^r F d r=\int_{R_0}^r \frac{x R_0^\delta}{r^{2+\delta}} d r\)
\(
\begin{aligned}
U & =+x R_0^\delta \int_{R_0}^r \frac{d r}{r^{2+\delta}}=+\frac{x R_0^\delta}{-1-\delta}\left[\frac{1}{r^{1+\delta}}\right]_{R_0}^r \\
U & =\frac{x R_0^\delta}{1+\delta}\left[\frac{1}{r^{1+\delta}}-\frac{1}{R_0^{1+\delta}}\right]=-\frac{x}{1+\delta}\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}\right] \\
& \cdot \\
U & =-x\left[\frac{R_0^\delta}{r^{1+\delta}}-\frac{1}{R_0}+\frac{1+\delta}{R_0}\right] \\
U & =-x\left[\frac{R_0^{-19}}{r^{-0.9}}-\frac{1.9}{R_0}\right] \\
& =\frac{2.3}{0.9} \times 10^{-18}\left[(0.8)^{0.9}-1.9\right] \mathrm{J}=-17.3 \mathrm{eV}
\end{aligned}
\)
Hence total energy of electron in ground state \(=(-17.3+5.9)=-11.4 \mathrm{eV}\)