Bohr's Model of Hydrogen Atom Questions in English

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For a hydrogen atom,the atomic number $Z = 1$,so $E_n = -13.6 \times \frac{1^2}{n^2} = -\frac{13.6}{n^2}$.
For a helium ion $(He^+)$,the atomic number $Z = 2$,so the energy $E'_{n}$ is given by: $E'_{n} = -13.6 \times \frac{2^2}{n^2} = -13.6 \times \frac{4}{n^2}$.
Comparing the two expressions,we get: $E'_{n} = 4 \times (-13.6 \times \frac{1}{n^2}) = 4E_n$.
Therefore,the energy of the $n^{th}$ orbit in a helium ion is $4E_n$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
The change in angular momentum $\Delta L$ when the electron transitions from orbit $n_1 = 4$ to $n_2 = 5$ is:
$\Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi}(n_2 - n_1)$.
Substituting the given values:
$\Delta L = \frac{6.6 \times 10^{-34}}{2 \times 3.14} \times (5 - 4)$.
$\Delta L = \frac{6.6 \times 10^{-34}}{6.28} \times 1$.
$\Delta L \approx 1.05 \times 10^{-34} \, J \cdot s$.

(C) According to Bohr's theory,the time period $T$ of an electron in an orbit is given by $T = \frac{2\pi r}{v}$.
Since $r \propto n^2$ and $v \propto \frac{1}{n}$,we have $T \propto \frac{n^2}{1/n} = n^3$.
Therefore,the ratio of the time periods for $n_2 = 2$ and $n_1 = 1$ is:
$\frac{T_2}{T_1} = \left( \frac{n_2}{n_1} \right)^3 = \left( \frac{2}{1} \right)^3 = \frac{8}{1}$.
Thus,the ratio is $8:1$.

(A) The energy of a photon emitted during a transition between energy levels $n_2$ and $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
Since the transition is from $n = 2$ to $n = 1$ for all cases,the term $\left( \frac{1}{1^2} - \frac{1}{2^2} \right)$ is constant.
Therefore,$\Delta E \propto Z^2$.
Since $\Delta E = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} \propto Z^2$,which implies $\lambda Z^2 = \text{constant}$.
For $H$ $(Z=1)$,$D$ $(Z=1)$,$He^+$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$:
$\lambda_1 (1)^2 = \lambda_2 (1)^2 = \lambda_3 (2)^2 = \lambda_4 (3)^2$.
Thus,$\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

(D) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a given transition $(n_1=2, n_2=4)$,the wave number is proportional to the square of the atomic number: $\bar{\nu} \propto Z^2$.
For Hydrogen $(H)$,$Z_1 = 1$. For Helium ion $(He^+)$,$Z_2 = 2$.
Therefore,$\frac{\bar{\nu}_{He^+}}{\bar{\nu}_{H}} = \left( \frac{Z_2}{Z_1} \right)^2 = \left( \frac{2}{1} \right)^2 = 4$.
Thus,$\bar{\nu}_{He^+} = 4 \times \bar{\nu}_{H} = 4 \times 20,397 \, cm^{-1} = 81,588 \, cm^{-1}$.

(D) According to Bohr's model,the radius of the $n^{th}$ orbit is given by $R_n \propto n^2$.
Given $R_1 = R$ and $R_2 = 4R$,we have $\frac{R_1}{R_2} = \frac{R}{4R} = \frac{1}{4}$.
Since $R_n \propto n^2$,we have $\frac{n_1^2}{n_2^2} = \frac{1}{4}$,which implies $\frac{n_1}{n_2} = \frac{1}{2}$.
The time period $T$ of an electron in the $n^{th}$ orbit is given by $T_n \propto n^3$.
Therefore,the ratio of the time periods is $\frac{T_1}{T_2} = \left( \frac{n_1}{n_2} \right)^3$.
Substituting the values,$\frac{T_1}{T_2} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}$.

(A) The energy of an electron in a hydrogen-like atom is given by the formula $E_n = E_0 \cdot \frac{Z^2}{n^2}$,where $E_0$ is the ground state energy of the hydrogen atom $(-13.6 \ eV)$.
For the ground state of a hydrogen atom,$n = 1$ and $Z = 1$. Thus,$E = E_0 \cdot \frac{1^2}{1^2} = E_0$.
For the $2^{nd}$ excited state of $Li^{++}$,the principal quantum number $n = 3$ (since $n=1$ is ground,$n=2$ is $1^{st}$ excited,and $n=3$ is $2^{nd}$ excited). The atomic number of Lithium is $Z = 3$.
Substituting these values into the formula: $E' = E_0 \cdot \frac{Z^2}{n^2} = E_0 \cdot \frac{3^2}{3^2} = E_0$.
Since $E = E_0$,we find that $E' = E$.

(C) The ionization potential of a hydrogen atom is $13.6 \ eV$. The energy of the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
When the hydrogen atom in the ground state $(n=1)$ absorbs a photon of energy $12.1 \ eV$,it is excited to a higher energy level $n$. The energy difference is given by:
$E = E_n - E_1$
$12.1 = -\frac{13.6}{n^2} - (-13.6)$
$12.1 = 13.6 - \frac{13.6}{n^2}$
$\frac{13.6}{n^2} = 13.6 - 12.1 = 1.5$
$n^2 = \frac{13.6}{1.5} \approx 9.06 \approx 9$
$n = 3$
The number of spectral lines emitted when the electron transitions from the $n^{th}$ state back to the ground state is given by the formula $\frac{n(n-1)}{2}$.
For $n=3$,the number of spectral lines = $\frac{3(3-1)}{2} = \frac{3 \times 2}{2} = 3$.

(A) The energy $E$ of a hydrogen-like atom with atomic number $Z$ and principal quantum number $n$ is given by the formula:
$E_n = -13.6 \times \frac{Z^2}{n^2} \, eV$
For the $He^+$ ion,the atomic number $Z = 2$.
For the first excited state,the principal quantum number $n = 2$.
Substituting these values into the formula:
$E_2 = -13.6 \times \frac{2^2}{2^2} \, eV$
$E_2 = -13.6 \times \frac{4}{4} \, eV$
$E_2 = -13.6 \, eV$.

(C) The energy of the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \; eV$.
The energy levels are:
$E_1 = -13.6 \; eV$
$E_2 = -\frac{13.6}{4} = -3.4 \; eV$
$E_3 = -\frac{13.6}{9} \approx -1.51 \; eV$
$E_4 = -\frac{13.6}{16} = -0.85 \; eV$
The energy of an emitted photon is the difference between two energy levels: $\Delta E = E_{n_2} - E_{n_1}$.
Checking the options:
$1$. $E_4 - E_3 = -0.85 - (-1.51) = 0.66 \; eV$ (approx $0.65 \; eV$). This is possible.
$2$. $E_3 - E_2 = -1.51 - (-3.4) = 1.89 \; eV$ (approx $1.9 \; eV$). This is possible.
$3$. $E_1$ is the ground state energy,and transitions to $E_1$ can release $13.6 \; eV$ (e.g.,from infinity to $n=1$). This is possible.
$4$. $11.1 \; eV$ cannot be expressed as the difference between any two energy levels of the hydrogen atom.
Therefore,$11.1 \; eV$ is not a possible energy for an emitted photon.

(B) According to Bohr's theory,the velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$v_n = v_0 \times \frac{Z}{n}$
where $v_0 = 2.18 \times 10^6 \; m/s$ is the velocity of an electron in the first orbit of the hydrogen atom.
For $He^{+}$,$Z = 2$ and for the $3^{rd}$ orbit,$n = 3$.
Substituting these values:
$v_3 = (2.18 \times 10^6) \times \frac{2}{3} \; m/s$
$v_3 = 2.18 \times 10^6 \times 0.666... \; m/s$
$v_3 \approx 1.453 \times 10^6 \; m/s \approx 1.46 \times 10^6 \; m/s$.

(B) In a Bohr orbit of the hydrogen atom,the total energy $E$ is given by $E = -K$,where $K$ is the kinetic energy.
This relationship arises because the total energy is the sum of kinetic energy and potential energy,$E = K + U$,and for a Bohr orbit,$U = -2K$.
Therefore,$E = K + (-2K) = -K$.
Thus,the ratio of kinetic energy to total energy is $K/E = K/(-K) = 1:-1$.

(B) According to the Bohr model,the electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$.
In the Bohr model,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
Substituting this into the force equation: $F \propto \frac{1}{r_n^2} \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.
Therefore,the force on the electron depends on the principal quantum number as $F \propto 1/n^4$.

(A) The energy of the $n$-th orbit for a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \, \text{eV}$.
From the given energy level diagram,for the ground state $(n = 1)$,the energy is $E_1 = -54.4 \, \text{eV}$.
Substituting these values into the formula: $-54.4 = -13.6 \frac{Z^2}{1^2}$.
Solving for $Z^2$: $Z^2 = \frac{54.4}{13.6} = 4$,which gives $Z = 2$.
The radius of the $n$-th Bohr orbit is given by $r_n = 0.53 \frac{n^2}{Z} \, \mathring{A}$.
For the first Bohr orbit $(n = 1)$ and $Z = 2$,the radius is $r_1 = 0.53 \frac{1^2}{2} = 0.265 \, \mathring{A}$.
Thus,the correct option is $A$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -\frac{13.6}{n^2} \ eV$.
Given $E_n = -0.544 \ eV$,we set up the equation: $-0.544 = -\frac{13.6}{n^2}$.
Solving for $n^2$: $n^2 = \frac{13.6}{0.544} = 25$.
Thus,$n = 5$.
According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by: $L = n\frac{h}{2\pi}$.
Substituting $n = 5$,we get: $L = 5\frac{h}{2\pi} = \frac{5h}{2\pi}$.

(B) The acceleration $a$ of an electron in a circular orbit is given by $a = \frac{v^2}{r}$.
From Bohr's theory,the velocity $v \propto \frac{Z}{n}$ and the radius $r \propto \frac{n^2}{Z}$.
Substituting these into the acceleration formula: $a \propto \frac{(Z/n)^2}{n^2/Z} = \frac{Z^2/n^2}{n^2/Z} = \frac{Z^3}{n^4}$.
Since both atoms are in the ground state,$n = 1$ for both.
Therefore,$a \propto Z^3$.
The ratio of acceleration for singly ionized helium $(Z=2)$ to hydrogen $(Z=1)$ is $\frac{a_{He^+}}{a_H} = \left( \frac{Z_{He^+}}{Z_H} \right)^3 = \left( \frac{2}{1} \right)^3 = 8$.

(A) According to Bohr's quantization condition,the angular momentum $J_n$ of an electron in the $n^{th}$ orbit is given by $J_n = \frac{n h}{2 \pi}$.
This implies $J_n \propto n$.
According to the de-Broglie hypothesis,the wavelength $\lambda_n$ is given by $\lambda_n = \frac{h}{p} = \frac{h}{mv_n}$.
For the $n^{th}$ orbit,the circumference is $2 \pi r_n = n \lambda_n$.
Since $r_n \propto n^2$ and $v_n \propto \frac{1}{n}$,the angular momentum $J_n = m v_n r_n$ is proportional to $n$.
From the relation $2 \pi r_n = n \lambda_n$,we have $\lambda_n = \frac{2 \pi r_n}{n}$.
Substituting $r_n \propto n^2$,we get $\lambda_n \propto \frac{n^2}{n} = n$.
Since $J_n \propto n$ and $\lambda_n \propto n$,it follows that $J_n \propto \lambda_n$.

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
Given $L = \frac{3h}{2\pi}$,comparing the two expressions,we get $n = 3$.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \text{ eV}$.
For $n = 3$,$E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \text{ eV}$.
The kinetic energy $K$ of an electron in an orbit is equal to the magnitude of its total energy,so $K = |E_n| = 1.51 \text{ eV}$.

(A) We know that for a hydrogen atom,the kinetic energy $(K)$ is equal to the binding energy $(E_n)$:
$K = \frac{1}{2} m v^2 = E_n$ --- $(i)$
According to Bohr's quantization condition of angular momentum:
$mvr = \frac{nh}{2\pi}$ --- $(ii)$
From $(i)$,we have $v^2 = \frac{2E_n}{m}$,so $v = \sqrt{\frac{2E_n}{m}}$.
From $(ii)$,the radius $r = \frac{nh}{2\pi mv}$.
The frequency of revolution $(f)$ is given by $f = \frac{v}{2\pi r}$.
Substituting the value of $r$ from $(ii)$ into the frequency formula:
$f = \frac{v}{2\pi (nh / 2\pi mv)} = \frac{v^2 m}{nh} = \frac{(2E_n / m) m}{nh} = \frac{2E_n}{nh}$.
Therefore,the frequency of revolution is $\frac{2E_n}{nh}$.

(D) The Bohr radius for the $n^{th}$ orbit is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius constant.
According to the problem,the difference between the $(n+1)^{th}$ and $n^{th}$ radius is equal to the $(n-1)^{th}$ radius:
$r_{n+1} - r_n = r_{n-1}$
Substituting the formula $r_n = a_0 n^2$:
$a_0(n+1)^2 - a_0 n^2 = a_0(n-1)^2$
Dividing by $a_0$:
$(n+1)^2 - n^2 = (n-1)^2$
$(n^2 + 2n + 1) - n^2 = n^2 - 2n + 1$
$2n + 1 = n^2 - 2n + 1$
$n^2 - 4n = 0$
$n(n - 4) = 0$
Since $n$ must be a positive integer greater than $1$ for the $(n-1)^{th}$ orbit to exist,we have $n = 4$.

(C) The number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula $N = \frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2}$.
Given that $N = 6$,we have $\frac{(n_2 - n_1 + 1)(n_2 - n_1)}{2} = 6$,which simplifies to $(n_2 - n_1 + 1)(n_2 - n_1) = 12$.
Let $x = n_2 - n_1$. Then $(x+1)x = 12$,which gives $x^2 + x - 12 = 0$. Solving this quadratic equation,we get $(x+4)(x-3) = 0$,so $x = 3$.
Thus,$n_2 - n_1 = 3$. Possible pairs are $(4, 1)$ or $(5, 2)$.
However,the problem states that the energy of the emitted photons can be equal to,less than,or greater than the absorbed photon energy. The absorbed photon energy corresponds to the transition from $n_1$ to $n_2$. If $n_1 = 2$ and $n_2 = 4$,the transitions include $4 \rightarrow 3, 4 \rightarrow 2, 4 \rightarrow 1, 3 \rightarrow 2, 3 \rightarrow 1, 2 \rightarrow 1$. Here,$E_{4 \rightarrow 2}$ equals the absorbed energy,$E_{4 \rightarrow 3}$ is less than the absorbed energy,and $E_{4 \rightarrow 1}, E_{3 \rightarrow 1}, E_{2 \rightarrow 1}$ are greater than the absorbed energy. This satisfies all conditions. Therefore,$n_1 = 2$ and $n_2 = 4$.

(D) The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
The velocity of the electron in the $n^{th}$ orbit is given by $v_n \propto \frac{1}{n}$.
The centripetal acceleration $a_n$ is given by $a_n = \frac{v_n^2}{r_n}$.
Substituting the proportionalities: $a_n \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
For the $M$ shell, $n = 3$. Thus, $a_M \propto \frac{1}{3^4} = \frac{1}{81}$.
For the $L$ shell, $n = 2$. Thus, $a_L \propto \frac{1}{2^4} = \frac{1}{16}$.
The ratio of the initial ($M$ shell) to the final ($L$ shell) centripetal acceleration is $\frac{a_M}{a_L} = \frac{1/81}{1/16} = \frac{16}{81}$.

(B) The frequency of orbital motion $f$ of an electron in a hydrogen-like atom is given by $f \propto \frac{Z^2}{n^3}$.
Since the atom is hydrogen,$Z = 1$,so $f \propto \frac{1}{n^3}$.
Let $f_1$ be the frequency in the initial state $n_1$ and $f_2$ be the frequency in the final state $n_2$.
Given that $f_1 = \frac{1}{27} f_2$,we have $\frac{1}{n_1^3} = \frac{1}{27} \cdot \frac{1}{n_2^3}$.
Rearranging the terms,we get $\frac{n_2^3}{n_1^3} = \frac{1}{27}$.
Taking the cube root of both sides,$\frac{n_2}{n_1} = \frac{1}{3}$,which implies $n_1 = 3n_2$.
Checking the options:
For option $B$,$n_1 = 3$ and $n_2 = 1$,which satisfies $n_1 = 3(1) = 3$.
Thus,the correct values are $n_1 = 3$ and $n_2 = 1$.

(A) The Bohr radius for the $K$ shell is given by the formula:
$a_{0} = \frac{h^{2} \varepsilon_{0}}{\pi m e^{2}}$
From this expression,it is clear that the Bohr radius is inversely proportional to the mass $(m)$ of the orbiting particle.
Therefore,the ratio of the radius of the electron orbit $(a_{0e})$ to the radius of the muon orbit $(a_{0m})$ is:
$\frac{a_{0m}}{a_{0e}} = \frac{m_{e}}{m_{m}}$
Given that $m_{m} = 207 m_{e}$,we substitute this into the equation:
$a_{0m} = \frac{m_{e}}{207 m_{e}} \times a_{0e} = \frac{a_{0e}}{207}$
The standard Bohr radius for a hydrogen atom is $a_{0e} \approx 0.529 \, \mathring{A} = 0.529 \times 10^{-10} \, m$.
Calculating the new radius:
$a_{0m} = \frac{0.529 \, \mathring{A}}{207} \approx 0.002555 \, \mathring{A} \approx 2.56 \times 10^{-3} \, \mathring{A}$.

(A) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $N = \frac{n(n-1)}{2}$.
Given $N = 10$,we have $\frac{n(n-1)}{2} = 10$,which implies $n^2 - n - 20 = 0$.
Solving the quadratic equation,$(n-5)(n+4) = 0$,we get $n = 5$ (since $n$ must be positive).
The energy of the incident photon corresponds to the transition from the ground state $(n=1)$ to the $n=5$ state.
The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} \right)$.
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{25} \right) = R \left( \frac{24}{25} \right)$.
Using $R \approx 1.097 \times 10^7 \ m^{-1}$,we get $\lambda = \frac{25}{24 \times 1.097 \times 10^7} \approx 9.5 \times 10^{-8} \ m = 95 \ nm$.

(C) Given:
Initial state $n_2 = 5$,final state $n_1 = 1$.
Mass of hydrogen atom $m \approx 1.67 \times 10^{-27} \, kg$.
Rydberg constant $R = 1.097 \times 10^7 \, m^{-1}$.
Planck's constant $h = 6.63 \times 10^{-34} \, J \cdot s$.
Using the Rydberg formula for the wavelength $\lambda$ of the emitted photon:
$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{5^2} \right) = 1.097 \times 10^7 \left( 1 - 0.04 \right) = 1.053 \times 10^7 \, m^{-1}$.
$\lambda = \frac{1}{1.053 \times 10^7} \approx 9.496 \times 10^{-8} \, m$.
By conservation of momentum,the momentum of the atom $p_{atom}$ must equal the momentum of the photon $p_{photon} = \frac{h}{\lambda}$.
$m v = \frac{h}{\lambda} \implies v = \frac{h}{m \lambda}$.
$v = \frac{6.63 \times 10^{-34}}{(1.67 \times 10^{-27}) \times (9.496 \times 10^{-8})} \approx \frac{6.63 \times 10^{-34}}{1.586 \times 10^{-34}} \approx 4.18 \, m/s$.
Rounding to the nearest provided option,the recoil speed is approximately $4.3 \, m/s$.

(C) The frequency of revolution of an electron in the $n^{th}$ Bohr orbit is given by $v_n \propto \frac{1}{n^3}$.
Therefore,$v_n = \frac{k}{n^3}$ for some constant $k$.
For the first orbit $(n=1)$,$v_1 = k$.
Thus,the ratio is $\frac{v_n}{v_1} = \frac{k/n^3}{k} = \frac{1}{n^3} = n^{-3}$.
Taking the logarithm on both sides,we get $\log (v_n / v_1) = \log (n^{-3}) = -3 \log n$.
This equation is of the form $y = mx$,where $y = \log (v_n / v_1)$,$x = \log n$,and the slope $m = -3$.
Since the slope is negative,the graph is a straight line passing through the origin with a negative slope.

(D) The Rydberg formula for the wavelength of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.
For the transition $n = 2 \rightarrow n = 1$,the term in the brackets is constant.
Therefore,$\frac{1}{\lambda} \propto Z^2$,which implies $\lambda \propto \frac{1}{Z^2}$.
To obtain the shortest wavelength,we need the largest atomic number $Z$.
The atomic numbers are: Hydrogen $(Z=1)$,Deuterium $(Z=1)$,Singly ionized Helium $(Z=2)$,and Doubly ionized Lithium $(Z=3)$.
Since Lithium has the highest atomic number $(Z=3)$,it produces the shortest wavelength.

(C) According to Bohr's theory for an electron in a circular orbit,the centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{ke^2}{r^2}$,which implies $v \propto r^{-1/2}$.
The time period $T$ is given by $T = \frac{2\pi r}{v}$.
Substituting $v \propto r^{-1/2}$,we get $T \propto \frac{r}{r^{-1/2}} = r^{3/2}$.
Given radii are $r_1 = R$ and $r_2 = 4R$.
Therefore,the ratio of time periods is $\frac{T_1}{T_2} = \left(\frac{r_1}{r_2}\right)^{3/2} = \left(\frac{R}{4R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$.
Thus,the ratio is $1 : 8$.

(B) According to Bohr's model for a hydrogen atom:
$1$. Angular momentum $L = \frac{nh}{2\pi}$,so $L \propto n$.
$2$. Orbit radius $r = a_0 \frac{n^2}{Z}$,so $r \propto n^2$ (for a fixed $Z$).
$3$. Frequency of revolution $f = \frac{v}{2\pi r}$. Since $v \propto \frac{Z}{n}$ and $r \propto \frac{n^2}{Z}$,we have $f \propto \frac{Z/n}{n^2/Z} = \frac{Z^2}{n^3}$,so $f \propto n^{-3}$.
Now,consider the product $frL$:
$frL \propto (n^{-3}) \cdot (n^2) \cdot (n) = n^0 = 1$.
Therefore,$frL$ is independent of the quantum number $n$ and is constant for all orbits.

(B) The radius of the $n^{th}$ Bohr orbit is given by the formula: $r_n = \frac{0.529 \times n^2}{Z} \mathring A$.
For a singly ionized helium atom $(He^+)$,the atomic number $Z = 2$.
For the second Bohr orbit,$n = 2$.
Substituting these values into the formula:
$r_2 = \frac{0.529 \times 2^2}{2} \mathring A$
$r_2 = \frac{0.529 \times 4}{2} \mathring A$
$r_2 = 0.529 \times 2 \mathring A$
$r_2 = 1.06 \mathring A$.

(A) The energy of an electron in an orbit of principal quantum number $n$ is given by the formula:
$E = -13.6 / n^2 \ eV$
Given $E = -3.4 \ eV$,we have:
$-3.4 = -13.6 / n^2$
$n^2 = 13.6 / 3.4 = 4$
$n = 2$
According to Bohr's postulate,the angular momentum $L$ of an electron in the $n$-th orbit is given by:
$L = nh / 2\pi$
Substituting $n = 2$:
$L = 2h / 2\pi = h / \pi$

(B) The energy of a photon emitted during a transition is given by $E = \Delta E = 13.6 \text{ eV} \times Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first transition (second excited state $n=3$ to first excited state $n=2$):
$E_1 = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 13.6 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 \left( \frac{5}{36} \right)$.
For the second transition (first excited state $n=2$ to ground state $n=1$):
$E_2 = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \left( \frac{3}{4} \right) = 13.6 \left( \frac{27}{36} \right)$.
Ratio of energies $z = \frac{E_1}{E_2} = \frac{5/36}{27/36} = \frac{5}{27}$.
Since $E = \frac{hc}{\lambda}$,we have $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1} = z$. Thus,$x = \frac{\lambda_1}{\lambda_2} = \frac{1}{z} = \frac{27}{5}$.
Since $p = \frac{E}{c}$,the ratio of momenta $y = \frac{p_1}{p_2} = \frac{E_1/c}{E_2/c} = \frac{E_1}{E_2} = z = \frac{5}{27}$.
Comparing with options:
$(A)$ $z = 1/x$ is correct because $z = \frac{E_1}{E_2}$ and $x = \frac{\lambda_1}{\lambda_2} = \frac{E_2}{E_1} = 1/z$.
$(B)$ $x = 9/4$ is incorrect (actual value is $27/5$).
$(C)$ $y = 5/27$ is correct.
$(D)$ $z = 5/27$ is correct.
Therefore,the wrong answer is $(B)$.

(A) When considering the motion of the nucleus,the mass of the electron $m_e$ is replaced by the reduced mass $\mu = \frac{m_e M}{m_e + M}$,where $M$ is the mass of the nucleus.
Since the mass of the deuterium nucleus $(M_D \approx 2M_H)$ is greater than that of the hydrogen nucleus,the reduced mass $\mu_D$ is greater than $\mu_H$.
The radius of the $n^{th}$ Bohr orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi \mu e^2}$. Since $\mu_D > \mu_H$,the radius $r_D < r_H$. Thus,option $A$ is correct.
The speed of the electron is given by $v_n = \frac{e^2}{2 n h \epsilon_0}$. This expression does not depend on the mass of the nucleus,so $v_D = v_H$. Option $B$ is incorrect.
The wavelength of the emitted photon is given by $\frac{1}{\lambda} = R_{\mu} Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$,where $R_{\mu} = \frac{\mu e^4}{8 \epsilon_0^2 h^3 c}$. Since $\mu_D > \mu_H$,$R_D > R_H$. Consequently,$\lambda_D < \lambda_H$. Option $C$ is incorrect.
The angular momentum of the electron in the $n^{th}$ orbit is $L = n \frac{h}{2\pi}$,which is independent of the mass of the nucleus. Thus,$L_D = L_H$. Option $D$ is incorrect.

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = n^2 a_0$.
The area of the $n^{th}$ orbit is given by $A_n = \pi r_n^2 = \pi (n^2 a_0)^2 = \pi n^4 a_0^2$.
For the first orbit $(n=1)$,the area is $A_1 = \pi a_0^2$.
Taking the ratio,we get $\frac{A_n}{A_1} = \frac{\pi n^4 a_0^2}{\pi a_0^2} = n^4$.
Taking the natural logarithm on both sides,we get $\ln \left( \frac{A_n}{A_1} \right) = \ln (n^4) = 4 \ln (n)$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \ln (A_n/A_1)$,$x = \ln (n)$,$m = 4$,and $c = 0$,we conclude that the graph is a straight line passing through the origin with a slope of $4$.

(C) The force acting on the electron is $F = \frac{k}{r}$.
For circular motion,this force provides the necessary centripetal force: $\frac{k}{r} = \frac{mv^2}{r}$.
This implies $mv^2 = k$,which means the kinetic energy $K_n = \frac{1}{2}mv^2 = \frac{k}{2}$ is constant and independent of $n$.
According to the Bohr quantization condition,$mvr = \frac{nh}{2\pi}$.
Since $v = \sqrt{\frac{k}{m}}$ is constant,we have $m \sqrt{\frac{k}{m}} r_n = \frac{nh}{2\pi}$.
Therefore,$r_n \propto n$.

(B) The energy required for an electron transition in a hydrogen-like ion is given by the formula:
$\Delta E = 13.6 Z^{2} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right) \text{eV}$
For $Li^{++}$ ion,the atomic number $Z = 3$.
The transition is from the first orbit $(n_{1} = 1)$ to the third orbit $(n_{2} = 3)$.
Substituting the values into the formula:
$\Delta E = 13.6 \times (3)^{2} \left( \frac{1}{1^{2}} - \frac{1}{3^{2}} \right) \text{eV}$
$\Delta E = 13.6 \times 9 \left( 1 - \frac{1}{9} \right) \text{eV}$
$\Delta E = 13.6 \times 9 \left( \frac{8}{9} \right) \text{eV}$
$\Delta E = 13.6 \times 8 = 108.8 \text{eV}$

(A) The rotational energy of a system is given by $E = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
According to Bohr's quantization rule,$L = \frac{nh}{2\pi}$.
The moment of inertia $I$ for a diatomic molecule about its center of mass is $I = \mu r^2$,where $\mu = \frac{m_1 m_2}{m_1 + m_2}$ is the reduced mass.
Substituting $L$ and $I$ into the energy formula:
$E = \frac{(nh/2\pi)^2}{2(\frac{m_1 m_2}{m_1 + m_2})r^2}$
$E = \frac{n^2 h^2}{4\pi^2 \cdot 2 \cdot \frac{m_1 m_2}{m_1 + m_2} r^2}$
$E = \frac{(m_1 + m_2)n^2 h^2}{8\pi^2 m_1 m_2 r^2}$.

(A) The frequency $\nu$ of the emitted radiation is given by the Rydberg formula: $\nu = c R Z^2 [1/(n-1)^2 - 1/n^2]$.
Simplifying the expression inside the brackets: $[n^2 - (n-1)^2] / [n^2(n-1)^2] = (n^2 - (n^2 - 2n + 1)) / [n^2(n-1)^2] = (2n - 1) / [n^2(n-1)^2]$.
Since $n >> 1$,we can approximate $(2n - 1) \approx 2n$ and $(n-1) \approx n$.
Substituting these approximations: $\nu \approx c R Z^2 [2n / (n^2 \cdot n^2)] = c R Z^2 [2n / n^4] = 2 c R Z^2 / n^3$.
Therefore,the frequency $\nu$ is proportional to $1/n^3$.

(B) The wavelength $\lambda$ of emitted radiation for a hydrogen-like atom is given by the Rydberg formula:
$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Since the transition is from $n_2 = 2$ to $n_1 = 1$ for all ions,the term in the bracket is constant: $\left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \frac{3}{4}$.
Thus,$\lambda \propto \frac{1}{Z^2}$.
For Hydrogen $(Z=1)$,$\lambda_1 \propto \frac{1}{1^2} = 1$.
For Deuterium $(Z=1)$,$\lambda_2 \propto \frac{1}{1^2} = 1$.
For Helium $(Z=2)$,$\lambda_3 \propto \frac{1}{2^2} = \frac{1}{4}$.
For Lithium $(Z=3)$,$\lambda_4 \propto \frac{1}{3^2} = \frac{1}{9}$.
Comparing these,we get $\lambda_1 = \lambda_2 = 4\lambda_3 = 9\lambda_4$.

(D) The wavelength of the emitted photon for a transition from the $n^{th}$ state to the ground state is given by the Rydberg formula: $\frac{1}{\Lambda_n} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$.
For large $n$,we can approximate this as $\Lambda_n = \frac{1}{R} (1 - \frac{1}{n^2})^{-1} \approx \frac{1}{R} (1 + \frac{1}{n^2}) = \frac{1}{R} + \frac{1}{R n^2}$.
The de Broglie wavelength of an electron in the $n^{th}$ orbit is $\lambda_n = \frac{h}{p} = \frac{h}{m v_n}$. Since $v_n \propto \frac{1}{n}$,we have $\lambda_n \propto n$,which implies $n^2 \propto \lambda_n^2$.
Substituting this into the expression for $\Lambda_n$,we get $\Lambda_n = A + \frac{B}{\lambda_n^2}$,where $A = \frac{1}{R}$ and $B$ is a constant related to the physical parameters of the atom.

(B) The binding energy of a hydrogen-like atom in the ground state is given by the formula $E_n = -\frac{m e^4}{8 \epsilon_0^2 h^2 n^2}$,where $m$ is the mass of the orbiting particle.
Since the charge of the muon is the same as the electron,the binding energy is directly proportional to the mass of the particle $(E \propto m)$.
Given that the mass of the muon $m_{\mu} = 207 m_e$,the binding energy of the muonic hydrogen atom is $E_{B\mu} = 207 E_{Be}$.
Therefore,$|E_{B\mu}| \approx 207 |E_{Be}|$. The closest option provided is $207|E_{Be}|$,which corresponds to option $B$.

(A) The frequency of a photon emitted during a transition is given by Rydberg's formula: $f = R c Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a hydrogen atom $(Z=1)$,the transition from the $2^{nd}$ excited state $(n_2=3)$ to the ground state $(n_1=1)$ gives frequency $f$:
$f = R c (1)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = R c \left( 1 - \frac{1}{9} \right) = \frac{8}{9} R c$.
Thus,$R c = \frac{9}{8} f$.
For $Li^{++}$ $(Z=3)$,the transition from the $1^{st}$ excited state $(n_2=2)$ to the ground state $(n_1=1)$ gives frequency $f'$:
$f' = R c (3)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R c (9) \left( 1 - \frac{1}{4} \right) = R c (9) \left( \frac{3}{4} \right) = \frac{27}{4} R c$.
Substituting $R c = \frac{9}{8} f$ into the equation for $f'$:
$f' = \frac{27}{4} \times \left( \frac{9}{8} f \right) = \frac{243}{32} f$.

(A) For a system following Bohr's model,the electrostatic force provides the necessary centripetal force: $\frac{1}{4\pi\varepsilon_0} \frac{(3q)(q)}{r^2} = \frac{mv^2}{r} \implies \frac{3q^2}{4\pi\varepsilon_0 r} = mv^2$ .......$(1)$
According to Bohr's quantization condition for the ground state $(n=1)$: $mvr = \frac{h}{2\pi} \implies r = \frac{h}{2\pi mv}$ .......$(2)$
Substitute $r$ from $(2)$ into $(1)$:
$\frac{3q^2}{4\pi\varepsilon_0} \left( \frac{2\pi mv}{h} \right) = mv^2$
$\frac{3q^2}{2\varepsilon_0 h} = v$
Thus,the orbital velocity is $v = \frac{3q^2}{2\varepsilon_0 h}$.

(A) For a hydrogen-like atom,the wavelength $\lambda$ of the emitted radiation for a transition from $n_2$ to $n_1$ is given by the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the transition $n = 2$ to $n = 1$,we have $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R Z^2 \left( 1 - \frac{1}{4} \right) = \frac{3R Z^2}{4}$.
Thus,$\lambda = \frac{4}{3RZ^2}$,which implies $\lambda Z^2 = \text{constant}$.
Given the atomic numbers: $Z_1 = 1$ (for $H$),$Z_2 = 1$ (for $D$),$Z_3 = 2$ (for $He^+$),and $Z_4 = 3$ (for $Li^{++}$).
Substituting these values: $\lambda_1 (1)^2 = \lambda_2 (1)^2 = \lambda_3 (2)^2 = \lambda_4 (3)^2$.
This simplifies to $\lambda_1 = \lambda_2 = 4 \lambda_3 = 9 \lambda_4$.

(B) In a lithium atom $(Z=3)$,there are $3$ electrons. Two electrons are in the $K$-shell $(n=1)$.
According to the problem,these two electrons screen the nucleus such that the effective charge seen by the outermost electron is $Z_{eff} = 3 - 2 = 1$.
The outermost electron is in the $L$-shell $(n=2)$.
The energy of an electron in a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z_{eff}^2}{n^2} \text{ eV}$.
For the outermost electron,$n=2$ and $Z_{eff}=1$.
$E_2 = -13.6 \times \frac{1^2}{2^2} = -13.6 \times \frac{1}{4} = -3.4 \text{ eV}$.
The ionization energy is the energy required to remove the electron from $n=2$ to $n=\infty$,which is $E_{\infty} - E_2 = 0 - (-3.4) = 3.4 \text{ eV}$.

(C) The wavelength of the spectral lines is given by $\lambda = \frac{1500p^2}{p^2 - 1}$.
To find the ionization energy,we consider the transition from $n = \infty$ to $n = 1$,which corresponds to the series limit.
As $p \to \infty$,the wavelength $\lambda_{\min} = \lim_{p \to \infty} \frac{1500p^2}{p^2 - 1} = 1500\ \mathring{A}$.
The energy corresponding to this wavelength is the ionization energy $E_i$,given by $E_i = \frac{hc}{\lambda_{\min}}$.
Substituting the given values,$E_i = \frac{12420\ eV\cdot\mathring{A}}{1500\ \mathring{A}} = 8.28\ eV$.
Since the ionization energy is $8.28\ eV$,the ionization potential is $8.28\ V$.

(D) The magnetic moment $M$ of an orbital electron is given by $M = \frac{qL}{2m}$,where $L$ is the orbital angular momentum.
According to Bohr's quantization condition,$L = \frac{nh}{2\pi}$,where $n$ is the principal quantum number.
Thus,$M \propto L \propto n$.
For the ground state,$n_1 = 1$,so $\mu_1 \propto 1$.
For the $3^{rd}$ excited state,$n_2 = 1 + 3 = 4$,so $\mu_2 \propto 4$.
Taking the ratio,$\frac{\mu_2}{\mu_1} = \frac{n_2}{n_1} = \frac{4}{1} = 4$.
Therefore,$\mu_2 = 4\mu_1$.

(C) The number of spectral lines emitted for transitions from level $n$ to lower levels is given by the formula $N = \frac{n(n-1)}{2}$.
Given $N = 10$,we have $\frac{n(n-1)}{2} = 10$,which implies $n^2 - n - 20 = 0$.
Solving the quadratic equation,$(n-5)(n+4) = 0$,we get $n = 5$ (since $n$ must be positive).
The angular momentum of an electron in the $n$-th orbit is given by $L = \frac{nh}{2\pi}$.
The change in angular momentum $\Delta L$ for a transition from $n_2$ to $n_1$ is $\Delta L = \frac{(n_2 - n_1)h}{2\pi}$.
To maximize $\Delta L$,we choose the transition from the highest level $n_2 = 5$ to the lowest level $n_1 = 1$.
Thus,$\Delta L_{max} = \frac{(5 - 1)h}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}$.