Bohr's Model of Hydrogen Atom Questions in English

(A) The total energy of the $n$-th orbit is given by $E_n = -\frac{m e^4}{8 \varepsilon_0^2 h^2 n^2}$.
Since $E_n \propto m$,the ratio of the energy of the $\mu^{-}$-system to the hydrogen atom is $\frac{E_{\mu}}{E_e} = \frac{m_{\mu}}{m_e} = 207$.
Thus,$E_{\mu} = 207 \times E_e = 207 \times (-13.6 \text{ eV}) = -13.6 \times 207 \text{ eV}$.
The radius of the $n$-th orbit is given by $r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$.
Since $r_n \propto \frac{1}{m}$,the ratio of the radius of the $\mu^{-}$-system to the hydrogen atom is $\frac{r_{\mu}}{r_H} = \frac{m_e}{m_{\mu}} = \frac{1}{207}$.
Thus,$r_{\mu} = \frac{r_H}{207}$.

(B) The potential energy $U$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $U = -27.2 / n^2 \text{ eV}$.
Given $U = -6.8 \text{ eV}$,we have $-27.2 / n^2 = -6.8$,which implies $n^2 = 27.2 / 6.8 = 4$,so $n = 2$.
The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_0$. For $n = 2$,$r_2 = 2^2 r_0 = 4 r_0$.
According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the de Broglie wavelength: $2 \pi r_n = n \lambda$.
Substituting the values,$2 \pi (4 r_0) = 2 \lambda$.
Solving for $\lambda$,we get $\lambda = (8 \pi r_0) / 2 = 4 \pi r_0$.

(B) According to the de-Broglie hypothesis,a revolving electron in a circular orbit exhibits wave nature.
For a stable circular orbit,the circumference of the orbit must be an integral multiple of the de-Broglie wavelength,which is given by the condition:
$2 \pi r_n = n \lambda$
where $r_n$ is the radius of the $n$th orbit,$n$ is the principal quantum number,and $\lambda$ is the de-Broglie wavelength.
For the first Bohr orbit,$n = 1$.
Substituting $n = 1$ into the equation,we get:
$2 \pi r_1 = 1 \cdot \lambda$
$\lambda = 2 \pi r_1$
Thus,the de-Broglie wavelength of the electron in the first Bohr orbit is equal to the circumference of the first orbit.

(C) The orbital angular momentum of an electron in a hydrogen atom is given by $L = \frac{nh}{2 \pi}$.
Initially, the electron is in the ground state, so $n_1 = 1$. The initial angular momentum is $L_1 = \frac{1 \cdot h}{2 \pi} = \frac{h}{2 \pi}$.
After absorbing energy $\Delta E$, the angular momentum increases by $\frac{h}{2 \pi}$.
Thus, the new angular momentum is $L_2 = L_1 + \frac{h}{2 \pi} = \frac{h}{2 \pi} + \frac{h}{2 \pi} = \frac{2h}{2 \pi} = \frac{2h}{2 \pi}$.
Comparing this with $L = \frac{nh}{2 \pi}$, we find the new principal quantum number is $n_2 = 2$.
The energy of an electron in the $n$-th state is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
For $n_1 = 1$, $E_1 = -13.6 \ eV$.
For $n_2 = 2$, $E_2 = -\frac{13.6 \ eV}{2^2} = -\frac{13.6 \ eV}{4} = -3.4 \ eV$.
The energy absorbed is $\Delta E = E_2 - E_1 = -3.4 \ eV - (-13.6 \ eV) = 10.2 \ eV$.

(B) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
Here, the electron jumps from $n_i = 2$ to $n_f = 1$.
Substituting the values: $\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right)$.
$\frac{1}{\lambda} = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right)$.
Therefore, $\lambda = \frac{4}{3R}$.

(D) In the given hypothetical world, the angular momentum is $L = 2n' \frac{h}{2\pi} = n' \frac{h}{\pi}$, where $n' = 1, 2, 3, \dots$.
Comparing this with the standard Bohr quantization $L = n \frac{h}{2\pi}$, we see that the allowed orbits correspond to $n = 2n'$.
The energy of an orbit is $E_n = -\frac{13.6}{n^2} \text{ eV}$. Substituting $n = 2n'$, we get $E_{n'} = -\frac{13.6}{(2n')^2} = -\frac{13.6}{4n'^2} = -\frac{3.4}{n'^2} \text{ eV}$.
For the visible range, the transition must end at the first excited state of this system. The ground state is $n'=1$ $(n=2)$, and the first excited state is $n'=2$ $(n=4)$.
The transition for the largest wavelength (smallest energy) in the visible range is from $n'=2$ to $n'=1$.
The energy difference is $\Delta E = E_2 - E_1 = -\frac{3.4}{2^2} - (-\frac{3.4}{1^2}) = -0.85 + 3.4 = 2.55 \text{ eV}$.
The wavelength is $\lambda = \frac{hc}{\Delta E} = \frac{1242 \text{ eV-nm}}{2.55 \text{ eV}} \approx 487 \text{ nm}$.

(A, B) The energy required for a transition in $He^{+}$ is given by $E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For $He^{+}$, $Z=2$, so $E = 13.6 \times 4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 54.4 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) eV$.
For a transition from $n=2$ to $n=4$, $E = 54.4 \left( \frac{1}{4} - \frac{1}{16} \right) = 54.4 \left( \frac{3}{16} \right) = 10.2 eV$.
Since the incident photon energy is $10.2 eV$, the $He^{+}$ electron can be excited from $n=2$ to $n=4$.
Once in the $n=4$ state, the number of spectral lines emitted is given by $\frac{n(n-1)}{2} = \frac{4(4-1)}{2} = 6$.
Therefore, statements $A$ and $B$ are correct.

(A) According to Bohr's postulate of quantization of angular momentum, the angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi}$
From this equation, we can express the quantum number $n$ as:
$n = \frac{2\pi m}{h} \cdot (vr)$
Since $m$, $h$, and $\pi$ are constants, we have:
$n \propto vr$
Therefore, the quantity $vr$ is proportional to the quantum number $n$.

(D) According to the Bohr's atomic model, the angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi} \dots (i)$
Since angular velocity $\omega = \frac{v}{r}$, we have $v = r\omega$.
Substituting $v = r\omega$ into equation $(i)$:
$m(r\omega)r = \frac{nh}{2\pi} \Rightarrow m\omega r^2 = \frac{nh}{2\pi} \Rightarrow \omega = \frac{nh}{2\pi mr^2} \dots (ii)$
The radius of the electron in the $n$-th orbit is given by:
$r = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \dots (iii)$
Substituting the expression for $r$ from equation $(iii)$ into equation $(ii)$:
$\omega = \frac{nh}{2\pi m} \left( \frac{\pi m Z e^2}{n^2 h^2 \epsilon_0} \right)^2$
$\omega = \frac{nh}{2\pi m} \cdot \frac{\pi^2 m^2 Z^2 e^4}{n^4 h^4 \epsilon_0^2}$
$\omega = \frac{\pi m Z^2 e^4}{2 h^3 \epsilon_0^2} \cdot \frac{1}{n^3}$
Therefore, $\omega \propto \frac{1}{n^3}$.

(A) The linear velocity $v$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$v = \frac{Z e^2}{2 \epsilon_0 n h}$
Where $Z$ is the atomic number, $e$ is the charge of the electron, $\epsilon_0$ is the permittivity of free space, $n$ is the principal quantum number, and $h$ is Planck's constant.
From this expression, it is clear that all terms except $n$ are constants for a given atom.
Therefore, the relationship is $v \propto \frac{1}{n}$.

(A-D) According to Bohr's model for a hydrogen atom:
$v_{n} \propto \frac{1}{n}$
$E_{n} \propto \frac{1}{n^{2}}$
$r_{n} \propto n^{2}$
For option $A$: $\frac{E_{n} r_{n}}{E_{1} r_{1}} \propto \frac{(1/n^{2}) \cdot n^{2}}{1} = 1$. This is a constant, so the slope is $0$.
For option $B$: $\frac{r_{n} v_{n}}{r_{1} v_{1}} \propto \frac{n^{2} \cdot (1/n)}{1} = n$. This is a straight line with slope $1$.
For option $C$: $\frac{r_{n}}{r_{1}} = n^{2}$. Taking natural log on both sides: $\ln \left(\frac{r_{n}}{r_{1}}\right) = 2 \ln(n)$. This is a straight line with slope $2$.
For option $D$: $\frac{r_{n}}{E_{n}} \propto \frac{n^{2}}{1/n^{2}} = n^{4}$. Thus, $\frac{r_{n} E_{1}}{E_{n} r_{1}} = n^{4}$. Taking natural log: $\ln \left(\frac{r_{n} E_{1}}{E_{n} r_{1}}\right) = 4 \ln(n)$. This is a straight line with slope $4$.
All options $A, B, C,$ and $D$ are mathematically correct based on Bohr's model.

(B) According to Bohr's quantization condition, the angular momentum of an electron in an orbit is given by $L = mvr = \frac{nh}{2\pi}$.
From the de-Broglie hypothesis, the wavelength of an electron is $\lambda = \frac{h}{mv}$.
Substituting $mv = \frac{h}{\lambda}$ into the quantization condition, we get $\frac{h}{\lambda} r = \frac{nh}{2\pi}$.
This simplifies to $2\pi r = n\lambda$, where $2\pi r$ is the circumference of the $n^{th}$ orbit.
For the second Bohr orbit, $n = 2$.
Therefore, the circumference of the second orbit is $2\pi r = 2\lambda$.
This implies that the number of de-Broglie wavelengths contained in the second Bohr orbit is $2$.

(D) The energy of the photon is given by $E = \frac{hc}{\lambda}$.
Substituting the given values:
$E = \frac{4 \times 10^{-15} \ eV \cdot s \times 3 \times 10^{8} \ m/s}{300 \times 10^{-9} \ m} = \frac{12 \times 10^{-7}}{300 \times 10^{-9}} \ eV = \frac{1200}{300} \ eV = 4 \ eV$.
The energy required to excite a hydrogen atom from the ground state $(n=1)$ to the first excited state $(n=2)$ is $\Delta E = 13.6 \ eV \times (1 - \frac{1}{4}) = 13.6 \times 0.75 = 10.2 \ eV$.
Since the energy of the photon $(4 \ eV)$ is less than the energy required for the first excitation $(10.2 \ eV)$ and also less than the ionization energy $(13.6 \ eV)$, the electron cannot absorb this energy to transition to a higher state.
Therefore, the electron will remain in the ground state.

(A) For a hydrogen-like atom, the velocity of an electron in the $n^{th}$ orbit is given by $v \propto \frac{Z}{n}$.
The radius of the $n^{th}$ orbit is given by $r \propto \frac{n^2}{Z}$.
From these two relations, we can write $r \propto \frac{n}{v}$ (since $Z \propto \frac{n}{v} \cdot n = \frac{n^2}{r}$, substituting $Z$ into the velocity relation).
Given that the radii are nearly the same $(r_1 \approx r_2)$, we have $\frac{n_1}{v_1} = \frac{n_2}{v_2}$.
Therefore, $\frac{n_1}{n_2} = \frac{v_1}{v_2} = \frac{3 \times 10^5}{2.5 \times 10^5} = \frac{3}{2.5} = \frac{6}{5}$.
Thus, the possible order of energy states ($n_1$ and $n_2$) is $6$ and $5$.

(C) According to Bohr's postulate, the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$, where $n$ is the principal quantum number.
Rearranging this, we get $\frac{2\pi L}{h} = n$.
The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{E_0}{n^2}$, where $E_0$ is the magnitude of the ground state energy.
Given that $E_n = -0.04 E_0$, we have $-\frac{E_0}{n^2} = -0.04 E_0$.
Dividing both sides by $-E_0$, we get $\frac{1}{n^2} = 0.04$.
$n^2 = \frac{1}{0.04} = \frac{100}{4} = 25$.
Therefore, $n = 5$.
Since $\frac{2\pi L}{h} = n$, the value is $5$.

(A) According to Bohr's model, the orbital angular momentum $L$ is given by the formula $L = n \frac{h}{2\pi}$.
The ground state corresponds to $n = 1$.
The first excited state is $n = 2$, the second excited state is $n = 3$, and the third excited state is $n = 4$.
Substituting $n = 4$ into the formula:
$L = 4 \times \frac{h}{2\pi} = 2 \times \frac{h}{\pi}$.
Given $h = 6.63 \times 10^{-34} \text{ Js}$ and $\pi \approx 3.14$:
$L = 2 \times \frac{6.63 \times 10^{-34}}{3.14} \approx 2 \times 2.111 \times 10^{-34} = 4.222 \times 10^{-34} \text{ kg m}^2\text{s}^{-1}$.

(D) For a hydrogen atom in the ground state $(n=1)$, Bohr's quantization condition for angular momentum is given by $mvr = \frac{nh}{2\pi}$.
Here, $m$ is the mass of the electron $(9.1 \times 10^{-31} \text{ kg})$, $v$ is the velocity, $r$ is the orbital radius $(5.3 \times 10^{-11} \text{ m})$, and $h$ is Planck's constant $(6.63 \times 10^{-34} \text{ Js})$.
Rearranging the formula for velocity $v$, we get $v = \frac{h}{2\pi mr}$.
Substituting the values: $v = \frac{6.63 \times 10^{-34}}{2 \times 3.14159 \times 9.1 \times 10^{-31} \times 5.3 \times 10^{-11}}$.
Calculating this, $v \approx 2.18 \times 10^6 \text{ ms}^{-1}$.
Rounding to two significant figures, we get $v \approx 2.2 \times 10^6 \text{ ms}^{-1}$.

(C) In the Bohr model, the Total Energy $(TE)$ of an electron in the ground state is $-13.6 \text{ eV}$.
Kinetic Energy $(KE)$ is given by the relation $KE = -TE$. Therefore, $KE = -(-13.6 \text{ eV}) = +13.6 \text{ eV}$.
Potential Energy $(PE)$ is given by the relation $PE = 2 \times TE$. Therefore, $PE = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$.
Thus, the potential energy is $-27.2 \text{ eV}$ and the kinetic energy is $+13.6 \text{ eV}$.

(A) In a hydrogen-like atom, the total energy $E$ of an electron is related to its kinetic energy $K$ and potential energy $U$ by the following relations:
$E = -K$
$U = 2E$
Given that the total energy $E = -13.6 \text{ eV}$, we can find $K$ and $U$:
$K = -E = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$
$U = 2E = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{13.6 \text{ eV}}{-27.2 \text{ eV}} = -\frac{1}{2}$.
Thus, the correct option is $A$.

(D) The radius of the $n$-th orbit in a hydrogen atom is given by the formula $r_n = n^2 r_1$, where $r_1$ is the radius of the first orbit (Bohr radius).
Given $r_1 = 5.3 \times 10^{-11} \text{ m}$ and $n = 3$.
Substituting these values into the formula:
$r_3 = 3^2 \times (5.3 \times 10^{-11} \text{ m})$
$r_3 = 9 \times 5.3 \times 10^{-11} \text{ m}$
$r_3 = 47.7 \times 10^{-11} \text{ m}$
$r_3 = 4.77 \times 10^{-10} \text{ m}$.
Thus, the radius of the $n = 3$ orbit is $4.77 \times 10^{-10} \text{ m}$.
Therefore, the correct option is $D$.

(C) According to Bohr's quantization condition, the angular momentum $L$ of an electron in an orbit is given by $L = \frac{nh}{2\pi}$.
Given that $L = \frac{3h}{\pi}$, we equate the two expressions: $\frac{nh}{2\pi} = \frac{3h}{\pi}$.
Solving for $n$, we get $n = 6$.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = -\frac{13.6}{n^2} \text{ eV}$.
Substituting $n = 6$ into the formula: $E_6 = -\frac{13.6}{6^2} = -\frac{13.6}{36}$.
Calculating the value: $E_6 \approx -0.377 \text{ eV}$, which rounds to $-0.38 \text{ eV}$.

(C) The radius of the $n$-th orbit in a hydrogen atom is given by $r_n = a_0 n^2$, where $a_0$ is the Bohr radius.
Therefore, the ratio of the radii is $r_i / r_f = n_i^2 / n_f^2 = 16 / 4 = 4$.
Taking the square root, we get $n_i / n_f = 2$, which implies $n_i = 2n_f$.
For the simplest transition, we take $n_f = 2$ and $n_i = 4$.
Using the Rydberg formula for the wavelength $\lambda$:
$1/\lambda = R(1/n_f^2 - 1/n_i^2)$
$1/\lambda = R(1/2^2 - 1/4^2) = R(1/4 - 1/16) = R(3/16)$.
Substituting $R = 1.0973 \times 10^7 \text{ m}^{-1}$:
$1/\lambda = 1.0973 \times 10^7 \times (3/16) \approx 0.20574 \times 10^7 \text{ m}^{-1}$.
$\lambda = 1 / (0.20574 \times 10^7) \approx 4.86 \times 10^{-7} \text{ m} = 486 \text{ nm}$.

(C) The magnetic field $B$ at the center of the orbit due to an electron moving in a circular path is given by $B = \frac{\mu_0 I}{2r}$.
The current $I$ is given by $I = \frac{ev}{2\pi r}$, where $e$ is the charge of the electron, $v$ is the velocity, and $r$ is the radius.
According to Bohr's model, $v \propto \frac{1}{n}$ and $r \propto n^2$.
Substituting these into the expression for current: $I \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
Now, substituting $I$ and $r$ into the expression for $B$: $B \propto \frac{I}{r} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.
Therefore, the ratio of the magnetic fields for the $2^{nd}$ and $4^{th}$ orbits is $\frac{B_2}{B_4} = \left( \frac{4}{2} \right)^5 = 2^5 = 32$.
Thus, the ratio is $32:1$.

(B) For a hydrogen atom, the radius of the $n^{th}$ orbit is given by the formula $r_n = a_0 \times n^2$, where $a_0 = 0.529 \text{ Å}$ is the Bohr radius.
For the first excited state, the principal quantum number is $n = 2$.
Substituting the value of $n$ into the formula, we get $r_2 = 0.529 \times (2)^2 \text{ Å}$.
$r_2 = 0.529 \times 4 \text{ Å} = 2.116 \text{ Å}$.
Since $1 \text{ Å} = 10^{-10} \text{ m}$, we have $r_2 = 2.116 \times 10^{-10} \text{ m}$.
Rounding to two significant figures, the radial distance is approximately $2.1 \times 10^{-10} \text{ m}$.

(D) According to Bohr's theory, for an electron in a hydrogen atom, the electrostatic force provides the necessary centripetal force:
$\frac{1}{4\pi \epsilon_0} \frac{e^2}{r^2} = \frac{m v^2}{r}$
Rearranging for radius $r$:
$r = \frac{1}{4\pi \epsilon_0} \frac{e^2}{m v^2}$
Given values:
$v = \sqrt{25.6} \times 10^5 \text{ m s}^{-1} \implies v^2 = 25.6 \times 10^{10} \text{ m}^2 \text{ s}^{-2}$
$m = 9 \times 10^{-31} \text{ kg}$
$e = 1.6 \times 10^{-19} \text{ C}$
$\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$
Substituting these values:
$r = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{9 \times 10^{-31} \times 25.6 \times 10^{10}}$
$r = \frac{9 \times 10^9 \times 2.56 \times 10^{-38}}{9 \times 25.6 \times 10^{-21}}$
$r = \frac{2.56 \times 10^{-29}}{25.6 \times 10^{-21}} = 0.1 \times 10^{-8} \text{ m} = 1 \times 10^{-9} \text{ m}$
Comparing with $x \times 10^{-9} \text{ m}$, we get $x = 1$.

(D) According to the Bohr model, the radius of the $n^{th}$ orbit is given by $r_n = n^2 a_0$, where $a_0$ is the Bohr radius.
The circumference of the $n^{th}$ orbit is $2\pi r_n = 2\pi n^2 a_0$.
According to the de Broglie hypothesis, the condition for a stable orbit is $n\lambda = 2\pi r_n$.
Substituting the expression for $r_n$, we get $n\lambda_n = 2\pi n^2 a_0$, which simplifies to $\lambda_n = 2\pi n a_0$.
For the ground state $(n = 1)$, $\lambda_1 = 2\pi(1)a_0 = 2\pi a_0$.
For the $n = 3$ level, $\lambda_3 = 2\pi(3)a_0 = 3(2\pi a_0) = 3\lambda_1$.
Therefore, $\lambda_3 = 3\lambda_1$.

(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the first case: The electron jumps from the second excited state $(n_i = 3)$ to the first excited state $(n_f = 2)$.
$\frac{1}{\lambda_0} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right)$.
So, $\lambda_0 = \frac{36}{5R}$.
For the second case: The electron jumps from the third excited state $(n_i = 4)$ to the second orbit $(n_f = 2)$.
$\frac{1}{\lambda} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{3}{16} \right)$.
So, $\lambda = \frac{16}{3R}$.
Now, express $\lambda$ in terms of $\lambda_0$:
$\lambda = \frac{16}{3R} = \frac{16}{3} \times \frac{5\lambda_0}{36} = \frac{80\lambda_0}{108} = \frac{20\lambda_0}{27}$.
Comparing this with the given expression $\frac{20\lambda_0}{x}$, we get $x = 27$.

(C) The energy of the emitted photon $(E)$ is given by the sum of the work function $(\phi)$ and the maximum kinetic energy $(K_{max})$:
$E = \phi + K_{max} = 2.65 \text{ eV} + 10.1 \text{ eV} = 12.75 \text{ eV}$.
In a hydrogen atom, the energy of an electron in the $n^{th}$ state is $E_n = -13.6 / n^2 \text{ eV}$.
The energy of the photon emitted when jumping from state $n$ to the ground state $(n=1)$ is:
$E = E_n - E_1 = -13.6 / n^2 - (-13.6 / 1^2) = 13.6 (1 - 1/n^2) \text{ eV}$.
Equating the two expressions:
$12.75 = 13.6 (1 - 1/n^2)$
$12.75 / 13.6 = 1 - 1/n^2$
$0.9375 = 1 - 1/n^2$
$1/n^2 = 1 - 0.9375 = 0.0625$
$n^2 = 1 / 0.0625 = 16$
$n = 4$.

(A) According to Bohr's quantization postulate, the angular momentum $L$ of an electron in an orbit is given by $L = mvr = nh / 2\pi$.
For the first orbit, $n = 1$, so $mvr = h / 2\pi$.
The velocity $v$ is given by $v = h / (2\pi mr)$.
The number of revolutions per second $f$ is the frequency, which is defined as $f = v / (2\pi r)$.
Substituting the value of $v$ into the frequency formula: $f = (h / 2\pi mr) / (2\pi r) = h / 4\pi^2mr^2$.
Thus, the correct option is $A$.

(B) In a hydrogen atom, the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$) provides the necessary centripetal force for the electron to move in a circular orbit of radius $r$.
According to Coulomb's law, the electrostatic force is $F = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
The centripetal force required for circular motion is $F = \frac{mv^2}{r}$.
Equating these two forces: $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
Multiplying both sides by $r/2$, we get $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{2r}$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\epsilon_0 r}$.
Since $e$, $\pi$, and $\epsilon_0$ are constants, the kinetic energy $K$ is proportional to $e^2/r$.

(A) According to Bohr's theory for a hydrogen atom:
$1$. The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
$2$. The speed of the electron in the $n^{th}$ orbit is given by $V_n \propto 1/n$.
Now, let us analyze the product $Vr$:
$Vr \propto (1/n) \times n^2 = n$.
This is directly proportional to $n$.
Let us analyze $1/Vr$:
$1/Vr \propto 1/n$.
Thus, the quantity $1/Vr$ is inversely proportional to the principal quantum number '$n$'.

(A) The magnetic dipole moment '$M$' of an electron revolving in a circular orbit of radius '$r$' with velocity '$v$' is given by $M = IA$, where '$I$' is the current and '$A$' is the area of the orbit.
$I = e/T = ev / (2\pi r)$
$A = \pi r^2$
So, $M = (ev / 2\pi r) \times \pi r^2 = evr / 2$.
The angular momentum '$L$' of the electron is $L = mvr$.
From the expression for '$M$',we have $M = (e/2m) \times (mvr) = (e/2m) \times L$.
Rearranging for '$L$',we get $L = 2mM / e$.

(C) The centripetal acceleration $a_c$ of an electron in a Bohr orbit is given by $a_c = \frac{v^2}{r}$.
For a hydrogen atom, the velocity $v$ in the $n^{th}$ orbit is proportional to $\frac{1}{n}$ $(v \propto \frac{1}{n})$ and the radius $r$ is proportional to $n^2$ $(r \propto n^2)$.
Substituting these into the formula for centripetal acceleration:
$a_c \propto \frac{(1/n)^2}{n^2} = \frac{1/n^2}{n^2} = \frac{1}{n^4}$.
Therefore, the ratio of centripetal acceleration for the $3^{rd}$ and $5^{th}$ orbits is:
$\frac{a_3}{a_5} = \frac{n_5^4}{n_3^4} = \left(\frac{5}{3}\right)^4 = \frac{625}{81}$.
Thus, the ratio is $625 : 81$.

(C) In the Bohr model of the hydrogen atom, the total energy $E$ of an electron in the $n^{th}$ orbit is given by $E = -\frac{13.6}{n^2} \text{ eV}$.
This implies $E \propto n^{-2}$.
The angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
This implies $L \propto n$, or $n \propto L$.
Substituting $n \propto L$ into the energy relation $E \propto n^{-2}$, we get $E \propto (L)^{-2}$, which means $E \propto L^{-2}$.

(B) The de-Broglie wavelength $\lambda$ associated with an electron in the $n^{th}$ orbit of a hydrogen atom is given by the relation $\lambda = \frac{h}{p} = \frac{2\pi r_n}{n}$, where $r_n$ is the radius of the $n^{th}$ orbit.
Since $r_n \propto n^2$, we have $\lambda \propto \frac{n^2}{n} = n$.
In the ground state, $n_1 = 1$.
The third excited state corresponds to $n_2 = 4$.
Since $n_2 > n_1$, the de-Broglie wavelength $\lambda$ increases as the electron moves to a higher energy state.

(A) The radius of the $n^{th}$ Bohr orbit in a hydrogen atom is given by the formula:
$r_n = n^2 a_0$
where $n$ is the principal quantum number $(n = 1, 2, 3, 4, ...)$ and $a_0$ is the Bohr radius.
For the first four orbits:
For $n = 1$, $r_1 = 1^2 a_0 = 1 a_0$
For $n = 2$, $r_2 = 2^2 a_0 = 4 a_0$
For $n = 3$, $r_3 = 3^2 a_0 = 9 a_0$
For $n = 4$, $r_4 = 4^2 a_0 = 16 a_0$
Thus, the ratio of the radii is $r_1 : r_2 : r_3 : r_4 = 1 : 4 : 9 : 16$.

(A) According to Bohr's model, the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2h^2\epsilon_0}{\pi me^2}$.
The velocity of the electron in the $n^{th}$ orbit is given by $v_n = \frac{e^2}{2\epsilon_0nh}$.
The orbital period $T$ is the time taken to complete one revolution, which is $T = \frac{2\pi r_n}{v_n}$.
Substituting the expressions for $r_n$ and $v_n$:
$T = \frac{2\pi (\frac{n^2h^2\epsilon_0}{\pi me^2})}{(\frac{e^2}{2\epsilon_0nh})}$
$T = \frac{2n^2h^2\epsilon_0}{me^2} \times \frac{2\epsilon_0nh}{e^2}$
$T = \frac{4\epsilon_0^2n^3h^3}{me^4}$.
Thus, the correct option is $A$.

(B) According to Bohr's theory, 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 Bohr's model, the radius of the $n^{th}$ orbit is proportional to $n^2$, i.e.,$r \propto n^2$.
Substituting this into the force equation: $F \propto \frac{1}{r^2} \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.
Therefore, $F \propto n^{-4}$.

(B) The orbital magnetic moment $(m_{orb})$ of an electron revolving in an orbit is given by the formula: $m_{orb} = \frac{e}{2m_e} L$, where $L$ is the orbital angular momentum.
According to Bohr's quantization condition, the orbital angular momentum of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
Substituting this value into the expression for $m_{orb}$, we get: $m_{orb} = \frac{e}{2m_e} \times \frac{nh}{2\pi}$.
Since $e$, $m_e$, $h$, and $\pi$ are constants, we can see that $m_{orb} \propto n$.
Therefore, the orbital magnetic moment is directly proportional to the principal quantum number $n$.

(B) 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 the second orbit of a hydrogen atom $(Z=1, n=2)$: $E = -13.6 \times \frac{1^2}{2^2} = -13.6 \times \frac{1}{4} \text{ eV}$.
Thus, $-13.6 \text{ eV} = 4E$.
For the third orbit of a helium atom $(Z=2, n=3)$: $E' = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} \text{ eV}$.
Substituting $-13.6 = 4E$ into the equation for $E'$:
$E' = (4E) \times \frac{4}{9} = \frac{16E}{9}$.

(C) According to Bohr's quantization postulate, the angular momentum $L$ of an electron in an orbit with quantum number $n$ is given by $L = n(h / 2\pi)$.
For the third Bohr orbit $(n_1 = 3)$, the angular momentum is $l = 3(h / 2\pi)$.
For the fourth Bohr orbit $(n_2 = 4)$, the angular momentum is $L' = 4(h / 2\pi)$.
Dividing the two expressions: $L' / l = (4(h / 2\pi)) / (3(h / 2\pi)) = 4/3$.
Therefore, $L' = (4/3)l$.

(C) The radius of an electron in a hydrogen-like atom is given by the formula $r_n = a_0 \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the principal quantum number, and $Z$ is the atomic number.
For the ground state of hydrogen, $n_H = 1$ and $Z_H = 1$. Thus, $r_H = a_0 \frac{1^2}{1} = a_0$.
For triply ionized beryllium $(Be^{3+})$, $Z_{Be} = 4$. Let the energy state be $n_{Be}$. The radius is $r_{Be} = a_0 \frac{n_{Be}^2}{4}$.
Given that $r_{Be} = r_H$, we have $a_0 \frac{n_{Be}^2}{4} = a_0$.
This simplifies to $\frac{n_{Be}^2}{4} = 1$, which means $n_{Be}^2 = 4$.
Therefore, $n_{Be} = 2$.

(A) 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.
Given the condition: $r_{n+1} - r_n = r_{n-1}$.
Substituting the formula: $a_0(n+1)^2 - a_0 n^2 = a_0(n-1)^2$.
Dividing by $a_0$: $(n+1)^2 - n^2 = (n-1)^2$.
Expanding the terms: $(n^2 + 2n + 1) - n^2 = n^2 - 2n + 1$.
Simplifying: $2n + 1 = n^2 - 2n + 1$.
Rearranging the equation: $n^2 - 4n = 0$.
Factoring: $n(n - 4) = 0$.
Since $n$ represents the orbit number, $n$ must be greater than $1$ for the $(n-1)^{th}$ orbit to exist. Thus, $n = 4$.

(A) According to Bohr's theory, the orbital angular momentum $L$ of an electron in the $n^{th}$ orbit is given by $L = \frac{nh}{2\pi}$.
The magnetic moment $\mu_l$ associated with the orbital motion of an electron is given by $\mu_l = \frac{e}{2m} L$.
Substituting the value of $L$, we get $\mu_l = \frac{e}{2m} \left( \frac{nh}{2\pi} \right)$.
Since $e$, $m$, $h$, and $\pi$ are constants, we can see that $\mu_l \propto n$.
Therefore, the magnetic moment is proportional to the principal quantum number $n$.

(C) According to Bohr's quantization condition, the angular momentum of an electron in the $n^{th}$ orbit is given by $mvr = \frac{nh}{2\pi}$.
From the de-Broglie relation, the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
Rearranging the Bohr quantization condition, we get $mv = \frac{nh}{2\pi r}$.
Substituting this into the de-Broglie wavelength formula: $\lambda = \frac{h}{(nh / 2\pi r)}$.
Simplifying the expression, we get $\lambda = \frac{2\pi r}{n}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = -13.6/n^2 \text{ eV}$.
Given $E_n = -0.544 \text{ eV}$.
So, $-13.6/n^2 = -0.544$.
$n^2 = 13.6 / 0.544 = 25$.
Therefore, $n = 5$.
The angular momentum $(L)$ of an electron in the $n^{th}$ orbit is given by Bohr's quantization condition: $L = nh/2\pi$.
Substituting $n = 5$, we get $L = 5h/2\pi$.

(D) According to Bohr's postulate for the hydrogen atom, an electron can revolve only in those orbits for which its angular momentum $(L)$ is an integral multiple of $h/2\pi$.
This is expressed by the quantization condition: $L = n(h/2\pi)$, where $n = 1, 2, 3, ...$ is the principal quantum number, and $h$ is Planck's constant.
Therefore, the angular momentum is the quantity that is an integral multiple of $h/2\pi$.

(C) In a hydrogen atom, the electrostatic force of attraction between the nucleus (charge $+e$) and the electron (charge $-e$) provides the necessary centripetal force for the electron to move in a circular orbit of radius $r$.
According to Coulomb's law, the electrostatic force is $F = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
The centripetal force required for circular motion is $F = \frac{mv^2}{r}$.
Equating these, we get $\frac{mv^2}{r} = \frac{1}{4\pi\epsilon_0} \frac{e^2}{r^2}$.
Multiplying both sides by $r/2$, we get $\frac{1}{2}mv^2 = \frac{1}{4\pi\epsilon_0} \frac{e^2}{2r}$.
The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2 = \frac{e^2}{8\pi\epsilon_0 r}$.
Since $\epsilon_0$ is a constant, the kinetic energy is proportional to $e^2/r$.

(B) For a hydrogen atom in the first orbit $(n=1)$:
$1$. The total energy is given by $E_T = -13.6 \text{ eV}$.
$2$. The kinetic energy is related to the total energy by $E_K = -E_T$, therefore $E_K = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$.
$3$. The potential energy is related to the total energy by $E_P = 2E_T$, therefore $E_P = 2 \times (-13.6 \text{ eV}) = -27.2 \text{ eV}$.

(C) Bohr's second postulate states that an electron can revolve only in those orbits for which its angular momentum $L$ is an integral multiple of $\frac{h}{2\pi}$.
Mathematically, $L = n \frac{h}{2\pi}$, where $n = 1, 2, 3, ...$ is the principal quantum number and $h$ is Planck's constant.
This condition is known as the quantisation of angular momentum.