Bohr's Model of Hydrogen Atom Questions in English

(B) From Bohr's quantization postulate,the angular momentum is given by $mvr = \frac{nh}{2\pi}$.
For the first orbit $(n=1)$,the velocity $v$ is $v = \frac{h}{2\pi mr}$.
The centripetal acceleration $a_c$ is given by $a_c = \frac{v^2}{r}$.
Substituting the expression for $v$ into the acceleration formula:
$a_c = \frac{(\frac{h}{2\pi mr})^2}{r} = \frac{h^2}{4\pi^2 m^2 r^2 \cdot r} = \frac{h^2}{4\pi^2 m^2 r^3}$.
Since $4\pi^2$ is a constant,we have $a_c \propto \frac{h^2}{m^2 r^3}$.

(D) According to de Broglie's hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
According to Bohr's postulate for the quantization of angular momentum,$L = mvr_n = \frac{nh}{2\pi}$.
Rearranging this equation to solve for the momentum $mv$,we get $mv = \frac{nh}{2\pi r_n}$.
Substituting this expression for $mv$ into the de Broglie wavelength formula:
$\lambda = \frac{h}{mv} = \frac{h}{(nh / 2\pi r_n)} = \frac{2\pi r_n}{n}$.
Thus,the de Broglie wavelength of an electron in the $n^{\text{th}}$ orbit is $\frac{2\pi r}{n}$.

(B) The centripetal acceleration $a$ is given by $a = \frac{v^2}{r}$.
In Bohr's model,the velocity $v \propto \frac{1}{n}$ and the radius $r \propto n^2$.
Substituting these into the expression for acceleration: $a \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
Therefore,the ratio of centripetal acceleration for the $3^{\text{rd}}$ orbit $(n_1 = 3)$ to the $5^{\text{th}}$ orbit $(n_2 = 5)$ is:
$\frac{a_3}{a_5} = \frac{n_2^4}{n_1^4} = \frac{5^4}{3^4} = \frac{625}{81}$.

(D) According to Bohr's theory,the angular momentum of an electron in the ground state $(n=1)$ is given by $L = mvr = \frac{h}{2 \pi}$.
From this,the velocity $v$ is $v = \frac{h}{2 \pi mR}$.
The time period $T$ of the electron revolving in the orbit is $T = \frac{2 \pi R}{v} = \frac{2 \pi R}{h / (2 \pi mR)} = \frac{4 \pi^2 mR^2}{h}$.
The equivalent current $I$ due to the revolving electron is $I = \frac{e}{T} = \frac{e}{4 \pi^2 mR^2 / h} = \frac{eh}{4 \pi^2 mR^2}$.
The orbital magnetic moment $M$ is given by $M = I \times A$,where $A = \pi R^2$ is the area of the orbit.
Substituting the values,$M = \left( \frac{eh}{4 \pi^2 mR^2} \right) \times (\pi R^2) = \frac{eh}{4 \pi m}$.
Thus,the correct option is $D$.

(D) The magnetic moment $M_0$ of an electron due to its orbital motion is given by the formula: $M_0 = \frac{e}{2m_e} L_0$,where $e$ is the charge of the electron,$m_e$ is the mass of the electron,and $L_0$ is the orbital angular momentum.
According to Bohr's quantization condition,the orbital angular momentum $L_0$ is given by $L_0 = \frac{nh}{2\pi}$,where $n$ is the principal quantum number and $h$ is Planck's constant.
Substituting this into the magnetic moment expression,we get $M_0 = \frac{e}{2m_e} \times \frac{nh}{2\pi}$.
Since $e$,$m_e$,$h$,and $\pi$ are constants,it follows that $M_0 \propto n$.

(D) According to Bohr's model,the velocity of an electron in the $n^{\text{th}}$ orbit is $v_n \propto \frac{1}{n}$.
The radius of the $n^{\text{th}}$ orbit is $r_n \propto n^2$.
The frequency of revolution $f$ is given by $f = \frac{v}{2\pi r}$.
Substituting the proportionalities,we get $f \propto \frac{(1/n)}{n^2} = \frac{1}{n^3}$.
Therefore,the frequency of revolution is inversely proportional to $n^3$.

(A) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.
The velocity $v$ of an electron in the $n^{\text{th}}$ orbit is given by $v = \frac{v_0}{n}$,where $v_0$ is a constant.
We need to find the ratio $\frac{L}{v}$.
Substituting the expressions,we get $\frac{L}{v} = \frac{nh/2\pi}{v_0/n}$.
Simplifying this,we get $\frac{L}{v} = \frac{h}{2\pi v_0} \cdot n^2$.
Since $h$,$\pi$,and $v_0$ are constants,the ratio $\frac{L}{v}$ is proportional to $n^2$.

(D) The magnetic field $B$ at the centre of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$,where $I$ is the current and $r$ is the radius of the orbit.
For an electron revolving in an orbit,the current $I = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the velocity and $T$ is the time period.
Substituting $I$ into the magnetic field formula: $B = \frac{\mu_0 ev}{4\pi r^2}$.
According to Bohr's model,the radius $r \propto n^2$ and the velocity $v \propto n^{-1}$.
Substituting these proportionalities: $B \propto \frac{v}{r^2} \propto \frac{n^{-1}}{(n^2)^2} = \frac{n^{-1}}{n^4} = n^{-5}$.
Therefore,the magnetic field at the centre is proportional to $n^{-5}$.

(D) In a hydrogen atom,the orbital velocity $v$ of an electron in the $n^{th}$ orbit is given by $v_n = \frac{v_1}{n}$,where $v_1$ is the velocity in the ground state $(n=1)$.
Given that the new velocity $v_n = \frac{1}{3} v_1$,we have $\frac{v_1}{n} = \frac{v_1}{3}$,which implies $n = 3$.
The radius of the $n^{th}$ orbit is given by $r_n = n^2 r_1$.
Substituting $n = 3$ into the formula,we get $r_3 = (3)^2 r_1 = 9 r_1$.
Therefore,the radius of the orbit is $9 r_1$.

(A) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = n \frac{h}{2 \pi}$.
For the second orbit,$n = 2$,so $L = 2 \frac{h}{2 \pi} = \frac{h}{\pi}$.
The rotational energy $E$ of a diatomic molecule is given by $E = \frac{L^2}{2I}$.
Substituting the value of $L$,we get $E = \frac{(\frac{h}{\pi})^2}{2I} = \frac{h^2}{\pi^2 \cdot 2I} = \frac{h^2}{2 I \pi^2}$.
Thus,the correct option is $A$.

(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 $\mu$ associated with an orbital electron is given by $\mu = \frac{e}{2m} L$.
Substituting the value of $L$,we get $\mu = \frac{e}{2m} \left( \frac{nh}{2\pi} \right)$.
Since $e$,$m$,$h$,and $\pi$ are constants,we have $\mu \propto n$.
Therefore,the magnetic moment is proportional to the principal quantum number $n$.

(A) According to Bohr's model, the radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n = n^2 r_1$, where $r_1$ is the radius of the innermost orbit (ground state).
Given $r_1 = 5.3 \times 10^{-11} \ m = 0.53 \ Å$.
For the fourth orbit, $n = 4$.
Therefore, $r_4 = (4)^2 \times r_1 = 16 \times 0.53 \ Å$.
$r_4 = 8.48 \ Å$.

(D) In the Bohr model,the potential energy $(P.E.)$ of an electron in the $n^{\text{th}}$ orbit is given by $P.E. = -\frac{kZe^2}{r_n}$. Since $r_n \propto n^2$,we have $P.E. \propto -\frac{1}{n^2}$.
As $n$ increases,the magnitude of the negative potential energy decreases,meaning the value becomes less negative (i.e.,it increases).
For $n=1$,$P.E. = -27.2 \text{ eV}$.
For $n=2$,$P.E. = -6.8 \text{ eV}$.
Since $-6.8 \text{ eV} > -27.2 \text{ eV}$,the potential energy in the $n=2$ orbit is greater than in the $n=1$ orbit.
Therefore,the statement in option $D$ is false.

(D) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2 \pi}$.
Since $L = I \omega$,we have $I \omega = \frac{nh}{2 \pi}$,which implies $\omega = \frac{nh}{2 \pi I}$.
The rotational kinetic energy $E_r$ is given by $E_r = \frac{1}{2} I \omega^2$.
Substituting the value of $\omega$,we get $E_r = \frac{1}{2} I \left(\frac{nh}{2 \pi I}\right)^2$.
Simplifying this expression,$E_r = \frac{1}{2} I \left(\frac{n^2 h^2}{4 \pi^2 I^2}\right) = \frac{n^2 h^2}{8 \pi^2 I}$.

(C) The radius of the $n^{\text{th}}$ Bohr orbit is given by the formula $r_n = n^2 r_0$,where $r_0$ is the radius of the first Bohr orbit $(n=1)$.
For the first Bohr orbit $(n=1)$,$r_1 = 1^2 r_0 = r_0$.
For the second Bohr orbit $(n=2)$,$r_2 = 2^2 r_0 = 4 r_0$.
The ratio of the radius of the first Bohr orbit to the second Bohr orbit is $\frac{r_1}{r_2} = \frac{r_0}{4 r_0} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(B) The orbital velocity of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by the formula: $V_n = \frac{e^2}{2 \varepsilon_0 h n}$.
From this expression,it is clear that the velocity is inversely proportional to the principal quantum number $n$,i.e.,$V_n \propto \frac{1}{n}$.
Therefore,for the $p^{\text{th}}$ and $n^{\text{th}}$ orbits,we have the ratio:
$\frac{V_p}{V_n} = \frac{1/p}{1/n} = \frac{n}{p}$.
Thus,the ratio $V_p : V_n$ is $n : p$.

(C) In the Bohr model,the centripetal force required for the circular motion of the electron is provided by the electrostatic Coulomb force of attraction between the nucleus (proton) and the electron.
The condition for equilibrium is:
$\text{Centripetal Force} = \text{Coulomb Force}$
$\frac{mv^2}{r_0} = \frac{1}{4 \pi \varepsilon_0} \frac{e \cdot e}{r_0^2}$
Simplifying the equation for $v^2$:
$v^2 = \frac{1}{4 \pi \varepsilon_0} \frac{e^2}{r_0 m}$
Taking the square root on both sides:
$v = \sqrt{\frac{e^2}{4 \pi \varepsilon_0 r_0 m}} = \frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$
Thus,the speed of the electron is $\frac{e}{\sqrt{4 \pi \varepsilon_0 r_0 m}}$.

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{th}$ orbit is given by:
$L = \frac{nh}{2\pi}$
This implies that $L \propto n$.
For the third Bohr orbit $(n_1 = 3)$,the angular momentum is $L_1 = l$.
For the fourth Bohr orbit $(n_2 = 4)$,let the angular momentum be $L_2 = L'$.
Using the proportionality $L \propto n$,we have:
$\frac{L_1}{L_2} = \frac{n_1}{n_2}$
$\frac{l}{L'} = \frac{3}{4}$
$L' = \frac{4}{3} l$

(C) The orbital speed of an electron in the $n^{th}$ Bohr orbit is given by $v_n \propto \frac{1}{n}$.
Given that the initial state is the ground state $(n_1 = 1)$ with speed $v_1$,and the final speed is $v_2 = \frac{v_1}{3}$.
Using the relation $\frac{v_1}{v_2} = \frac{n_2}{n_1}$,we get $\frac{v_1}{v_1/3} = \frac{n_2}{1}$,which implies $n_2 = 3$.
The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
Therefore,$\frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$.
Thus,the radius of the orbit in the excited state is $r_2 = 9 r_1$.

(D) For a hydrogen atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
Since the area of the orbit $A_n = \pi r_n^2$,we have $A_n \propto (n^2)^2 = n^4$.
The first excited state corresponds to $n = 2$,and the second excited state corresponds to $n = 3$.
Therefore,the ratio of the area of the second excited state $(A_3)$ to the first excited state $(A_2)$ is:
$\frac{A_3}{A_2} = \left(\frac{3}{2}\right)^4 = \frac{81}{16}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
Since $E \propto \frac{Z^2}{n^2}$,we can write the ratio for the two cases.
For the hydrogen atom $(Z_H = 1)$ in the second orbit $(n_H = 2)$: $E_H = E \propto \frac{1^2}{2^2} = \frac{1}{4}$.
For the helium atom $(Z_{He} = 2)$ in the third orbit $(n_{He} = 3)$: $E_{He} \propto \frac{2^2}{3^2} = \frac{4}{9}$.
Now,taking the ratio: $\frac{E_{He}}{E_H} = \frac{4/9}{1/4} = \frac{4}{9} \times 4 = \frac{16}{9}$.
Therefore,$E_{He} = \frac{16}{9} E$.

(B) The centripetal force $F$ acting on the electron is given by $F = \frac{mv^2}{r}$.
According to Bohr's theory,the velocity $v$ of the electron in the $n^{th}$ orbit is proportional to $\frac{1}{n}$ $(v \propto \frac{1}{n})$.
The radius $r$ of the $n^{th}$ orbit is proportional to $n^2$ $(r \propto n^2)$.
Substituting these proportionalities into the force equation:
$F \propto \frac{v^2}{r}$
$F \propto \frac{(1/n)^2}{n^2}$
$F \propto \frac{1/n^2}{n^2}$
$F \propto \frac{1}{n^4}$ or $F \propto n^{-4}$.

(B) The velocity of an electron in the $n^{th}$ Bohr orbit is given by the formula $v_n = \frac{Ze^2}{2 \varepsilon_0 nh}$.
From this expression,we can see that the velocity is inversely proportional to the principal quantum number $n$,i.e.,$v_n \propto \frac{1}{n}$.
For the first orbit,$n_1 = 1$,and for the second orbit,$n_2 = 2$.
Therefore,the ratio of the velocities is $\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(C) The moment of inertia $I$ of an electron of mass $m$ moving in an orbit of radius $r$ is given by $I = mr^2$.
The radius of the $n^{\text{th}}$ Bohr orbit is given by the formula:
$r_n = \frac{\varepsilon_0 h^2 n^2}{\pi m e^2}$
For the second Bohr orbit,$n = 2$. Substituting $n = 2$ into the formula:
$r_2 = \frac{\varepsilon_0 h^2 (2)^2}{\pi m e^2} = \frac{4 \varepsilon_0 h^2}{\pi m e^2}$
Now,calculate the moment of inertia $I$ for the second orbit:
$I = m \times (r_2)^2$
$I = m \times \left( \frac{4 \varepsilon_0 h^2}{\pi m e^2} \right)^2$
$I = m \times \frac{16 \varepsilon_0^2 h^4}{\pi^2 m^2 e^4}$
$I = \frac{16 \varepsilon_0^2 h^4}{\pi^2 m e^4}$

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.
Change in angular momentum $\Delta L = L_2 - L_1$.
Here,$n_1 = 4$ and $n_2 = 5$.
$\Delta L = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{h}{2\pi} (n_2 - n_1)$.
Substituting the values: $\Delta L = \frac{6.63 \times 10^{-34}}{2 \times 3.14} (5 - 4)$.
$\Delta L = \frac{6.63 \times 10^{-34}}{6.28} \approx 1.055 \times 10^{-34} \text{ J s}$.
Rounding to the nearest given option,the change is approximately $1 \times 10^{-34} \text{ J s}$.

(D) Concept: Bohr orbit radius.
In Bohr's model, the radius of the $n^{th}$ orbit is given by the formula:
$r_n = a_0 \cdot \frac{n^2}{Z}$
For a hydrogen atom, the atomic number $Z = 1$.
Therefore, the radius of the $n^{th}$ orbit is $r_n = a_0 n^2$.
For the smallest orbit (ground state), $n = 1$, so $r_1 = a_0 (1)^2 = a_0$.
For the third orbit, $n = 3$, so the radius is:
$r_3 = a_0 (3)^2 = 9 a_0$.
Thus, the correct option is $D$.

(D) The orbital magnetic moment is defined as $m_{orb} = iA$,where $i = \frac{e}{T}$ is the current,$A = \pi r^2$ is the area,and $T$ is the time period of the electron's revolution.
Substituting $i$ and $A$,we get $m_{orb} = e \left( \frac{\pi r^2}{T} \right) \dots (1)$.
According to Bohr's quantization condition for angular momentum,$mvr = \frac{nh}{2\pi} \dots (2)$.
Also,the time period $T$ is related to velocity $v$ and radius $r$ by $T = \frac{2\pi r}{v}$,which implies $\frac{r}{T} = \frac{v}{2\pi}$.
Substituting this into the expression for $m_{orb}$,we get $m_{orb} = e \pi r \left( \frac{r}{T} \right) = e \pi r \left( \frac{v}{2\pi} \right) = \frac{evr}{2}$.
From equation $(2)$,$vr = \frac{nh}{2\pi m}$.
Substituting this into the expression for $m_{orb}$,we get $m_{orb} = \frac{e}{2} \left( \frac{nh}{2\pi m} \right) = n \left( \frac{eh}{4\pi m} \right)$.
Since $e, h, \pi,$ and $m$ are constants,we conclude that $m_{orb} \propto n$.

(B) In a hydrogen atom,the electrostatic force provides the necessary centripetal force for the electron to move in a circular orbit of radius $r$:
$\frac{m v^2}{r} = \frac{e^2}{r^2}$
Multiplying both sides by $\frac{r}{2}$,we get:
$\frac{1}{2} m v^2 = \frac{e^2}{2 r}$
Since the kinetic energy $K$ is defined as $\frac{1}{2} m v^2$,we have:
$K = \frac{e^2}{2 r}$
Therefore,the kinetic energy is proportional to $\frac{e^2}{2 r}$.

(C) The centripetal force is provided by the electrostatic force: $\frac{mv^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies mv^2 = \frac{e^2}{4\pi\varepsilon_0 r}$.
From Bohr's quantization condition: $mvr = \frac{nh}{2\pi}$. For ground state $(n=1)$,$mvr = \frac{h}{2\pi} \implies v = \frac{h}{2\pi mr}$.
Substituting $v$ into the force equation: $m(\frac{h}{2\pi mr})^2 = \frac{e^2}{4\pi\varepsilon_0 r} \implies \frac{h^2}{4\pi^2 mr^2} = \frac{e^2}{4\pi\varepsilon_0 r} \implies r = \frac{\varepsilon_0 h^2}{\pi m e^2}$.
The current $I$ due to the orbiting electron is $I = \frac{e}{T} = \frac{ev}{2\pi r}$.
The magnetic field at the center is $B = \frac{\mu_0 I}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.
Substituting $v = \frac{h}{2\pi mr}$ and $r = \frac{\varepsilon_0 h^2}{\pi m e^2}$:
$B = \frac{\mu_0 e}{4\pi r^2} \cdot \frac{h}{2\pi mr} = \frac{\mu_0 e h}{8\pi^2 m r^3}$.
Substituting $r^3 = (\frac{\varepsilon_0 h^2}{\pi m e^2})^3 = \frac{\varepsilon_0^3 h^6}{\pi^3 m^3 e^6}$:
$B = \frac{\mu_0 e h}{8\pi^2 m} \cdot \frac{\pi^3 m^3 e^6}{\varepsilon_0^3 h^6} = \frac{\mu_0 e^7 \pi m^2}{8 \varepsilon_0^3 h^5}$.

(A) According to Bohr's theory,the total energy $E$ of an electron in the $n^{\text{th}}$ orbit is given by $E = -\frac{13.6 Z^2}{n^2} \text{ eV}$,which implies $E \propto \frac{1}{n^2}$.
The angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$,which implies $L \propto n$.
From the relation $L \propto n$,we have $n \propto L$.
Substituting this into the energy relation: $E \propto \frac{1}{n^2} \implies E \propto \frac{1}{L^2}$.
Taking the square root of both sides,we get $\sqrt{E} \propto \frac{1}{L}$,which rearranges to $L \propto \frac{1}{\sqrt{E}}$.

(B) The centripetal acceleration is given by $a = \frac{v^2}{r}$.
In Bohr's model,the velocity of an electron in the $n^{\text{th}}$ orbit is $v \propto \frac{1}{n}$ and the radius is $r \propto n^2$.
Substituting these proportionalities into the expression for acceleration,we get $a \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.
Therefore,the ratio of centripetal acceleration for the $3^{\text{rd}}$ orbit $(n_1 = 3)$ and $5^{\text{th}}$ orbit $(n_2 = 5)$ is $\frac{a_3}{a_5} = \frac{n_2^4}{n_1^4} = \frac{5^4}{3^4}$.
Calculating the values,we get $\frac{625}{81}$.

(A) The orbital period of revolution of an electron in the $n^{\text{th}}$ orbit is $T_n = \frac{2 \pi r_n}{v_n}$.
According to Bohr's model,the radius of the $n^{\text{th}}$ orbit is $r_n = \left(\frac{h^2 \varepsilon_0}{\pi m e^2}\right) n^2$ (for $Z=1$).
The velocity of the electron in the $n^{\text{th}}$ orbit is $v_n = \left(\frac{e^2}{2 h \varepsilon_0}\right) \frac{1}{n}$.
Substituting these values into the expression for $T_n$:
$T_n = 2 \pi \left(\frac{h^2 \varepsilon_0 n^2}{\pi m e^2}\right) \div \left(\frac{e^2}{2 h \varepsilon_0 n}\right)$
$T_n = 2 \pi \left(\frac{h^2 \varepsilon_0 n^2}{\pi m e^2}\right) \times \left(\frac{2 h \varepsilon_0 n}{e^2}\right)$
$T_n = \frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$.

(B) The gyromagnetic ratio is defined as the ratio of the magnetic moment $\mu$ to the angular momentum $L$ of the electron.
$\text{Gyromagnetic ratio} = \frac{\mu}{L} = \frac{iA}{mvr}$
Using the Bohr model,the current $i = \frac{e}{T} = \frac{e \omega}{2 \pi}$ and the area $A = \pi r^2$.
The angular momentum $L = mvr = mr(r \omega) = mr^2 \omega$.
Substituting these values:
$\frac{\mu}{L} = \frac{(\frac{e \omega}{2 \pi})(\pi r^2)}{mr^2 \omega} = \frac{e}{2m}$
Since $e$ and $m$ are constants,the gyromagnetic ratio is independent of the orbit $n$ of the electron.

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

(B) In the Bohr model of the atom,the kinetic energy $(K)$ of an electron in the $n^{\text{th}}$ orbit is given by $K = \frac{kZe^2}{2r_n}$.
The total energy $(E)$ of the electron in the $n^{\text{th}}$ orbit is given by $E = -\frac{kZe^2}{2r_n}$.
Comparing these two expressions,we find that $K = -E$.
Therefore,the ratio of the kinetic energy to the total energy is $\frac{K}{E} = \frac{-E}{E} = -1$.
This can be expressed as the ratio $1:-1$.

(C) In the Bohr model of the hydrogen atom,the total energy $E$ of an electron in the $n^{\text{th}}$ orbit is given by $E \propto \frac{1}{n^2}$.
According to Bohr's quantization postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$,which implies $L \propto n$.
Substituting $n \propto L$ into the energy expression,we get $E \propto \frac{1}{L^2}$.
Therefore,$E \propto L^{-2}$.

(A) According to the de-Broglie hypothesis,the wavelength $\lambda$ is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.
For an electron in the $n^{th}$ Bohr orbit,the quantization condition is $mvr = \frac{nh}{2\pi}$,which implies $mv = \frac{nh}{2\pi r}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.
For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n = a_0 n^2$,where $a_0$ is the Bohr radius.
Thus,$\lambda_n = \frac{2\pi (a_0 n^2)}{n} = 2\pi a_0 n$.
Therefore,the ratio of wavelengths for the first $(n=1)$ and second $(n=2)$ orbits is $\frac{\lambda_1}{\lambda_2} = \frac{2\pi a_0 (1)}{2\pi a_0 (2)} = \frac{1}{2}$.

(D) For an electron revolving in a Bohr orbit of a hydrogen atom at a distance $r$ from the nucleus:
Kinetic energy $(K.E.)$ is given by $K = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{2r} = \frac{e^{2}}{8 \pi \epsilon_{0} r}$.
Potential energy $(P.E.)$ is given by $P = -\frac{1}{4 \pi \epsilon_{0}} \cdot \frac{e^{2}}{r}$.
Taking the ratio of $K.E.$ to $P.E.$:
$\frac{K}{P} = \frac{\frac{e^{2}}{8 \pi \epsilon_{0} r}}{-\frac{e^{2}}{4 \pi \epsilon_{0} r}} = -\frac{4 \pi \epsilon_{0} r}{8 \pi \epsilon_{0} r} = -\frac{1}{2}$.

(D) Given: Radius $r = 0.53 \text{ Å} = 0.53 \times 10^{-10} \text{ m}$.
Time period $T = 1.571 \times 10^{-16} \text{ s}$.
The velocity $v$ of the electron is given by the formula $v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \pi r}{T}$.
Substituting the values: $v = \frac{2 \times 3.142 \times 0.53 \times 10^{-10}}{1.571 \times 10^{-16}}$.
Since $2 \times 3.142 = 6.284$,we have $v = \frac{6.284 \times 0.53 \times 10^{-10}}{1.571 \times 10^{-16}}$.
Note that $\frac{6.284}{1.571} = 4$.
So,$v = 4 \times 0.53 \times 10^{6} \text{ m/s}$.
$v = 2.12 \times 10^{6} \text{ m/s}$.

(C) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_{n} = -\frac{13.6}{n^{2}} \text{ eV}$.
Thus,$E_{n} \propto \frac{1}{n^{2}}$.
The angular momentum of an electron in the $n^{th}$ orbit is given by $L_{n} = \frac{nh}{2\pi}$.
Thus,$L_{n} \propto n$,which implies $n \propto L_{n}$.
Substituting $n \propto L_{n}$ into the energy relation,we get $E_{n} \propto \frac{1}{L_{n}^{2}}$.
Therefore,the correct relation is $E_{n} \propto \frac{1}{L_{n}^{2}}$.

(A) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
For the second orbit,$n = 2$,so $L = \frac{2h}{2\pi} = \frac{h}{\pi}$.
The rotational kinetic energy $E$ is given by $E = \frac{L^{2}}{2I}$.
Substituting the value of $L$,we get $E = \frac{(h/\pi)^{2}}{2I} = \frac{h^{2}}{2I\pi^{2}}$.

(C) According to Bohr's theory,the centripetal force acting on the electron is given by $F = \frac{mv^2}{r}$.
From Bohr's model,the velocity of the electron is $v \propto \frac{1}{n}$ and the radius of the orbit is $r \propto n^2$.
Substituting these proportionalities into the force equation:
$F \propto \frac{(1/n)^2}{n^2} = \frac{1/n^2}{n^2} = \frac{1}{n^4}$.
Therefore,the force is related to the principal quantum number as $F \propto n^{-4}$.

(C) According to Bohr's theory,the velocity $(v)$ of an electron in the $n^{th}$ orbit is given by $v = \frac{e^{2}}{2 \epsilon_{0} n h}$.
For the first orbit,$n = 1$,so the velocity is $v = \frac{e^{2}}{2 \epsilon_{0} h}$.
The ratio of the velocity of the electron $(v)$ to the velocity of light $(c)$ is given by $\frac{v}{c} = \frac{e^{2}}{2 \epsilon_{0} h c}$.

(B) The orbital period $T$ is given by $T = \frac{2 \pi r}{v}$.
According to Bohr's model,the radius of the $n^{\text{th}}$ orbit is $r = \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2}}$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v = \frac{e^{2}}{2 \epsilon_{0} n h}$.
Substituting these expressions into the formula for $T$:
$T = \frac{2 \pi \left( \frac{n^{2} h^{2} \epsilon_{0}}{\pi m e^{2}} \right)}{\left( \frac{e^{2}}{2 \epsilon_{0} n h} \right)}$
$T = \frac{2 n^{2} h^{2} \epsilon_{0}}{m e^{2}} \times \frac{2 \epsilon_{0} n h}{e^{2}}$
$T = \frac{4 \epsilon_{0}^{2} n^{3} h^{3}}{m e^{4}}$.

(B) According to Bohr's model, the radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = a_0 n^2$, where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
This implies that the radius $r$ is directly proportional to the square of the principal quantum number, i.e., $r \propto n^2$.
For the first four orbits $(n = 1, 2, 3, 4)$, the radii are proportional to $1^2, 2^2, 3^2, 4^2$.
Calculating these values, we get $1: 4: 9: 16$.
Therefore, the correct option is $B$.

(C) The de-Broglie wavelength $\lambda$ associated with an electron in an orbit of radius $r$ is given by the relation $2\pi r = n\lambda$,where $n$ is the principal quantum number.
For a hydrogen atom,the radius of the $n^{th}$ orbit is $r_n \propto n^2$.
Substituting this into the de-Broglie relation: $2\pi (k n^2) = n\lambda$,which simplifies to $\lambda \propto n$.
The ground state corresponds to $n = 1$.
The third excited state corresponds to $n = 1 + 3 = 4$.
Since the principal quantum number $n$ increases from $1$ to $4$,the de-Broglie wavelength $\lambda$ will increase.

(B) The speed of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v_n = \frac{v_1}{n}$,where $v_1$ is the speed in the ground state $(n=1)$.
Given,$v_1 = 2.2 \times 10^6 \ m/s$.
The third excited state corresponds to $n = 4$ (since ground state is $n=1$,first excited is $n=2$,second is $n=3$,and third is $n=4$).
Using the relation $v_n = \frac{v_1}{n}$,we get:
$v_4 = \frac{v_1}{4} = \frac{2.2 \times 10^6 \ m/s}{4}$.
$v_4 = 0.55 \times 10^6 \ m/s = 5.5 \times 10^5 \ m/s$.

(A) According to Bohr's postulates for the Hydrogen atom,an electron revolves around the nucleus in specific orbits called stationary orbits.
In these stationary orbits,the electron does not radiate energy,even though it is accelerating due to its circular motion (its velocity vector changes continuously).
Therefore,the electron does not radiate light while revolving in a stationary orbit.

(A) According to Bohr's theory,the wavelength $\lambda$ of the emitted radiation is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{p^2} - \frac{1}{n^2} \right)$.
Since $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is inversely proportional to the energy difference $\Delta E$ between the orbits.
To obtain the maximum wavelength,the energy difference $\Delta E$ must be minimum.
The energy difference between consecutive orbits decreases as the principal quantum number increases.
Comparing the energy gaps: $(E_5 - E_4) < (E_4 - E_3) < (E_3 - E_2) < (E_2 - E_1)$.
Therefore,the transition from $n=5$ to $p=4$ results in the smallest energy difference and consequently the maximum wavelength.

(A) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.
Here,$I$ is the current,which is $I = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the orbital velocity and $r$ is the radius of the orbit.
According to Bohr's theory,$r \propto n^2$ and $v \propto \frac{1}{n}$.
Substituting these into the current expression: $I \propto \frac{(1/n)}{n^2} = n^{-3}$.
Now,substituting $I$ and $r$ into the magnetic field formula: $B \propto \frac{I}{r} \propto \frac{n^{-3}}{n^2} = n^{-5}$.
Therefore,the magnetic field at the center is proportional to $n^{-5}$.