Bohr's Model of Hydrogen Atom Questions in English

(A) According to Bohr's quantization condition,the angular momentum of an electron is given by: $mvr = \frac{nh}{2\pi}$.
From this,the velocity $v$ is: $v = \frac{nh}{2\pi mr}$.
The orbital acceleration (centripetal acceleration) of the electron is given by: $a = \frac{v^2}{r}$.
Substituting the expression for $v$: $a = \frac{(\frac{nh}{2\pi mr})^2}{r} = \frac{n^2 h^2}{4\pi^2 m^2 r^2 \cdot r} = \frac{n^2 h^2}{4\pi^2 m^2 r^3}$.

(B) The time period $T$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $T = \frac{4 \varepsilon_{0}^{2} n^{3} h^{3}}{m Z^{2} e^{4}}$.
Since the orbital frequency $f$ is the reciprocal of the time period $(f = 1/T)$, we have:
$f \propto \frac{1}{n^{3}}$.
Therefore, the orbital frequency is proportional to $n^{-3}$.

(A) The linear momentum $p$ of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $p = mv = \frac{mZe^2}{2 \epsilon_0 nh}$. Since $Z=1$ for hydrogen,$p \propto \frac{1}{n}$.
The angular momentum $L$ of an electron in the $n^{th}$ orbit is given by Bohr's quantization condition: $L = \frac{nh}{2\pi}$. Thus,$L \propto n$.
The product of linear momentum and angular momentum is $p \times L \propto \left(\frac{1}{n}\right) \times n = n^0$.
Comparing this with $n^x$,we get $x = 0$.

(A) According to Bohr's quantization condition,the angular momentum of an electron is given by $mvr = \frac{nh}{2\pi}$.
Rearranging this,we get $2\pi r = n \left( \frac{h}{mv} \right)$.
Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we can substitute this into the equation to get $2\pi r = n\lambda$.
For the first orbit,$n = 1$.
Therefore,the wavelength $\lambda = 2\pi r$.

(B) The radius of an orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius $(5.3 \times 10^{-11} \ m)$ and $n$ is the principal quantum number.
Since $r \propto n^2$,we have the ratio:
$\frac{r_f}{r_i} = \left(\frac{n_f}{n_i}\right)^2$
Given $r_i = 5.3 \times 10^{-11} \ m$ (for $n_i = 1$) and $r_f = 21.2 \times 10^{-11} \ m$:
$\frac{21.2 \times 10^{-11}}{5.3 \times 10^{-11}} = \left(\frac{n}{1}\right)^2$
$4 = n^2$
$n = 2$
Therefore,the principal quantum number of the final state is $n=2$.

(D) The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p$ is the momentum of the electron.
For an electron in the $n^{th}$ orbit of a hydrogen atom,the velocity $v$ is inversely proportional to the principal quantum number $n$ $(v \propto \frac{1}{n})$.
Since momentum $p = mv$,we have $p \propto \frac{1}{n}$.
Substituting this into the de-Broglie relation,we get $\lambda = \frac{h}{p} \propto n$.
The ground state corresponds to $n = 1$. The third excited state corresponds to $n = 4$.
Since the principal quantum number $n$ increases from $1$ to $4$,the de-Broglie wavelength $\lambda$ will increase.

(B) The radius of the $n^{\text{th}}$ orbit is given by $r_n = a_0 n^2$.
For the third orbit $(n=3)$,the radius is $r_3 = a_0 \times 3^2 = 9 a_0$.
According to Bohr's quantization condition,the angular momentum is $mvr = \frac{nh}{2\pi}$.
Rearranging for momentum $mv$,we get $mv = \frac{nh}{2\pi r}$.
Substituting $n=3$ and $r=9a_0$:
$mv = \frac{3h}{2\pi(9a_0)} = \frac{h}{6\pi a_0}$.
The de-Broglie wavelength is $\lambda = \frac{h}{mv}$.
Substituting the value of $mv$:
$\lambda = \frac{h}{h / (6\pi a_0)} = 6\pi a_0$.

(D) The velocity of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $v \propto \frac{1}{n}$.
Since the momentum $p = mv$, we have $p \propto \frac{1}{n}$.
The de-Broglie wavelength is given by the relation $\lambda = \frac{h}{p}$.
Substituting the proportionality for momentum, we get $\lambda \propto n$.
As the electron jumps to a higher excited state, the principal quantum number $n$ increases.
Therefore, the de-Broglie wavelength $\lambda$ will increase.

(C) According to Bohr's second postulate,the angular momentum is given by $\frac{nh}{2 \pi} = mvr_n$.
From this,the de-Broglie wavelength $\lambda_n$ is $\lambda_n = \frac{h}{mv} = \frac{2 \pi r_n}{n}$.
We know that the radius of the $n^{\text{th}}$ Bohr orbit is given by $r_n = r_1 \times n^2$,where $r_1 = r$.
For the $4^{\text{th}}$ orbit $(n = 4)$,the radius is $r_4 = r \times 4^2 = 16r$.
Substituting these values into the wavelength formula:
$\lambda_4 = \frac{2 \pi r_4}{4} = \frac{2 \pi \times (16r)}{4} = 8 \pi r$.

(B) The de-Broglie wavelength is given by $\lambda = \frac{h}{p}$,where $h$ is Planck's constant and $p$ is the momentum of the electron.
Since $p = mv$ and kinetic energy $E_K = \frac{p^2}{2m}$,we have $p = \sqrt{2mE_K}$.
In a hydrogen atom,the kinetic energy of an electron in the $n^{th}$ orbit is proportional to $\frac{1}{n^2}$,i.e.,$E_K \propto \frac{1}{n^2}$.
Therefore,the momentum $p = \sqrt{2mE_K} \propto \sqrt{\frac{1}{n^2}} = \frac{1}{n}$.
Since $\lambda = \frac{h}{p}$,it follows that $\lambda \propto n$.
When the electron jumps from the third excited state $(n = 4)$ to the ground state $(n = 1)$,the ratio of the new wavelength $\lambda_{final}$ to the initial wavelength $\lambda_{initial}$ is $\frac{\lambda_{final}}{\lambda_{initial}} = \frac{n_{final}}{n_{initial}} = \frac{1}{4}$.
Thus,the de-Broglie wavelength becomes $\frac{1}{4}$ of its initial value.

(B) According to the de-Broglie hypothesis,the wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{mv}$,where $h$ is Planck's constant,$m$ is the mass,and $v$ is the velocity of the electron.
Rearranging this,we get $mv = \frac{h}{\lambda} \quad --- (1)$.
According to Bohr's postulate for the quantization of angular momentum,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = mvr = \frac{nh}{2\pi}$.
Rearranging this for $mv$,we get $mv = \frac{nh}{2\pi r} \quad --- (2)$.
Equating the expressions for $mv$ from equation $(1)$ and $(2)$:
$\frac{h}{\lambda} = \frac{nh}{2\pi r}$.
Canceling $h$ from both sides,we get $\frac{1}{\lambda} = \frac{n}{2\pi r}$.
Therefore,the de-Broglie wavelength is $\lambda = \frac{2\pi r}{n}$.

(B) According to Bohr's quantization postulate,the angular momentum of an electron in a stationary orbit is given by:
$mvr = \frac{nh}{2\pi}$
Rearranging this equation,we get:
$2\pi r = \frac{nh}{mv}$
Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we can substitute this into the equation:
$2\pi r = n\lambda$
For the ground state of the hydrogen atom,the principal quantum number is $n = 1$.
Substituting $n = 1$ into the equation,we get:
$\lambda = 2\pi r$

(B) Using Rydberg's formula for the hydrogen atom transition to the ground state $(m=1)$:
$\frac{1}{\lambda} = R \left[ \frac{1}{1^2} - \frac{1}{n^2} \right]$
$\frac{1}{\lambda R} = 1 - \frac{1}{n^2}$
$\frac{1}{n^2} = 1 - \frac{1}{\lambda R}$
$\frac{1}{n^2} = \frac{\lambda R - 1}{\lambda R}$
Taking the reciprocal on both sides:
$n^2 = \frac{\lambda R}{\lambda R - 1}$
Therefore,the quantum number of the excited state is:
$n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$

(B) The energy of the electron in the ground state $(n=1)$ is $E_1 = -13.6 \text{ eV}$.
When the atom absorbs a photon of energy $12.1 \text{ eV}$,the new energy level $E_n$ is given by:
$E_n = E_1 + 12.1 \text{ eV} = -13.6 \text{ eV} + 12.1 \text{ eV} = -1.5 \text{ eV}$.
Using the Bohr formula $E_n = -\frac{13.6}{n^2} \text{ eV}$:
$-1.5 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{1.5} \approx 9.06 \approx 9$.
Thus,$n = 3$.
The electron is excited to the $n=3$ state. The possible transitions for the electron to return to the ground state are:
$1$. From $n=3$ to $n=2$
$2$. From $n=3$ to $n=1$
$3$. From $n=2$ to $n=1$
The number of spectral lines is given by the formula $N = \frac{n(n-1)}{2} = \frac{3(3-1)}{2} = 3$.
Therefore,$3$ spectral lines are emitted.

(D) The electrostatic force between the nucleus and the electron provides the necessary centripetal force for circular motion.
Electrostatic force $=$ Centripetal force
$\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{r^{2}} = \frac{m v^{2}}{r}$
Solving for velocity $v$:
$v = \sqrt{\frac{1}{4 \pi \varepsilon_{0}} \frac{Z e^{2}}{m r}}$
Given values: $Z = 1$, $e = 1.6 \times 10^{-19} \,C$, $r = 0.1 \times 10^{-9} \,m$, $m = 9.1 \times 10^{-31} \,kg$, and $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \,N \cdot m^{2} \cdot C^{-2}$.
Substituting these values:
$v = \sqrt{\frac{9 \times 10^{9} \times 1 \times (1.6 \times 10^{-19})^{2}}{9.1 \times 10^{-31} \times 0.1 \times 10^{-9}}}$
$v = \sqrt{\frac{9 \times 10^{9} \times 2.56 \times 10^{-38}}{9.1 \times 10^{-41}}}$
$v = \sqrt{\frac{23.04 \times 10^{-29}}{9.1 \times 10^{-41}}} = \sqrt{2.53 \times 10^{12}} \approx 1.59 \times 10^{6} \,ms^{-1}$.

(D) The radius of the $n^{\text{th}}$ Bohr orbit is given by $r_n \propto n^2$.
The velocity of the electron in the $n^{\text{th}}$ orbit is $v_n \propto \frac{1}{n}$.
The angular velocity is $\omega_n = \frac{v_n}{r_n} \propto \frac{1/n}{n^2} = \frac{1}{n^3}$.
The current $I_n$ produced by the revolving electron is $I_n = \frac{e}{T_n} = \frac{e \omega_n}{2 \pi} \propto \omega_n \propto \frac{1}{n^3}$.
The magnetic field at the centre of the orbit is $B_n = \frac{\mu_0 I_n}{2 r_n}$.
Substituting the proportionalities: $B_n \propto \frac{I_n}{r_n} \propto \frac{1/n^3}{n^2} = \frac{1}{n^5}$.
Therefore,$B_n \propto n^{-5}$.

(A) The magnetic field $B$ at the center of a circular current loop is given by $B = \frac{\mu_0 I}{2r}$.
Since the electron revolves with period $T$, the equivalent current is $I = \frac{e}{T}$.
Given $T = \frac{2\pi r}{v}$, we have $I = \frac{ev}{2\pi r}$.
Substituting this into the magnetic field formula: $B = \frac{\mu_0 (ev/2\pi r)}{2r} = \frac{\mu_0 ev}{4\pi r^2}$.
For a hydrogen atom in the ground state $(n=1)$, the velocity $v$ and radius $r$ are given by Bohr's theory:
$v = \frac{e^2}{2\epsilon_0 h}$ and $r = \frac{\epsilon_0 h^2}{\pi m e^2}$.
Substituting these into the expression for $B$:
$B = \frac{\mu_0 e}{4\pi} \cdot \left( \frac{e^2}{2\epsilon_0 h} \right) \cdot \left( \frac{\pi m e^2}{\epsilon_0 h^2} \right)^2$
$B = \frac{\mu_0 e^3}{8\pi \epsilon_0 h} \cdot \frac{\pi^2 m^2 e^4}{\epsilon_0^2 h^4} = \frac{\mu_0 e^7 \pi m^2}{8 \epsilon_0^3 h^5}$.

(B) According to Bohr's theory,the angular momentum '$L$' of an electron in the '$n^{th}$' orbit is given by $L = \frac{nh}{2\pi}$.
The magnetic moment '$M$' associated with a revolving electron is given by $M = \frac{e}{2m_e} L$.
Substituting the value of '$L$' into the equation for '$M$':
$M = \frac{e}{2m_e} \left( \frac{nh}{2\pi} \right) = \frac{enh}{4\pi m_e}$.
Since '$e$','$h$','$m_e$',and '$\pi$' are constants,we have $M \propto n$.

(D) The current $i$ produced by an electron moving in a circular orbit is given by $i = \frac{e}{T}$,where $T$ is the time period of revolution. Since $T = \frac{2 \pi r}{v}$,we have $i = \frac{ev}{2 \pi r}$.
The magnetic moment $M$ of a current loop is $M = iA$,where $A = \pi r^2$ is the area of the orbit.
Substituting the values,$M = \left(\frac{ev}{2 \pi r}\right) (\pi r^2) = \frac{evr}{2}$.
Multiplying and dividing by the mass of the electron $m_{e}$,we get $M = \frac{e}{2 m_{e}} (m_{e}vr)$.
According to Bohr's quantization condition,the angular momentum $L = m_{e}vr = \frac{nh}{2 \pi}$.
Substituting this into the expression for $M$,we get $M = \frac{e}{2 m_{e}} \left(\frac{nh}{2 \pi}\right)$.

(D) The total energy $E$ of an electron in the $n^{\text{th}}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For the hydrogen atom $(_1H^1)$,$Z_H = 1$ and $n = 2$. Thus,$E_H = -13.6 \frac{1^2}{2^2} = -13.6 \times \frac{1}{4} \text{ eV}$.
For the helium ion $(He^+)$,$Z_{He} = 2$ and $n = 2$. Thus,$E_{He} = -13.6 \frac{2^2}{2^2} = -13.6 \times 1 \text{ eV}$.
The ratio of the total energy of the hydrogen atom to that of the helium ion is: $\frac{E_H}{E_{He}} = \frac{-13.6 \times (1/4)}{-13.6 \times 1} = \frac{1}{4}$.

(D) The radius of the $n^{th}$ orbit is given by $r_{n} \propto \frac{n^{2}}{Z}$.
For an $H$-atom,the atomic number $Z = 1$.
Therefore,the radius is proportional to the square of the principal quantum number: $r_{n} \propto n^{2}$.
For the second orbit $(n_{2} = 2)$ and the third orbit $(n_{3} = 3)$:
$\frac{r_{2}}{r_{3}} = \left(\frac{n_{2}}{n_{3}}\right)^{2} = \left(\frac{2}{3}\right)^{2} = \frac{4}{9}$.
Thus,the ratio is $4: 9$.

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in the $n^{\text{th}}$ orbit is given by $L = \frac{nh}{2\pi}$.
For the $5^{\text{th}}$ orbit $(n_1 = 5)$,the angular momentum is $L_5 = \frac{5h}{2\pi}$.
For the $3^{\text{rd}}$ orbit $(n_2 = 3)$,the angular momentum is $L_3 = \frac{3h}{2\pi}$.
The change in angular momentum $\Delta L$ is given by:
$\Delta L = L_5 - L_3$
$\Delta L = \frac{5h}{2\pi} - \frac{3h}{2\pi}$
$\Delta L = \frac{h}{2\pi} (5 - 3)$
$\Delta L = \frac{2h}{2\pi} = \frac{h}{\pi}$.

(A) The radius of an electron orbit in a hydrogen-like atom is given by the formula: $r_n = n^2 a_0$,where $n$ is the principal quantum number and $a_0$ is the Bohr radius $(5.3 \times 10^{-11} \ m)$.
For the $n=4$ state,we substitute the values into the formula:
$r_4 = (4)^2 \times (5.3 \times 10^{-11} \ m)$
$r_4 = 16 \times 5.3 \times 10^{-11} \ m$
$r_4 = 84.8 \times 10^{-11} \ m$
$r_4 = 8.48 \times 10^{-10} \ m$
Therefore,the correct option is $A$.

(C) The total energy $E$ of an electron in a hydrogen atom is given by the sum of its kinetic energy $K$ and potential energy $U$.
For a hydrogen atom, the total energy $E$ is related to the kinetic energy $K$ by the relation $E = -K$.
Given that the ground state energy $E = -13.6 \text{ eV}$.
Therefore, the kinetic energy $K = -E = -(-13.6 \text{ eV}) = 13.6 \text{ eV}$.
Thus, the correct option is $C$.

(B) For an electron in a hydrogen atom,the electrostatic potential energy $U$ and kinetic energy $K$ are related by the virial theorem.
$U = -2K$
Given that the kinetic energy $K = \frac{e^{2}}{8 \pi \varepsilon_{0} r}$.
Substituting this value into the relation:
$U = -2 \times \left( \frac{e^{2}}{8 \pi \varepsilon_{0} r} \right)$
$U = -\frac{e^{2}}{4 \pi \varepsilon_{0} r}$
Thus,the potential energy is $-\frac{e^{2}}{4 \pi \varepsilon_{0} r}$.

(A) The total energy $(E)$ of an electron in a hydrogen atom is the sum of its kinetic energy $(K)$ and potential energy $(U)$.
For a hydrogen atom,the relationship between total energy $(E)$,kinetic energy $(K)$,and potential energy $(U)$ is given by $E = -K = U/2$.
Given the ground state energy $E = -13.6 \ eV$.
Therefore,the potential energy $U = 2 \times E$.
$U = 2 \times (-13.6 \ eV) = -27.2 \ eV$.
Thus,the correct option is $A$.

(A) In the Bohr model of a hydrogen atom, the kinetic energy $(K)$, potential energy $(U)$, and total energy $(E)$ of an electron in an orbit are related as follows:
$K = -E$
$U = 2E$
$K = -\frac{U}{2}$
Given that the kinetic energy is $K = x$.
Since $K = -E$, we have $x = -E$, which implies $E = -x$.
Therefore, the total energy of the electron is $-x$.

(C) According to Bohr's quantization condition for angular momentum,the orbital angular momentum $L$ is given by:
$L = mvr = \frac{nh}{2\pi}$
Where:
$m = 6.0 \times 10^{24} \ kg$ (Mass of the Earth)
$v = 3 \times 10^{4} \ m/s$ (Orbital speed)
$r = 1.5 \times 10^{11} \ m$ (Orbital radius)
$h = 6.625 \times 10^{-34} \ J \ s$ (Planck's constant)
Rearranging the formula to solve for the quantum number $n$:
$n = \frac{2\pi mvr}{h}$
Substituting the values:
$n = \frac{2 \times 3.14159 \times (6.0 \times 10^{24}) \times (3 \times 10^{4}) \times (1.5 \times 10^{11})}{6.625 \times 10^{-34}}$
$n = \frac{169.646 \times 10^{39}}{6.625 \times 10^{-34}}$
$n \approx 25.6 \times 10^{73} = 2.56 \times 10^{74} \approx 2.6 \times 10^{74}$
Thus,the quantum number is $2.6 \times 10^{74}$.

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_{n} = n^{2} a_{0}$,where $a_{0}$ is the Bohr radius.
For the first orbit,$n = 1$,so $r_{1} = (1)^{2} a_{0} = a_{0}$.
The third excited state corresponds to $n = 4$ (since ground state is $n=1$,first excited is $n=2$,second excited is $n=3$,and third excited is $n=4$).
Therefore,the radius in the third excited state is $r_{4} = (4)^{2} a_{0} = 16 a_{0}$.

(B) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit $n$ is given by the formula:
$L = \frac{nh}{2\pi}$
Given that $n = 5$ and Planck's constant $h \approx 6.63 \times 10^{-34} \text{ J s}$:
$L = \frac{5 \times 6.63 \times 10^{-34}}{2 \times 3.14}$
$L = \frac{33.15 \times 10^{-34}}{6.28}$
$L \approx 5.278 \times 10^{-34} \text{ J s}$
Rounding to two significant figures,we get $L \approx 5.3 \times 10^{-34} \text{ J s}$.

(C) According to Bohr's postulate,the angular momentum $L$ of an electron in an orbit $n$ is given by $L = n \frac{h}{2 \pi}$.
For the initial state $n_i = 4$,the angular momentum is $L_i = 4 \frac{h}{2 \pi}$.
For the final state $n_f = 1$,the angular momentum is $L_f = 1 \frac{h}{2 \pi}$.
The change in angular momentum $\Delta L$ is given by $\Delta L = L_i - L_f$.
Substituting the values,$\Delta L = \frac{4 h}{2 \pi} - \frac{1 h}{2 \pi} = \frac{3 h}{2 \pi}$.

(A) The wavelength $\lambda$ of the radiation absorbed when an electron transitions from orbit $n_1$ to $n_2$ is given by the Rydberg formula:
$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Given $n_1 = 2$ and $n_2 = 4$:
$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$
$\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$
$\frac{1}{\lambda} = R \left[ \frac{4 - 1}{16} \right]$
$\frac{1}{\lambda} = \frac{3 R}{16}$
Therefore,the wavelength is $\lambda = \frac{16}{3 R}$.

(A) The number of spectral lines emitted when an electron transitions from energy level $n$ to lower levels is given by $N = \frac{n(n-1)}{2}$.
For the first case,$N = 3$:
$3 = \frac{n_{1}(n_{1}-1)}{2} \Rightarrow n_{1}^2 - n_{1} - 6 = 0 \Rightarrow (n_{1}-3)(n_{1}+2) = 0$.
Since $n_{1} > 0$,we have $n_{1} = 3$.
For the second case,$N = 6$:
$6 = \frac{n_{2}(n_{2}-1)}{2} \Rightarrow n_{2}^2 - n_{2} - 12 = 0 \Rightarrow (n_{2}-4)(n_{2}+3) = 0$.
Since $n_{2} > 0$,we have $n_{2} = 4$.
The orbital speed of an electron in the $n$-th orbit is given by $v_n = \frac{Ze^2}{2\varepsilon_0 hn}$,which implies $v \propto \frac{1}{n}$.
Therefore,the ratio of the speeds is $\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{4}{3}$.

(A) The ionization energy of a hydrogen-like ion is given by the formula $E = R c h Z^2$,where $R$ is the Rydberg constant,$c$ is the speed of light,$h$ is Planck's constant,and $Z$ is the atomic number.
For the lithium ion $Li^{2+}$,the atomic number $Z = 3$.
Substituting the value of $Z$ into the formula:
$E = R c h (3)^2 = 9 R c h$.
Therefore,the ionization energy of $Li^{2+}$ is $9 h c R$.

(B) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = n^2 r_1$, where $r_1$ is the radius of the first orbit $(r_1 = r)$.
For the second Bohr orbit, $n = 2$.
Substituting the value of $n$ into the formula:
$r_2 = (2)^2 \times r$
$r_2 = 4 r$
Therefore, the radius of the second Bohr orbit is $4 r$.

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by $r_n \propto n^2$.
Since the area $A$ of the orbit is $A = \pi r_n^2$,we have $A_n \propto (n^2)^2 = n^4$.
For the ground state,$n_1 = 1$.
For the first excited state,$n_2 = 2$.
The ratio of the area of the first excited state $(A_2)$ to the ground state $(A_1)$ is:
$\frac{A_2}{A_1} = \left(\frac{n_2}{n_1}\right)^4 = \left(\frac{2}{1}\right)^4 = 16$.
Therefore,the ratio is $16: 1$.

(A) The radius of a hydrogen atom is given by the formula $r_n = a_0 n^2$,where $a_0 = 0.53 \ Å$ is the Bohr radius and $n$ is the principal quantum number.
Given,the ground state radius $r_1 = 0.53 \ Å$ (for $n_1 = 1$).
The radius of the excited state is $r_2 = 2.12 \ Å$ (for $n_2 = n$).
Using the relation $r \propto n^2$,we have:
$\frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2$
Substituting the values:
$\frac{2.12}{0.53} = \left(\frac{n}{1}\right)^2$
$4 = n^2$
$n = \sqrt{4} = 2$
Therefore,the principal quantum number of the final state is $n=2$.

(C) Given:
Orbital speed $v = 3 \times 10^4 \ m/s$
Radius $r = 1.5 \times 10^{11} \ m$
Mass of Earth $m_e = 6 \times 10^{24} \ kg$
Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s$
According to Bohr's quantization condition for angular momentum:
$L = m_e v r = \frac{n h}{2 \pi}$
Rearranging for the quantum number $n$:
$n = \frac{2 \pi m_e v r}{h}$
Substituting the values:
$n = \frac{2 \times 3.14159 \times (6 \times 10^{24}) \times (3 \times 10^4) \times (1.5 \times 10^{11})}{6.626 \times 10^{-34}}$
$n = \frac{169.646 \times 10^{39}}{6.626 \times 10^{-34}}$
$n \approx 25.603 \times 10^{73} \approx 2.56 \times 10^{74}$
Rounding to the nearest provided option,$n = 2.57 \times 10^{74}$.

(B) The radius of an electron in the $n$th orbit of a hydrogen-like atom is given by the formula $r_n = n^2 r_1$,where $r_1$ is the Bohr radius (radius of the first orbit).
Given,$r_1 = 0.529 Å$.
We need to find the radius of the third orbit,so $n = 3$.
Substituting the values into the formula:
$r_3 = 3^2 \times r_1$
$r_3 = 9 \times 0.529 Å$
$r_3 = 4.761 Å$.

(B) According to Bohr's postulate, the angular momentum $L$ is given by $L = \frac{n h}{2 \pi}$.
Given $L = \frac{3 h}{2 \pi}$, we get $n = 3$.
The de-Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.
From the angular momentum quantization, $mvr = \frac{nh}{2\pi} = \frac{3h}{2\pi}$, so $mv = \frac{3h}{2\pi r}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{mv} = \frac{h \cdot 2\pi r}{3h} = \frac{2}{3} \pi r$.
The radius of the $n$-th orbit for a hydrogen-like ion is $r = a_{0} \frac{n^2}{Z}$.
For $Li^{2+}$, $Z = 3$ and $n = 3$, so $r = a_{0} \frac{3^2}{3} = 3 a_{0}$.
Substituting $r$ into the expression for $\lambda$: $\lambda = \frac{2}{3} \pi (3 a_{0}) = 2 \pi a_{0}$.
Comparing this with the given form $p \pi a_{0}$, we find $p = 2$.

(B) The angular momentum of an electron in an orbit of radius $r$ is given by Bohr's quantization condition:
$mvr = \frac{nh}{2\pi}$
For the ground state $(n=1)$,the momentum $p = mv$ is:
$p = \frac{h}{2\pi r}$
The de-Broglie wavelength $\lambda$ is defined as:
$\lambda = \frac{h}{p}$
Substituting the value of $p$ from the angular momentum equation:
$\lambda = \frac{h}{(h / 2\pi r)} = 2\pi r$
In the ground state of a hydrogen atom,the Bohr radius $r = 0.53 \text{ Å}$.
Therefore,$\lambda = 2 \times 3.14 \times 0.53 \text{ Å} \approx 3.33 \text{ Å}$.
Thus,the correct option is $3.3 \text{ Å}$.

(C) The time period of revolution $T$ of an electron in the $n^{th}$ orbit is defined as the ratio of the circumference of the orbit to the orbital velocity of the electron: $T = \frac{2 \pi r_{n}}{v_{n}}$.
According to Bohr's theory,the radius of the $n^{th}$ orbit is $r_{n} \propto n^{2}$ and the velocity of the electron in the $n^{th}$ orbit is $v_{n} \propto \frac{1}{n}$.
Substituting these proportionalities into the formula for the time period:
$T_{n} \propto \frac{r_{n}}{v_{n}} \propto \frac{n^{2}}{1/n} = n^{3}$.
Thus,the time period of revolution is proportional to $n^{3}$.

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

(B) The energy of the $ n $-th orbit in a hydrogen atom is given by $ E_n = -\frac{13.6}{n^2} \text{ eV} $.
For the ground state $( n_1 = 1 )$,$ E_1 = -13.6 \text{ eV} $.
When the atom absorbs $ 10.2 \text{ eV} $,the new energy is $ E_2 = -13.6 + 10.2 = -3.4 \text{ eV} $.
Since $ E_2 = -\frac{13.6}{n_2^2} = -3.4 \text{ eV} $,we find $ n_2^2 = 4 $,so $ n_2 = 2 $.
The orbital angular momentum is given by $ L = \frac{nh}{2\pi} $.
The increase in angular momentum is $ \Delta L = L_2 - L_1 = \frac{n_2 h}{2\pi} - \frac{n_1 h}{2\pi} = \frac{(n_2 - n_1)h}{2\pi} $.
Substituting $ n_2 = 2 $,$ n_1 = 1 $,and $ h = 6.626 \times 10^{-34} \text{ Js} $:
$ \Delta L = \frac{(2 - 1) \times 6.626 \times 10^{-34}}{2 \times 3.14159} \approx 1.054 \times 10^{-34} \text{ Js} $.
Thus,the increase is $ 1.05 \times 10^{-34} \text{ Js} $.

(D) The time period of revolution of an electron in a Bohr orbit is given by $T = \frac{2\pi r}{v}$.
We know that the radius of the orbit $r \propto n^2$ and the velocity of the electron $v \propto \frac{1}{n}$.
Substituting these relations into the time period formula: $T \propto \frac{n^2}{1/n} = n^3$.
For the ground state $(n_1 = 1)$,the time period is $T_1 = T$.
For the first excited state $(n_2 = 2)$,the time period is $T_2$.
Using the proportionality $T \propto n^3$,we have $\frac{T_2}{T_1} = \left(\frac{n_2}{n_1}\right)^3$.
$\frac{T_2}{T} = \left(\frac{2}{1}\right)^3 = 8$.
Therefore,$T_2 = 8T$.

(C) According to Bohr's theory,the angular momentum of an electron in an orbit is quantized as $mvr = \frac{nh}{2\pi}$.
This implies that the orbital velocity $v$ is inversely proportional to the radius $r$,i.e.,$v \propto \frac{1}{r}$.
Let $v_1$ and $r_1$ be the velocity and radius in the ground state,and $v_2$ and $r_2$ be the velocity and radius in the excited state.
Given $v_1 = v$,$r_1 = R$,and $v_2 = \frac{v}{3}$.
Using the relation $\frac{v_1}{v_2} = \frac{r_2}{r_1}$,we get $\frac{v}{v/3} = \frac{r_2}{R}$.
Solving for $r_2$,we find $3 = \frac{r_2}{R}$,which gives $r_2 = 3R$.

(D) The number of spectral lines $N$ obtained due to transitions of electrons from the $n$th orbit to lower orbits is given by the formula: $N = \frac{n(n-1)}{2}$.
For the first case,$N = 6$:
$6 = \frac{n_1(n_1-1)}{2} \Rightarrow n_1^2 - n_1 - 12 = 0 \Rightarrow (n_1-4)(n_1+3) = 0$. Since $n > 0$,we have $n_1 = 4$.
For the second case,$N = 3$:
$3 = \frac{n_2(n_2-1)}{2} \Rightarrow n_2^2 - n_2 - 6 = 0 \Rightarrow (n_2-3)(n_2+2) = 0$. Since $n > 0$,we have $n_2 = 3$.
The velocity of an electron in the $n$th orbit of a hydrogen atom is given by $v_n \propto \frac{1}{n}$.
Therefore,the ratio of the velocities is:
$\frac{v_1}{v_2} = \frac{n_2}{n_1} = \frac{3}{4}$.

(A) The frequency of revolution of the electron is given by $f = \frac{v}{2 \pi r}$.
Substituting the given values: $f = \frac{2.2 \times 10^{6}}{2 \pi (5 \times 10^{-11})} \approx 7.0 \times 10^{15} \, Hz$.
The current $i$ associated with the motion of the electron is $i = qf$, where $q$ is the charge of the electron $(1.6 \times 10^{-19} \, C)$.
$i = (1.6 \times 10^{-19} \, C) \times (7.0 \times 10^{15} \, Hz) = 11.2 \times 10^{-4} \, A$.
Converting to milliamperes: $i = 1.12 \times 10^{-3} \, A = 1.12 \, mA$.