Bohr's Model of Hydrogen Atom Questions in English

(A) The radius of the $n^{\text{th}}$ orbit in a hydrogen atom is given by the formula $r_n = r_1 \times n^2$,where $r_1$ is the radius of the first orbit (Bohr radius).
Given $r_1 = 5.3 \times 10^{-11} \ m = 0.53 \ \mathring{A}$.
For the third orbit,$n = 3$.
Therefore,$r_3 = r_1 \times (3)^2$.
$r_3 = 0.53 \ \mathring{A} \times 9 = 4.77 \ \mathring{A}$.

(B) According to Bohr's model,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$ for a given atom (where $Z$ is constant).
Given that the radius of the third orbit $(n=3)$ is $R$,we have $r_3 = R$.
The radius of the fourth orbit $(n=4)$ is $r_4$.
Using the proportionality $r_n \propto n^2$,we can write:
$\frac{r_4}{r_3} = \frac{4^2}{3^2}$
$\frac{r_4}{R} = \frac{16}{9}$
Therefore,$r_4 = \frac{16}{9} R$.

(B) The magnetic moment $\mu$ of an electron revolving in an orbit is given by $\mu = iA$,where $i$ is the current and $A$ is the area of the orbit.
$i = \frac{e}{T} = \frac{ev}{2\pi r}$,where $v$ is the velocity and $r$ is the radius.
$A = \pi r^2$.
Therefore,$\mu = \left(\frac{ev}{2\pi r}\right) \pi r^2 = \frac{evr}{2}$.
According to Bohr's theory,$r \propto n^2$ and $v \propto \frac{1}{n}$.
Substituting these relations: $\mu \propto \left(\frac{1}{n}\right) \cdot n^2 = n^1$.
Comparing $\mu \propto n^1$ with $\mu \propto n^x$,we get $x = 1$.

(A) The frequency of emitted radiation is given by $\nu = R c Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For the transition $n=2$ to $n=1$:
$\nu_1 = K \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = K \left( 1 - \frac{1}{4} \right) = K \left( \frac{3}{4} \right) = 3 \times 10^{15} \,Hz$.
For the transition $n=3$ to $n=1$:
$\nu_2 = K \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = K \left( 1 - \frac{1}{9} \right) = K \left( \frac{8}{9} \right)$.
Taking the ratio:
$\frac{\nu_2}{\nu_1} = \frac{K(8/9)}{K(3/4)} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}$.
Therefore, $\nu_2 = \frac{32}{27} \times 3 \times 10^{15} \,Hz = \frac{32}{9} \times 10^{15} \,Hz$.
Comparing this with $\frac{x}{9} \times 10^{15} \,Hz$, we get $X = 32$.

(B) The energy difference for the transition $n=3$ to $n=2$ is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.9 \text{ eV}$.
The emitted photon has wavelength $\lambda = \frac{hc}{\Delta E}$.
Due to conservation of momentum,the atom recoils with velocity $v$,so $mv = \frac{h}{\lambda'}$,where $\lambda'$ is the wavelength of the emitted photon considering recoil.
The energy balance is $\Delta E = \frac{1}{2}mv^2 + \frac{hc}{\lambda'}$.
Substituting $v = \frac{h}{m\lambda'}$,we get $\Delta E = \frac{h^2}{2m\lambda'^2} + \frac{hc}{\lambda'}$.
Solving for $\lambda'$,we find $\lambda' \approx \lambda \left(1 + \frac{\Delta E}{2mc^2}\right)$.
The fractional change in wavelength is $\frac{\Delta \lambda}{\lambda} = \frac{\lambda' - \lambda}{\lambda} = \frac{\Delta E}{2mc^2}$.
Substituting the values: $\Delta E = 1.9 \times 1.6 \times 10^{-19} \text{ J}$,$m = 1.67 \times 10^{-27} \text{ kg}$,$c = 3 \times 10^8 \text{ m/s}$.
$\frac{\Delta \lambda}{\lambda} = \frac{1.9 \times 1.6 \times 10^{-19}}{2 \times 1.67 \times 10^{-27} \times (3 \times 10^8)^2} \approx \frac{3.04 \times 10^{-19}}{3.006 \times 10^{-10}} \approx 10^{-9}$.
Since the question asks for percentage change,$\% \text{ change} = \frac{\Delta \lambda}{\lambda} \times 100 \approx 10^{-9} \times 10^2 = 10^{-7}$.
Thus,$n = 7$.

(C) According to Bohr's postulate,the angular momentum (moment of momentum) of an electron in the $n^{\text{th}}$ orbit is given by the formula:
$L = \frac{nh}{2\pi}$
Given that the electron is in the $4^{\text{th}}$ orbit,we have $n = 4$.
Substituting the value of $n$ into the formula:
$L = \frac{4h}{2\pi}$
$L = \frac{2h}{\pi}$
Therefore,the moment of momentum is $\frac{2h}{\pi}$.

(D) The electrostatic force provides the necessary centripetal force for the electron in a circular orbit:
$F = \frac{k(Ze)(e)}{r^2} = \frac{mv^2}{r}$
From this,the kinetic energy $(KE)$ is:
$KE = \frac{1}{2} mv^2 = \frac{kZe^2}{2r}$
The potential energy $(U)$ is given by:
$U = -\frac{kZe^2}{r}$
The total energy $(E)$ is the sum of kinetic and potential energy:
$E = KE + U = \frac{kZe^2}{2r} - \frac{kZe^2}{r} = -\frac{kZe^2}{2r}$
Comparing $E$ and $U$:
$E = \frac{1}{2} \left( -\frac{kZe^2}{r} \right) = \frac{U}{2}$
Therefore,$2E = U$.

(A) The centripetal force required for the electron to orbit the nucleus is provided by the electrostatic force of attraction: $F_{C} = F_{e}$.
Using the formula for centripetal force $F_{C} = \frac{mv^2}{r}$ and Coulomb's law $F_{e} = \frac{kZe^2}{r^2}$,we have: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$.
Multiplying both sides by $mr^2$,we get: $m^2v^2r^2 = mkZe^2r$.
Since angular momentum $L = mvr$,we can write: $L^2 = mkZe^2r$.
Taking the square root of both sides,we find: $L = \sqrt{mkZe^2r}$.
Since $m$,$k$,$Z$,and $e$ are constants,the angular momentum is proportional to the square root of the radius: $L \propto \sqrt{r}$.

(C) The radius of the $n^{\text{th}}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0 = 0.529 \mathring{A}$.
Given $r_n = 8.48 \mathring{A}$,we have $8.48 = 0.529 \times n^2$.
$n^2 = \frac{8.48}{0.529} \approx 16$.
Thus,$n = 4$.
The energy of an electron in the $n^{\text{th}}$ orbit is given by $E_n = \frac{E}{n^2}$,where $E$ is the ground state energy $(-13.6 \text{ eV})$.
Substituting $n = 4$,we get $E_4 = \frac{E}{4^2} = \frac{E}{16}$.
Comparing this with $E/x$,we find $x = 16$.

(A) The energy required to excite a hydrogen atom to the first excitation level is given by the potential at which the first dip occurs, which is $E = 10.2 \, eV$.
The energy of the emitted photon is related to its wavelength $\lambda$ by the equation $E = \frac{hc}{\lambda}$.
Substituting the given values: $10.2 \, eV = \frac{1245 \, eV \cdot nm}{\lambda}$.
Solving for $\lambda$: $\lambda = \frac{1245}{10.2} \, nm \approx 122.06 \, nm$.
Rounding to the nearest integer, the wavelength is $122 \, nm$.

(C) Part $1$: Energy of $H$ atom in $n=2$ state is $E_H = -13.6 \times (1^2/2^2) = -3.4 \ eV$. Ground state is $-13.6 \ eV$. Excitation energy $\Delta E_H = -3.4 - (-13.6) = 10.2 \ eV$. $He^{+}$ in $n=2$ state has energy $E_{He^+} = -13.6 \times (2^2/2^2) = -13.6 \ eV$. Ground state is $-54.4 \ eV$. Total energy of $He^{+}$ after transfer = $-13.6 + 10.2 = -3.4 \ eV$. Since $E_n = -13.6 \times (Z^2/n^2) = -13.6 \times (4/n^2)$,we have $-3.4 = -54.4/n^2 \implies n^2 = 16 \implies n = 4$. Correct option is $(C)$.
Part $2$: For $He^{+}$,transitions to visible region $(n=2)$ from $n=4$ is $4 \to 2$. $\frac{1}{\lambda} = R Z^2 (\frac{1}{2^2} - \frac{1}{4^2}) = 1.097 \times 10^7 \times 4 \times (\frac{1}{4} - \frac{1}{16}) = 1.097 \times 10^7 \times 4 \times \frac{3}{16} = 0.82275 \times 10^7 \ m^{-1}$. $\lambda \approx 1.215 \times 10^{-7} \ m$. Wait,checking $n=4 \to 3$ or $n=3 \to 2$. For $He^{+}$,$n=3 \to 2$ gives $\frac{1}{\lambda} = R(4)(\frac{1}{4} - \frac{1}{9}) = R(4)(\frac{5}{36}) = R(\frac{5}{9}) \approx 1.097 \times 10^7 \times 0.555 \approx 0.61 \times 10^7 \implies \lambda \approx 1.64 \times 10^{-7} \ m$. Re-evaluating: $n=4 \to 2$ is $4.68 \times 10^{-7} \ m$. Correct option is $(C)$.
Part $3$: $KE = |E| = 13.6 \frac{Z^2}{n^2}$. For $H$ $(Z=1, n=2)$,$KE_H = 13.6/4 = 3.4 \ eV$. For $He^{+}$ $(Z=2, n=2)$,$KE_{He^+} = 13.6 \times (4/4) = 13.6 \ eV$. Ratio $3.4/13.6 = 1/4$. Correct option is $(A)$.

(D) For a $1s$ orbital of a hydrogen atom, the radial wave function is given by $R(r) = 2(Z/a_0)^{3/2} e^{-Zr/a_0}$.
The radial probability distribution function $P(r)$ is defined as the probability of finding the electron in a spherical shell of thickness $dr$ at a distance $r$ from the nucleus.
$P(r) = 4\pi r^2 R^2(r) dr$.
Substituting the expression for $R(r)$, we get $P(r) = 4\pi r^2 [2(Z/a_0)^{3/2} e^{-Zr/a_0}]^2 dr = 16\pi (Z/a_0)^3 r^2 e^{-2Zr/a_0} dr$.
At $r = 0$, $P(r) = 0$ because of the $r^2$ term.
As $r \to \infty$, $P(r) \to 0$ because of the exponential decay term $e^{-2Zr/a_0}$.
The function $P(r)$ starts from zero, increases to a maximum value at $r = a_0/Z$, and then decreases asymptotically to zero.
This corresponds to the shape shown in Graph $D$.

(B) For a single electron system, the radius of the $n^{th}$ orbit is given by $r_n = a_0 \times \frac{n^2}{Z}$.
Given $Z=2$ for $He^{+}$.
For the initial orbit, $r_2 = 105.8 \ pm$:
$105.8 = 52.9 \times \frac{n_2^2}{2} \implies n_2^2 = \frac{105.8 \times 2}{52.9} = 4 \implies n_2 = 2$.
For the final orbit, $r_1 = 26.45 \ pm$:
$26.45 = 52.9 \times \frac{n_1^2}{2} \implies n_1^2 = \frac{26.45 \times 2}{52.9} = 1 \implies n_1 = 1$.
The transition is from $n=2$ to $n=1$.
The energy of the emitted photon is $\Delta E = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
$\Delta E = 2.2 \times 10^{-18} \times (2)^2 \times \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 2.2 \times 10^{-18} \times 4 \times \left( 1 - 0.25 \right) = 8.8 \times 10^{-18} \times 0.75 = 6.6 \times 10^{-18} \ J$.
Using $\Delta E = \frac{hc}{\lambda}$, we get $\lambda = \frac{hc}{\Delta E} = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{6.6 \times 10^{-18}} = 3 \times 10^{-8} \ m = 30 \ nm$.

(A) The energy of a photon emitted during a transition from $n_2$ to $n_1$ in a hydrogen-like atom is given by $\Delta E = 13.6 Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right] \ eV$.
For the transition $n = 2$ to $n = 1$:
$E_1 = 13.6 Z^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 13.6 Z^2 \left( 1 - \frac{1}{4} \right) = 13.6 Z^2 \left( \frac{3}{4} \right)$.
For the transition $n = 3$ to $n = 2$:
$E_2 = 13.6 Z^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = 13.6 Z^2 \left( \frac{1}{4} - \frac{1}{9} \right) = 13.6 Z^2 \left( \frac{5}{36} \right)$.
According to the problem,$E_1 - E_2 = 74.8 \ eV$:
$13.6 Z^2 \left( \frac{3}{4} - \frac{5}{36} \right) = 74.8$.
Calculating the term in the bracket:
$\frac{3}{4} - \frac{5}{36} = \frac{27 - 5}{36} = \frac{22}{36} = \frac{11}{18}$.
Substituting back:
$13.6 Z^2 \times \frac{11}{18} = 74.8$.
$Z^2 = \frac{74.8 \times 18}{13.6 \times 11} = 5.5 \times \frac{18}{11} = 0.5 \times 18 = 9$.
$Z = 3$.

(A) The energy of the absorbed photon is $\frac{hc}{\lambda_{a}} = 13.6 \text{ eV} \times \left[\frac{1}{1^2} - \frac{1}{4^2}\right] = 13.6 \times \frac{15}{16} \text{ eV} \quad (i)$
The energy of the emitted photon is $\frac{hc}{\lambda_{e}} = 13.6 \text{ eV} \times \left[\frac{1}{m^2} - \frac{1}{4^2}\right] \quad (ii)$
Given $\frac{\lambda_{a}}{\lambda_{e}} = \frac{1}{5}$,we divide $(ii)$ by $(i)$:
$\frac{\lambda_{a}}{\lambda_{e}} = \frac{13.6 \times [1 - 1/16]}{13.6 \times [1/m^2 - 1/16]} = 5 \implies \frac{15/16}{1/m^2 - 1/16} = 5$
$\frac{15}{16} = 5 \times \left(\frac{1}{m^2} - \frac{1}{16}\right) \implies \frac{3}{16} = \frac{1}{m^2} - \frac{1}{16}$
$\frac{1}{m^2} = \frac{4}{16} = \frac{1}{4} \implies m = 2$. Thus,option $(3)$ is correct.
For kinetic energy,$K_n \propto \frac{1}{n^2}$. So,$\frac{K_{m=2}}{K_{n=1}} = \frac{1/2^2}{1/1^2} = \frac{1}{4}$. Thus,option $(2)$ is correct.
For momentum,$\Delta p = \frac{h}{\lambda}$. So,$\frac{\Delta p_{a}}{\Delta p_{e}} = \frac{\lambda_{e}}{\lambda_{a}} = 5$. Thus,option $(4)$ is incorrect.

(B) The potential energy is $V(r) = Fr$. The magnitude of the centripetal force is $F_c = |-\frac{dV}{dr}| = F$.
For circular motion,$F = \frac{mv^2}{R} \implies v^2 = \frac{FR}{m}$.
Using Bohr's quantization condition,$mvr = \frac{nh}{2\pi} \implies v = \frac{nh}{2\pi mR}$.
Substituting $v$ into the force equation: $F = \frac{m}{R} \left( \frac{n^2 h^2}{4 \pi^2 m^2 R^2} \right) = \frac{n^2 h^2}{4 \pi^2 m R^3}$.
Thus,$R^3 = \frac{n^2 h^2}{4 \pi^2 mF} \implies R \propto n^{2/3}$.
From $v = \frac{nh}{2\pi mR}$,since $R \propto n^{2/3}$,we get $v \propto \frac{n}{n^{2/3}} = n^{1/3}$. So,$(B)$ is correct.
The total energy $E = K.E. + P.E. = \frac{1}{2}mv^2 + FR$.
Since $mv^2 = FR$,$E = \frac{1}{2}FR + FR = \frac{3}{2}FR$.
Substituting $R = \left( \frac{n^2 h^2}{4 \pi^2 mF} \right)^{1/3}$,we get $E = \frac{3}{2} F \left( \frac{n^2 h^2}{4 \pi^2 mF} \right)^{1/3} = \frac{3}{2} \left( \frac{n^2 h^2 F^3}{4 \pi^2 mF} \right)^{1/3} = \frac{3}{2} \left( \frac{n^2 h^2 F^2}{4 \pi^2 m} \right)^{1/3}$. So,$(C)$ is correct.

(D) At $d = d_0$, the electron-electron repulsion and nucleus-nucleus repulsion are absent. The potential energy is primarily due to the attraction between the proton and electron in each $H$ atom.
The potential energy $(P.E.)$ for one $H$ atom is given by:
$P.E. = \frac{-K q_1 q_2}{r} = \frac{-(9 \times 10^9) \times (1.6 \times 10^{-19})^2}{0.529 \times 10^{-10}} \ J$
$P.E. = -4.355 \times 10^{-18} \ J$
To convert this to $kJ \ mol^{-1}$ for one mole of $H$ atoms:
$E_0 = (-4.355 \times 10^{-18} \ J) \times (6.023 \times 10^{23} \ mol^{-1}) \times 10^{-3} \ kJ/J$
$E_0 = -2623.249 \ kJ \ mol^{-1}$
Since the question asks for the magnitude of the potential energy $E_0$ as indicated in the figure, the value is $2623.249 \ kJ \ mol^{-1}$.

(C) The radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z} = 4.5 a_0$.
Given orbital angular momentum $L = \frac{nh}{2\pi} = \frac{3h}{2\pi}$,so $n = 3$.
Substituting $n=3$ into the radius formula: $4.5 = \frac{3^2}{Z} = \frac{9}{Z}$,which gives $Z = 2$.
The atom is a Helium ion $(He^+)$.
When the atom de-excites from $n=3$,the possible transitions are $3 \rightarrow 2$,$3 \rightarrow 1$,and $2 \rightarrow 1$.
The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$.
For $3 \rightarrow 1$: $\frac{1}{\lambda_{3 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{3^2} \right) = 4R \left( 1 - \frac{1}{9} \right) = 4R \left( \frac{8}{9} \right) = \frac{32R}{9} \Rightarrow \lambda_{3 \rightarrow 1} = \frac{9}{32R}$.
For $3 \rightarrow 2$: $\frac{1}{\lambda_{3 \rightarrow 2}} = R(2^2) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{9} \right) = 4R \left( \frac{5}{36} \right) = \frac{5R}{9} \Rightarrow \lambda_{3 \rightarrow 2} = \frac{9}{5R}$.
For $2 \rightarrow 1$: $\frac{1}{\lambda_{2 \rightarrow 1}} = R(2^2) \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R \Rightarrow \lambda_{2 \rightarrow 1} = \frac{1}{3R}$.
The possible wavelengths are $\frac{9}{32R}$ and $\frac{9}{5R}$. Thus,the correct option is $(C)$.

(B) The energy of the incident photon is given by $E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1242 \ eV \ nm}{90 \ nm} = 13.8 \ eV$.
The energy required to ionize a hydrogen atom from the $n^{\text{th}}$ orbital is $E_n = \frac{13.6 \ eV}{n^2}$.
According to the law of conservation of energy,the energy of the photon is equal to the sum of the ionization energy and the kinetic energy of the ejected electron:
$E_{\text{photon}} = E_n + K.E.$
Substituting the given values:
$13.8 \ eV = \frac{13.6 \ eV}{n^2} + 10.4 \ eV$.
Rearranging the equation:
$\frac{13.6}{n^2} = 13.8 - 10.4 = 3.4$.
Solving for $n^2$:
$n^2 = \frac{13.6}{3.4} = 4$.
Therefore,$n = 2$.

(B) The central force provides the necessary centripetal force: $kr = \frac{mv^2}{r} \implies kr^2 = mv^2$ $(1)$.
By Bohr's quantization rule: $L = mvr = n\hbar \implies v = \frac{n\hbar}{mr}$ $(2)$.
Substitute $(2)$ into $(1)$: $kr^2 = m(\frac{n\hbar}{mr})^2 = \frac{n^2\hbar^2}{mr^2}$.
Rearranging gives $r^4 = \frac{n^2\hbar^2}{mk}$,so $r^2 = n\hbar \sqrt{\frac{1}{mk}}$. Thus,$(A)$ is correct.
From $(1)$,$v^2 = \frac{kr^2}{m} = \frac{k}{m} (n\hbar \sqrt{\frac{1}{mk}}) = n\hbar \sqrt{\frac{k}{m^2}} = n\hbar \sqrt{\frac{k}{m^3}}$. Thus,$(B)$ is correct.
From $(1)$,$\frac{v^2}{r^2} = \frac{k}{m}$,so $\frac{v}{r} = \sqrt{\frac{k}{m}}$. Since $L = mvr$,$\frac{L}{mr^2} = \frac{mvr}{mr^2} = \frac{v}{r} = \sqrt{\frac{k}{m}}$. Thus,$(C)$ is correct.
Total energy $E = K + V = \frac{1}{2}mv^2 + \frac{1}{2}kr^2 = \frac{1}{2}(kr^2) + \frac{1}{2}kr^2 = kr^2 = k(n\hbar \sqrt{\frac{1}{mk}}) = n\hbar \sqrt{\frac{k}{m}}$. Thus,$(D)$ is incorrect.

(B) The kinetic energy $(KE)$ of an electron in the $n^{th}$ orbit of a hydrogen-like atom with atomic number $Z$ is given by the formula: $KE = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
Calculating for each option:
$(A)$ For the first orbit of $H$ atom $(Z=1, n=1)$: $KE = 13.6 \times \frac{1^2}{1^2} = 13.6 \text{ eV}$.
$(B)$ For the first orbit of $He^{+}$ $(Z=2, n=1)$: $KE = 13.6 \times \frac{2^2}{1^2} = 13.6 \times 4 = 54.4 \text{ eV}$.
$(C)$ For the second orbit of $He^{+}$ $(Z=2, n=2)$: $KE = 13.6 \times \frac{2^2}{2^2} = 13.6 \text{ eV}$.
$(D)$ For the second orbit of $Li^{2+}$ $(Z=3, n=2)$: $KE = 13.6 \times \frac{3^2}{2^2} = 13.6 \times 2.25 = 30.6 \text{ eV}$.
Comparing the values, the highest kinetic energy is $54.4 \text{ eV}$, which corresponds to the first orbit of $He^{+}$.

(B) The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
According to Bohr's quantization condition,$mvr = \frac{nh}{2\pi}$,which implies $mv = \frac{nh}{2\pi r}$.
Substituting this into the wavelength formula: $\lambda = \frac{h}{nh / (2\pi r)} = \frac{2\pi r}{n}$.
Thus,$\lambda \propto \frac{r}{n}$.
For the ground state,$n_1 = 1$ and $r_1 = 5.3 \times 10^{-11} \ m$.
For the third excited state,$n_4 = 4$ and $r_4 = 8.48 \times 10^{-10} \ m = 84.8 \times 10^{-11} \ m$.
The ratio is $\frac{\lambda_1}{\lambda_4} = \frac{r_1}{n_1} \times \frac{n_4}{r_4} = \frac{5.3 \times 10^{-11}}{1} \times \frac{4}{84.8 \times 10^{-11}} = \frac{5.3 \times 4}{84.8} = \frac{21.2}{84.8} = \frac{1}{4}$.

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

(B) The energy of an electron in a hydrogen-like atom is given by $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For $H$ atom $(Z=1)$ in ground state $(n=1)$: $E = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV}$.
For $He^{+}$ ion $(Z=2)$:
Ground state $(n=1)$: $E = -13.6 \times \frac{2^2}{1^2} = -54.4 \text{ eV}$.
First excited state $(n=2)$: $E = -13.6 \times \frac{2^2}{2^2} = -13.6 \text{ eV}$.
For $Li^{++}$ ion $(Z=3)$:
Ground state $(n=1)$: $E = -13.6 \times \frac{3^2}{1^2} = -122.4 \text{ eV}$.
First excited state $(n=2)$: $E = -13.6 \times \frac{3^2}{2^2} = -30.6 \text{ eV}$.
Second excited state $(n=3)$: $E = -13.6 \times \frac{3^2}{3^2} = -13.6 \text{ eV}$.
Comparing the values:
Statement $(A)$: $H$ (ground) = $-13.6 \text{ eV}$,$He^{+}$ (1st excited) = $-13.6 \text{ eV}$. (Correct)
Statement $(B)$: $H$ (ground) = $-13.6 \text{ eV}$,$Li^{++}$ (2nd excited) = $-13.6 \text{ eV}$. (Correct)
Statement $(C)$: $H$ (ground) = $-13.6 \text{ eV}$,$He^{+}$ (ground) = $-54.4 \text{ eV}$. (Incorrect)
Statement $(D)$: $He^{+}$ (1st excited) = $-13.6 \text{ eV}$,$Li^{++}$ (ground) = $-122.4 \text{ eV}$. (Incorrect)
Thus,only $(A)$ and $(B)$ are correct.

(C) According to Bohr's model,the radius of an orbit for a hydrogen-like ion is given by the formula: $r_n = a_0 \frac{n^2}{Z}$.
For the $Li^{++}$ ion,the atomic number $Z = 3$.
For the ground state,the principal quantum number $n = 1$.
Substituting these values into the formula:
$r = a_0 \frac{1^2}{3} = \frac{1}{3} a_0$.
Comparing this with the given expression $\frac{1}{X} a_0$,we find that $X = 3$.

(C) The Bohr model is based on the assumption of a single electron revolving around a nucleus. It considers only the electrostatic force of attraction between the nucleus and the electron. It does not account for the electron-electron repulsion that occurs in multi-electron atoms. Therefore,the model is strictly applicable only to hydrogen-like atoms (atoms with one electron,such as $H$,$He^+$,$Li^{2+}$). Since the reason correctly explains why the model is limited to single-electron systems,both Assertion $A$ and Reason $R$ are true,and $R$ is the correct explanation of $A$.

(D) The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6 \ eV}{n^2}$.
The fourth excited state corresponds to $n_1 = 5$ (since ground state is $n=1$,first excited is $n=2$,...,fourth excited is $n=5$).
The energy emitted during a transition from $n_1$ to $n$ is given by $\Delta E = 13.6 \left( \frac{1}{n^2} - \frac{1}{n_1^2} \right) \ eV$.
Given $\Delta E = 2.86 \ eV$ and $n_1 = 5$,we have:
$2.86 = 13.6 \left( \frac{1}{n^2} - \frac{1}{5^2} \right)$
$\frac{2.86}{13.6} = \frac{1}{n^2} - \frac{1}{25}$
$0.2103 \approx \frac{1}{n^2} - 0.04$
$\frac{1}{n^2} = 0.2103 + 0.04 = 0.2503 \approx \frac{1}{4}$
$n^2 = 4 \implies n = 2$.

(D) The radius of the $n^{\text{th}}$ orbit for a hydrogen-like atom is given by the formula: $r_n = a_0 \cdot \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For $Li^{2+}$,the atomic number $Z_1 = 3$ and the orbit number $n = 5$. Thus,$r_{Li^{2+}} = a_0 \cdot \frac{5^2}{3} = a_0 \cdot \frac{25}{3}$.
For $He^{+}$,the atomic number $Z_2 = 2$ and the orbit number $n = 5$. Thus,$r_{He^{+}} = a_0 \cdot \frac{5^2}{2} = a_0 \cdot \frac{25}{2}$.
The ratio of the radii is: $\frac{r_{Li^{2+}}}{r_{He^{+}}} = \frac{a_0 \cdot \frac{25}{3}}{a_0 \cdot \frac{25}{2}} = \frac{25}{3} \cdot \frac{2}{25} = \frac{2}{3}$.

(C) Statement $(I):$ The energy of a photon is given by $E = hf$. The dimensions of Planck's constant $h$ are $[h] = [E]/[f] = [ML^2T^{-2}]/[T^{-1}] = [ML^2T^{-1}]$. The dimensions of angular momentum $L = mvr$ are $[L] = [M][LT^{-1}][L] = [ML^2T^{-1}]$. Thus,Statement $(I)$ is correct.
Statement $(II):$ According to Bohr's postulate,the angular momentum $L$ of an electron in a stable orbit is an integral multiple of $h/(2\pi)$,i.e.,$L = nh/(2\pi)$,where $n$ is an integer. The statement claims it is an integral multiple of $h$,which is incorrect because the factor $1/(2\pi)$ is missing. Thus,Statement $(II)$ is incorrect.

(D) The energy of an electron in a hydrogen-like ion is given by $E_n = -13.6 \frac{Z^2}{n^2} \ eV$.
For the ground state,$n_1 = 1$. For the $2^{\text{nd}}$ excitation state,$n_2 = 3$ (since $1^{\text{st}}$ excitation is $n=2$ and $2^{\text{nd}}$ excitation is $n=3$).
The energy difference is given by $\Delta E = E_3 - E_1 = 13.6 Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the given values: $108.8 = 13.6 Z^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$.
$108.8 = 13.6 Z^2 \left( 1 - \frac{1}{9} \right) = 13.6 Z^2 \left( \frac{8}{9} \right)$.
$Z^2 = \frac{108.8 \times 9}{13.6 \times 8} = 8 \times 9 = 9$.
$Z = 3$.

(C) The centripetal force is provided by the constant force $F$ directed towards the origin:
$\frac{m v^2}{r} = F \quad ....(1)$
According to Bohr's quantization condition for angular momentum:
$mvr = \frac{nh}{2\pi} \Rightarrow v = \frac{nh}{2\pi mr} \quad ....(2)$
Substitute $v$ from equation $(2)$ into equation $(1)$:
$\frac{m}{r} \left( \frac{nh}{2\pi mr} \right)^2 = F$
$\frac{m}{r} \cdot \frac{n^2 h^2}{4\pi^2 m^2 r^2} = F$
$\frac{n^2 h^2}{4\pi^2 m r^3} = F$
Rearranging for $r^3$:
$r^3 = \frac{n^2 h^2}{4\pi^2 m F}$
Since $h, m, F$ are constants,$r^3 \propto n^2$,which implies $r \propto n^{2/3}$.
Now,substitute $r \propto n^{2/3}$ into the expression for $v$ from equation $(2)$:
$v \propto \frac{n}{r} \propto \frac{n}{n^{2/3}} \propto n^{1 - 2/3} \propto n^{1/3}$.
Thus,$r \propto n^{2/3}$ and $v \propto n^{1/3}$.

(B) According to Bohr's quantization condition,the circumference of the orbit is an integer multiple of the De-Broglie wavelength: $2 \pi r_n = n \lambda$.
The radius of the $n$-th orbit is given by $r_n = a_0 \frac{n^2}{Z}$,where $a_0 = 0.052 \ nm$ and $Z = 1$ for hydrogen.
For $n=2$,the radius is $r_2 = 0.052 \times \frac{2^2}{1} = 0.052 \times 4 = 0.208 \ nm$.
Substituting this into the quantization condition: $2 \pi (0.208) = 2 \lambda$.
Solving for $\lambda$: $\lambda = \pi \times 0.208 \approx 3.14159 \times 0.208 \approx 0.653 \ nm$.
Rounding to the nearest given option,the value is $0.67 \ nm$.

(B) The magnetic moment $M$ of an electron revolving in the $n$-th orbit is given by the formula: $M = \frac{n e h}{4 \pi m}$.
From this expression, we can see that the magnetic moment is directly proportional to the principal quantum number $n$, i.e., $M \propto n$.
For the first orbit $(n = 1)$, the magnetic moment is $M_1 = \mu_{B}$.
For the second orbit $(n = 2)$, the magnetic moment is $M_2 = 2 \times M_1$.
Therefore, $M_2 = 2 \mu_{B}$.

(D) The number of spectral lines emitted when an electron transitions from an excited state $n_1$ to the ground state $n=1$ is given by the formula $N = \frac{n_1(n_1-1)}{2}$.
Given that the number of emitted wavelengths is $6$,we have $\frac{n_1(n_1-1)}{2} = 6$,which simplifies to $n_1^2 - n_1 - 12 = 0$.
Solving this quadratic equation,we get $(n_1-4)(n_1+3) = 0$. Since $n_1$ must be positive,$n_1 = 4$.
The electron was initially in state $n_2$ and absorbed a photon to reach $n_1 = 4$. After reaching $n_1$,it returned to the ground state $(n=1)$ emitting $6$ lines. This implies the transition sequence covers all levels from $4$ down to $1$.
Since the atom was in an excited state $n_2$ and jumped to $n_1=4$,and the total emission spectrum corresponds to transitions from $n=4$ to $n=1$,the initial state must have been $n_2=2$ to allow for the absorption of a photon to reach $n_1=4$.

(B) The electrostatic force $F$ between the nucleus and the electron is given by Coulomb's law: $F = \frac{1}{4\pi\epsilon_0} \frac{Ze^2}{r^2}$,which implies $F \propto \frac{1}{r^2}$.
According to the Bohr model,the radius of the $n^{th}$ orbit is given by $r_n = a_0 n^2$,which implies $r \propto n^2$.
Substituting the relation $r \propto n^2$ into the force equation,we get $F \propto \frac{1}{(n^2)^2} = \frac{1}{n^4}$.
Therefore,the force on the electron is inversely proportional to the fourth power of the principal quantum number $n$.

(D) According to Bohr's quantization condition,the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$.
For the ground state of a hydrogen atom,$n = 1$.
Thus,$mvr = \frac{h}{2\pi}$,which implies $2\pi r = \frac{h}{mv}$.
Since the de-Broglie wavelength is defined as $\lambda = \frac{h}{mv}$,we have $\lambda = 2\pi r$.
For the ground state,the radius of the first orbit is $r = 0.53 \mathring A$.
Substituting the values,we get $\lambda = 2 \times 3.14 \times 0.53 \mathring A = 3.33 \mathring A$.

(B) The radius of an orbit in a hydrogen atom is given by $R_n = n^2 a_0$,where $a_0 = 0.53 \mathring{A}$.
For the initial state: $2.12 = n_1^2 \times 0.53 \implies n_1^2 = 4 \implies n_1 = 2$.
For the final state: $4.77 = n_2^2 \times 0.53 \implies n_2^2 = 9 \implies n_2 = 3$.
Since the radius increases,the electron has transitioned from a lower energy level to a higher energy level,meaning the atom has absorbed a photon.
The energy of the absorbed photon is $\Delta E = E_3 - E_2 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} \right) \ eV$.
$\Delta E = -13.6 \left( \frac{1}{9} - \frac{1}{4} \right) = -13.6 \left( \frac{4-9}{36} \right) = -13.6 \left( -\frac{5}{36} \right) \ eV$.
$\Delta E = \frac{68}{36} \ eV \approx 1.89 \ eV$.

(B) According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
Given $L = \frac{3h}{2\pi}$,we find that the principal quantum number $n = 3$.
The radius of the $n$-th orbit is given by $r_n = n^2 r_0$,where $r_0 = 0.529 \mathring{A}$.
For $n = 3$,$r_3 = 3^2 \times 0.529 \mathring{A} = 9 \times 0.529 \mathring{A} = 4.761 \mathring{A}$.
According to the de Broglie hypothesis,the circumference of the orbit is equal to $n$ times the wavelength: $2\pi r_n = n\lambda$.
Therefore,$\lambda = \frac{2\pi r_n}{n} = \frac{2\pi \times 9r_0}{3} = 6\pi r_0$.
Substituting $r_0 = 0.529 \mathring{A}$,we get $\lambda = 6 \times 3.1416 \times 0.529 \mathring{A} \approx 9.97 \mathring{A} \approx 10 \mathring{A}$.

(A) The velocity of an electron in the $n$-th orbit is given by $v = \frac{Z e^2}{2 \epsilon_0 n h}$.
For an electron,the de Broglie wavelength is $\lambda_e = \frac{h}{m v} = \frac{2 \epsilon_0 n h^2}{m Z e^2}$.
For a neutron with thermal energy $E = k_B T$,the de Broglie wavelength is $\lambda_n = \frac{h}{\sqrt{2 m_N k_B T}}$.
Equating $\lambda_e = \lambda_n$,we get $\frac{h}{m v} = \frac{h}{\sqrt{2 m_N k_B T}}$,which implies $m^2 v^2 = 2 m_N k_B T$.
Substituting $v = \frac{Z e^2}{2 \epsilon_0 n h}$,we get $T = \frac{m^2 Z^2 e^4}{8 \epsilon_0^2 n^2 h^2 m_N k_B}$.
Using $n=3$,$T = \frac{m^2 Z^2 e^4}{72 \epsilon_0^2 h^2 m_N k_B}$.
Since $a_0 = \frac{h^2 \epsilon_0}{\pi m e^2}$,we have $a_0^2 = \frac{h^4 \epsilon_0^2}{\pi^2 m^2 e^4}$.
Substituting $\frac{m^2 e^4}{\epsilon_0^2} = \frac{h^4}{\pi^2 a_0^2}$,we get $T = \frac{Z^2 h^2}{72 \pi^2 a_0^2 m_N k_B}$.
Comparing this with the given expression $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$,we find $\alpha = 72$.

(B) $1$. Energy of the incident photon $h v_1$ is used to ionize the hydrogen atom and provide kinetic energy to the electron:
$h v_1 = E_{ionization} + KE_{electron} = 13.6 eV + 10 eV = 23.6 eV$.
$2$. The positronium atom is a system of an electron and a positron. Its reduced mass $\mu$ is given by $\mu = \frac{m_e m_p}{m_e + m_p} = \frac{m_e^2}{2m_e} = \frac{m_e}{2}$.
$3$. The ground state energy of the positronium atom is $E_p = -13.6 eV \times \frac{\mu}{m_e} = -13.6 eV \times \frac{1}{2} = -6.8 eV$.
$4$. When the electron (with $10 eV$ kinetic energy) combines with a stationary positron to form a positronium atom,the total energy available is $10 eV$. This energy is used to form the positronium atom in its ground state (releasing $6.8 eV$ as a photon) and to provide kinetic energy to the center of mass of the positronium atom $(5 eV)$.
$5$. By conservation of energy: $10 eV = h v_2 + |E_p| + KE_{COM} = h v_2 + 6.8 eV + 5 eV$.
$h v_2 = 10 eV - 11.8 eV = -1.8 eV$.
Wait,re-evaluating: The energy released as a photon $h v_2$ is the difference between the initial kinetic energy of the electron and the sum of the binding energy and the kinetic energy of the center of mass of the positronium atom.
$h v_2 = KE_{electron} - (|E_p| + KE_{COM}) = 10 eV - (6.8 eV + 5 eV) = -1.8 eV$.
Actually,the energy balance is: $KE_{electron} + |E_p| = h v_2 + KE_{COM}$.
$10 eV + 6.8 eV = h v_2 + 5 eV \implies h v_2 = 11.8 eV$.
$6$. The difference between the two photon energies is $|h v_1 - h v_2| = |23.6 eV - 11.8 eV| = 11.8 eV$.

(A) According to Rydberg's formula for hydrogen-like ions,the wavelength $\lambda$ of the emitted radiation is given by:
$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Since the transition is from $n_2 = 2$ to $n_1 = 1$ for both ions,the term $\left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = \left[ 1 - \frac{1}{4} \right] = \frac{3}{4}$ is constant.
Therefore,$\lambda \propto \frac{1}{Z^2}$.
For Helium $(He^+)$,$Z_{He} = 2$,so $\lambda_{He} \propto \frac{1}{2^2} = \frac{1}{4}$.
For Lithium $(Li^{++})$,$Z_{Li} = 3$,so $\lambda_{Li} \propto \frac{1}{3^2} = \frac{1}{9}$.
The ratio of wavelengths is $\frac{\lambda_{He}}{\lambda_{Li}} = \frac{1/4}{1/9} = \frac{9}{4}$.
Thus,the ratio is $9:4$.

(B) The velocity of an electron in the $n^{\text{th}}$ orbit is given by $v_n = \frac{e^2}{2 \varepsilon_0 nh}$.
This implies $v_n \propto \frac{1}{n}$.
For the ground state,$n_1 = 1$,and for the first excited state,$n_2 = 2$.
Thus,the ratio of velocities is $\frac{v_2}{v_1} = \frac{n_1}{n_2} = \frac{1}{2}$,which means $v_2 = 0.5 v_1$.
The change in velocity is $\Delta v = |v_2 - v_1| = |0.5 v_1 - v_1| = 0.5 v_1$.
The percentage change is $\frac{\Delta v}{v_1} \times 100\% = \frac{0.5 v_1}{v_1} \times 100\% = 50\%$.

(A) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
For the $2^{\text{nd}}$ 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 $E = \frac{(\frac{h}{\pi})^2}{2I} = \frac{h^2}{2I\pi^2}$.

(A) The magnetic field at the center of a circular current loop is given by $B = \frac{\mu_{0} I}{2r}$.
For an electron revolving in an orbit,the equivalent current is $I = ef = e \left( \frac{v}{2 \pi r} \right) = \frac{ev}{2 \pi r}$.
Substituting this into the magnetic field formula: $B = \frac{\mu_{0} ev}{4 \pi r^{2}}$.
For the ground state of a hydrogen atom,the velocity $v = \frac{e^{2}}{2 \epsilon_{0} h}$ and the radius $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$.
Substituting $v$ and $r$ into the expression for $B$:
$B = \frac{\mu_{0} e}{4 \pi r^{2}} \left( \frac{e^{2}}{2 \epsilon_{0} h} \right) = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h r^{2}}$.
Now,substituting $r = \frac{h^{2} \epsilon_{0}}{\pi m e^{2}}$:
$B = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \left( \frac{\pi m e^{2}}{h^{2} \epsilon_{0}} \right)^{2} = \frac{\mu_{0} e^{3}}{8 \pi \epsilon_{0} h} \cdot \frac{\pi^{2} m^{2} e^{4}}{h^{4} \epsilon_{0}^{2}} = \frac{\mu_{0} \pi m^{2} e^{7}}{8 \epsilon_{0}^{3} h^{5}}$.

(A) According to Bohr's quantization condition,the angular momentum $L$ is given by $L = n \frac{h}{2 \pi}$,where $n$ is the orbit number.
For the third orbit,$n = 3$,so $L = 3 \frac{h}{2 \pi} = \frac{3h}{2 \pi}$.
The rotational kinetic energy $K$ of a diatomic molecule with moment of inertia $I$ is given by $K = \frac{L^2}{2I}$.
Substituting the value of $L$ into the energy formula:
$K = \frac{(\frac{3h}{2 \pi})^2}{2I} = \frac{\frac{9h^2}{4 \pi^2}}{2I} = \frac{9h^2}{8 \pi^2 I}$.
Thus,the correct option is $A$.

(D) According to Bohr's postulate for the quantization of angular momentum,$mvr = \frac{nh}{2\pi}$.
From the de-Broglie hypothesis,the wavelength is given by $\lambda = \frac{h}{mv}$.
Rearranging the angular momentum equation,we get $mv = \frac{nh}{2\pi r}$.
Substituting this into the de-Broglie 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 $r_n \propto n^2$.
Substituting $r \propto n^2$ into the wavelength expression: $\lambda \propto \frac{n^2}{n} = n$.
Therefore,$\lambda \propto n$.

(A) According to Bohr's first postulate,the centripetal force required for the circular motion of the electron is provided by the electrostatic force of attraction between the nucleus and the electron.
The electrostatic force is given by $F_e = \frac{1}{4 \pi \epsilon_0} \frac{(Ze)(e)}{r^2} = K \frac{Ze^2}{r^2}$.
The centripetal force is given by $F_c = \frac{mv^2}{r}$.
Equating the two forces: $\frac{mv^2}{r} = K \frac{Ze^2}{r^2}$.
Multiplying both sides by $\frac{r}{2}$,we get: $\frac{1}{2} mv^2 = \frac{1}{2} K \frac{Ze^2}{r}$.
Since kinetic energy $E_k = \frac{1}{2} mv^2$,we have $E_k = \frac{Ze^2}{2r} K$.

(C) The radius of the $n^{\text{th}}$ orbit is given by $r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2 Z}$.
The velocity of the electron in the $n^{\text{th}}$ orbit is given by $v_n = \frac{Z e^2}{2 \varepsilon_0 n h}$.
The orbital period $T$ is defined as $T = \frac{2 \pi r_n}{v_n}$.
Substituting the expressions for $r_n$ and $v_n$:
$T = \frac{2 \pi (n^2 h^2 \varepsilon_0 / \pi m e^2 Z)}{(Z e^2 / 2 \varepsilon_0 n h)}$.
$T = \frac{2 n^2 h^2 \varepsilon_0}{m e^2 Z} \times \frac{2 \varepsilon_0 n h}{Z e^2} = \frac{4 \varepsilon_0^2 n^3 h^3}{m Z^2 e^4}$.
For a hydrogen atom,the atomic number $Z = 1$.
Therefore,$T = \frac{4 \varepsilon_0^2 n^3 h^3}{m e^4}$.

(B) The magnetic dipole moment $(m)$ of an electron revolving in a circular orbit is given by $m = -\frac{e}{2m_e} L$,where $e$ is the charge of the electron,$m_e$ is the mass of the electron,and $L$ is the angular momentum.
The gyromagnetic ratio $(R)$ is defined as the ratio of the magnetic dipole moment to the angular momentum:
$R = \left| \frac{m}{L} \right| = \frac{e}{2m_e}$.
Therefore,the relationship between the magnetic dipole moment $(m)$,the gyromagnetic ratio $(R)$,and the angular momentum $(L)$ is given by:
$m = -RL$.