Atomic models and Planck's quantum theory Questions in English

(C) The kinetic energy $(K.E.)$ of an electron in the $n^{th}$ Bohr orbit is given by the formula: $K.E. = \frac{kZe^2}{2r}$.
Since the radius of the $n^{th}$ orbit is $r_n \propto n^2$,we can substitute this into the expression for kinetic energy.
Therefore,$K.E. \propto \frac{1}{n^2}$.
This relationship indicates that as the principal quantum number $(n)$ increases,the kinetic energy decreases following an inverse square curve. Thus,the correct graph is the one showing $K.E.$ decreasing as $n$ increases.

(D) According to Planck's quantum theory,the emission or absorption of energy occurs in the form of discrete packets called quanta.
When an electron transitions from a higher energy level $(n_{high})$ to a lower energy level $(n_{low})$,it emits energy in the form of a photon (quantum).
Since all the given transitions ($n = 4 \rightarrow n = 2$,$n = 3 \rightarrow n = 1$,and $n = 5 \rightarrow n = 3$) involve an electron moving from a higher energy state to a lower energy state,each transition results in the emission of one quantum of energy.

(A) The frequency of emitted radiation is directly proportional to the energy difference between the energy levels involved in the transition.
Emission occurs when an electron jumps from a higher energy level $(n_2)$ to a lower energy level $(n_1)$.
The frequency $\nu$ is given by the Rydberg formula: $\nu = R_{H} c z^2 \left[\frac{1}{n_1^2} - \frac{1}{n_2^2}\right]$.
For option $A$ ($n = 2$ to $n = 1$): $\Delta E \propto \left[\frac{1}{1^2} - \frac{1}{2^2}\right] = 1 - 0.25 = 0.75$.
For option $B$ ($n = 6$ to $n = 2$): $\Delta E \propto \left[\frac{1}{2^2} - \frac{1}{6^2}\right] = 0.25 - 0.027 = 0.223$.
Comparing the values,the transition from $n = 2$ to $n = 1$ results in the largest energy difference and thus the maximum frequency of emission.

(A) The radius of the $n^{th}$ orbit of the $H$-atom is given by the formula $r_n = 0.529 \times n^2 \mathring{A}$.
For the $2^{nd}$ orbit $(n=2)$: $r_2 = 0.529 \times (2)^2 = 0.529 \times 4 = 2.116 \mathring{A}$.
For the $3^{rd}$ orbit $(n=3)$: $r_3 = 0.529 \times (3)^2 = 0.529 \times 9 = 4.761 \mathring{A}$.
The distance between the $2^{nd}$ and $3^{rd}$ orbit is $r_3 - r_2 = 4.761 \mathring{A} - 2.116 \mathring{A} = 2.645 \mathring{A} \approx 2.65 \mathring{A}$.
Thus,the correct option is $A$.

(B) The longest wavelength corresponds to the minimum energy transition in the Lyman series,which occurs from $n_2 = 2$ to $n_1 = 1$.
Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the $H$ atom,$Z = 1$,$n_1 = 1$,and $n_2 = 2$.
$\frac{1}{\lambda} = R \times 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R$.
$\lambda = \frac{4}{3R}$.
Given $\frac{1}{R} \approx 911.6 \mathring{A}$,we have $\lambda = \frac{4}{3} \times 911.6 \mathring{A} \approx 1215.5 \mathring{A}$.

(B) The total energy $E$ of $n$ photons is given by $E = \frac{nhc}{\lambda}$.
Given: $E = 1 \ J$,$\lambda = 4000 \ \mathring{A} = 4000 \times 10^{-10} \ m = 4 \times 10^{-7} \ m$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$.
Rearranging for $n$: $n = \frac{E \lambda}{hc}$.
Substituting the values: $n = \frac{1 \times 4 \times 10^{-7}}{6.626 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{4 \times 10^{-7}}{19.878 \times 10^{-26}} \approx 0.2012 \times 10^{19} = 2.012 \times 10^{18}$.
Rounding to the nearest option,$n \approx 2 \times 10^{18}$.

(C) For a hydrogen atom,the electronic transition from $n=4$ to $n=2$ corresponds to the Balmer series,which falls in the visible region. Thus,the Assertion is True.
However,the Lyman series corresponds to transitions ending at $n=1$,which falls in the ultraviolet $(UV)$ region,not the visible region. Thus,the Reason is False.

(B) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the minimum wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.
For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R_H (\frac{1}{1^2} - 0) = R_H$,which means $\lambda_L = \frac{1}{R_H}$.
For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R_H (\frac{1}{2^2} - 0) = \frac{R_H}{4}$,which means $\lambda_B = \frac{4}{R_H}$.
The ratio of minimum wavelengths $\frac{\lambda_L}{\lambda_B} = \frac{1/R_H}{4/R_H} = \frac{1}{4} = 0.25$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ eV / n^2$.
The energy difference between $n = 3$ and $n = 2$ is $\Delta E = E_3 - E_2 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} \right) = -13.6 \left( \frac{1}{9} - \frac{1}{4} \right) = -13.6 \left( \frac{4 - 9}{36} \right) = -13.6 \left( -\frac{5}{36} \right) = \frac{13.6 \times 5}{36} \ eV$.
Given $\Delta E = E$,so $E = \frac{68}{36} \ eV = \frac{17}{9} \ eV$.
The ionization energy of the $H$-atom is the energy required to move an electron from $n = 1$ to $n = \infty$,which is $E_{\infty} - E_1 = 0 - (-13.6 \ eV) = 13.6 \ eV$.
Substituting $13.6 = E \times \frac{36}{5}$,we get $13.6 = E \times 7.2$.
Therefore,the ionization energy is $7.2 \ E$.

(A) The formula for the radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For the $B^{4+}$ ion, the atomic number $Z = 5$ and the orbit number $n = 4$.
Substituting these values into the formula:
$r_4 = \frac{52.9 \times (4)^2}{5} \ pm$
$r_4 = \frac{52.9 \times 16}{5} \ pm$
$r_4 = \frac{846.4}{5} \ pm$
$r_4 = 169.28 \ pm \approx 169.3 \ pm$.

(D) Rutherford's atomic model proposed that electrons revolve around the nucleus,but it could not explain the stability of the atom or the distribution of electrons. It also failed to describe the energies of electrons. Bohr's atomic model was the one that successfully described the energies of electrons.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$.
For $He^{+}$,the atomic number $Z = 2$.
The orbit number $n = 3$.
Substituting these values into the formula:
$E_3 = -2.18 \times 10^{-18} \times \frac{2^2}{3^2} \ J$
$E_3 = -2.18 \times 10^{-18} \times \frac{4}{9} \ J$
$E_3 = -2.18 \times 10^{-18} \times 0.4444 \ J$
$E_3 \approx -0.9688 \times 10^{-18} \ J$
$E_3 \approx -9.69 \times 10^{-19} \ J$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula:
$E_n = -R_H \times \frac{Z^2}{n^2} \ J$
For a hydrogen atom,the atomic number $Z = 1$.
For the fourth orbit,$n = 4$.
Substituting the values:
$E_4 = -2.18 \times 10^{-18} \times \frac{1^2}{4^2} \ J$
$E_4 = -2.18 \times 10^{-18} \times \frac{1}{16} \ J$
$E_4 = -0.13625 \times 10^{-18} \ J \approx -0.136 \times 10^{-18} \ J$

(D) For the first orbit of a monopositive $He^{+}$ ion,the atomic number $Z = 2$ and the orbit number $n = 1$.
The energy of an electron in the $n^{th}$ orbit is given by the formula:
$E_n = -2.18 \times 10^{-18} \left( \frac{Z^2}{n^2} \right) \ J$
Substituting the values $Z = 2$ and $n = 1$:
$E_1 = -2.18 \times 10^{-18} \left( \frac{2^2}{1^2} \right) \ J$
$E_1 = -2.18 \times 10^{-18} \times 4 \ J$
$E_1 = -8.72 \times 10^{-18} \ J$

(D) For the first orbit of the monopositive $He^{+}$ ion,the atomic number $Z = 2$ and the principal quantum number $n = 1$.
The energy of an orbit is given by the formula: $E_n = -2.18 \times 10^{-18} \left(\frac{Z^2}{n^2}\right) \ J$.
Substituting the values: $E_1 = -2.18 \times 10^{-18} \left(\frac{2^2}{1^2}\right) \ J$.
$E_1 = -2.18 \times 10^{-18} \times 4 \ J = -8.72 \times 10^{-18} \ J$.

(C) The radius of the $n^{\text{th}}$ orbit is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For $He^{+}$, the atomic number $Z = 2$ and for the first orbit $n = 1$.
Substituting these values: $r_1 = \frac{52.9 \times (1)^2}{2} \ pm = 26.45 \ pm$.

(B) According to Bohr's model of the atom,the angular momentum $(L)$ of an electron in a stationary orbit is quantized and is given by the equation:
$L = mvr = \frac{nh}{2 \pi}$
where $m$ is the mass of the electron,$v$ is the velocity,$r$ is the radius of the orbit,$n$ is the principal quantum number $(n = 1, 2, 3, ...)$,and $h$ is Planck's constant.

(C) The Bohr model of the atom could not explain the splitting of spectral lines in the presence of a magnetic field,which is known as the Zeeman effect. Therefore,the statement that it explains the Zeeman effect is incorrect.

(C) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For the $Li^{2+}$ ion, the atomic number $Z = 3$.
For the first orbit, the principal quantum number $n = 1$.
Substituting these values into the formula: $r_1 = \frac{52.9 \times (1)^2}{3} \ pm = 17.63 \ pm$.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For $He^{+}$, the atomic number $Z = 2$ and for the first orbit, $n = 1$.
Substituting these values: $r_1 = \frac{52.9 \times (1)^2}{2} \ pm = 26.45 \ pm$.

(A) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = 52.9 \times \frac{n^2}{Z} \ pm$.
For $Be^{3+}$, the atomic number $Z = 4$ and the orbit number $n = 4$.
Substituting these values into the formula:
$r_4 = 52.9 \times \frac{4^2}{4} \ pm$
$r_4 = 52.9 \times \frac{16}{4} \ pm$
$r_4 = 52.9 \times 4 \ pm = 211.6 \ pm$.

(D) The $Bohr$ atomic model is based on the quantization of energy and angular momentum for a single-electron species like the hydrogen atom.
It successfully explains the stability of the atom and the line spectra of hydrogen.
However,it fails to explain the ability of atoms to form molecules by chemical bonds,as it does not account for the interaction between multiple atoms or the nature of chemical bonding.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $r_n = 52.9 \times \frac{n^2}{Z} \ pm$.
For a hydrogen atom, the atomic number $Z = 1$.
For the fourth orbit, $n = 4$.
Substituting these values into the formula: $r_4 = 52.9 \times \frac{4^2}{1} \ pm$.
$r_4 = 52.9 \times 16 \ pm = 846.4 \ pm$.

(D) The radius of the $n^{th}$ orbit of a hydrogen atom is given by the formula $r_n = n^2 a_0$, where $a_0$ is the radius of the first orbit.
Given that the radius of the first orbit is $R \text{ pm}$, we have $a_0 = R \text{ pm}$.
For the fourth orbit $(n = 4)$, the radius is $r_4 = (4)^2 \times R \text{ pm} = 16 \ R \text{ pm}$.

(C) The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula: $r_n = \frac{52.9 \times n^2}{Z} \ pm$.
For $He^{+}$, the atomic number $Z = 2$.
For the third orbit, $n = 3$.
Substituting these values into the formula:
$r_3 = \frac{52.9 \times (3)^2}{2} \ pm$
$r_3 = \frac{52.9 \times 9}{2} \ pm$
$r_3 = \frac{476.1}{2} \ pm = 238.05 \ pm \approx 238.1 \ pm$.

(C) The wave number $\bar{\nu}$ is calculated using the Rydberg formula: $\bar{\nu} = R_{H} (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
Given $n_1 = 1$,$n_2 = 2$,and $R_{H} = 109677 \ cm^{-1}$.
Substituting the values: $\bar{\nu} = 109677 \times (\frac{1}{1^2} - \frac{1}{2^2})$
$\bar{\nu} = 109677 \times (1 - 0.25) = 109677 \times 0.75$
$\bar{\nu} = 82257.75 \ cm^{-1} \approx 82258 \ cm^{-1}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -R_H \times \frac{Z^2}{n^2}$.
For $Li^{2+}$,the atomic number $Z = 3$ and for the first orbit,$n = 1$.
Substituting the values: $E_1 = -2.18 \times 10^{-18} \times \frac{3^2}{1^2} \ J$.
$E_1 = -2.18 \times 10^{-18} \times 9 \ J$.
$E_1 = -19.62 \times 10^{-18} \ J$.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For $Be^{3+}$, the atomic number $Z = 4$ and for the first orbit, $n = 1$.
Substituting these values: $r_1 = 0.529 \times \frac{1^2}{4} \ \mathring{A}$.
$r_1 = 0.529 \times 0.25 \ \mathring{A} = 0.13225 \ \mathring{A}$.
Since $1 \ \mathring{A} = 100 \ pm$, we have $r_1 = 0.13225 \times 100 \ pm = 13.225 \ pm$, which is approximately $13.23 \ pm$.

$(A)$ The radius of the $n^{th}$ orbit for a hydrogen-like species is given by the formula:
$r_n = a_0 \frac{n^2}{z}$
where $a_0$ is the Bohr radius $(0.529 \ \mathring{A})$, $n$ is the principal quantum number (order of orbit), and $z$ is the atomic number (nuclear charge).

(A) According to Bohr's postulate,the frequency $(v)$ of the emitted or absorbed radiation during a transition between two stationary states with energy difference $\Delta E$ is given by the equation:
$h v = \Delta E$
Where $h$ is Planck's constant.
Rearranging for frequency:
$v = \frac{\Delta E}{h}$

(B) The energy of an electron in a stationary state is given by the formula: $E_n = -R_H \times \frac{1}{n^2}$
Where $R_H$ is the Rydberg constant,which is $2.18 \times 10^{-18} \ J$.
For $n=2$:
$E_2 = -2.18 \times 10^{-18} \times \frac{1}{2^2}$
$E_2 = -2.18 \times 10^{-18} \times \frac{1}{4}$
$E_2 = -0.545 \times 10^{-18} \ J$

(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Lyman series,$n_1 = 1$.
The longest wavelength corresponds to the smallest energy transition,which occurs from $n_2 = 2$ to $n_1 = 1$.
Substituting the values: $\frac{1}{\lambda} = 109677 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$.
$\frac{1}{\lambda} = 109677 \left( 1 - \frac{1}{4} \right) = 109677 \times \frac{3}{4} \ cm^{-1}$.
$\frac{1}{\lambda} = 82257.75 \ cm^{-1}$.
$\lambda = \frac{1}{82257.75} \ cm \approx 1.216 \times 10^{-5} \ cm$.

(A) The wavenumber $(\bar{\nu})$ for a transition in a hydrogen atom is given by the Rydberg formula: $\bar{\nu} = R_{H} \times Z^2 \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Here,$R_{H} = 109677 \ cm^{-1}$,$Z = 1$ (for hydrogen atom),$n_1 = 2$,and $n_2 = 5$.
Substituting the values: $\bar{\nu} = 109677 \times 1^2 \times \left( \frac{1}{2^2} - \frac{1}{5^2} \right)$.
$\bar{\nu} = 109677 \times \left( \frac{1}{4} - \frac{1}{25} \right)$.
$\bar{\nu} = 109677 \times \left( \frac{25 - 4}{100} \right) = 109677 \times \frac{21}{100}$.
$\bar{\nu} = 109677 \times 0.21 = 23032.17 \ cm^{-1}$.
Rounding to the nearest integer,we get $23032 \ cm^{-1}$.

(B) The lowest energy transition in the Balmer series occurs from $n_2 = 3$ to $n_1 = 2$.
Using the Rydberg formula for wave number: $\overline{v} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Substituting the values: $\overline{v} = R_H \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$.
$\overline{v} = R_H \left[ \frac{1}{4} - \frac{1}{9} \right] = R_H \left[ \frac{9-4}{36} \right] = R_H \left( \frac{5}{36} \right)$.

(A) For the Lyman series,the electron transitions to the ground state,so $n_1 = 1$.
The lowest energy transition in the Lyman series corresponds to the transition from the nearest higher energy level,which is $n_2 = 2$.
The Rydberg formula for the wave number $\bar{v}$ is given by $\bar{v} = R_{H} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Substituting the values $n_1 = 1$ and $n_2 = 2$:
$\bar{v} = R_{H} \left( \frac{1}{1^2} - \frac{1}{2^2} \right) \ cm^{-1}$
$\bar{v} = R_{H} \left( 1 - \frac{1}{4} \right) \ cm^{-1}$
$\bar{v} = R_{H} \left( \frac{3}{4} \right) \ cm^{-1}$

(D) For a hydrogen atom,the wavenumber $\bar{\nu}$ is given by the Rydberg formula:
$\bar{\nu} = R_H \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right] \ cm^{-1}$
Given $n_i = 2$ and $n_f = 1$:
$\bar{\nu} = 109677 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] \ cm^{-1}$
$\bar{\nu} = 109677 \left[ 1 - \frac{1}{4} \right] \ cm^{-1}$
$\bar{\nu} = 109677 \left[ \frac{3}{4} \right] \ cm^{-1}$
$\bar{\nu} = 82257.75 \ cm^{-1} \approx 82257.8 \ cm^{-1}$

(A) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = R_{H} [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$
Given $n_1 = 2$,$n_2 = 4$,and $R_{H} = 109677 \ cm^{-1}$.
Substituting the values:
$\bar{\nu} = 109677 [\frac{1}{2^2} - \frac{1}{4^2}] \ cm^{-1}$
$= 109677 [\frac{1}{4} - \frac{1}{16}] \ cm^{-1}$
$= 109677 [\frac{4-1}{16}] \ cm^{-1}$
$= 109677 [\frac{3}{16}] \ cm^{-1}$
$= 20564.44 \ cm^{-1}$

(A) The formula for the wavenumber $(\bar{\nu})$ is given by the Rydberg equation: $\bar{\nu} = R_{H} \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right] \ cm^{-1}$
Given $n_f = 2$,$n_i = 3$,and $R_{H} = 109677 \ cm^{-1}$.
Substituting the values: $\bar{\nu} = 109677 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] \ cm^{-1}$
$= 109677 \left[ \frac{1}{4} - \frac{1}{9} \right] \ cm^{-1}$
$= 109677 \left[ \frac{9-4}{36} \right] \ cm^{-1}$
$= 109677 \left[ \frac{5}{36} \right] \ cm^{-1}$
$= 15232.9 \ cm^{-1}$

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \ eV$.
Given $E_n = -3.4 \ eV$,we have $-3.4 = -13.6 / n^2$,which gives $n^2 = 4$,so $n = 2$.
According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{n h}{2 \pi}$.
Substituting $n = 2$,we get $L = \frac{2 h}{2 \pi} = \frac{h}{\pi}$.

(A) According to Bohr's postulate,the angular momentum $(L)$ of an electron in an orbit is given by the formula: $L = \frac{nh}{2\pi}$.
For the first orbit,$n = 1$.
Substituting the values: $L = \frac{1 \times 6.626 \times 10^{-34}}{2 \times 3.14159}$.
$L = \frac{6.626 \times 10^{-34}}{6.28318} \approx 1.0545 \times 10^{-34} \ J \ s$.
Thus,the correct option is $A$.

(A) According to Bohr's postulate for the hydrogen atom,the electron revolves only in those orbits for which the angular momentum is an integral multiple of $\frac{h}{2 \pi}$.
Mathematically,this is expressed as: $mvr = \frac{nh}{2 \pi}$,where $m$ is the mass of the electron,$v$ is the velocity,$r$ is the radius of the orbit,$n$ is the principal quantum number $(n = 1, 2, 3, ...)$,and $h$ is Planck's constant.

(C) The angular momentum $(L)$ of an electron in a stationary orbit is given by Bohr's postulate: $L = mvr = \frac{nh}{2 \pi}$.
For the fourth orbit,the principal quantum number $n = 4$.
Substituting the value of $n$ in the formula: $L = \frac{4h}{2 \pi} = \frac{2h}{\pi}$.

(B) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula:
$E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J$
For $He^{+}$,the atomic number $Z = 2$ and for the first orbit $n = 1$.
Substituting these values into the formula:
$E_1 = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} \ J$
$E_1 = -2.18 \times 10^{-18} \times 4 \ J$
$E_1 = -8.72 \times 10^{-18} \ J$

(A) The energy of an electron in the $n^{\text{th}}$ Bohr orbit of a hydrogen atom is given by the formula:
$E_n = -\frac{13.6 \times Z^2}{n^2} \text{ eV}$
For a hydrogen atom,the atomic number $Z = 1$.
Substituting $Z = 1$ into the formula,we get:
$E_n = -\frac{13.6 \times (1)^2}{n^2} \text{ eV} = -\frac{13.6}{n^2} \text{ eV}$.

(A) The wave number $\bar{\nu}$ of a spectral line in the hydrogen emission spectrum is given by the Rydberg formula: $\bar{\nu} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given that $\bar{\nu} = \frac{8}{9} R_H$,we substitute this into the formula:
$\frac{8}{9} R_H = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Dividing both sides by $R_H$,we get $\frac{8}{9} = \frac{1}{n_1^2} - \frac{1}{n_2^2}$.
For the emission spectrum,$n_1 < n_2$. Since the value is $\frac{8}{9}$,we test $n_1 = 1$:
$\frac{8}{9} = 1 - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 1 - \frac{8}{9} = \frac{1}{9}$.
Thus,$n_2^2 = 9$,which means $n_2 = 3$.
Therefore,the electron jumps from $n=3$ to $n=1$.

(A) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For the ground state of $H$ $(Z=1, n=1)$,$E_1 = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV}$.
Now,calculate the energy for the options:
$A$: $He^{+}$ $(Z=2)$,first excited state $(n=2)$: $E_2 = -13.6 \times \frac{2^2}{2^2} = -13.6 \text{ eV}$.
$B$: $Be^{3+}$ $(Z=4)$,ground state $(n=1)$: $E_1 = -13.6 \times \frac{4^2}{1^2} = -217.6 \text{ eV}$.
$C$: $Li^{2+}$ $(Z=3)$,first excited state $(n=2)$: $E_2 = -13.6 \times \frac{3^2}{2^2} = -30.6 \text{ eV}$.
$D$: $Li^{2+}$ $(Z=3)$,ground state $(n=1)$: $E_1 = -13.6 \times \frac{3^2}{1^2} = -122.4 \text{ eV}$.
Thus,the ground state energy of $H$ is equal to the first excited state energy of $He^{+}$.

(C) Rutherford's $\alpha-$particle scattering experiment provided evidence for the existence of a small,dense,positively charged nucleus at the center of the atom.
It concluded that most of the space in the atom is empty and that the size of the nucleus is very small compared to the size of the atom.
However,the concept that electrons move in circular paths of fixed energy (orbits) was proposed by Niels Bohr,not Rutherford.
Rutherford's model could not explain the stability of the atom,as classical physics predicted that revolving electrons would lose energy and spiral into the nucleus.

(B) $J.J.$ Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.
Thomson proposed the plum pudding model of the atom,which states that the positive charge is uniformly distributed and electrons are embedded into it.
Hence,the correct option is $(B)$.

(B) $I$) The energy of the hydrogen atom in its ground state $(n=1)$ is given by $E_n = -13.6 \ eV / n^2$. For $n=1$,$E_1 = -13.6 \ eV$. Thus,statement $I$ is correct.
$II$) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = 52.9 \times n^2 / Z \ pm$. For $H$ $(Z=1)$ and $n=3$,$r_3 = 52.9 \times 3^2 / 1 = 476.1 \ pm$. Thus,statement $II$ is incorrect.
$III$) The radius of the first orbit $(n=1)$ is $r_1 = 52.9 \times (1^2 / Z) \ pm$. As $Z$ increases,$r_1$ decreases. The atomic numbers are $H(1), He^{+}(2), Li^{2+}(3), Be^{3+}(4)$. Therefore,the order of radii is $H > He^{+} > Li^{2+} > Be^{3+}$. Thus,statement $III$ is correct.
Conclusion: $I$ and $III$ are correct.

(B) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by $r_n = a_0 \times \frac{n^2}{Z}$, where $a_0$ is the Bohr radius $(52.9 \text{ pm})$.
For $H$ atom $(Z=1)$: $x = r_3 - r_2 = a_0 \times (3^2 - 2^2) / 1 = a_0 \times (9 - 4) = 5a_0$.
For $Li^{2+}$ ion $(Z=3)$: $y = r_4 - r_3 = a_0 \times (4^2 - 3^2) / 3 = a_0 \times (16 - 9) / 3 = \frac{7}{3}a_0$.
The ratio $y:x = (\frac{7}{3}a_0) : (5a_0) = \frac{7}{3} : 5 = 7 : 15$.