Atomic models and Planck's quantum theory Questions in English

(A) The energy change for the transition from $n_1 = 1$ to $n_2 = 2$ is calculated as:
$\Delta E = - 2.0 \times 10^{-18} \times \left( \frac{1}{2^2} - \frac{1}{1^2} \right)$
$\Delta E = - 2.0 \times 10^{-18} \times \left( \frac{1}{4} - 1 \right) = - 2.0 \times 10^{-18} \times \left( - \frac{3}{4} \right) = 1.5 \times 10^{-18} \, J$
Using the relation $\Delta E = \frac{hc}{\lambda}$,the wavelength $\lambda$ is:
$\lambda = \frac{hc}{\Delta E} = \frac{6.625 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-18}}$
$\lambda = 1.325 \times 10^{-7} \, m$

(D) The total energy $(E)$ of an electron in a hydrogen atom is the sum of its kinetic energy $(K.E.)$ and potential energy $(P.E.)$.
For an electron in an orbit of radius $r$,the electrostatic potential energy is given by $P.E. = -\frac{e^2}{r}$.
The kinetic energy is given by $K.E. = \frac{1}{2} \frac{e^2}{r}$.
Therefore,the total energy is $E = K.E. + P.E. = \frac{e^2}{2r} - \frac{e^2}{r} = -\frac{e^2}{2r}$.

(C) The energy of a photon is given by the equation $E = h \nu$.
Given:
Planck's constant,$h = 6.63 \times 10^{-34} \ J \cdot s$
Frequency,$\nu = 2.47 \times 10^{15} \ Hz$
Substituting the values:
$E = (6.63 \times 10^{-34} \ J \cdot s) \times (2.47 \times 10^{15} \ s^{-1})$
$E = 16.3761 \times 10^{-19} \ J$
$E = 1.63761 \times 10^{-18} \ J \approx 1.640 \times 10^{-18} \ J$.

(A) The wave number $\bar{\nu}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{\nu} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
The first emission line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$.
For the $H$ atom,$Z = 1$.
Substituting these values: $\bar{\nu} = R (1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right)$.
$\bar{\nu} = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{9 - 4}{36} \right) = \frac{5}{36} R$.

(A) The frequency $\nu$ is given by the formula $\nu = R_{\infty} \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$ and $n_2 = 3, 4, 5, \dots, \infty$.
The limiting line corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$.
Substituting the values: $\nu = 3.29 \times 10^{15} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right)$.
$\nu = 3.29 \times 10^{15} \times \frac{1}{4} = 8.225 \times 10^{14} \ s^{-1}$.

(C) For an emission line,$n_f < n_i$.
The Rydberg formula for the wave number is $\bar{\nu} = R_H Z^2 \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$.
For atomic hydrogen,$Z = 1$,$n_i = 8$,and $n_f = n$.
Substituting these values: $\bar{\nu} = R_H \left[ \frac{1}{n^2} - \frac{1}{8^2} \right] = R_H \left( \frac{1}{n^2} \right) - \frac{R_H}{64}$.
Comparing this with the equation of a straight line $y = mx + c$,where $y = \bar{\nu}$ and $x = \frac{1}{n^2}$:
The slope $m = R_H$ and the intercept $c = -\frac{R_H}{64}$.
Thus,the plot is linear with a slope of $R_H$.

(C) According to Einstein's photoelectric equation,the kinetic energy $(K.E.)$ of the ejected electron is given by: $K.E. = h\nu - h\nu_0$,where $h\nu$ is the energy of the incident photon and $h\nu_0$ is the work function of the metal.
$1$. The relationship between $K.E.$ and energy of light $(h\nu)$ is a straight line with a positive intercept $(-h\nu_0)$,which matches graph $(A)$.
$2$. The number of ejected electrons depends on the intensity of light,not the frequency (above the threshold),so graph $(B)$ is a valid representation.
$3$. The relationship between $K.E.$ and frequency $(
u)$ is $K.E. = h\nu - h\nu_0$. This is a straight line with a negative intercept on the $K.E.$ axis,not one passing through the origin. Therefore,graph $(C)$ is incorrect.
$4$. $K.E.$ is independent of the intensity of incident light,so graph $(D)$ is a valid representation.
Thus,the graph that does not represent the relationship is $(C)$.

(C) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For the $He^{+}$ ion,the atomic number $Z = 2$.
The ground state is $n = 1$,the first excited state is $n = 2$,and the second excited state is $n = 3$.
Substituting $Z = 2$ and $n = 3$ into the formula:
$E_3 = -13.6 \times \frac{2^2}{3^2} \ eV$
$E_3 = -13.6 \times \frac{4}{9} \ eV$
$E_3 = -13.6 \times 0.4444 \ eV = -6.04 \ eV$.

(A) The wavelength is given as $\lambda = 900 \ nm = 900 \times 10^{-9} \ m = 9 \times 10^{-7} \ m$.
Using the Rydberg formula for the hydrogen atom: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Paschen series,$n_1 = 3$. Substituting the values: $\frac{1}{9 \times 10^{-7}} = 1.097 \times 10^7 \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right)$.
$0.101 = \frac{1}{9} - \frac{1}{n_2^2} \implies \frac{1}{n_2^2} = 0.111 - 0.101 = 0.01$.
$n_2^2 = 100 \implies n_2 = 10$.
Since $900 \ nm$ falls in the infrared region,it corresponds to the Paschen series $(n_1 = 3)$. Among the given options,the Paschen series is the correct spectral region.

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.
$E = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4000 \times 10^{-10}} = 4.9695 \times 10^{-19} \, J$.
The kinetic energy of the emitted photoelectron is $K.E. = \frac{1}{2}mv^2$.
$K.E. = \frac{1}{2} \times 9 \times 10^{-31} \times (6 \times 10^5)^2 = 1.62 \times 10^{-19} \, J$.
According to Einstein's photoelectric equation,$E = \phi + K.E.$,where $\phi$ is the work function.
$\phi = E - K.E. = 4.9695 \times 10^{-19} - 1.62 \times 10^{-19} = 3.3495 \times 10^{-19} \, J$.
To convert the work function into $eV$,divide by the charge of an electron $(1.6 \times 10^{-19} \, C)$:
$\phi = \frac{3.3495 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 2.093 \, eV \approx 2.1 \, eV$.

(C) The wavenumber is given by $\bar{\nu} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For any series,$\Delta \bar{\nu} = \bar{\nu}_{\max} - \bar{\nu}_{\min} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_1+1^2} \right] - R_H \left[ \frac{1}{n_1^2} - \frac{1}{\infty^2} \right] = R_H \left[ \frac{1}{\infty^2} - \frac{1}{(n_1+1)^2} \right] = \frac{R_H}{(n_1+1)^2}$.
For the Lyman series,$n_1 = 1$,so $\Delta \bar{\nu}_{\text{Lyman}} = \frac{R_H}{(1+1)^2} = \frac{R_H}{4}$.
For the Balmer series,$n_1 = 2$,so $\Delta \bar{\nu}_{\text{Balmer}} = \frac{R_H}{(2+1)^2} = \frac{R_H}{9}$.
Therefore,the ratio is $\frac{\Delta \bar{\nu}_{\text{Lyman}}}{\Delta \bar{\nu}_{\text{Balmer}}} = \frac{R_H/4}{R_H/9} = \frac{9}{4}$.

(D) For a hydrogen atom,the total energy $(E)$ of an electron in an orbital is constant and depends only on the principal quantum number $(n)$.
$E = -K.E = \frac{P.E}{2}$,which implies $|P.E| = 2 \times |K.E|$. Thus,option $C$ is correct.
The probability density for a $1s$ orbital is given by $\psi^2 = \frac{1}{\pi a_0^3} e^{-2r/a_0}$. This value is maximum at $r = 0$ (the nucleus),so option $B$ is correct.
The electron can be found at any distance $r$ from the nucleus,including $2a_0$,although the probability density decreases exponentially with distance. Thus,option $A$ is correct.
The total energy of the electron in a specific orbital is constant and does not change with its instantaneous distance from the nucleus. Therefore,the statement that the total energy is maximum at $a_0$ is incorrect.

(A) The shortest wavelength $(\lambda)$ for a spectral series is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the shortest wavelength,$n_2 = \infty$,so $\frac{1}{\lambda} = \frac{R_H}{n_1^2}$.
Thus,$\lambda = \frac{n_1^2}{R_H}$.
The ratio of the shortest wavelengths of two series with lower energy levels $n_1$ and $n_2$ is $\frac{\lambda_2}{\lambda_1} = \frac{n_2^2}{n_1^2} = 9$.
This implies $\frac{n_2}{n_1} = 3$.
For the Lyman series,$n_1 = 1$. For the Paschen series,$n_2 = 3$.
Therefore,the ratio is $\frac{3^2}{1^2} = 9$. The series are Lyman and Paschen.

(A) The energy of a photon is given by $\Delta E = \frac{hc}{\lambda}$.
Rearranging for wavelength,we get $\lambda = \frac{hc}{\Delta E}$.
Given $\Delta E$ is in electron volts $(eV)$,we use $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^{8} \ m/s$,and $1 \ eV = 1.602 \times 10^{-19} \ J$.
$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{\Delta E \times 1.602 \times 10^{-19}} \ m$.
$\lambda \approx \frac{12395}{\Delta E} \times 10^{-10} \ m$.

(A) The frequency of a spectral line in the $H$ atom is given by $\nu = R_H c (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For a Lyman series line,$n_1 = 1$. Let the transition be $n_2 = n$. Then $\nu = R_H c (1 - \frac{1}{n^2})$.
Condition $(i)$ states: $\nu(n \to 1) = \nu(n' \to 1) + \nu(n'' \to 2)$.
Substituting the Rydberg formula: $(1 - \frac{1}{n^2}) = (1 - \frac{1}{n'^2}) + (\frac{1}{4} - \frac{1}{n''^2})$.
For $n=3$,$\nu(3 \to 1) = R_H c (1 - \frac{1}{9}) = \frac{8}{9} R_H c$.
Checking the conditions for $n=3$:
$(i)$ $\nu(3 \to 1) = \nu(2 \to 1) + \nu(3 \to 2) = R_H c (1 - \frac{1}{4}) + R_H c (\frac{1}{4} - \frac{1}{9}) = R_H c (1 - \frac{1}{9}) = \frac{8}{9} R_H c$. This matches.
Thus,the transition corresponds to $n_2 = 3$ to $n_1 = 1$.

(B) The Brackett series corresponds to electronic transitions ending at the $n = 4$ energy level.
The first line of the Brackett series occurs for the transition from $n = 5$ to $n = 4$.
To emit this line,the hydrogen atom must first be excited from the ground state $(n = 1)$ to the $n = 5$ state.
The energy required for this excitation is given by $\Delta E = E_5 - E_1$.
Using the formula $E_n = -13.6 / n^2 \ eV$,we have:
$E_1 = -13.6 \ eV$
$E_5 = -13.6 / 5^2 = -13.6 / 25 = -0.544 \ eV$.
Therefore,$\Delta E = -0.544 - (-13.6) = 13.056 \ eV \approx 13.06 \ eV$.
Thus,the potential required to accelerate the electron is $13.06 \ V$,which corresponds to an energy of $13.06 \ eV$.

(B) For the Lyman series,the transition occurs from higher energy levels to $n_1 = 1$.
For maximum wavelength,the energy difference must be minimum,which corresponds to the transition from $n_2 = 2$ to $n_1 = 1$.
The Rydberg formula is $\frac{1}{\lambda} = R \cdot Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $He^+$ ion,$Z = 2$,so $Z^2 = 4$.
Substituting the values: $\frac{1}{\lambda} = R \times 4 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R$.
Therefore,$\lambda = \frac{1}{3R}$.

(C) The energy difference $\Delta E$ for a transition in a hydrogen atom is given by $\Delta E = 13.6 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
The wavelength $\lambda$ is inversely proportional to the energy difference: $\lambda = \frac{hc}{\Delta E}$.
Thus,a larger $\Delta E$ corresponds to a smaller $\lambda$.
Calculating relative energy differences:
$(a) \ \Delta E_1 \propto (1/1^2 - 1/3^2) = 1 - 0.111 = 0.889$
$(b) \ \Delta E_3 \propto (1/10^2 - 1/12^2) = 0.01 - 0.0069 = 0.0031$
$(c) \ \Delta E_2 \propto (1/3^2 - 1/5^2) = 0.111 - 0.04 = 0.071$
$(d) \ \Delta E_4 \propto (1/20^2 - 1/22^2) = 0.0025 - 0.00206 = 0.00044$
Comparing energy differences: $\Delta E_1 > \Delta E_2 > \Delta E_3 > \Delta E_4$.
Therefore,the wavelengths follow the order: $\lambda_1 < \lambda_2 < \lambda_3 < \lambda_4$.

(C) The Bohr model of the atom proposed that electrons move in well-defined circular orbits around the nucleus. However,this was later proven to be incorrect by the Heisenberg Uncertainty Principle and the wave-particle duality of matter,which suggest that electrons exist in three-dimensional orbitals rather than fixed circular paths. Therefore,the statement that an electron moves in a circular orbit around the nucleus is considered false in the context of modern quantum mechanics.

(C) The radius of an orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For $r_{2H}$,$n=2, Z=1 \implies r_{2H} = 0.529 \times 4 = 2.116 \ \mathring{A}$.
For $r_{1He^{+}}$,$n=1, Z=2 \implies r_{1He^{+}} = 0.529 \times \frac{1}{2} = 0.2645 \ \mathring{A}$.
For $r_{1H}$,$n=1, Z=1 \implies r_{1H} = 0.529 \ \mathring{A}$.
Thus,$r_{2H} > r_{1H} > r_{1He^{+}}$,so option $A$ is incorrect.
Ionisation energy is given by $I.E. = 13.6 \times \frac{Z^2}{n^2} \ \text{eV}$.
$I.E._{H} (Z=1, n=1) = 13.6 \ \text{eV}$.
$I.E._{He^{+}} (Z=2, n=1) = 13.6 \times 4 = 54.4 \ \text{eV}$.
$I.E._{Li^{+2}} (Z=3, n=1) = 13.6 \times 9 = 122.4 \ \text{eV}$.
Thus,$I.E._{Li^{+2}} > I.E._{He^{+}} > I.E._{H}$,so option $B$ is incorrect.
Total energy is $E_n = -13.6 \times \frac{Z^2}{n^2} \ \text{eV}$.
$E_{1H} = -13.6 \ \text{eV}$,$E_{2H} = -3.4 \ \text{eV}$,$E_{3H} = -1.51 \ \text{eV}$.
Since $-13.6 < -3.4 < -1.51$,$E_{1H} < E_{2H} < E_{3H}$ is correct.
Option $C$ is correct.

(B) Let the electron be excited from state $n_1$ to $n_2$ by absorbing a photon of energy $E = E_{n_2} - E_{n_1}$.
When returning to the ground state $(n=1)$ from $n_2$,the total number of photons emitted is given by $\frac{n_2(n_2-1)}{2} = 10$.
Solving for $n_2$: $n_2^2 - n_2 - 20 = 0 \implies (n_2-5)(n_2+4) = 0$. Since $n_2 > 0$,we have $n_2 = 5$.
The energy of the incident photon is $E_{incident} = E_5 - E_{n_1}$.
There are $7$ transitions that have higher energy than the incident photon. These transitions must be from levels $n > n_1$ to levels $n \le n_1$.
The number of photons emitted with energy less than or equal to the incident photon is $10 - 7 = 3$. These correspond to transitions from levels $n \le n_1$ to $n=1$.
The number of such transitions is $\frac{n_1(n_1-1)}{2} = 3$.
Solving for $n_1$: $n_1^2 - n_1 - 6 = 0 \implies (n_1-3)(n_1+2) = 0$. Thus,$n_1 = 3$.
Since $n_1 = 3$,the electron was initially in the $2^{nd}$ excited state ($n=3$ is the $2^{nd}$ excited state).

(D) According to Bohr's postulate,the angular momentum $(mvr)$ of an electron in the $n^{th}$ shell is given by the formula: $mvr = \frac{nh}{2 \pi}$.
For the second shell,$n = 2$.
The value of Planck's constant $h$ in $CGS$ units is $6.626 \times 10^{-27} \ erg \ s$.
Substituting the values: $mvr = \frac{2 \times 6.626 \times 10^{-27}}{2 \times 3.14159}$.
$mvr = \frac{6.626 \times 10^{-27}}{3.14159} \approx 2.109 \times 10^{-27} \ erg \ s$.
Thus,the correct option is $D$.

(C) For a hydrogen-like species,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \, eV$.
For $He^{+}$ ion,the atomic number $Z = 2$.
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 $n=4$).
$E_4 = -13.6 \times \frac{2^2}{4^2} = -13.6 \times \frac{4}{16} = -13.6 \times 0.25 = -3.4 \, eV$.
The potential energy $(P.E.)$ is related to total energy $(E)$ by the relation $P.E. = 2 \times E$.
Therefore,$P.E. = 2 \times (-3.4 \, eV) = -6.8 \, eV$.

(B) The energy of the third Balmer transition in $He^{\oplus}$ $(Z=2)$ corresponds to the transition from $n=5$ to $n=2$.
$\Delta E = 13.6 \times Z^{2} \times (\frac{1}{n_1^{2}} - \frac{1}{n_2^{2}}) = 13.6 \times 4 \times (\frac{1}{2^{2}} - \frac{1}{5^{2}}) = 54.4 \times (0.25 - 0.04) = 54.4 \times 0.21 = 11.424 \, eV$.
$(A)$ First excitation energy of $He^{\oplus}$ ($n=1$ to $n=2$): $\Delta E = 13.6 \times 4 \times (1 - \frac{1}{4}) = 54.4 \times 0.75 = 40.8 \, eV$.
$(B)$ Third separation energy of $Li^{2+}$ $(Z=3)$ corresponds to $n=4$ to $n=\infty$: $\Delta E = 13.6 \times 9 \times (\frac{1}{4^{2}}) = 122.4 \times 0.0625 = 7.65 \, eV$.
$(C)$ Fourth excitation energy of $H$ atom ($n=1$ to $n=5$): $\Delta E = 13.6 \times 1 \times (1 - \frac{1}{25}) = 13.6 \times 0.96 = 13.056 \, eV$.
$(D)$ Ionisation energy of $Be^{3+}$ $(Z=4)$: $\Delta E = 13.6 \times 16 = 217.6 \, eV$.
Comparing the values,$7.65 \, eV < 11.424 \, eV$. Thus,option $B$ is correct.

(A) The frequency of a spectral line is given by the formula: $v = RC Z^{2} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.
For the Paschen series,the transition is to the $n_{1} = 3$ energy level. The first line corresponds to the transition from $n_{2} = 4$ to $n_{1} = 3$.
For the $Be^{+3}$ ion,the atomic number $Z = 4$. Substituting these values into the formula:
$v = RC \times (4)^{2} \times \left( \frac{1}{3^{2}} - \frac{1}{4^{2}} \right)$
$v = RC \times 16 \times \left( \frac{1}{9} - \frac{1}{16} \right)$
$v = 16 RC \times \left( \frac{16 - 9}{144} \right)$
$v = 16 RC \times \frac{7}{144}$
$v = \frac{7 RC}{9}$

(D) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the Balmer series,$n_1 = 2$.
For the third line of the Balmer series,$n_2 = 2 + 3 = 5$. For $He^{+}$,$Z = 2$. So,$\frac{1}{\lambda_1} = R \cdot 2^2 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{25} \right) = 4R \left( \frac{21}{100} \right) = \frac{21R}{25}$.
For the first line of the Balmer series,$n_2 = 2 + 1 = 3$. For $Li^{2+}$,$Z = 3$. So,$\frac{1}{\lambda_2} = R \cdot 3^2 \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = 9R \left( \frac{1}{4} - \frac{1}{9} \right) = 9R \left( \frac{5}{36} \right) = \frac{5R}{4}$.
The ratio $\frac{\lambda_1}{\lambda_2} = \frac{\lambda_1}{1} \times \frac{1}{\lambda_2} = \frac{25}{21R} \times \frac{5R}{4} = \frac{125}{84}$.

(B) The energy of an electron in a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
As the principal quantum number $n$ increases,the energy levels become closer to each other.
The energy difference between two successive levels $n$ and $n+1$ is given by $\Delta E = E_{n+1} - E_n = 13.6 \times Z^2 \times (\frac{1}{n^2} - \frac{1}{(n+1)^2})$.
As $n$ increases,the term $(\frac{1}{n^2} - \frac{1}{(n+1)^2})$ decreases,leading to a decrease in the energy gap.
Therefore,the energy gap decreases as $n$ increases.

(B) The spectrum of helium $(He)$ is similar to that of $Li^{+}$ because both are $2$-electron systems. Helium has $2$ electrons,and the lithium ion $(Li^{+})$ also has $3 - 1 = 2$ electrons. According to Bohr's theory,spectra are similar for species having the same number of electrons.

(B) The energy of an electron in the $n$-th orbit of a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$. For $He^{+}$,$Z = 2$,so $E_n = -13.6 \times \frac{4}{n^2} = -\frac{54.4}{n^2} \ eV$.
To ionize the electron,the energy supplied must be at least equal to the ionization energy,which is $|E_n| = \frac{54.4}{n^2} \ eV$.
Given that the collision with an electron of $2.5 \ eV$ energy causes ionization,the energy of the electron in the excited state must be greater than or equal to $-2.5 \ eV$.
Thus,$\frac{54.4}{n^2} \leq 2.5$.
$n^2 \geq \frac{54.4}{2.5} = 21.76$.
Since $n$ must be an integer,the minimum value of $n$ is $\sqrt{21.76} \approx 4.66$.
Therefore,the minimum integer value for $n$ is $5$.

(A) The angular momentum $J$ of an electron of mass $m$ moving in an orbit of radius $r$ with velocity $v$ is given by $J = mvr$.
Kinetic energy $(KE)$ is given by $KE = \frac{1}{2} mv^2$.
We can express $KE$ in terms of $J$ by substituting $v = \frac{J}{mr}$ into the $KE$ formula:
$KE = \frac{1}{2} m \left( \frac{J}{mr} \right)^2 = \frac{1}{2} m \left( \frac{J^2}{m^2r^2} \right) = \frac{J^2}{2mr^2}$.
Thus,the correct option is $(A)$.

(B) The wavelength of $H$-like atoms is given by the Rydberg formula:
$\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
For the shortest wavelength $x$ of the Balmer series of $He^{+}$ $(Z=2)$: $n_1 = 2$ and $n_2 = \infty$.
$\frac{1}{x} = R(2)^2 \left[ \frac{1}{2^2} - \frac{1}{\infty^2} \right] = 4R \left[ \frac{1}{4} \right] = R$
$\Rightarrow x = \frac{1}{R}$
For the longest wavelength $\lambda$ of the Paschen series of $Li^{2+}$ $(Z=3)$: $n_1 = 3$ and $n_2 = 4$.
$\frac{1}{\lambda} = R(3)^2 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]$
$\frac{1}{\lambda} = 9R \left[ \frac{1}{9} - \frac{1}{16} \right] = 9R \left[ \frac{16-9}{144} \right] = 9R \left[ \frac{7}{144} \right] = \frac{7R}{16}$
$\Rightarrow \lambda = \frac{16}{7R} = \frac{16x}{7}$
Hence,the correct option is $B$.

(D) According to Kepler's third law for circular orbits,$T^2 \propto r^3$,which implies $T \propto r^{3/2}$.
Given the distances $r_1 = r$ and $r_2 = 4r$,the ratio of their time periods is calculated as:
$\frac{T_1}{T_2} = (\frac{r_1}{r_2})^{3/2} = (\frac{r}{4r})^{3/2} = (\frac{1}{4})^{3/2} = \frac{1}{8}$.
Therefore,the ratio of their time periods is $1:8$.

(C) The area of an orbit is given by $A = \pi r^2$. Since $r \propto n^2$,we have $A \propto (n^2)^2 = n^4$.
Given $\frac{A_1}{A_2} = \frac{4}{1}$,then $(\frac{n_1}{n_2})^4 = 4$,which implies $\frac{n_1}{n_2} = (4)^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$.
The frequency of an electron in an orbit is given by $f \propto \frac{1}{n^3}$.
Therefore,$\frac{f_1}{f_2} = (\frac{n_2}{n_1})^3 = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.
Thus,the ratio is $1 : 2\sqrt{2}$.

(A) The hydrogen emission spectrum consists of several series of lines based on the energy level transitions of the electron.
The series are:
$1$. Lyman series: $n_1 = 1$ (Ultraviolet region)
$2$. Balmer series: $n_1 = 2$ (Visible region)
$3$. Paschen series: $n_1 = 3$ (Infrared region)
$4$. Brackett series: $n_1 = 4$ (Infrared region)
$5$. Pfund series: $n_1 = 5$ (Infrared region)
Therefore,the series observed in the visible region is the Balmer series.

(C) The velocity of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $v_n = \frac{v_0}{n}$,where $v_0$ is a constant.
This implies that velocity is inversely proportional to the orbit number $n$,i.e.,$v \propto \frac{1}{n}$.
For the $1^{st}$,$2^{nd}$,and $3^{rd}$ orbits,the ratio of velocities is:
$v_1 : v_2 : v_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$.
Therefore,the correct option is $C$.

(B) The Rydberg formula for the Balmer series $(n_1 = 2)$ is given by $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{2^2} - \frac{1}{n_2^2} \right)$.
For the $H_{\alpha}$ line,$n_2 = 3$: $\frac{1}{X} = RZ^2 \left( \frac{1}{4} - \frac{1}{9} \right) = RZ^2 \left( \frac{5}{36} \right) \implies RZ^2 = \frac{36}{5X}$.
For the $H_{\beta}$ line,$n_2 = 4$: $\frac{1}{\lambda_{\beta}} = RZ^2 \left( \frac{1}{4} - \frac{1}{16} \right) = RZ^2 \left( \frac{3}{16} \right)$.
Substituting $RZ^2$: $\frac{1}{\lambda_{\beta}} = \left( \frac{36}{5X} \right) \left( \frac{3}{16} \right) = \frac{108}{80X}$.
Therefore,$\lambda_{\beta} = X \left( \frac{80}{108} \right) \ \mathring{A}$.

(C) The ionization potential $(I.P.)$ of a hydrogen-like species is given by $E_n = -13.6 \times Z^2 / n^2 \ eV$. For the ground state $(n=1)$,$I.P. = 13.6 \times Z^2 = 36 \ eV$.
The energy of any state $n$ is $E_n = -36 / n^2 \ eV$.
The second excitation energy corresponds to the transition from the ground state $(n=1)$ to the third energy level $(n=3)$.
The energy of the third level is $E_3 = -36 / 3^2 = -36 / 9 = -4 \ eV$.
The energy of the ground state is $E_1 = -36 / 1^2 = -36 \ eV$.
The second excitation energy is $\Delta E = E_3 - E_1 = -4 - (-36) = 32 \ eV$.

(A) For a hydrogen-like species,the energy of an orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For $n=1, 2, 3, 4$,the energy values are $E_1 = -13.6 Z^2, E_2 = -3.4 Z^2, E_3 = -1.51 Z^2, E_4 = -0.85 Z^2$.
Calculating the energy differences:
$(E_2 - E_1) = (-3.4 - (-13.6)) Z^2 = 10.2 Z^2$
$(E_3 - E_2) = (-1.51 - (-3.4)) Z^2 = 1.89 Z^2$
$(E_4 - E_3) = (-0.85 - (-1.51)) Z^2 = 0.66 Z^2$
Comparing these values,we find that $(E_2 - E_1) > (E_3 - E_2) > (E_4 - E_3)$.

(C) When an electron transitions from an excited state $x$ to lower energy levels,the total number of spectral lines is calculated by summing the possible transitions from each level.
For an atom with $x$ energy levels,the transitions are from level $x$ to $(x-1), (x-2), \dots, 1$.
The number of lines is given by the sum: $(x-1) + (x-2) + (x-3) + \dots + 2 + 1$.
This is equivalent to the sum of the first $(x-1)$ natural numbers,which is $1 + 2 + 3 + \dots + (x - 1)$.

(D) The number of emission lines when an electron transitions from an excited state $n$ to a lower state is given by the formula $\frac{n(n-1)}{2}$,where $n$ is the principal quantum number of the highest energy level involved.
For transitions from $n = 4$ to lower energy levels,the number of possible emission lines is calculated as:
$\text{Number of lines} = \frac{4(4-1)}{2} = \frac{4 \times 3}{2} = 6$.
The possible transitions are:
$4 \to 3, 4 \to 2, 4 \to 1, 3 \to 2, 3 \to 1, 2 \to 1$.
Thus,there are $6$ possible emission lines.

(B) The speed of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v_n = v_1 \times \frac{Z}{n}$.
For a hydrogen atom,$Z = 1$.
In the first orbit $(n = 1)$,the speed is $v_1 = x$.
In the third orbit $(n = 3)$,the speed is $v_3 = v_1 \times \frac{1}{3} = \frac{x}{3}$.

(A) The Lyman series corresponds to transitions ending at $n_1 = 1$. The $1^{st}$ Lyman line is $n_2 = 2 \rightarrow n_1 = 1$. The $3^{rd}$ Lyman line is $n_2 = 4 \rightarrow n_1 = 1$.
Given the transition is from the $3^{rd}$ Lyman line $(n_2 = 4)$ to the $1^{st}$ Lyman line $(n_2 = 2)$:
Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
For hydrogen,$Z = 1$,$n_1 = 2$,and $n_2 = 4$:
$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$
$\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$
$\frac{1}{\lambda} = R \left[ \frac{4-1}{16} \right] = R \left[ \frac{3}{16} \right]$
$\lambda = \frac{16}{3R}$

(C) The energy difference $\Delta E$ for one mole of photons is given by $\Delta E = \frac{N_A hc}{\lambda}$.
Given $\lambda = 121.5 \ nm = 121.5 \times 10^{-9} \ m$,$h = 6.626 \times 10^{-34} \ J \cdot s$,$c = 3 \times 10^8 \ m/s$,and $N_A = 6.022 \times 10^{23} \ mol^{-1}$.
$\Delta E = \frac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{121.5 \times 10^{-9}} \ J/mol$.
$\Delta E \approx \frac{11.97 \times 10^1}{121.5 \times 10^{-9}} \times 10^{-11} \approx 984,000 \ J/mol$.
Rounding to the nearest provided option,$\Delta E \approx 982 \ kJ/mol$.

(C) According to Einstein's photoelectric equation: $h v = h v_0 + K.E.$
Therefore,$K.E. = h v - h v_0$.
Given the ratio of kinetic energies for frequencies $v_1$ and $v_2$ is $1:2$:
$\frac{K.E._1}{K.E._2} = \frac{h v_1 - h v_0}{h v_2 - h v_0} = \frac{1}{2}$
Cross-multiplying gives:
$2(h v_1 - h v_0) = 1(h v_2 - h v_0)$
$2h v_1 - 2h v_0 = h v_2 - h v_0$
Rearranging for the threshold frequency $v_0$:
$2h v_1 - h v_2 = 2h v_0 - h v_0$
$v_0 = 2v_1 - v_2$

(B) Given: Wavelength $\lambda = 4500 \ \mathring{A} = 4500 \times 10^{-10} \ m = 4.5 \times 10^{-7} \ m$.
Power of bulb $P = 150 \ W = 150 \ J/s$.
Efficiency of light emission = $8 \ \%$.
Energy emitted as light per second $E = 150 \times \frac{8}{100} = 12 \ J/s$.
Let $n$ be the number of photons emitted per second.
The energy of one photon is $E_{photon} = \frac{hc}{\lambda}$.
Total energy $E = n \times \frac{hc}{\lambda} \implies n = \frac{E \lambda}{hc}$.
Substituting the values: $n = \frac{12 \times 4.5 \times 10^{-7}}{6.626 \times 10^{-34} \times 3 \times 10^8} \approx \frac{54 \times 10^{-7}}{19.878 \times 10^{-26}} \approx 2.716 \times 10^{19}$.
Thus,the number of photons is approximately $2.72 \times 10^{19}$.

(C) The power of the bulb is $P = 200 \ W = 200 \ J \ s^{-1}$.
The wavelength of the light is $\lambda = 200 \ nm = 200 \times 10^{-9} \ m = 2 \times 10^{-7} \ m$.
The energy of one photon is $E = \frac{hc}{\lambda}$.
Let $n$ be the number of photons emitted per second.
The total energy emitted per second is $P = n \times E = n \times \frac{hc}{\lambda}$.
Rearranging for $n$: $n = \frac{P \times \lambda}{h \times c}$.
Substituting the values: $n = \frac{200 \times 2 \times 10^{-7}}{6.626 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{400 \times 10^{-7}}{19.878 \times 10^{-26}} \approx 2.012 \times 10^{20}$ photons per second.

(D) The Einstein energy $E$ is given by the formula: $E = \frac{N_A hc}{\lambda}$.
Given: $E = 1.995 \times 10^{12} \ erg \ mol^{-1} = 1.995 \times 10^5 \ J \ mol^{-1}$ (since $1 \ J = 10^7 \ erg$).
Using $N_A = 6.022 \times 10^{23} \ mol^{-1}$,$h = 6.626 \times 10^{-34} \ J \ s$,and $c = 3 \times 10^8 \ m \ s^{-1}$:
$\lambda = \frac{N_A hc}{E} = \frac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{1.995 \times 10^5}$.
$\lambda \approx \frac{1.197 \times 10^{-1}}{1.995 \times 10^5} \approx 6 \times 10^{-7} \ m$.
Converting to $\mathring{A}: 6 \times 10^{-7} \ m = 6000 \ \mathring{A}$.

(A) The energy of a photon is given by the equation $E = h \nu$,where $h$ is Planck's constant and $\nu$ is the frequency.
Given $E_1 = h \times (2 \times 10^{12}) = E$.
For the second photon,$E_2 = h \times (4 \times 10^{12})$.
Taking the ratio: $\frac{E_2}{E_1} = \frac{h \times 4 \times 10^{12}}{h \times 2 \times 10^{12}} = 2$.
Therefore,$E_2 = 2E$.

(D) Thomson's atomic model proposed that the atom consists of a positively charged sphere with electrons embedded in it like seeds in a watermelon or plums in a pudding. Due to these analogies,it is commonly referred to as the $Plum \ pudding$ model,$Raisin \ pudding$ model,or $Watermelon$ model. Therefore,all the given options are correct.