Atomic models and Planck's quantum theory Questions in English

(D) Energy of one photon $E = \frac{hc}{\lambda} = \frac{12400 \ eV \ \mathring{A}}{6200 \ \mathring{A}} = 2 \ eV$.
Total energy emitted by the bulb in $20 \ s$ with $80 \%$ efficiency is $E_{total} = P \times t \times \text{efficiency} = 40 \ W \times 20 \ s \times 0.8 = 640 \ J$.
Convert total energy to $eV$: $E_{total} = \frac{640}{1.6 \times 10^{-19}} \ eV = 400 \times 10^{19} \ eV = 4 \times 10^{21} \ eV$.
Number of photons $n = \frac{E_{total}}{E_{photon}} = \frac{4 \times 10^{21} \ eV}{2 \ eV} = 2 \times 10^{21}$.

(C) According to Bohr's theory,the velocity of an electron in a stationary orbit is given by $v = \frac{2 \pi Z e^2}{n h}$.
This expression shows that the velocity $v$ is independent of the mass of the electron $(m_e)$. Thus,doubling the mass does not change the velocity.
However,the kinetic energy $(KE)$ of an electron is given by $KE = \frac{1}{2} m_e v^2$.
Since $KE \propto m_e$,if the mass of the electron is doubled,the kinetic energy will also double,meaning it increases.

(B) The radius of the $n^{th}$ Bohr orbit is given by $r_n = a_0 n^2$.
Given the ratio of radii is $\frac{r_{n_1}}{r_{n_2}} = \frac{n_1^2}{n_2^2} = \frac{1}{4}$,which implies $\frac{n_1}{n_2} = \frac{1}{2}$.
Since the orbits are among the first four $(n = 1, 2, 3, 4)$,the possible pairs $(n_1, n_2)$ are $(1, 2)$ or $(2, 4)$.
The energy of the $n^{th}$ orbit is $E_n = -\frac{13.6}{n^2} \ eV$.
For the pair $(1, 2)$,the energy difference is $\Delta E = E_2 - E_1 = -\frac{13.6}{4} - (-13.6) = -3.4 + 13.6 = 10.2 \ eV$.
For the pair $(2, 4)$,the energy difference is $\Delta E = E_4 - E_2 = -\frac{13.6}{16} - (-\frac{13.6}{4}) = -0.85 + 3.4 = 2.55 \ eV$.

(D) The velocity of an electron in an orbit is given by the formula $v = v_0 \times \frac{Z}{n}$,where $v_0$ is a constant,$Z$ is the atomic number,and $n$ is the principal quantum number.
From this relation,we can conclude:
$1$. $v \propto Z$ (at constant $n$),which represents a straight line passing through the origin.
$2$. $v \propto \frac{1}{n}$ (at constant $Z$),which represents a straight line passing through the origin when plotted against $\frac{1}{n}$.
$3$. $v \propto \frac{1}{n}$ (at constant $Z$),which represents a rectangular hyperbola when plotted against $n$.
Comparing these with the given options,the graph of $v$ vs. $n$ should be a rectangular hyperbola,not a straight line with a negative slope. Therefore,the linear graph of $v$ vs. $n$ is incorrect.

(A) According to the Bohr model,the energy of an electron in the $n^{th}$ orbit of a hydrogen-like species with atomic number $Z$ is given by $E_n = -13.6 \frac{Z^2}{n^2} \text{ eV}$.
For hydrogen $(Z=1)$,the energy is $E_n(H) = -13.6 \frac{1^2}{n^2} = -13.6 \frac{1}{n^2} \text{ eV}$.
Comparing these two expressions,we get $E_n = Z^2 \times E_n(H)$.

(C) According to Bohr's theory,the radius of an orbit is $r \propto n^{2}$ and the velocity of an electron is $v \propto \frac{1}{n}$.
Time period $T$ is given by $T = \frac{2 \pi r}{v}$.
Substituting the proportionality relations,$T \propto \frac{n^{2}}{1/n} = n^{3}$.
Since $r \propto n^{2}$,we have $n \propto r^{1/2}$,so $T \propto (r^{1/2})^{3} = r^{3/2}$.
Therefore,the ratio of time periods is $\frac{T_{1}}{T_{2}} = \left(\frac{R_{1}}{R_{2}}\right)^{3/2}$.
Given $R_{1} = R$ and $R_{2} = 4R$,we get $\frac{T_{1}}{T_{2}} = \left(\frac{R}{4R}\right)^{3/2} = \left(\frac{1}{4}\right)^{3/2} = \frac{1}{8}$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{E_o}{n^2}$,where $E_o = 13.6 \ eV$.
The energy difference between $n = 2$ and $n = 3$ is given as $E = E_3 - E_2 = -\frac{E_o}{3^2} - (-\frac{E_o}{2^2}) = E_o(\frac{1}{4} - \frac{1}{9}) = \frac{5E_o}{36}$.
Therefore,$E_o = \frac{36E}{5} = 7.2E$.
The ionization potential of the $H$ atom is the energy required to remove an electron from $n = 1$ to $n = \infty$,which is $E' = E_{\infty} - E_1 = 0 - (-\frac{E_o}{1^2}) = E_o$.
Substituting the value of $E_o$,we get $E' = 7.2E$.

(NONE) The equation $E = -2.178 \times 10^{-18} \ J \ \frac{Z^2}{n^2}$ describes the energy of an electron in a hydrogen-like atom.
Option $A$ is correct because as $n$ increases,energy becomes less negative (closer to zero),meaning the electron is less tightly bound.
Option $B$ is correct as the negative sign indicates a stable bound state relative to the zero-energy state at infinity.
Option $C$ is correct because the radius of the orbit is given by $r_n = a_0 \frac{n^2}{Z}$,so $r \propto n^2$.
Option $D$ is correct as $\Delta E = E_f - E_i$ can be calculated using this formula.
Since all statements $A, B, C,$ and $D$ are scientifically correct based on the Bohr model,there is no incorrect statement provided in the options.

(C) The ultraviolet region of the hydrogen spectrum corresponds to the Lyman series,where transitions occur to the ground state $(n_1 = 1)$.
Four lines in the ultraviolet region imply transitions from $n_2 = 2, 3, 4, 5$ to $n_1 = 1$.
This indicates the electron is initially in the $n = 5$ excited state.
The possible transitions from $n = 5$ to lower energy levels are given by the formula $\frac{n(n-1)}{2} = \frac{5(5-1)}{2} = 10$ total lines.
The lines in the infrared region correspond to the Paschen series $(n_1 = 3)$ and Brackett series $(n_1 = 4)$.
For $n = 5$,the possible transitions are:
$5 \rightarrow 4$ (Brackett series,$IR$),
$5 \rightarrow 3$ (Paschen series,$IR$),
$4 \rightarrow 3$ (Paschen series,$IR$).
Thus,there are $3$ lines in the infrared region.
Therefore,the correct option is $C$.

(B) The $5^{th}$ excited state of a hydrogen atom corresponds to the principal quantum number $n = 6$.
The ground state corresponds to $n = 1$.
The Balmer series consists of spectral lines produced when an electron transitions from any higher energy level $(n_2 > 2)$ to the $n_1 = 2$ energy level.
Given the electron transitions from $n = 6$ to $n = 1$,the transitions that result in the Balmer series are those ending at $n = 2$.
These transitions are: $6 \to 2$,$5 \to 2$,$4 \to 2$,and $3 \to 2$.
Thus,the total number of different spectral lines in the Balmer series is $4$.

(C) According to Bohr's theory,the energy of an electron in a state is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ \text{eV}$,which implies $E \propto Z^2$. If $Z$ is doubled,$E$ becomes $4$ times the original value.
The radius of an orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$,which implies $r \propto \frac{1}{Z}$. If $Z$ is doubled,$r$ becomes half of the original value.
The velocity of an electron in an orbit is given by $v_n = 2.18 \times 10^6 \times \frac{Z}{n} \ \text{m/s}$,which implies $v \propto Z$. If $Z$ is doubled,$v$ becomes $2$ times the original value.
Therefore,the statement that the velocity of the electron in an orbit is doubled is consistent with Bohr's theory.

(D) The number of spectral lines emitted when an electron jumps from $n_2$ to $n_1$ is given by the formula $\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
For $n_2 = 5$ and $n_1 = 2$,the total number of lines is $\frac{(5 - 2)(5 - 2 + 1)}{2} = \frac{3 \times 4}{2} = 6$.
The possible transitions are:
$5 \to 4, 5 \to 3, 5 \to 2, 4 \to 3, 4 \to 2, 3 \to 2$.
$1$. Ultraviolet region (Lyman series,$n_1 = 1$): None of these transitions end at $n = 1$,so the number of lines is $0$.
$2$. Visible region (Balmer series,$n_1 = 2$): The transitions ending at $n = 2$ are $5 \to 2, 4 \to 2, 3 \to 2$. Total lines = $3$.
$3$. Infrared region (Paschen series,$n_1 = 3$): The transitions ending at $n = 3$ are $5 \to 3, 4 \to 3$. Total lines = $2$.
Since none of the individual statements $A, B, C$ are correct,and option $D$ claims all are correct,there is an error in the provided options. However,based on standard textbook problems of this type,if the question implies transitions ending at $n=2$,the visible lines are $3$. Given the options,the question is flawed.

(B) For a hydrogen-like species,the energy of an electron in the $n^{th}$ shell is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $He^{+}$,the atomic number $Z = 2$.
The second excited state corresponds to $n = 3$ (ground state $n=1$,first excited state $n=2$,second excited state $n=3$).
Substituting the values: $E_3 = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} \ eV$.
$E_3 = -6.044 \ eV$.
The kinetic energy $(KE)$ of an electron is equal to the magnitude of its total energy,i.e.,$KE = |E_n|$.
Therefore,$KE = 6.044 \ eV \approx 6.04 \ eV$.

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R_{H} Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Lyman series of $H$ atom $(Z=1)$,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$.
$\frac{1}{x} = R_{H} (1)^2 \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right] = R_{H} \implies R_{H} = \frac{1}{x}$.
For the Balmer series of $He^{+}$ ion $(Z=2)$,the longest wavelength occurs at $n_1 = 2$ and $n_2 = 3$.
$\frac{1}{\lambda_{\max}} = R_{H} (2)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$.
Substituting $R_{H} = \frac{1}{x}$:
$\frac{1}{\lambda_{\max}} = \frac{1}{x} \times 4 \left[ \frac{1}{4} - \frac{1}{9} \right] = \frac{4}{x} \left[ \frac{5}{36} \right] = \frac{5}{9x}$.
Therefore,$\lambda_{\max} = \frac{9x}{5}$.

(B) The velocity of an electron in a hydrogen-like species is given by the formula: $v = v_0 \times \frac{Z}{n}$,where $v_0$ is the velocity of an electron in the ground state of hydrogen $(2.18 \times 10^6 \ m/s)$,$Z$ is the atomic number,and $n$ is the principal quantum number.
For the ground state of a hydrogen atom $(H)$: $Z_1 = 1$,$n_1 = 1$. So,$v_1 = v_0 \times \frac{1}{1} = v_0$.
For the second excited state of a $He^{+}$ ion: $Z_2 = 2$,and the second excited state corresponds to $n_2 = 3$. So,$v_2 = v_0 \times \frac{2}{3}$.
The ratio of the velocities is $\frac{v_1}{v_2} = \frac{v_0}{v_0 \times (2/3)} = \frac{3}{2} = 1.5$.

(A) The relationship between total energy $(E_n)$ and potential energy $(PE)$ is given by $E_n = \frac{1}{2} PE$.
Substituting the given value: $E_n = \frac{-6.8 \ eV}{2} = -3.4 \ eV$.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is $E_n = \frac{-13.6 \ eV}{n^2}$.
Equating the two: $\frac{-13.6}{n^2} = -3.4$,which gives $n^2 = 4$,so $n = 2$.
The state $n = 1$ is the ground state,and $n = 2$ corresponds to the first excited state.

(B) According to Bohr's model,the radius of the $n$-th orbit is given by $r_n = a_0 \frac{n^2}{Z}$,where $a_0$ is the Bohr radius.
For the ground state of hydrogen atom,$n = 1$ and $Z = 1$,so $r_H = a_0 \frac{1^2}{1} = a_0$.
For $Be^{+3}$ ion,$Z = 4$. We need to find the orbit $n$ such that $r_{Be} = r_H$.
$a_0 \frac{n^2}{4} = a_0$.
$n^2 = 4$.
$n = 2$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$ or $E_n = -K \times \frac{Z^2}{n^2} \ J$.
For $He^{+}$ $(Z=2)$,the ionization energy is the energy required to remove the electron from the ground state $(n=1)$ to infinity $(n=\infty)$.
$IE = E_{\infty} - E_1 = 0 - (-K \times \frac{2^2}{1^2}) = 4K = 8.68 \times 10^{-18} \ J$.
Therefore,$K = \frac{8.68 \times 10^{-18}}{4} = 2.17 \times 10^{-18} \ J$.
Now,for $Be^{3+}$ $(Z=4)$ in the second orbit $(n=2)$:
$E_2 = -K \times \frac{Z^2}{n^2} = -(2.17 \times 10^{-18}) \times \frac{4^2}{2^2} = -(2.17 \times 10^{-18}) \times \frac{16}{4} = -(2.17 \times 10^{-18}) \times 4 = -8.68 \times 10^{-18} \ J$.

(C) The energy of a transition in a hydrogen-like atom is given by $\Delta E = 13.6 \ Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.
For absorption,the electron moves from a lower energy level $(n_1)$ to a higher energy level $(n_2)$.
For $1$ to $2$: $\Delta E = 13.6 (1 - \frac{1}{4}) = 10.2 \ eV$.
For $2$ to $3$: $\Delta E = 13.6 (\frac{1}{4} - \frac{1}{9}) \approx 1.89 \ eV$.
For $1$ to $\infty$: $\Delta E = 13.6 (1 - 0) = 13.6 \ eV$.
$\infty$ to $1$ represents emission,not absorption.
Thus,the transition from $1$ to $\infty$ involves the maximum energy absorption.

(B) The visible region corresponds to the Balmer series $(n_f = 2)$.
For $n$ transitions resulting in $4$ lines in the Balmer series,the electron must jump from $n_i = 6$ to $n_f = 2$ (since the number of lines $= n_i - n_f = 6 - 2 = 4$).
The infrared $(IR)$ region includes the Paschen $(n_f = 3)$,Brackett $(n_f = 4)$,and Pfund $(n_f = 5)$ series.
For transitions from $n_i = 6$ to lower energy levels:
Paschen series $(n_f = 3)$: $6$ $\rightarrow 3, 5$ $\rightarrow 3, 4$ $\rightarrow 3$ ($3$ lines).
Brackett series $(n_f = 4)$: $6$ $\rightarrow 4, 5$ $\rightarrow 4$ ($2$ lines).
Pfund series $(n_f = 5)$: $6 \rightarrow 5$ ($1$ line).
Total lines in the $IR$ region $= 3 + 2 + 1 = 6$.

(D) According to Bohr's postulate,the energy difference between two stationary states is given by $\Delta E = E_2 - E_1$.
This energy difference is equal to the energy of the photon absorbed or emitted,which is given by $E = h v$.
Equating the two,we get $h v = E_2 - E_1$.
Therefore,the frequency $v$ is given by $v = \frac{E_2 - E_1}{h}$.

(A) The radius of the $n^{th}$ Bohr orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
For a hydrogen atom,$Z = 1$.
Radius of the $3^{rd}$ orbit $(r_3)$ = $0.529 \times 3^2 = 0.529 \times 9 = 4.761 \mathring{A}$.
Radius of the $2^{nd}$ orbit $(r_2)$ = $0.529 \times 2^2 = 0.529 \times 4 = 2.116 \mathring{A}$.
The distance between the $3^{rd}$ and $2^{nd}$ orbits is $r_3 - r_2 = 4.761 \mathring{A} - 2.116 \mathring{A} = 2.645 \mathring{A}$.
Since $1 \mathring{A} = 10^{-8} \ cm$,the distance is $2.645 \times 10^{-8} \ cm$.

(D) The Lyman series in the atomic spectrum of the $H$ atom corresponds to electronic transitions from higher energy levels $(n_2 = 2, 3, 4, \dots, \infty)$ to the ground state $(n_1 = 1)$.
The wave number $(\bar{\nu})$ is given by the Rydberg formula: $\bar{\nu} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
The 'last line' of a spectral series corresponds to the transition from the highest possible energy level,which is $n_2 = \infty$,to the series limit level,which is $n_1 = 1$.
Therefore,the transition is from $n_2 = \infty$ to $n_1 = 1$.

(D) For an electron in a hydrogen-like atom,the relationship between kinetic energy $(K.E.)$ and potential energy $(P.E.)$ is given by $P.E. = -2 \times K.E.$
Initially,$K.E._1 = y$,so $P.E._1 = -2y$.
Finally,$K.E._2 = y/4$,so $P.E._2 = -2 \times (y/4) = -y/2$.
The change in potential energy is $\Delta P.E. = P.E._2 - P.E._1 = (-y/2) - (-2y) = (-y/2) + 2y = \frac{3}{2}y$.

(C) The Rydberg formula for the wavelength of a spectral line is given by: $\frac{1}{\lambda} = R Z^{2} \left( \frac{1}{n_1^{2}} - \frac{1}{n_2^{2}} \right)$.
For the Lyman series,$n_1 = 1$. For maximum wavelength,the transition must be from the nearest energy level,so $n_2 = 2$.
For $He^{+}$ ion,the atomic number $Z = 2$.
Substituting these values: $\frac{1}{\lambda_{\max}} = R (2)^{2} \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right)$.
$\frac{1}{\lambda_{\max}} = R \times 4 \times \left( 1 - \frac{1}{4} \right) = R \times 4 \times \frac{3}{4} = 3R$.
Therefore,$\lambda_{\max} = \frac{1}{3R}$.

(C) $Bohr's$ theory is applicable only to single-electron species (hydrogen-like atoms).
$H$ has $1$ electron.
$Li^{2+}$ has $3 - 2 = 1$ electron.
$Be^{3+}$ has $4 - 3 = 1$ electron.
$He^{2+}$ has $2 - 2 = 0$ electrons.
Since $He^{2+}$ has no electrons,$Bohr's$ theory is not applicable to it.

(A) In Bohr's atomic model,the radius of the $n^{th}$ orbit is given by $r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2}$,which implies $r \propto n^2$.
The total energy of an electron in the $n^{th}$ orbit is $E_n = -\frac{m Z^2 e^4}{8 \epsilon_0^2 h^2 n^2}$,which implies $E \propto -1/n^2$.
Considering the magnitude of energy,the relationship is $r \propto n^2$ and $E \propto 1/n^2$.

(D) For a single $He^{+}$ ion in the $n = 5$ state,the number of emission lines is $n - 1 = 4$. Thus,option $A$ is incorrect.
The ionization energy for $n = 5$ is given by $I.E. = \frac{13.6 \times Z^2}{n^2} = \frac{13.6 \times 2^2}{5^2} = \frac{13.6 \times 4}{25} = 2.176 \, eV$. Thus,option $B$ is incorrect.
The longest wavelength corresponds to the smallest energy transition,which is $n = 5 \to n = 4$. The formula is $\frac{1}{\lambda} = R \times Z^2 \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \times 4 \times \left(\frac{1}{4^2} - \frac{1}{5^2}\right) = 4R \times \left(\frac{25-16}{400}\right) = 4R \times \frac{9}{400} = \frac{9R}{100}$. Therefore,$\lambda = \frac{100}{9R} \approx \frac{11.11}{R}$. Since $11.11/R > 10/R$,option $C$ is incorrect.
The radius of the orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \, \mathring{A}$. For $n = 5$ and $Z = 2$,$r_5 = 0.529 \times \frac{25}{2} = 0.529 \times 12.5 = 6.6125 \, \mathring{A}$. Since $6.6125 \, \mathring{A} > 6 \, \mathring{A}$,option $D$ is correct.

(A) The power of the bulb is $P = 150 \ W$.
Only $8 \%$ of the energy is emitted as light,so the power emitted as light is $P_{light} = 150 \times \frac{8}{100} = 12 \ W$.
The energy of one photon is given by $E = \frac{hc}{\lambda}$.
The number of photons emitted per second $(n)$ is given by $n = \frac{P_{light}}{E} = \frac{P_{light} \times \lambda}{hc}$.
Substituting the values: $n = \frac{12 \times 6600 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^{8}}$.
Using $h = 6.6 \times 10^{-34} \ J \cdot s$ for simplicity as per the provided solution: $n = \frac{12 \times 6600 \times 10^{-10}}{6.6 \times 10^{-34} \times 3 \times 10^{8}} = \frac{12 \times 6600 \times 10^{-10}}{19.8 \times 10^{-26}} = 4 \times 10^{19}$ photons per second.

(C) The angular momentum of an electron in a given shell is given by the formula: $L = n \frac{h}{2\pi}$.
For the $M$-shell,the principal quantum number $n = 3$.
Given Planck's constant $h = 6.626 \times 10^{-34} \ J \cdot s = 6.626 \times 10^{-27} \ erg \cdot s$.
Substituting the values: $L = 3 \times \frac{6.626 \times 10^{-27}}{2 \times 3.14159}$.
$L = 3 \times 1.054 \times 10^{-27} \ erg \cdot s = 3.162 \times 10^{-27} \ erg \cdot s$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
The energy difference between adjacent levels $n$ and $n+1$ is $\Delta E = E_{n+1} - E_n = 13.6 \times Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right) = 13.6 \times Z^2 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.
As $n$ increases,the term $\frac{2n+1}{n^2(n+1)^2}$ decreases rapidly.
For example:
For $n=1$ to $2$,$\Delta E = 13.6 \times (1 - 0.25) = 10.2 \text{ eV}$.
For $n=2$ to $3$,$\Delta E = 13.6 \times (0.25 - 0.111) = 1.89 \text{ eV}$.
Thus,the energy difference decreases as the principal quantum number increases.

(A) The radius of the $n^{\text{th}}$ Bohr orbit is given by the formula $r_n = 0.529 \frac{n^2}{Z} \, \mathring{A}$.
Since $Z$ is constant for a given atom,the relationship is $r_n \propto n^2$.
For the first three orbits $(n = 1, 2, 3)$,the ratio is $r_1 : r_2 : r_3 = 1^2 : 2^2 : 3^2$.
Therefore,the ratio is $1 : 4 : 9$.

(B) The radius of an orbit in a hydrogen-like species is given by the formula $r_n = a_0 \frac{n^2}{Z}$.
For a given ion,$Z$ is constant,so $r_n \propto n^2$.
Therefore,the ratio $\frac{r_2}{r_5} = \frac{(2)^2}{(5)^2} = \frac{4}{25}$.

(B) 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 shortest wavelength in the Balmer series of $He^{+}$ $(Z=2)$: $n_1 = 2$,$n_2 = \infty$.
$\frac{1}{X} = R \times 2^2 \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) = R \times 4 \times \frac{1}{4} = R$.
So,$R = \frac{1}{X}$.
For the longest wavelength in the Paschen series of $Li^{2+}$ $(Z=3)$: $n_1 = 3$,$n_2 = 4$.
$\frac{1}{\lambda_{Li^{2+}}} = R \times 3^2 \left(\frac{1}{3^2} - \frac{1}{4^2}\right) = R \times 9 \left(\frac{1}{9} - \frac{1}{16}\right) = R \times 9 \left(\frac{16-9}{144}\right) = R \times 9 \times \frac{7}{144} = R \times \frac{7}{16}$.
Substituting $R = \frac{1}{X}$:
$\frac{1}{\lambda_{Li^{2+}}} = \frac{1}{X} \times \frac{7}{16} \implies \lambda_{Li^{2+}} = \frac{16}{7} \ X$.

(C) According to Bohr's postulate,the angular momentum of an electron is quantized as $mvr = \frac{nh}{2\pi}$.
According to De Broglie's hypothesis,$\lambda = \frac{h}{mv}$,which implies $mv = \frac{h}{\lambda}$.
Substituting this into the Bohr's postulate: $(\frac{h}{\lambda})r = \frac{nh}{2\pi}$.
This simplifies to $2\pi r = n\lambda$,where $n$ is the principal quantum number.
For the $3^{rd}$ orbit,$n = 3$,so the number of waves is $3$.

(A) For a hydrogen-like atom,the radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.
Given $r_2 = x$,we have $r_n = k \cdot n^2$.
So,$x = k \cdot (2)^2 = 4k$,which implies $k = x/4$.
For the $4^{th}$ orbit,$r_4 = k \cdot (4)^2 = 16k = 16 \cdot (x/4) = 4x$.
According to the Bohr quantization condition,$mvr_n = n \frac{h}{2 \pi}$.
For the $4^{th}$ orbit,$mvr_4 = 4 \frac{h}{2 \pi}$.
The de-Broglie wavelength is given by $\lambda = \frac{h}{mv}$.
Rearranging the quantization condition,$\frac{h}{mv} = \frac{2 \pi r_4}{n}$.
Substituting $n = 4$ and $r_4 = 4x$,we get $\lambda = \frac{2 \pi (4x)}{4} = 2 \pi x$.

(C) The energy of an electron in the $n^{th}$ Bohr orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For $n=1$,$E_1 = -13.6 \ eV$.
For $n=2$,$E_2 = -\frac{13.6}{4} = -3.4 \ eV$.
For $n=3$,$E_3 = -\frac{13.6}{9} \approx -1.51 \ eV$.
The difference between $1^{st}$ and $2^{nd}$ orbits is $\Delta E_{2-1} = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
The difference between $2^{nd}$ and $3^{rd}$ orbits is $\Delta E_{3-2} = E_3 - E_2 = -1.51 - (-3.4) = 1.89 \ eV$.
The ratio is $\frac{\Delta E_{2-1}}{\Delta E_{3-2}} = \frac{10.2}{1.89} \approx 5.4$.

(D) For $He^{+}$ ion,the atomic number $Z = 2$.
The Rydberg formula for the wave number is $\bar{\nu} = \frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$.
For the maximum wavelength $(\lambda_{\max})$,the energy difference must be minimum,which corresponds to the transition from the nearest higher energy level,$n_2 = 2$.
Substituting these values: $\frac{1}{\lambda_{\max}} = R (2)^2 \left[ \frac{1}{1^2} - \frac{1}{2^2} \right] = 4R \left[ 1 - \frac{1}{4} \right] = 4R \left( \frac{3}{4} \right) = 3R$.
Therefore,$\lambda_{\max} = \frac{1}{3R}$.

(B) The Rydberg formula is given by $\bar{v} = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_2 > n_1$.
Given that the sum of the energy levels is $n_2 + n_1 = 4$ and their difference is $n_2 - n_1 = 2$.
Adding these two equations: $2n_2 = 6 \implies n_2 = 3$.
Subtracting the equations: $2n_1 = 2 \implies n_1 = 1$.
For the $He^{+}$ ion,the atomic number $Z = 2$.
Substituting these values into the Rydberg formula:
$\bar{v} = R \times (2)^2 \left( \frac{1}{1^2} - \frac{1}{3^2} \right)$
$\bar{v} = R \times 4 \left( 1 - \frac{1}{9} \right)$
$\bar{v} = 4R \left( \frac{8}{9} \right) = \frac{32R}{9}$.
Thus,the correct option is $B$.

(D) Given equations:
$2n_2 + 3n_1 = 18$ $(I)$
$2n_2 - 3n_1 = 6$ $(II)$
Adding equations $(I)$ and $(II)$:
$4n_2 = 24 \implies n_2 = 6$ $(III)$
Substituting the value of $(III)$ into $(II)$:
$2(6) - 3n_1 = 6 \implies 12 - 3n_1 = 6 \implies 3n_1 = 6 \implies n_1 = 2$
Thus,the transition occurs between orbits $6 \rightarrow 2$.
The formula to find the total number of emitted photons is:
$\text{Total photons} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
$= \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 \times 5}{2} = 10$
Therefore,the correct option is $D$.

(D) For $Be^{3+}$,the atomic number $Z = 4$.
The first excited state corresponds to $n = 2$.
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 $n = 2$,$E_2 = -13.6 \times \frac{4^2}{2^2} = -13.6 \times \frac{16}{4} = -13.6 \times 4 = -54.4 \ eV$.
The separation energy (or ionization energy from the excited state) is the energy required to move the electron from $n = 2$ to $n = \infty$.
$E_{separation} = E_{\infty} - E_2 = 0 - (-54.4 \ eV) = 54.4 \ eV$.

(B) The time period $(T)$ of an electron in the $n^{th}$ orbit is given by the formula $T \propto \frac{n^3}{Z^2}$.
Therefore,the ratio of time periods for two different species is given by $\frac{T_1}{T_2} = \frac{n_1^3}{Z_1^2} \times \frac{Z_2^2}{n_2^3}$.
For the second orbit of hydrogen atom $(n_1 = 2, Z_1 = 1)$ and the third orbit of $He^{+}$ ion $(n_2 = 3, Z_2 = 2)$:
$\frac{T_1}{T_2} = \frac{2^3}{1^2} \times \frac{2^2}{3^3} = \frac{8}{1} \times \frac{4}{27} = \frac{32}{27}$.
Thus,the ratio is $32/27$.

(A) The energy required is the difference between the energy level of the $n=4$ state and the ground state $(n=1)$.
The formula for energy difference is: $E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.
For $He^{+}$,$Z = 2$. The ground state is $n_1 = 1$. The second line of the Balmer series corresponds to a transition from $n=4$ to $n=2$,so the electron must be excited to the $n=4$ state.
$E = 13.6 \times 2^2 \times (\frac{1}{1^2} - \frac{1}{4^2}) \ eV$
$E = 13.6 \times 4 \times (1 - \frac{1}{16}) \ eV$
$E = 54.4 \times (\frac{15}{16}) \ eV$
$E = 51 \ eV$.

(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 $H$ atom $(Z=1)$ in the second quantum state $(n=2)$: $E_2 = -13.6 \times \frac{1^2}{2^2} = -\frac{13.6}{4} \ eV$.
For the $He^{+}$ ion $(Z=2)$ in the third quantum state $(n=3)$: $E_3 = -13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} \ eV$.
Comparing the two expressions: $\frac{E_3}{E_2} = \frac{-13.6 \times (4/9)}{-13.6 \times (1/4)} = \frac{4}{9} \times 4 = \frac{16}{9}$.
Wait,let us re-evaluate: $E_2 = -\frac{13.6}{4}$ and $E_3 = -\frac{13.6 \times 4}{9}$.
Thus,$E_3 = E_2 \times \frac{4/9}{1/4} = E_2 \times \frac{16}{9}$.
Since the question defines the energy as $-E_2$,the magnitude is $E_2$. The correct value is $-\frac{16}{9}E_2$.

(C) The energy of the incident radiation is given by $E = \frac{hc}{\lambda}$.
Using $hc = 12400\, eV\, \mathring{A}$ and $\lambda = 250\, nm = 2500\, \mathring{A}$,we get $E = \frac{12400}{2500} = 4.96\, eV$.
According to the photoelectric equation,$E = W_0 + K.E._{max}$,where $W_0$ is the work function and $K.E._{max}$ is the maximum kinetic energy.
The stopping potential is $0.5\, V$,so $K.E._{max} = 0.5\, eV$.
Substituting the values: $4.96 = W_0 + 0.5$.
Therefore,$W_0 = 4.96 - 0.5 = 4.46\, eV$,which is approximately $4.5\, eV$.

(D) The statement in option $B$ is false because as the temperature of a black body increases,the intensity of radiation increases and the peak of the emission curve shifts to shorter wavelengths (higher frequency).
The statement in option $D$ is also false because the $Rydberg$ constant $(R_H)$ has the unit of reciprocal length,i.e.,$cm^{-1}$ or $m^{-1}$.
However,in the context of standard multiple-choice questions,option $D$ is the most fundamentally incorrect regarding physical units.

(D) For the Lyman series,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$.
Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For hydrogen $(Z=1)$: $\frac{1}{A} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,so $R = \frac{1}{A}$.
For the Paschen series,the longest wavelength occurs at the first line,where $n_1 = 3$ and $n_2 = 4$.
For $He^{+}$ $(Z=2)$: $\frac{1}{\lambda} = R(2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$.
$\frac{1}{\lambda} = 4R \left( \frac{1}{9} - \frac{1}{16} \right) = 4R \left( \frac{16-9}{144} \right) = 4R \left( \frac{7}{144} \right) = \frac{7R}{36}$.
Substituting $R = \frac{1}{A}$: $\frac{1}{\lambda} = \frac{7}{36A}$.
Therefore,$\lambda = \frac{36A}{7}$.

$(B)$ The radius of an orbit in a hydrogen-like atom is given by $r = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
Given $r = 211.6 \text{ pm} = 2.116 \mathring{A}$ and for hydrogen $Z = 1$.
Substituting the values: $2.116 = 0.529 \times n^2$.
$n^2 = \frac{2.116}{0.529} = 4$.
$n = 2$.
Since the electron transitions from higher orbitals to the $n = 2$ orbit, this corresponds to the Balmer series.

(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 $Li^{2+}$ ion,the atomic number $Z = 3$.
The first excited state corresponds to $n = 2$.
Substituting these values into the formula:
$E_2 = -13.6 \times \frac{3^2}{2^2} \ eV$
$E_2 = -13.6 \times \frac{9}{4} \ eV$
$E_2 = -13.6 \times 2.25 \ eV = -30.6 \ eV$.

(C) According to Einstein's photoelectric equation,the kinetic energy of the ejected electron is given by:
$E = W + K.E.$
$\frac{hc}{\lambda} = \frac{hc}{\lambda_0} + \frac{1}{2}mv^2$
Rearranging for the kinetic energy term:
$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}$
$\frac{1}{2}mv^2 = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_0}\right) = hc \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$
Solving for velocity $v$:
$v^2 = \frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)$
$v = \sqrt{\frac{2hc}{m} \left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)}$