Atomic models and Planck's quantum theory Questions in English

(A) For a hydrogen-like atom,the energy of an electron in the $n^{th}$ shell is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
As the shell number $n$ increases from $K$ $(n=1)$ to $N$ $(n=4)$,the value of $n$ increases.
$1$. $K.E. = |E_n| = 13.6 \times \frac{Z^2}{n^2}$. As $n$ increases,$K.E.$ decreases.
$2$. Total energy $E_n = -13.6 \times \frac{Z^2}{n^2}$. As $n$ increases,the magnitude of the negative value decreases,meaning the total energy increases.
$3$. Potential energy $P.E. = 2 \times E_n = -27.2 \times \frac{Z^2}{n^2}$. As $n$ increases,the magnitude of the negative value decreases,meaning the potential energy increases.
Therefore,the kinetic energy decreases as the electron moves to a higher shell.

(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r_n = a_0 \times \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the orbit number,and $Z$ is the atomic number.
For $Li^{+2}$ ion $(Z=3)$: $n_1 = 2$,$Z_1 = 3$. So,$r_1 = a_0 \times \frac{2^2}{3} = a_0 \times \frac{4}{3}$.
For $Be^{+3}$ ion $(Z=4)$: $n_2 = 3$,$Z_2 = 4$. So,$r_2 = a_0 \times \frac{3^2}{4} = a_0 \times \frac{9}{4}$.
The ratio is $\frac{r_1}{r_2} = \frac{4/3}{9/4} = \frac{4}{3} \times \frac{4}{9} = \frac{16}{27}$.

(A) The energy of an electron in a hydrogen-like species is given by the formula: $E_n = E_0 \frac{Z^2}{n^2}$,where $E_0$ is the ground state energy of hydrogen,$Z$ is the atomic number,and $n$ is the principal quantum number.
For hydrogen in the ground state $(n=1, Z=1)$:
$E_H = E_0 \frac{1^2}{1^2} = E_0$
For $Be^{3+}$ in the first excited state $(n=2, Z=4)$:
$E_{Be^{3+}} = E_0 \frac{4^2}{2^2} = E_0 \frac{16}{4} = 4E_0$
The ratio of the energy of the electron in the ground state of hydrogen to the energy of the electron in the first excited state of $Be^{3+}$ is:
Ratio $= \frac{E_H}{E_{Be^{3+}}} = \frac{E_0}{4E_0} = \frac{1}{4} = 1:4$

(D) In an electronic transition,an electron jumps from a higher energy level to a lower energy level.
This process is extremely rapid.
The time taken for such an electronic transition is typically of the order of $10^{-8} \, sec$.

(B) The total energy of an electron in an atom is given by the expression $E_n = -\frac{13.6 \ Z^2}{n^2} \ eV$.
Since the electron is bound to the nucleus by electrostatic forces of attraction,the potential energy is negative and its magnitude is greater than the kinetic energy.
Therefore,the total energy (sum of kinetic and potential energy) is always negative,indicating that the electron is in a bound state.

(A) In the Bohr model,the energy of an electron in the $n^{th}$ orbit is given by the expression $E_{n} = -13.6 \times \frac{Z^{2}}{n^{2}} \text{ eV}$.
As the distance from the nucleus increases,the value of $n$ increases.
When the electron is at an infinite distance from the nucleus,$n = \infty$.
Substituting $n = \infty$ into the formula,we get $E_{\infty} = -13.6 \times \frac{Z^{2}}{\infty^{2}} = 0$.
Therefore,the energy of the electron at infinite distance from the nucleus is taken as zero.

(C) The energy of an electron at an infinite distance from the nucleus is defined as $0$.
As the electron approaches the nucleus,it becomes bound to the nucleus,and the system releases energy,resulting in a negative potential energy.
The energy of an electron in a hydrogen-like atom is given by $E_n = -\frac{13.6 \ Z^2}{n^2} \ eV$.
As the distance from the nucleus decreases (i.e.,$n$ decreases),the value of $E_n$ becomes more negative (i.e.,it decreases).
Therefore,the energy decreases to a greater negative value.

(A) The energy change for an electron transition in a hydrogen atom is given by $\Delta E = 13.6 \ \text{eV} \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For a transition from $n_1$ to $n_2$,the energy required is proportional to $\left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Comparing the options:
$A$: $n=1$ to $n=2$,$\Delta E \propto (1 - 0.25) = 0.75$.
$B$: $n=2$ to $n=3$,$\Delta E \propto (0.25 - 0.111) = 0.139$.
$C$: $n=\infty$ to $n=1$ represents an emission (energy release),not absorption.
$D$: $n=3$ to $n=5$,$\Delta E \propto (0.111 - 0.04) = 0.071$.
The transition from $n=1$ to $n=2$ requires the largest amount of energy among the given absorption transitions.

(D) Bohr’s model of the atom was successful in explaining the stability of the atom and the line spectra of hydrogen-like species.
However,it failed to explain the splitting of spectral lines in the presence of an external magnetic field (Zeeman effect) or an external electric field (Stark effect).
It also contradicts Heisenberg’s uncertainty principle,as it assumes definite paths for electrons.
Therefore,none of the given phenomena are explained by Bohr’s model.

(D) Bohr's atomic theory introduced the concept of stationary states or stable orbits in which electrons revolve around the nucleus without radiating energy. This concept successfully addressed the instability issue present in Rutherford's atomic model.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = \frac{-13.6}{n^2} \ eV$.
For the $n = 2$ state,the energy is $E_2 = \frac{-13.6}{2^2} = \frac{-13.6}{4} = -3.4 \ eV$.
To remove an electron from the $n = 2$ state to infinity ($n = \infty$,where $E = 0$),the energy required is $\Delta E = E_{\infty} - E_2 = 0 - (-3.4 \ eV) = 3.4 \ eV$.

(A) The total energy of an electron in the ground state of a hydrogen atom is $E = -13.6 \, eV$.
To remove the electron from the ground state (ionize it),we must provide energy equal to the binding energy,which is $|E| = 13.6 \, eV$.
This energy is provided as kinetic energy to the emitted electron.
The stopping potential $V_s$ is related to the maximum kinetic energy $K_{max}$ by the equation $K_{max} = e \times V_s$.
Since $K_{max} = 13.6 \, eV$,we have $13.6 \, eV = e \times V_s$.
Therefore,$V_s = 13.6 \, V$.

(D) Bohr's model is applicable only to hydrogen-like species,which are systems containing only one electron.
$H$ has $1$ electron,$He^{+}$ has $2-1 = 1$ electron,and $Li^{2+}$ has $3-2 = 1$ electron.
$Na$ (Sodium) has $11$ electrons.
Since $Na$ is a multi-electron system,Bohr's model cannot explain its emission spectrum.

(C) According to Bohr's postulate,the angular momentum $(L)$ of an electron revolving in the $n^{th}$ orbit is quantized.
The formula for angular momentum is given by $L = mvr = \frac{nh}{2\pi}$,where $n$ is the principal quantum number,$h$ is Planck's constant,and $m, v, r$ are the mass,velocity,and radius of the orbit respectively.
Thus,the correct expression is $\frac{nh}{2\pi}$.

(D) The energy of an electron in a hydrogen-like atom is given by the formula $E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$.
As the principal quantum number $n$ increases,the value of $E_n$ becomes less negative,which means the energy increases.
As $n \to \infty$,the energy $E_n$ approaches $0$,which is the maximum possible energy for a bound electron.
Therefore,an electron possesses maximum energy when it is at an infinite distance from the nucleus.

(C) The energy of an electron in the $n^{\text{th}}$ shell is given by the formula:
$E_n = -13.6 \frac{Z^2}{n^2} \ \text{eV}$
As we move away from the nucleus,the principal quantum number $n$ increases.
Since the energy expression has a negative sign,as $n$ increases,the value of $\frac{1}{n^2}$ decreases,making the total energy $E_n$ less negative (i.e.,it becomes more positive).
Therefore,the energy of the electron increases as it moves further from the nucleus.

(C) The energy difference for an electronic transition in a hydrogen atom is given by $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
For a hydrogen atom,$Z=1$,so $\Delta E = 13.6 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
Calculating the energy for each transition:
$A$: $n=1$ to $n=2$,$\Delta E = 13.6 \times (1 - 1/4) = 13.6 \times 0.75 = 10.2 \text{ eV}$.
$B$: $n=2$ to $n=3$,$\Delta E = 13.6 \times (1/4 - 1/9) = 13.6 \times 0.138 = 1.89 \text{ eV}$.
$C$: $n=1$ to $n=\infty$,$\Delta E = 13.6 \times (1 - 0) = 13.6 \text{ eV}$.
$D$: $n=3$ to $n=5$,$\Delta E = 13.6 \times (1/9 - 1/25) = 13.6 \times 0.071 = 0.97 \text{ eV}$.
Comparing the values,the transition from $n=1$ to $n=\infty$ requires the largest amount of energy.

(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{2.18 \times 10^{-11}}{n^2} \, ergs$.
For $n=1$,$E_1 = -2.18 \times 10^{-11} \, ergs$.
For $n=3$,$E_3 = -\frac{2.18 \times 10^{-11}}{3^2} = -\frac{2.18 \times 10^{-11}}{9} \approx -0.2422 \times 10^{-11} \, ergs$.
The energy absorbed is $\Delta E = E_3 - E_1 = (-0.2422 \times 10^{-11}) - (-2.18 \times 10^{-11}) = 1.9378 \times 10^{-11} \, ergs = 0.19378 \times 10^{-10} \, ergs$.
Comparing this with the given options,the closest value is $0.1911 \times 10^{-10} \, ergs$.

(B) Given,$E = -3.4 \ eV$.
The energy of an electron in the $n^{th}$ orbit is given by $E = -\frac{13.6 \ eV}{n^2}$.
Substituting the value: $-3.4 = -\frac{13.6}{n^2} \implies n^2 = \frac{13.6}{3.4} = 4 \implies n = 2$.
According to Bohr's postulate,the angular momentum $L$ is given by $L = \frac{nh}{2\pi}$.
For $n = 2$,$L = \frac{2h}{2\pi} = \frac{h}{\pi}$.
Using $h = 6.626 \times 10^{-34} \ J \ s$ and $\pi \approx 3.14$:
$L = \frac{6.626 \times 10^{-34}}{3.14} \approx 2.11 \times 10^{-34} \ kg \ m^2 \ s^{-1}$.
Thus,the correct option is $B$.

(B) The potential energy $(PE)$ of an electron at a distance $r$ from the nucleus of charge $Ze$ is given by the electrostatic potential energy formula:
$PE = \frac{k_e q_1 q_2}{r}$
Here,$q_1 = Ze$ (charge of the nucleus) and $q_2 = -e$ (charge of the electron).
Thus,$PE = \frac{k_e (Ze)(-e)}{r} = -\frac{k_e Z e^2}{r}$.
In Gaussian units where $k_e = 1$,the expression simplifies to $PE = -\frac{Z e^2}{r}$.

(B) According to Coulomb's law,the electrostatic force of attraction $(F)$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is given by the formula: $F = k \frac{|q_1 q_2|}{r^2}$.
In a hydrogen atom,the nucleus has a charge of $+e$ (where $Z=1$) and the electron has a charge of $-e$.
Substituting these values into the formula,we get: $F = k \frac{|(+e)(-e)|}{r^2} = k \frac{e^2}{r^2}$.
Assuming the constant $k$ is taken as unity in the given options,the expression simplifies to $\frac{e^2}{r^2}$.

(C) According to Bohr's model of the hydrogen atom:
$1$. 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,given by $\frac{mv^2}{r} = \frac{Ze^2}{r^2}$. Thus,option $A$ is correct.
$2$. Bohr postulated that the angular momentum of an electron in a stationary orbit is quantized,given by $mvr = \frac{nh}{2\pi}$. Thus,option $B$ is correct.
$3$. In the Bohr model,the nucleus is assumed to be stationary,but the mass of the proton (nucleus) is not ignored; rather,the model assumes the nucleus is infinitely heavy compared to the electron. However,stating that the 'mass of the proton is ignored' is technically incorrect as the model relies on the electrostatic interaction between the proton and electron.
Therefore,option $C$ is the incorrect statement.

(B) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by the formula: $v_n = 2.18 \times 10^8 \times \frac{Z}{n} \ \text{cm s}^{-1}$.
For the hydrogen atom,the atomic number $Z = 1$.
Thus,$v_n = 2.18 \times 10^8 \times \frac{1}{n} \ \text{cm s}^{-1}$.
For the speed $v_n$ to be maximum,the denominator $n$ must be the smallest possible integer.
Since the orbit number $n$ starts from $1$,the speed is maximum at $n = 1$.

(B) The energy difference for the transition from $C$ to $A$ is the sum of the energy differences for the transitions from $C$ to $B$ and $B$ to $A$.
$E_C - E_A = (E_C - E_B) + (E_B - E_A)$
Since energy $E = \frac{hc}{\lambda}$,we can write:
$\frac{hc}{\lambda_3} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$
Dividing both sides by $hc$:
$\frac{1}{\lambda_3} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$
$\frac{1}{\lambda_3} = \frac{\lambda_2 + \lambda_1}{\lambda_1 \lambda_2}$
Therefore,$\lambda_3 = \frac{\lambda_1 \lambda_2}{\lambda_1 + \lambda_2}$.

(C) The binding energy of a hydrogen-like species is given by the formula $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For a hydrogen atom,$Z = 1$ and $n = 1$,so $E_H = 13.6 \ eV$.
For a singly ionized helium atom $(He^+)$,$Z = 2$ and $n = 1$.
Therefore,$E_{He^+} = 13.6 \times \frac{2^2}{1^2} \ eV$.
$E_{He^+} = 13.6 \times 4 \ eV = 54.4 \ eV$.

(D) The wave number for the transition of $n=4$ to $n=2$ in the $He^{+}$ spectrum is given by the Rydberg formula: $\bar{\nu} = R Z^{2} ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} )$.
For $He^{+}$,$Z=2$,$n_{1}=2$,and $n_{2}=4$:
$\bar{\nu} = R(2)^{2} ( \frac{1}{2^{2}} - \frac{1}{4^{2}} ) = 4R ( \frac{1}{4} - \frac{1}{16} ) = 4R ( \frac{3}{16} ) = \frac{3R}{4}$.
Thus,the wavelength $\lambda = \frac{4}{3R}$.
For the hydrogen atom $(Z=1)$,we seek a transition $n_{2} \rightarrow n_{1}$ such that $\bar{\nu} = R(1)^{2} ( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ) = \frac{3R}{4}$.
This simplifies to $( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ) = \frac{3}{4}$.
Testing values,if $n_{1}=1$ and $n_{2}=2$,then $( \frac{1}{1^{2}} - \frac{1}{2^{2}} ) = 1 - \frac{1}{4} = \frac{3}{4}$.
Therefore,the transition $n=2 \rightarrow n=1$ in hydrogen has the same wavelength.

(C) The energy levels in a hydrogen atom are denoted by the principal quantum number $n$.
The $K$ shell corresponds to $n = 1$,the $L$ shell to $n = 2$,and the $M$ shell to $n = 3$.
When an electron transitions from any higher energy level $(n_2 > 3)$ to the $M$ shell $(n_1 = 3)$,the resulting spectral lines belong to the Paschen series.
Therefore,the correct option is $C$.

(B) The maximum number of electrons in a shell is given by $2n^2$. For the third shell $(n=3)$,the maximum number of electrons is $2(3)^2 = 18$. Thus,option $A$ is correct.
When an electron falls from a higher energy level $(n_2)$ to a lower energy level $(n_1)$,the energy emitted is given by $\Delta E = E_{n_2} - E_{n_1}$. Since the energy levels are different,the energy emitted depends on the initial level $(n_2)$. Therefore,falling to the same level from different higher levels results in different amounts of energy being emitted. Thus,option $B$ is incorrect.
The Balmer series corresponds to transitions where the electron falls to the $n=2$ level,which lies in the visible region. Thus,option $C$ is correct.
The ground state of a hydrogen atom is the lowest energy state,which corresponds to $n=1$. Thus,option $D$ is correct.

(C) Given: $n_2 = 8$ and $n_1 = 2$.
The total number of spectral lines emitted when an electron transitions from a higher energy level $n_2$ to a lower energy level $n_1$ is calculated using the formula:
$N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
Substituting the values:
$N = \frac{(8 - 2)(8 - 2 + 1)}{2}$
$N = \frac{6 \times 7}{2}$
$N = \frac{42}{2}$
$N = 21$
Thus,the total number of spectral lines is $21$.

(D) The number of possible spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula $\frac{n(n-1)}{2}$.
For $n = 4$,the number of lines is $\frac{4(4-1)}{2} = 6$.
The possible transitions are:
$1$. $4 \rightarrow 3$
$2$. $4 \rightarrow 2$
$3$. $4 \rightarrow 1$
$4$. $3 \rightarrow 2$
$5$. $3 \rightarrow 1$
$6$. $2 \rightarrow 1$
Since all the transitions listed in options $A$,$B$,and $C$ are valid individual steps that can occur during the cascade of an electron from $n=4$ to $n=1$,the correct answer is $D$.

(A) The Ritz combination principle states that the wave number of any spectral line in an atomic spectrum can be expressed as the difference between two spectral terms.
Mathematically,it is represented as $\bar{\nu} = T(n_1) - T(n_2)$,where $T(n) = \frac{R_H Z^2}{n^2}$.
This leads to the Rydberg formula: $\bar{\nu} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
Option $A$ represents the Rydberg formula,which is a direct application of the Ritz combination principle.
Therefore,$A$ is the correct answer.

(B) The wavelength of a spectral line for an electronic transition is inversely proportional to the difference in the energy levels involved in the transition.
$\Delta E = E_2 - E_1 = \frac{hc}{\lambda}$
Therefore,$\lambda = \frac{hc}{\Delta E}$,which implies $\lambda \propto \frac{1}{\Delta E}$.

(C) The Rydberg formula for the wavelength of spectral lines is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
As the wavelength $\lambda$ increases,the energy difference $\Delta E = \frac{hc}{\lambda}$ decreases.
In any spectral series,as we move to higher energy levels (higher $n$),the energy gap between successive levels decreases,meaning the spectral lines become closer together at shorter wavelengths (higher energy),not longer wavelengths.
Therefore,the statement that spectral lines are closer together at longer wavelengths is false.

(D) The line spectrum of the hydrogen atom provides evidence for the quantization of energy levels.
When an electron transitions from a higher energy level $(n_2)$ to a lower energy level $(n_1)$,it emits a photon.
The energy of this emitted photon is given by $\Delta E = E_2 - E_1 = h\nu$.
This indicates that the energy is not lost continuously,but rather in discrete bundles or packets of energy called quanta or photons.

(A) In the hydrogen spectrum,the series of lines produced when an electron jumps from higher energy levels $(n_2 = 3, 4, 5, ...)$ to the $L$ shell $(n_1 = 2)$ is known as the Balmer series.
This series lies in the visible region of the electromagnetic spectrum.

(D) The number of spectral lines produced when an electron transitions from an excited state $n_2$ to a lower energy state $n_1$ is given by the formula: $\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Here,the electron is excited to the $5^{th}$ level $(n_2 = 5)$ and returns to the ground state $(n_1 = 1)$.
Substituting the values: $\frac{(5 - 1)(5 - 1 + 1)}{2} = \frac{4 \times 5}{2} = 10$.
Therefore,a total of $10$ spectral lines are produced.

(A) The total number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $\text{Total lines} = \frac{n(n-1)}{2}$.
Given that the total number of lines is $10$,we have:
$\frac{n(n-1)}{2} = 10$
$n(n-1) = 20$
$n^2 - n - 20 = 0$
$(n-5)(n+4) = 0$
Since $n$ must be positive,$n = 5$.
The visible spectrum corresponds to the Balmer series,where transitions occur to the $n_1 = 2$ energy level.
For an electron falling from $n = 5$ to $n = 2$,the possible transitions are:
$5 \rightarrow 2$,$4 \rightarrow 2$,and $3 \rightarrow 2$.
Thus,there are $3$ lines in the visible region.

(C) The wave number $\bar{\nu}$ is given by the Rydberg formula: $\bar{\nu} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the first Lyman transition in $H$ atom $(Z=1)$: $n_1=1, n_2=2$. So,$\bar{\nu}_1 = R_H (1)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3}{4} R_H$.
For the second Balmer transition in a hydrogen-like ion $(Z)$: $n_1=2, n_2=4$. So,$\bar{\nu}_2 = R_H Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H Z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = R_H Z^2 \left( \frac{3}{16} \right)$.
Equating the two: $\frac{3}{4} R_H = R_H Z^2 \left( \frac{3}{16} \right)$.
$\frac{3}{4} = Z^2 \left( \frac{3}{16} \right) \implies Z^2 = \frac{3}{4} \times \frac{16}{3} = 4$.
Thus,$Z = 2$. The ion with $Z=2$ is $He^{+}$.

(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
For a hydrogen atom,$Z = 1$. The electron falls from infinity $(n_2 = \infty)$ to the stationary state $n_1 = 1$.
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \ m^{-1} \times (1)^2 \times (\frac{1}{1^2} - \frac{1}{\infty^2})$
$\frac{1}{\lambda} = 1.097 \times 10^7 \times (1 - 0) = 1.097 \times 10^7 \ m^{-1}$
$\lambda = \frac{1}{1.097 \times 10^7} \ m \approx 9.115 \times 10^{-8} \ m$
To convert meters to nanometers $(nm)$,multiply by $10^9$: $\lambda = 9.115 \times 10^{-8} \times 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

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

(A) The energy of a photon is given by $E = mc^2 = \frac{hc}{\lambda}$,which implies $m = \frac{h}{\lambda c}$. Thus,$m \propto \frac{1}{\lambda}$.
For the Lyman series,the first line corresponds to the transition from $n_2 = 2$ to $n_1 = 1$:
$\frac{1}{\lambda_1} = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R_H \left( 1 - \frac{1}{4} \right) = \frac{3R_H}{4}$.
For the Balmer series,the first line corresponds to the transition from $n_2 = 3$ to $n_1 = 2$:
$\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R_H}{36}$.
Since $m \propto \frac{1}{\lambda}$,the ratio of masses is $\frac{m_1}{m_2} = \frac{\lambda_2}{\lambda_1} = \frac{1/\lambda_1}{1/\lambda_2} = \frac{3R_H/4}{5R_H/36} = \frac{3}{4} \times \frac{36}{5} = \frac{27}{5}$.
Therefore,the ratio is $27 : 5$.

(C) The Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$.
For the series limit of the Lyman series,$n_1 = 1$ and $n_2 = \infty$.
Thus,$\frac{1}{\lambda_L} = R(1 - 0) = R$,which implies $\lambda_L = \frac{1}{R} = 912 \ \mathring{A}$.
For the series limit of the Balmer series,$n_1 = 2$ and $n_2 = \infty$.
Thus,$\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - 0 \right) = \frac{R}{4}$.
This gives $\lambda_B = \frac{4}{R} = 4 \lambda_L$.
Substituting the value of $\lambda_L$,we get $\lambda_B = 4 \times 912 \ \mathring{A}$.

(A) The wave number $\bar{\nu}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{\nu} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the first line of the Balmer series in the hydrogen atom $(Z=1)$,the transition is from $n_2=3$ to $n_1=2$: $\bar{\nu} = R (1)^2 (\frac{1}{2^2} - \frac{1}{3^2}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{5}{36})$.
For the $He^{+}$ ion $(Z=2)$,we need to find the transition $n_2 \rightarrow n_1$ such that $\bar{\nu} = R (2)^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = R \frac{5}{36}$.
Simplifying,$4 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{5}{36}$,which gives $(\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{5}{144}$.
Testing the options,for $n_2=4$ and $n_1=2$: $(\frac{1}{4} - \frac{1}{16}) = \frac{4-1}{16} = \frac{3}{16} \neq \frac{5}{144}$.
For $n_2=6$ and $n_1=4$: $(\frac{1}{16} - \frac{1}{36}) = \frac{9-4}{144} = \frac{5}{144}$.
Thus,the transition $6 \rightarrow 4$ in $He^{+}$ matches the wave number.

(A) According to Einstein's photoelectric equation,the kinetic energy of the emitted electron is given by:
$K_{max} = hv - hv_0$
Since $K_{max}$ must be positive for photoelectric emission to occur,we must have:
$hv > hv_0$ or $v > v_0$
Here,$v$ is the frequency of incident light and $v_0$ is the threshold frequency. Thus,the photoelectric effect occurs only when the incident light has a frequency greater than a certain minimum value,known as the threshold frequency.

(C) According to Einstein's photoelectric equation: $KE = h\nu - h\nu_0$,where $h\nu_0$ is the work function.
Initially: $KE_1 = h\nu - h\nu_0 \ldots(1)$
When frequency is doubled: $KE_2 = h(2\nu) - h\nu_0 = 2h\nu - h\nu_0 \ldots(2)$
From equation $(1)$,$h\nu = KE_1 + h\nu_0$. Substituting this into equation $(2)$:
$KE_2 = 2(KE_1 + h\nu_0) - h\nu_0$
$KE_2 = 2KE_1 + 2h\nu_0 - h\nu_0$
$KE_2 = 2KE_1 + h\nu_0$
Since $h\nu_0 > 0$,the new kinetic energy $KE_2$ is greater than $2KE_1$.

(B) The energy of a photon is given by $E = h\nu = \frac{hc}{\lambda}$.
As the wavelength increases,the energy of the photon decreases.
The order of wavelengths for the given colors is $\lambda_{\text{green}} < \lambda_{\text{yellow}} < \lambda_{\text{red}}$.
Therefore,the energy order is $E_{\text{green}} > E_{\text{yellow}} > E_{\text{red}}$.
Since the surface does not eject electrons with yellow light,the energy of yellow light is below the threshold energy (work function) of the surface.
Because the energy of red light is even lower than that of yellow light,it will also be below the threshold energy.
Thus,no electrons will be ejected by red light,regardless of intensity or duration.

(D) The photoelectric equation is given by $h\nu = \Phi + KE$,where $\Phi$ is the work function and $KE$ is the kinetic energy.
For zero velocity,$KE = 0$,so $h\nu = \Phi$.
Given $\Phi = 4 \, eV$,we have $h\nu = 4 \, eV$.
Using the relation $\lambda = \frac{hc}{\Phi}$,where $hc \approx 12400 \, eV \cdot \mathring{A}$:
$\lambda = \frac{12400 \, eV \cdot \mathring{A}}{4 \, eV} = 3100 \, \mathring{A}$.
Alternatively,using constants: $\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4 \times 1.602 \times 10^{-19}} \approx 3100 \, \mathring{A}$.
Hence,the correct option is $D$.

(C) According to the photoelectric effect,the kinetic energy $(KE)$ of the emitted photoelectrons is given by the equation: $KE = h\nu - \Phi$,where $h$ is Planck's constant,$\nu$ is the frequency of the incident light,and $\Phi$ is the work function of the metal.
Since $\Phi$ is constant for a given metal,the kinetic energy of the emitted electrons depends directly on the frequency of the incident light $(\nu)$.

(D) The $1s$ electronic level allows the hydrogen atom to absorb a photon but not to emit one.
On absorption of radiation,the electron is transferred from the $1s$ level to a higher energy level,such as $2s$ or $3s$.
However,to emit a photon,an electron must transition from a higher energy level to a lower energy level.
Since $1s$ is the ground state (the lowest energy level),the electron cannot transition to a lower energy level.
Therefore,emission is not possible from the $1s$ state.

(B) The energy difference between the ground state $(GS)$ and the first excited state $(1^{st} ES)$ is given by the difference in their respective ionisation energies: $\Delta E = IE_{GS} - IE_{1^{st} ES} = 1540 \ kJ \ mol^{-1} - 300 \ kJ \ mol^{-1} = 1240 \ kJ \ mol^{-1}$.
To convert this energy into electron volts $(eV)$,we use the conversion factor $1 \ eV \approx 96.5 \ kJ \ mol^{-1}$:
$\Delta E = \frac{1240}{96.5} \approx 12.85 \ eV$.
Using the relation $\lambda (nm) = \frac{1240}{\Delta E (eV)}$:
$\lambda = \frac{1240}{12.85} \approx 96.5 \ nm$.
Since the wavelength is approximately $96.5 \ nm$,which falls in the range of $10 \ nm$ to $400 \ nm$,the emitted light lies in the $UV$ region.