Atomic models and Planck's quantum theory Questions in English

(B) For the Lyman series,the minimum wavelength occurs when $n_1 = 1$ and $n_2 = \infty$. The formula is $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For Lyman series: $\frac{1}{\lambda_{L}} = R (\frac{1}{1^2} - 0) = R$,so $\lambda_{L} = \frac{1}{R}$.
For Balmer series: $n_1 = 2$ and $n_2 = \infty$. So,$\frac{1}{\lambda_{B}} = R (\frac{1}{2^2} - 0) = \frac{R}{4}$,so $\lambda_{B} = \frac{4}{R}$.
The ratio $\frac{\lambda_{L}}{\lambda_{B}} = \frac{1/R}{4/R} = \frac{1}{4} = 0.25$.

(C) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given: $n_1 = 1$ (first energy level),$n_2 = \infty$ (infinite energy level),and $R_H = 1.097 \times 10^{7} \ m^{-1}$.
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^{7} \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right)$.
Since $\frac{1}{\infty} = 0$,we have $\frac{1}{\lambda} = 1.097 \times 10^{7} \ m^{-1}$.
Therefore,$\lambda = \frac{1}{1.097 \times 10^{7}} \ m = 0.9115 \times 10^{-7} \ m$.
Converting to nanometers $(nm)$: $\lambda = 0.9115 \times 10^{-7} \times 10^{9} \ nm = 91.15 \ nm \approx 91 \ nm$.

(C) The radius of an orbit in a hydrogen-like species is given by $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the second excited state of $H$ $(Z=1)$,the electron is in the $n=3$ orbit:
$r_{H(n=3)} = 0.529 \times \frac{3^2}{1} = 0.529 \times 9 \ \mathring{A}$.
For the first excited state of $Li^{2+}$ $(Z=3)$,the electron is in the $n=2$ orbit:
$r_{Li^{2+}(n=2)} = 0.529 \times \frac{2^2}{3} = 0.529 \times \frac{4}{3} \ \mathring{A}$.
The ratio of the radius of the first excited state of $Li^{2+}$ to the radius of the second excited state of $H$ is:
$\frac{r_{Li^{2+}(n=2)}}{r_{H(n=3)}} = \frac{0.529 \times \frac{4}{3}}{0.529 \times 9} = \frac{4}{3 \times 9} = \frac{4}{27}$.
However,checking the provided options,the ratio is likely intended as the inverse,i.e.,the ratio of the radius of the second excited state of $H$ to the first excited state of $Li^{2+}$,which is $27:4$.

(B) The Rydberg formula for the spectral lines of hydrogen is given by $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the Balmer series,the transition ends at $n_1 = 2$.
The lines in the Balmer series are defined by transitions from $n_2 = 3, 4, 5, \dots$ to $n_1 = 2$.
- The first line $(H_\alpha)$ corresponds to $n = 3 \to n = 2$.
- The second line $(H_\beta)$ corresponds to $n = 4 \to n = 2$.
Therefore,the transition from $n = 4$ to $n = 2$ represents the second line of the Balmer series.

(B) The radius of 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 principal quantum number,and $Z$ is the atomic number.
For a hydrogen atom in the $n = 1$ state,$r = a_0 \times \frac{1^2}{1} = a_0$.
We need to find a species where $\frac{n^2}{Z} = 1$,which implies $n^2 = Z$.
For $Be^{3+}$ $(Z = 4)$,$n^2 = 4$,so $n = 2$.
Thus,for $Be^{3+}$ with $n = 2$,the radius is $a_0 \times \frac{2^2}{4} = a_0$.

(B) is true because,according to Bohr's model,electrons revolve in stationary orbits with fixed energy,preventing them from falling into the nucleus.
$R$ is true because electrons in these stationary states are indeed referred to as orbital electrons.
However,the fact that it is an orbital electron is not the direct physical explanation for why it does not fall into the nucleus (the explanation lies in the quantization of angular momentum and stationary states).
Therefore,both are true,but $R$ is not the correct explanation of $A$.

(D) The total number of spectral lines emitted when an electron transitions from $n_2$ to $n_1$ is given by the formula: $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Given $n_2 = 6$ and $n_1 = 2$,the total number of lines is $N = \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 \times 5}{2} = 10$.
Since the transition ends at $n_1 = 2$,all these lines belong to the Balmer series.
The question asks for the number of spectral lines excluding the Balmer series.
Therefore,the number of lines excluding the Balmer series is $10 - 10 = 0$.

(C) The energy of $n$ photons is given by the formula: $E = \frac{nhc}{\lambda}$.
Rearranging for $n$: $n = \frac{E \times \lambda}{hc}$.
Given: $E = 1 \, J$,$\lambda = 5000 \, \mathring{A} = 5000 \times 10^{-10} \, m$,$h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m/s$.
Substituting the values: $n = \frac{1 \times 5000 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^8}$.
$n = \frac{5 \times 10^{-7}}{1.9878 \times 10^{-25}} \approx 2.515 \times 10^{18}$.
Thus,the correct option is $C$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 \times 10^{-18} \ J \times \frac{Z^2}{n^2}$.
For $He^+$,$Z = 2$ and $n = 1$,so $E_{1, He^+} = E_{1, H} \times Z^2$.
Given $E_{1, He^+} = -871.6 \times 10^{-20} \ J$.
Therefore,$-871.6 \times 10^{-20} = E_{1, H} \times (2)^2$.
$-871.6 \times 10^{-20} = E_{1, H} \times 4$.
$E_{1, H} = \frac{-871.6 \times 10^{-20}}{4} = -217.9 \times 10^{-20} \ J$.

(B) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula $V_n = V_0 \times \frac{Z}{n}$.
Since $V_0$ and $Z$ are constant for a given species,the velocity is inversely proportional to the principal quantum number $n$ $(V \propto \frac{1}{n})$.
Therefore,the ratio of velocities for the first,second,and third orbits is $V_1 : V_2 : V_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For a hydrogen atom $(Z=1)$ in the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
The binding energy is the energy required to remove the electron,which is $|E_1| = 13.6 \ eV$.
For a $He^+$ ion,the atomic number $Z=2$.
In the ground state $(n=1)$,the energy is $E_1 = -13.6 \times \frac{2^2}{1^2} = -13.6 \times 4 = -54.4 \ eV$.
The energy required to remove the electron (binding energy) is $|-54.4 \ eV| = 54.4 \ eV$.
Thus,the correct option is $D$.

(A) For a hydrogen-like atom,the energy of the $n^{th}$ orbit is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
As $n$ increases,$E_n$ becomes less negative,meaning $E_1 < E_2 < E_3$.
Therefore,the statement $E_1 > E_2 > E_3$ is incorrect.
Potential energy $(PE)$ is given by $PE_n = -27.2 \times \frac{Z^2}{n^2} \text{ eV}$,so $PE_1 < PE_2 < PE_3$ is correct.
Kinetic energy $(KE)$ is given by $KE_n = 13.6 \times \frac{Z^2}{n^2} \text{ eV}$,so $KE_1 > KE_2 > KE_3$ is correct.

(B) 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 $H$ $(Z=1, n=1)$: $E_1 = -13.6 \times \frac{1^2}{1^2} = -13.6 \text{ eV}$.
For the $n=2$ state of $He^+$ $(Z=2, n=2)$: $E_2 = -13.6 \times \frac{2^2}{2^2} = -13.6 \text{ eV}$.
For the $n=3$ state of $Li^{2+}$ $(Z=3, n=3)$: $E_3 = -13.6 \times \frac{3^2}{3^2} = -13.6 \text{ eV}$.
For the $n=4$ state of $Be^{3+}$ $(Z=4, n=4)$: $E_4 = -13.6 \times \frac{4^2}{4^2} = -13.6 \text{ eV}$.
Thus,$E_1(H) = E_2(He^+) = E_3(Li^{2+}) = E_4(Be^{3+})$.

(B) The ultraviolet region corresponds to the Lyman series $(n_1 = 1)$.
For $n_2$ to $n_1 = 1$ transitions,the number of spectral lines is given by $n_2 - n_1 = 5$.
Therefore,$n_2 - 1 = 5$,which gives $n_2 = 6$.
Now,for transitions from $n_2 = 6$ to the infrared region (Paschen,Brackett,and Pfund series):
$1$. Paschen series $(n_1 = 3)$: Transitions are $6 \to 3, 5 \to 3, 4 \to 3$ ($3$ lines).
$2$. Brackett series $(n_1 = 4)$: Transitions are $6 \to 4, 5 \to 4$ ($2$ lines).
$3$. Pfund series $(n_1 = 5)$: Transition is $6 \to 5$ ($1$ line).
Total number of lines in the infrared region = $3 + 2 + 1 = 6$.

(B) The energy of an electron in a hydrogen atom is given by $E_n = \frac{-1312}{n^2} \ kJ \ mol^{-1}$.
Here,the constant value $1312 \ kJ \ mol^{-1}$ represents the energy required to remove an electron from the ground state $(n=1)$ to infinity,which is defined as the Ionization Energy of the hydrogen atom.
Since the constant in the equation is specific to the hydrogen atom,the Ionization Energy remains constant for the hydrogen atom.

(A) The radius of the $n^{th}$ Bohr orbit for a hydrogen-like atom is given by the formula $r_n = 0.529 \times \frac{n^2}{Z} \,\ \mathring{A}$.
For hydrogen,the atomic number $Z = 1$,so the formula simplifies to $r_n = 0.529 \times n^2 \,\mathring{A}$.
For $n = 1$: $r_1 = 0.529 \times 1^2 = 0.529 \,\mathring{A}$.
For $n = 2$: $r_2 = 0.529 \times 2^2 = 0.529 \times 4 = 2.116 \,\mathring{A}$.
For $n = 3$: $r_3 = 0.529 \times 3^2 = 0.529 \times 9 = 4.761 \,\mathring{A}$.
For $n = 4$: $r_4 = 0.529 \times 4^2 = 0.529 \times 16 = 8.464 \,\mathring{A}$.
Thus,the radii are $0.529 \,\mathring{A}, 2.116 \,\mathring{A}, 4.761 \,\mathring{A}, 8.464 \,\mathring{A}$.

(A) For the limiting line of the Lyman series,$n_1 = 1$ 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 $(Z=1)$,$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$.
The value of Rydberg constant $R \approx 1.097 \times 10^7 \, \text{m}^{-1} = 1.097 \times 10^5 \, \text{cm}^{-1}$.
Frequency $\nu = \frac{c}{\lambda} = c \times R$.
Using $c = 3 \times 10^8 \, \text{m/s} = 3 \times 10^{10} \, \text{cm/s}$:
$\nu = (3 \times 10^{10} \, \text{cm/s}) \times (1.097 \times 10^5 \, \text{cm}^{-1}) \approx 3.29 \times 10^{15} \, \text{sec}^{-1}$.

(D) The angular momentum of an electron in a Bohr orbit is given by the formula $L = \frac{nh}{2\pi}$.
For the $2^{nd}$ Bohr orbit of the $H$ atom,$n = 2$,so $x = \frac{2h}{2\pi} = \frac{h}{\pi}$.
The $1^{st}$ excited state of $Li^{+2}$ corresponds to $n = 2$ (since the ground state is $n = 1$).
Therefore,the angular momentum for $n = 2$ is $L = \frac{2h}{2\pi} = \frac{h}{\pi} = x$.
Thus,the angular momentum remains $x$.

(B) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = a_0 n^2$,where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
Given the ratio of the radii $r_1 : r_2 = 4 : 1$,we have $\frac{n_1^2}{n_2^2} = \frac{4}{1}$,which implies $\frac{n_1}{n_2} = \frac{2}{1}$.
Thus,$n_1 = 2$ and $n_2 = 1$.
The orbit with $n=1$ is designated as $K$ and the orbit with $n=2$ is designated as $L$.
Therefore,the designations are $L$ and $K$.

(C) In an absorption spectrum,transitions occur from the ground state (lowest energy level) to higher energy levels.
Looking at the provided energy level diagram,the transitions $1, 2, 3$ originate from the lowest energy level $A$ to higher levels $X, B, C$ respectively.
Therefore,transitions $1, 2, 3$ can appear in the absorption spectrum.
Transitions $4, 5, 6$ originate from higher energy levels to lower energy levels,which corresponds to emission,not absorption.
Thus,lines $4, 5, 6$ do not appear in the absorption spectrum.

(A) The first quantum mechanical model of the atom was proposed by $Niels \ Bohr$ in $1913$.
This model was based on the quantum theory of radiation,which suggests that electrons revolve in discrete orbits with quantized energy levels.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 / n^2 \, eV$.
For $n=3 \rightarrow n=2$: $\Delta E = E_3 - E_2 = -13.6(1/9 - 1/4) = -13.6(-5/36) \approx 1.89 \, eV$.
For $n=4 \rightarrow n=3$: $\Delta E = E_4 - E_3 = -13.6(1/16 - 1/9) = -13.6(-7/144) \approx 0.66 \, eV$.
For $n=5 \rightarrow n=4$: $\Delta E = E_5 - E_4 = -13.6(1/25 - 1/16) = -13.6(-9/400) \approx 0.31 \, eV$.
Comparing the values,the transition $n=3 \rightarrow n=2$ releases the maximum energy.

(D) The frequency of emitted radiation is given by $\nu = \frac{\Delta E}{h}$.
Emission occurs when an electron jumps from a higher energy level to a lower energy level.
The energy difference $\Delta E$ is proportional to the frequency $\nu$.
For emission,we look for transitions from higher to lower levels ($n=3 \to n=1$ or $n=5 \to n=2$).
The energy difference for $n=5 \to n=2$ is $\Delta E = 13.6 \times Z^2 \times (\frac{1}{2^2} - \frac{1}{5^2}) = 13.6 \times (0.25 - 0.04) = 13.6 \times 0.21 = 2.856 \ eV$.
The energy difference for $n=3 \to n=1$ is $\Delta E = 13.6 \times (1 - \frac{1}{9}) = 13.6 \times 0.888 = 12.08 \ eV$.
Since $2.856 \ eV < 12.08 \ eV$,the transition $n=5 \to n=2$ has the lowest energy change and therefore the lowest frequency.

(C) The energy of the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \, eV$.
For the ground state $(n=1)$,$E_1 = -13.6 \, eV$.
For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \, eV$.
The energy required for excitation is $\Delta E = E_2 - E_1$.
$\Delta E = -3.4 \, eV - (-13.6 \, eV) = 10.2 \, eV$.

(D) The relationship between wavelength $(\lambda)$,speed of light $(c)$,and frequency $(\nu)$ is given by $\lambda = \frac{c}{\nu}$.
Given: $c = 3.0 \times 10^8 \ m \ s^{-1}$ and $\nu = 8 \times 10^{15} \ s^{-1}$.
Substituting the values: $\lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}} = 0.375 \times 10^{-7} \ m = 3.75 \times 10^{-8} \ m$.
To convert meters to nanometers $(nm)$,multiply by $10^9$: $\lambda = 3.75 \times 10^{-8} \times 10^9 \ nm = 37.5 \ nm$.
The approximate value is $4 \times 10^1 \ nm$.

(B) The total number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula: $\frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Here,$n_2 = 6$ and $n_1 = 3$.
Substituting the values: $\frac{(6 - 3)(6 - 3 + 1)}{2} = \frac{3 \times 4}{2} = 6$.
Therefore,the total number of spectral lines is $6$.

(C) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} \ J \ atom^{-1}$.
For $He^+$ $(Z = 2)$,the ionization energy is the energy required to remove the electron from $n = 1$ to $n = \infty$,which is equal to $-E_1$. Given $IE = 19.6 \times 10^{-18} \ J \ atom^{-1}$,so $E_1(He^+) = -19.6 \times 10^{-18} \ J \ atom^{-1}$.
Using the formula $E_n = -R_H \times \frac{Z^2}{n^2}$,we can find the constant $R_H$ for this specific unit system: $19.6 \times 10^{-18} = R_H \times \frac{2^2}{1^2} \implies R_H = 4.9 \times 10^{-18} \ J \ atom^{-1}$.
Now,for $Li^{2+}$ $(Z = 3)$,the energy of the first stationary state $(n = 1)$ is: $E_1(Li^{2+}) = -R_H \times \frac{Z^2}{n^2} = -4.9 \times 10^{-18} \times \frac{3^2}{1^2} = -4.9 \times 10^{-18} \times 9 = -44.1 \times 10^{-18} \ J \ atom^{-1} = -4.41 \times 10^{-17} \ J \ atom^{-1}$.

(C) According to Planck's quantum theory,the energy $E$ emitted per second is given by $E = \frac{nhc}{\lambda}$,where $n$ is the number of photons emitted per second.
The power of the bulb is $150 \ \text{W}$,and $8\%$ of this is emitted as light energy:
$E = \frac{150 \times 8}{100} = 12 \ \text{J s}^{-1}$ (since $1 \ \text{W} = 1 \ \text{J s}^{-1}$).
Given values:
$\lambda = 4500 \ \mathring{A} = 4500 \times 10^{-10} \ \text{m}$
$c = 3 \times 10^8 \ \text{m s}^{-1}$
$h = 6.626 \times 10^{-34} \ \text{J s}$
Substituting these into the equation:
$12 = \frac{n \times (6.626 \times 10^{-34}) \times (3 \times 10^8)}{4500 \times 10^{-10}}$
Solving for $n$:
$n = \frac{12 \times 4500 \times 10^{-10}}{6.626 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{54000 \times 10^{-10}}{19.878 \times 10^{-26}}$
$n \approx 2.716 \times 10^{19} \ \text{s}^{-1} \approx 2.72 \times 10^{19} \ \text{s}^{-1}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the ground state,$n = 1$.
The second excited state corresponds to $n = 3$.
The energy of the electron in the second excited state is $E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \ eV$.
The energy required to ionize the atom from this state is the energy needed to bring the electron to $n = \infty$,which is $E_{\infty} - E_3 = 0 - (-1.51) = 1.51 \ eV$.

(D) According to $Bohr's$ postulate,the angular momentum $(L)$ of an electron in a stationary orbit is quantized and is an integral multiple of $\frac{h}{2 \pi}$.
Mathematically,this is expressed as: $L = mvr = \frac{n h}{2 \pi}$,where $n = 1, 2, 3, .....$ is the principal quantum number,$h$ is $Planck's$ constant,and $m$ is the mass of the electron.

(C) Given the formula: $E_n = -313.6/n^2$
Substituting the value of $E_n$: $-34.84 = -313.6/n^2$
Rearranging for $n^2$: $n^2 = -313.6 / -34.84$
$n^2 = 9$
Taking the square root: $n = 3$
Therefore,the correct option is $C$.

(D) The velocity of an electron in a hydrogen-like species is given by $v = 2.188 \times 10^6 \times \frac{Z}{n} \, m \cdot s^{-1}$.
For $Be^{3+}$,the atomic number $Z = 4$.
The $3^{rd}$ excited state corresponds to $n = 4$ (since ground state is $n=1$,$1^{st}$ excited is $n=2$,$2^{nd}$ is $n=3$,and $3^{rd}$ is $n=4$).
Substituting the values: $v = 2.188 \times 10^6 \times \frac{4}{4} = 2.188 \times 10^6 \, m \cdot s^{-1}$.
Converting to $km \cdot s^{-1}$ by dividing by $1000$ $(10^3)$: $v = 2.188 \times 10^3 \, km \cdot s^{-1}$.

(D) Given energy $E = 2 \ eV = 2 \times 1.602 \times 10^{-19} \ J = 3.204 \times 10^{-19} \ J$.
For wavelength $(\lambda)$: $E = \frac{hc}{\lambda} \implies \lambda = \frac{hc}{E}$.
$\lambda = \frac{6.626 \times 10^{-34} \ J \cdot s \times 3 \times 10^8 \ m \cdot s^{-1}}{3.204 \times 10^{-19} \ J} = 6.204 \times 10^{-7} \ m$.
For frequency $(\nu)$: $\nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \ m \cdot s^{-1}}{6.204 \times 10^{-7} \ m} = 0.4835 \times 10^{15} \ s^{-1} \approx 4.8 \times 10^{14} \ s^{-1}$.

(A) The ionization energy $(I.P.)$ of a $H$ atom is given by $I.P. = E_{\infty} - E_1 = 0 - E_1 = -E_1$.
The energy required to excite an electron from the first orbit $(n=1)$ to the second orbit $(n=2)$ is $\Delta E = E_2 - E_1$.
Since $E_n = E_1 / n^2$,we have $E_2 = E_1 / 4$.
Therefore,$\Delta E = (E_1 / 4) - E_1 = -3/4 E_1$.
Substituting $E_1 = -I.P.$,we get $\Delta E = 3/4 I.P.$

(B) The radius of an orbit in a hydrogen-like species is given by the formula $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the second orbit of hydrogen $(n=2, Z=1)$: $(r_2)_H = 0.529 \times \frac{2^2}{1} = 0.529 \times 4 \ \mathring{A}$.
For the third orbit of $Li^{+2}$ $(n=3, Z=3)$: $(r_3)_{Li^{+2}} = 0.529 \times \frac{3^2}{3} = 0.529 \times 3 \ \mathring{A}$.
The ratio is $\frac{(r_2)_H}{(r_3)_{Li^{+2}}} = \frac{0.529 \times 4}{0.529 \times 3} = \frac{4}{3}$.
Thus,the ratio is $4 : 3$.

(C) The wavelength $(\lambda)$ of an emitted photon is inversely proportional to the energy difference $(\Delta E)$ between the two energy levels,given by the relation $\Delta E = \frac{hc}{\lambda}$.
To obtain the shortest wavelength,the energy difference $(\Delta E)$ must be the maximum.
The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -13.6 \ \text{eV} / n^2$.
The energy difference for a transition from $n_2$ to $n_1$ is $\Delta E = 13.6 \ \text{eV} \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Comparing the given transitions:
$(A)$ $n = 2 \rightarrow n = 1$: $\Delta E \propto (1 - 0.25) = 0.75$
$(B)$ $n = 3 \rightarrow n = 1$: $\Delta E \propto (1 - 0.111) = 0.889$
$(C)$ $n = 4 \rightarrow n = 1$: $\Delta E \propto (1 - 0.0625) = 0.9375$
$(D)$ $n = 4 \rightarrow n = 3$: $\Delta E \propto (0.111 - 0.0625) = 0.0485$
The transition $n = 4 \rightarrow n = 1$ has the largest energy difference,and therefore,it corresponds to the shortest wavelength.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{13.6}{n^2} \ eV$.
For the ground state $(n=1)$,$E_1 = -13.6 \ eV$.
For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \ eV$.
The energy difference required for excitation is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
To find the energy per atom (or per mole basis converted to per atom),we divide by the Avogadro number:
Energy required = $\frac{10.2 \ eV}{6.022 \times 10^{23}} \approx 1.69 \times 10^{-23} \ eV$.

(B) The energy of an electron in the ground state of a hydrogen atom is $E_1 = -13.6 \, eV$.
The ionization energy of the excited state is given as $+0.85 \, eV$,which means the energy of the electron in that excited state is $E_n = -0.85 \, eV$.
When the electron returns from the excited state to the ground state,the energy of the emitted photon is given by the difference in energies:
$\Delta E = E_n - E_1 = -0.85 \, eV - (-13.6 \, eV) = 12.75 \, eV$.

(D) For an electron transitioning from an excited state $n$ to lower energy levels,the total number of possible emission lines is given by the formula: $\frac{n(n - 1)}{2}$.
Given that the highest energy level shown is $n = 4$,we substitute $n = 4$ into the formula:
$\text{Number of lines} = \frac{4(4 - 1)}{2} = \frac{4 \times 3}{2} = 6$.
Therefore,the total number of emission lines is $6$.

(B) The quantum theory was proposed by Max Planck. However,it was Niels Bohr who first applied the quantum theory to explain the structure of the atom and the stability of the hydrogen atom in his model. Therefore,the correct answer is $B$.

(B) In Rutherford's $\alpha$-particle scattering experiment,the nucleus is a very small,dense,and positively charged region at the center of the atom.
Because the nucleus is extremely small compared to the size of the atom,$\alpha$-particles do not pass through the nucleus.
Instead,they are either deflected by the electrostatic repulsion of the positive charge or pass through the empty space surrounding the nucleus.
Therefore,the observation that 'a few $\alpha$-particles pass through the nucleus' is incorrect.

(A) The energy of a photon emitted during an electronic transition is given by $\Delta E = h \nu = \frac{hc}{\lambda}$.
Since $\lambda = \frac{hc}{\Delta E}$,the shortest wavelength corresponds to the transition with the largest energy difference $(\Delta E)$.
Comparing the transitions:
$(A)$ $n_4 \to n_1$: Largest energy gap.
$(B)$ $n_2 \to n_1$: Smaller energy gap.
$(C)$ $n_4 \to n_2$: Smaller energy gap.
$(D)$ $n_3 \to n_1$: Smaller energy gap than $n_4 \to n_1$.
Therefore,the transition $n_4 \to n_1$ involves the maximum energy change,resulting in the shortest wavelength.

(C) The total number of spectral lines emitted when an electron transitions from an excited state $n_2$ to a lower state $n_1$ is given by the formula: $N = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Given that the electron transitions to the ground state $(n_1 = 1)$ and the total number of lines $N = 10$,we have:
$10 = \frac{(n_2 - 1)(n_2 - 1 + 1)}{2}$
$10 = \frac{(n_2 - 1)(n_2)}{2}$
$20 = n_2^2 - n_2$
$n_2^2 - n_2 - 20 = 0$
Solving the quadratic equation: $(n_2 - 5)(n_2 + 4) = 0$.
Since $n_2$ must be positive,$n_2 = 5$.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula $r_n = \frac{n^2 h^2}{4 \pi^2 m Z e^2 k}$,where $m$ is the mass of the electron.
Since both hydrogen and deuterium have the same atomic number $(Z = 1)$ and the same electron mass $(m)$,the radius of the first orbit is independent of the mass of the nucleus.
Therefore,the radius of the first orbit in hydrogen $(r_H)$ and deuterium $(r_D)$ are equal.
The ratio $r_H : r_D$ is $1 : 1$.

(C) The ground state energy of a hydrogen atom is $-13.6 \,eV$. The energy required to excite an electron from the $n=1$ to $n=2$ state is $\Delta E = E_2 - E_1 = -3.4 \,eV - (-13.6 \,eV) = 10.2 \,eV$.
Since the incident electron has $10.8 \,eV$ of energy,the $H$ atom absorbs $10.2 \,eV$ to undergo excitation,and the remaining energy $(10.8 \,eV - 10.2 \,eV = 0.6 \,eV)$ is retained by the electron as kinetic energy.

(D) For a hydrogen-like species,the radius of the $n^{th}$ orbit is given by $r_n = 0.529 \times \frac{n^2}{Z} \mathring{A}$.
The second excited state corresponds to the $n = 3$ energy level.
For $Li^{2+}$,the atomic number $Z = 3$.
Substituting the values: $r_3 = 0.529 \times \frac{3^2}{3} \mathring{A} = 0.529 \times 3 \mathring{A} = 1.587 \mathring{A}$.

(A) The visible region in the hydrogen spectrum corresponds to the Balmer series,where transitions occur to the $n = 2$ energy level.
Since the electron is transitioning from the $6^{th}$ orbit to the $3^{rd}$ orbit,it does not reach the $n = 2$ level.
Therefore,the number of spectral lines in the visible region is $0$.

(B) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $H$ atom $(Z=1)$,the energy difference $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = 13.6 \times 1^2 \times (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times \frac{3}{4} = 10.2 \ eV$.
For $Be^{3+}$ ion,the atomic number $Z = 4$.
The energy difference for the same transition ($n=1$ to $n=2$) is $\Delta E' = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
$\Delta E' = 13.6 \times 4^2 \times (\frac{1}{1^2} - \frac{1}{2^2}) = 13.6 \times 16 \times \frac{3}{4} = 13.6 \times 4 \times 3 = 163.2 \ eV$.

(C) The frequency $\nu$ is given by $\nu = \frac{c}{\lambda}$.
For the minimum frequency,the wavelength $\lambda$ must be maximum.
For the Lyman series $(n_1 = 1)$,the maximum wavelength corresponds to the transition from $n_2 = 2$ to $n_1 = 1$: $\frac{1}{\lambda_{L, \text{max}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( \frac{3}{4} \right) \implies \lambda_{L, \text{max}} = \frac{4}{3R}$.
For the Balmer series $(n_1 = 2)$,the maximum wavelength corresponds to the transition from $n_2 = 3$ to $n_1 = 2$: $\frac{1}{\lambda_{B, \text{max}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) \implies \lambda_{B, \text{max}} = \frac{36}{5R}$.
The ratio of minimum frequencies is $\frac{\nu_{L, \text{min}}}{\nu_{B, \text{min}}} = \frac{c / \lambda_{L, \text{max}}}{c / \lambda_{B, \text{max}}} = \frac{\lambda_{B, \text{max}}}{\lambda_{L, \text{max}}} = \frac{36 / 5R}{4 / 3R} = \frac{36}{5} \times \frac{3}{4} = \frac{27}{5} = 5.4$.

(D) The energy of an electron in the $n^{th}$ Bohr orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For the first Bohr orbit of $He^{+}$,we have $n = 1$ and the atomic number $Z = 2$.
Substituting these values into the formula: $E_1 = -13.6 \times \frac{2^2}{1^2} \ eV = -13.6 \times 4 \ eV = -54.4 \ eV$.