Atomic models and Planck's quantum theory Questions in English

(C) Rutherford's atomic model,based on his alpha-particle scattering experiment,proposed that the positive charge and most of the mass are concentrated in a small central region called the nucleus. He suggested that electrons revolve around the nucleus in circular paths and are held by electrostatic forces. However,Rutherford did not mention the presence of neutrons in the nucleus,as the neutron was discovered by James Chadwick in $1932$,long after Rutherford's model was proposed in $1911$.

(C) Niels Bohr proposed his atomic model in $1913$. He obtained detailed information about the structure of atoms by studying the similarity between the solar system and the central atom,where electrons revolve around the nucleus in fixed orbits,similar to planets revolving around the Sun.

(B) Bohr's atomic model was specifically developed to explain the line spectrum of the $H$ atom and other hydrogen-like species (ions containing only one electron,such as $He^+$,$Li^{2+}$,etc.).
It fails to explain the spectra of multi-electron atoms.

(B) The ground state of a hydrogen atom corresponds to $n = 1$.
Excited states are defined as $n > 1$.
For the third orbit $(n = 3)$,the excited state is calculated as $(n - 1) = 3 - 1 = 2$.
Therefore,the third orbit represents the $2^{nd}$ excited state.

(D) Bohr's atomic model is applicable only to hydrogen-like species,which are ions containing only one electron.
$He^+$ has $2-1 = 1$ electron.
$Li^{2+}$ has $3-1 = 2$ electrons (Wait,$Li^{2+}$ has $3-1=2$ is incorrect,$Li$ has $3$ electrons,so $Li^{2+}$ has $3-2=1$ electron).
$Be^{3+}$ has $4-3 = 1$ electron.
Since all the given species ($He^+$,$Li^{2+}$,$Be^{3+}$) contain only one electron,Bohr's model is applicable to all of them.

(D) According to Bohr's postulate,the angular momentum $(L)$ of an electron in the $n^{th}$ orbit is given by the formula: $L = \frac{nh}{2\pi}$.
For the $7^{th}$ orbit,$n = 7$.
Substituting the value of $n$ into the formula:
$L = \frac{7h}{2\pi} = 3.5 \times \frac{h}{\pi}$.

(D) The energy of an electron in the $n^{th}$ orbit of an $H$ atom is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $H$ atom,$Z = 1$,so $E_n = -13.6 / n^2 \ eV$.
Given $E_n = -3.4 \ eV$,we have $-3.4 = -13.6 / n^2$,which implies $n^2 = 4$,so $n = 2$.
The angular momentum $L$ is given by Bohr's postulate: $L = \frac{nh}{2\pi}$.
Substituting $n = 2$ and $h = 6.626 \times 10^{-34} \ J \cdot s$:
$L = \frac{2 \times 6.626 \times 10^{-34}}{2 \times 3.14} \approx 2.1 \times 10^{-34} \ J \cdot s$.

(C) According to Bohr's postulate,angular momentum $L = \frac{nh}{2\pi}$.
Given $L = 3.164 \times 10^{-34} \ kg \ m^2 \ s^{-1}$ and $h = 6.626 \times 10^{-34} \ J \ s$.
$n = \frac{L \times 2\pi}{h} = \frac{3.164 \times 10^{-34} \times 2 \times 3.1416}{6.626 \times 10^{-34}} \approx 3$.
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 $H$ atom,$Z = 1$ and $n = 3$.
$E_3 = -13.6 \times \frac{1^2}{3^2} = -13.6 \times \frac{1}{9} \approx -1.51 \ eV$.

(A) The energy difference for an electron transition in a hydrogen-like atom is given by the formula: $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \ eV$.
For a hydrogen atom,$Z = 1$,$n_1 = 1$,and $n_2 = 3$.
Substituting these values:
$\Delta E = 13.6 \times 1^2 \times (\frac{1}{1^2} - \frac{1}{3^2}) \ eV$.
$\Delta E = 13.6 \times (1 - \frac{1}{9}) \ eV$.
$\Delta E = 13.6 \times \frac{8}{9} \ eV$.
$\Delta E = 12.088 \approx 12.1 \ eV$.

(D) The ionization energy for a hydrogen-like species is given by the formula: $E_n = 13.6 \times Z^2 \text{ eV/atom}$.
For a hydrogen atom,$Z = 1$,so $E_H = 13.6 \times 1^2 = E$.
For $Li^{2+}$ ion,the atomic number $Z = 3$.
Therefore,the ionization energy for $Li^{2+}$ is $E_{Li^{2+}} = 13.6 \times 3^2 = 13.6 \times 9 = 9E$.

(B) According to Bohr's model,the kinetic energy $(KE)$ of an electron in a hydrogen-like species is given by the formula: $KE = \frac{1}{2} \frac{kZe^2}{r}$.
Substituting the Coulomb constant $k = \frac{1}{4\pi\varepsilon_0}$ and the atomic number $Z = 4$ for $Be^{3+}$:
$KE = \frac{1}{2} \times \frac{1}{4\pi\varepsilon_0} \times \frac{4e^2}{r}$.
Simplifying the expression:
$KE = \frac{4e^2}{8\pi\varepsilon_0 r} = \frac{e^2}{2\pi\varepsilon_0 r}$.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -R_H \times \frac{Z^2}{n^2}$.
The energy difference between two orbits $n_1$ and $n_2$ is $\Delta E = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the first and second orbits $(n_1=1, n_2=2)$: $\Delta E_1 = R_H \times Z^2 \times (1 - \frac{1}{4}) = R_H \times Z^2 \times \frac{3}{4} = \frac{3B}{4}$.
For the first and third orbits $(n_1=1, n_2=3)$: $\Delta E_2 = R_H \times Z^2 \times (1 - \frac{1}{9}) = R_H \times Z^2 \times \frac{8}{9}$.
Taking the ratio: $\frac{\Delta E_2}{\Delta E_1} = \frac{8/9}{3/4} = \frac{8}{9} \times \frac{4}{3} = \frac{32}{27}$.
Therefore,$\Delta E_2 = \frac{32}{27} \times \Delta E_1 = \frac{32}{27} \times \frac{3B}{4}$.

(B) The energy difference $\Delta E$ for an electron transition from the ground state $(n_1 = 1)$ to an excited state $(n_2 = n)$ is given by the Rydberg formula for energy:
$\Delta E = E_n - E_1 = Rhc \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
Substituting $n_1 = 1$ and $n_2 = n$:
$\Delta E = Rhc \left( \frac{1}{1^2} - \frac{1}{n^2} \right)$
$\Delta E = Rhc (1 - n^{-2})$

(C) The energy change for a transition in a hydrogen atom is given by the formula: $\Delta E = 2.18 \times 10^{-18} \ J \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Here,$n_1 = 2$ and $n_2 = 3$.
$\Delta E = 2.18 \times 10^{-18} \times (\frac{1}{2^2} - \frac{1}{3^2})$
$\Delta E = 2.18 \times 10^{-18} \times (\frac{1}{4} - \frac{1}{9})$
$\Delta E = 2.18 \times 10^{-18} \times (\frac{9-4}{36})$
$\Delta E = 2.18 \times 10^{-18} \times \frac{5}{36}$
$\Delta E = 0.3027 \times 10^{-18} \ J \ atom^{-1}$
$\Delta E = 3.03 \times 10^{-19} \ J \ atom^{-1}$.

(B) Given: $\lambda = 242 \, nm = 242 \times 10^{-9} \, m$.
The ionization energy $(IE)$ is given by the formula: $IE = \frac{N_A hc}{\lambda}$.
Substituting the values: $N_A = 6.022 \times 10^{23} \, mol^{-1}$,$h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^8 \, m \cdot s^{-1}$.
$IE = \frac{6.022 \times 10^{23} \times 6.626 \times 10^{-34} \times 3 \times 10^8}{242 \times 10^{-9}} \, J \cdot mol^{-1}$.
$IE \approx \frac{11.97 \times 10^{-2}}{242 \times 10^{-9}} \approx 4.945 \times 10^5 \, J \cdot mol^{-1}$.
Converting to $kJ \cdot mol^{-1}$: $IE = 494.5 \, kJ \cdot mol^{-1}$.

(C) For the Balmer series,the Rydberg formula is $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 2$.
For the first line,$n_2 = 3$: $\frac{1}{\lambda_1} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{9} \right) = R_H \left( \frac{5}{36} \right)$.
For the second line,$n_2 = 4$: $\frac{1}{\lambda_2} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R_H \left( \frac{1}{4} - \frac{1}{16} \right) = R_H \left( \frac{3}{16} \right)$.
Taking the ratio: $\frac{\lambda_2}{\lambda_1} = \frac{5/36}{3/16} = \frac{5}{36} \times \frac{16}{3} = \frac{80}{108} = \frac{20}{27}$.
Therefore,$\lambda_2 = \lambda_1 \times \frac{20}{27} = 656.1 \times \frac{20}{27} = 486 \ nm$.

(B) The Rydberg formula for the wavelength of a transition is given by $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For $He^{+}$ ion $(Z = 2)$,the transition is from $n_2 = 4$ to $n_1 = 2$:
$\frac{1}{\lambda} = R (2)^2 (\frac{1}{2^2} - \frac{1}{4^2}) = R (4) (\frac{1}{4} - \frac{1}{16}) = R (4) (\frac{3}{16}) = R (\frac{3}{4})$.
For $H$ atom $(Z = 1)$,we want the same wavelength $\lambda$ for a transition from $n_2$ to $n_1$:
$\frac{1}{\lambda} = R (1)^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) = R (\frac{3}{4})$.
Comparing the two,we get $(\frac{1}{n_1^2} - \frac{1}{n_2^2}) = \frac{3}{4} = (1 - \frac{1}{4}) = (\frac{1}{1^2} - \frac{1}{2^2})$.
Thus,the transition is from $n_2 = 2$ to $n_1 = 1$.

(A) For the Lyman series,the transition occurs to the ground state,so $n_1 = 1$. The electron transitions from the $4^{th}$ orbit,so $n_2 = 4$.
Using the Rydberg formula: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$
Given $R_H = 1.097 \times 10^7 \ m^{-1}$ and $Z = 1$ for hydrogen:
$\frac{1}{\lambda} = 1.097 \times 10^7 \times 1^2 \times (\frac{1}{1^2} - \frac{1}{4^2})$
$\frac{1}{\lambda} = 1.097 \times 10^7 \times (1 - \frac{1}{16})$
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \frac{15}{16} \ m^{-1}$
$\lambda = \frac{16}{1.097 \times 10^7 \times 15} \ m$
$\lambda \approx 9.7 \times 10^{-8} \ m$

(B) The Rydberg formula for wavelength is given by $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the highest energy transition in $H$ atom $(n_1=1, n_2=\infty)$,$\lambda_H = 91.2 \ nm$.
For $He^{+}$,$Z=2$. The transition corresponds to the same energy levels $(n_1=1, n_2=\infty)$.
Thus,$\lambda_{He^+} = \frac{\lambda_H}{Z^2} = \frac{91.2}{2^2} = \frac{91.2}{4} = 22.8 \ nm$.

(B) 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 limiting line of the Lyman series,$n_1 = 1$ and $n_2 = \infty$,so $\frac{1}{\lambda} = R Z^2$.
Let $\lambda_H$ be the wavelength for the hydrogen atom $(Z_H = 1)$ and $\lambda_X$ be the wavelength for the ion $X$ $(Z_X = Z)$.
Given $\lambda_H = 16 \lambda_X$,we have $\frac{1}{\lambda_H} = \frac{1}{16} \frac{1}{\lambda_X}$.
Substituting the Rydberg formula: $R(1)^2 = \frac{1}{16} R(Z)^2$.
$1 = \frac{Z^2}{16} \implies Z^2 = 16 \implies Z = 4$.
The ion with atomic number $Z = 4$ is $Be^{3+}$.

(B) The frequency of emitted radiation is given by the Rydberg formula: $\nu = R_H \times c \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For a hydrogen atom,$Z = 1$,$n_1 = 1$ (ground state),and $n_2 = 3$.
The Rydberg constant in terms of frequency is $R_H \times c \approx 3.29 \times 10^{15} \ s^{-1}$.
Substituting the values: $\nu = 3.29 \times 10^{15} \times (\frac{1}{1^2} - \frac{1}{3^2}) \times 1^2$.
$\nu = 3.29 \times 10^{15} \times (1 - \frac{1}{9}) = 3.29 \times 10^{15} \times \frac{8}{9}$.
$\nu = 2.924 \times 10^{15} \ s^{-1}$.

(B) The infrared region of the hydrogen spectrum corresponds to the Paschen series,Brackett series,and Pfund series. The first emission line in the infrared region is the first line of the Paschen series.
For the Paschen series,$n_1 = 3$ and for the first line,$n_2 = 4$.
The Rydberg formula is given by: $\bar{\nu} = R \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For hydrogen,$Z = 1$,so $\bar{\nu} = R \times (\frac{1}{3^2} - \frac{1}{4^2})$.
$\bar{\nu} = R \times (\frac{1}{9} - \frac{1}{16}) = R \times (\frac{16 - 9}{144}) = \frac{7R}{144} \ cm^{-1}$.

(A) For the maximum energy in the Balmer series,the transition occurs from $n_2 = \infty$ to $n_1 = 2$.
Using the Rydberg formula: $\bar{\nu} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \times Z^2$.
Substituting the values for Hydrogen $(Z = 1)$: $\bar{\nu} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \times 1^2$.
$\bar{\nu} = R \left( \frac{1}{4} - 0 \right) = \frac{R}{4}$.

(B) The radius of an orbit in a hydrogen-like species is given by the formula: $r = r_0 \times \frac{n^2}{Z}$.
For the $2^{nd}$ orbit of $H$ $(n_1 = 2, Z_1 = 1)$: $r_1 = r_0 \times \frac{2^2}{1} = 4r_0$.
For the $3^{rd}$ orbit of $He^{+}$ $(n_2 = 3, Z_2 = 2)$: $r_2 = r_0 \times \frac{3^2}{2} = 4.5r_0$.
The ratio is $\frac{r_1}{r_2} = \frac{4r_0}{4.5r_0} = \frac{4}{4.5} = \frac{8}{9}$.

(C) For an electron in a hydrogen-like atom,the electrostatic force provides the necessary centripetal force: $\frac{mv^2}{r} = \frac{kZe^2}{r^2}$.
Simplifying this,we get $mv^2 = \frac{kZe^2}{r}$.
Therefore,$v^2 = \frac{kZe^2}{mr}$.
Taking the square root,we get $v = \sqrt{\frac{kZe^2}{mr}}$.

(A) According to Bohr's postulate,the angular momentum of an electron in the $n^{th}$ orbit is given by:
$mvr = \frac{nh}{2\pi}$
Rearranging the formula to solve for $n$:
$n = \frac{2\pi rmv}{h}$

(C) The radius of an orbit in a hydrogen-like species is given by the formula $r = r_0 \times \frac{n^2}{Z}$.
For the second orbit,$n = 2$.
For $H$ $(Z=1)$,$He^{+}$ $(Z=2)$,and $Li^{2+}$ $(Z=3)$:
$r_H : r_{He^+} : r_{Li^{2+}} = \frac{2^2}{1} : \frac{2^2}{2} : \frac{2^2}{3}$
$= 4 : 2 : \frac{4}{3}$
Multiplying by $3$,we get $12 : 6 : 4$,which simplifies to $6 : 3 : 2$.

(B) According to Bohr's theory,the velocity of an electron in the $n^{th}$ orbit of a hydrogen-like atom is given by $v = \frac{2\pi Z e^2}{nh}$.
For a hydrogen atom,$Z = 1$,so the expression becomes $v = \frac{2\pi e^2}{nh}$.
Rearranging the formula to solve for $n$:
$n = \frac{2\pi e^2}{vh}$.

(A) The velocity of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $v = 2.18 \times 10^6 \times \frac{Z}{n} \, m \, s^{-1}$.
Given velocity $v = 0.0109 \times 10^{10} \, cm \, s^{-1}$.
Converting to $m \, s^{-1}$: $v = 0.0109 \times 10^{10} \times 10^{-2} \, m \, s^{-1} = 1.09 \times 10^6 \, m \, s^{-1}$.
For a hydrogen atom,$Z = 1$.
Substituting the values: $1.09 \times 10^6 = 2.18 \times 10^6 \times \frac{1}{n}$.
Therefore,$n = \frac{2.18 \times 10^6}{1.09 \times 10^6} = 2$.

(B) The velocity of an electron in a hydrogen-like species is given by $v = 2.18 \times 10^6 \times \frac{Z}{n} \ m/s$.
For the ground state of hydrogen $(H)$: $Z_1 = 1$,$n_1 = 1$.
For the second excited state of $He^{+}$: $Z_2 = 2$,$n_2 = 3$.
The ratio is $\frac{v_H}{v_{He^{+}}} = \frac{Z_1 / n_1}{Z_2 / n_2} = \frac{1 / 1}{2 / 3} = \frac{3}{2} = 1.5$.

(B) The time period of revolution $T$ of an electron in a Bohr orbit is given by the formula $T \propto \frac{n^3}{Z^2}$.
For the hydrogen atom $(H)$: $n_1 = 2$,$Z_1 = 1$. Thus,$T_H \propto \frac{2^3}{1^2} = 8$.
For the $He^{+}$ ion: $n_2 = 3$,$Z_2 = 2$. Thus,$T_{He^+} \propto \frac{3^3}{2^2} = \frac{27}{4}$.
The ratio $\frac{T_H}{T_{He^+}} = \frac{8}{27/4} = \frac{8 \times 4}{27} = \frac{32}{27}$.

(B) Given: $n_2 = 8$ and $n_1 = 1$.
The number of spectral lines emitted during the transition of an electron from $n_2$ to $n_1$ is given by the formula:
$\text{Number of spectral lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$
Substituting the values:
$\text{Number of spectral lines} = \frac{(8 - 1)(8 - 1 + 1)}{2} = \frac{7 \times 8}{2} = 28$.

(C) The number of spectral lines emitted when an electron transitions from orbit $n_2$ to $n_1$ is given by the formula: $\text{Number of lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Given $n_1 = 2$ and the number of lines $= 15$,we substitute these values:
$15 = \frac{(n - 2)(n - 2 + 1)}{2}$.
$30 = (n - 2)(n - 1)$.
$30 = n^2 - 3n + 2$.
$n^2 - 3n - 28 = 0$.
Factoring the quadratic equation: $(n - 7)(n + 4) = 0$.
Since $n$ must be a positive integer,$n = 7$.

(C) The number of spectral lines emitted during a transition from an excited state $n_2$ to a lower state $n_1$ is given by the formula: $\text{Number of lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Given $n_2 = 7$ and $n_1 = 2$.
Substituting the values: $\text{Number of lines} = \frac{(7 - 2)(7 - 2 + 1)}{2} = \frac{5 \times 6}{2} = 15$.

(C) According to the Bohr model,the circumference of the $n^{th}$ orbit is equal to an integral multiple of the de Broglie wavelength: $2\pi r = n\lambda$.
Also,the number of waves formed by an electron in an orbit is equal to the principal quantum number $n$.
For the $5^{th}$ orbit,$n = 5$.
Therefore,the number of waves formed is $5$.

(B) For a single-electron species like the $H$ atom,the energy of an orbital depends only on the principal quantum number $(n)$.
Since all orbitals with the same $n$ value have the same energy,they are degenerate.
For $n = 3$,the orbitals are $3s, 3p, 3d$,and they all have the same energy.
For $n = 4$,the orbitals are $4s, 4p, 4d, 4f$,and they all have the same energy.
Since $n = 3 < n = 4$,the energy of all $n = 3$ orbitals is less than the energy of all $n = 4$ orbitals.
Therefore,the correct order is $3s = 3p = 3d < 4s = 4p = 4d = 4f$.

(D) 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 $H^{-}$ ion,which is a two-electron system,the effective nuclear charge is not simply $Z=1$. However,in the context of standard textbook problems of this type,it is assumed to follow the Bohr model logic for hydrogen-like species.
Given for $H^{-}$ $(Z=1)$ in the second quantum state $(n=2)$: $E_{H^-} = -k \times \frac{1^2}{2^2} = -\frac{k}{4} = -E_2$,which implies $k = 4E_2$.
For the $He^{+}$ ion $(Z=2)$ in the third quantum state $(n=3)$: $E_{He^+} = -k \times \frac{2^2}{3^2} = -k \times \frac{4}{9}$.
Substituting $k = 4E_2$: $E_{He^+} = -(4E_2) \times \frac{4}{9} = -\frac{16}{9} E_2$.

(D) In Bohr's model,the time period of revolution $T$ is given by the relation $T \propto \frac{n^3}{Z^2}$.
For $T_{1,2}$,$n = 1$ and $Z = 2$,so $T_{1,2} \propto \frac{1^3}{2^2} = \frac{1}{4}$.
For $T_{2,1}$,$n = 2$ and $Z = 1$,so $T_{2,1} \propto \frac{2^3}{1^2} = \frac{8}{1} = 8$.
Therefore,the ratio $T_{1,2} : T_{2,1} = \frac{1}{4} : 8 = 1 : 32$.

(C) The Rydberg formula for a hydrogen-like ion is given by: $1 / \lambda = R Z^{2} [ 1 / n_{1}^{2} - 1 / n_{2}^{2} ]$
For the Paschen series,the transition ends at $n_{1} = 3$.
For the shortest wavelength,the transition starts from $n_{2} = \infty$.
For $Li^{2+}$ ion,the atomic number $Z = 3$.
Substituting these values: $1 / \lambda = R \times 3^{2} [ 1 / 3^{2} - 1 / \infty^{2} ]$
$1 / \lambda = R \times 9 [ 1 / 9 - 0 ]$
$1 / \lambda = R \times 9 \times (1 / 9) = R$
Therefore,$\lambda = 1 / R$.

(B) The spectral lines of the hydrogen spectrum are classified based on the lower energy level $(n_1)$ to which the electron transitions:
$n_1 = 1$: Lyman series
$n_1 = 2$: Balmer series
$n_1 = 3$: Paschen series
$n_1 = 4$: Brackett series
$n_1 = 5$: Pfund series
$n_1 = 6$: Humphrey series
For the Humphrey series,the transition must end at $n_1 = 6$. Therefore,the transition $n_2 = 10 \to n_1 = 6$ is correct.

(D) 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 a transition $n_2 \to n_1$,the energy emitted is proportional to $(\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Comparing the options:
$(A)$ $\infty \to 1$: $\Delta E \propto (1 - 0) = 1$
$(B)$ $2 \to 1$: $\Delta E \propto (1 - 0.25) = 0.75$
$(C)$ $3 \to 2$: $\Delta E \propto (0.25 - 0.111) = 0.139$
$(D)$ $n \to (n - 1)$ for $n \geq 4$: As $n$ increases,the energy gap between consecutive levels decreases. For example,if $n=4$,$4 \to 3$ gives $\Delta E \propto (0.111 - 0.0625) = 0.0485$.
Since the energy gap between consecutive energy levels decreases as the principal quantum number $n$ increases,the transition $n \to (n - 1)$ for large $n$ (where $n \geq 4$) results in the minimum energy emission compared to the other specific transitions listed.

(C) According to Bohr's model,the radius of the $n^{th}$ orbit is given by the formula $r_n = a_0 \frac{n^2}{Z}$,where $a_0$ is the Bohr radius,$n$ is the principal quantum number,and $Z$ is the atomic number.
For a given atom,$Z$ is constant,so $r_n \propto n^2$.
For the $3^{rd}$ orbit $(n=3)$ and the $1^{st}$ orbit $(n=1)$:
$\frac{r_3}{r_1} = \frac{3^2}{1^2} = \frac{9}{1}$.
Therefore,$r_3 = 9 r_1$.

(C) The $5^{th}$ excited state corresponds to the principal quantum number $n = 6$ (since ground state is $n = 1$).
For a sample containing only one hydrogen atom,the electron can only undergo transitions from $n = 6$ to $n = 1$ in a single path,but the question asks for the maximum number of spectral lines possible for a single atom transitioning from $n = 6$ to $n = 1$.
However,in a single atom,the electron can only occupy one energy level at a time and can only emit one photon per transition. If the electron transitions from $n = 6$ to $n = 1$ in a single atom,the maximum number of spectral lines it can emit is equal to the number of steps it takes to reach the ground state,which is $n - 1 = 6 - 1 = 5$ lines.
Thus,the maximum number of spectral lines for a single atom is $n - 1 = 5$.

(D) For a hydrogen-like species,the Rydberg formula is given by: $\frac{1}{\lambda} = R Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For the Lyman series,$n_1 = 1$. For $Li^{2+}$,$Z = 3$.
First line of Lyman series ($n_1 = 1$ to $n_2 = 2$): $\frac{1}{\lambda_1} = R (3)^2 (\frac{1}{1^2} - \frac{1}{2^2}) = 9R (1 - \frac{1}{4}) = 9R (\frac{3}{4}) = \frac{27R}{4}$. So,$\lambda_1 = \frac{4}{27R}$.
Second line of Lyman series ($n_1 = 1$ to $n_2 = 3$): $\frac{1}{\lambda_2} = R (3)^2 (\frac{1}{1^2} - \frac{1}{3^2}) = 9R (1 - \frac{1}{9}) = 9R (\frac{8}{9}) = 8R$. So,$\lambda_2 = \frac{1}{8R}$.
Given $R \approx 1.097 \times 10^7 \ m^{-1} = 1.097 \times 10^{-2} \ \mathring{A}^{-1}$.
$\lambda_1 = \frac{4}{27 \times 1.097 \times 10^{-2}} \approx 135.05 \ \mathring{A}$.
$\lambda_2 = \frac{1}{8 \times 1.097 \times 10^{-2}} \approx 113.95 \ \mathring{A}$.
Difference $\Delta \lambda = \lambda_1 - \lambda_2 = 135.05 - 113.95 = 21.1 \ \mathring{A}$.

(A) For an electron in a hydrogen-like atom,the relationship between kinetic energy $(KE)$ and potential energy $(PE)$ is given by $PE = -2 \times KE$.
Initial state: $KE_{1} = x$,so $PE_{1} = -2x$.
Final state: $KE_{2} = \frac{x}{4}$,so $PE_{2} = -2 \times (\frac{x}{4}) = -\frac{x}{2}$.
The change in potential energy is $\Delta PE = PE_{2} - PE_{1}$.
$\Delta PE = -\frac{x}{2} - (-2x) = -\frac{x}{2} + 2x = +\frac{3x}{2}$.

(B) For a hydrogen atom,$Z = 1$. The Rydberg formula is $\frac{1}{\lambda} = R_H \times (1)^2 \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $n_1 = 2$ and $n_2 = 4$,and $R_H = 1.097 \times 10^7 \ m^{-1}$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 1.097 \times 10^7 \times \left( \frac{1}{4} - \frac{1}{16} \right)$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \times \left( \frac{3}{16} \right) = 2.0568 \times 10^6 \ m^{-1}$.
$\lambda = \frac{1}{2.0568 \times 10^6} \approx 4.86 \times 10^{-7} \ m = 486 \ nm$.

(C) 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 principal quantum number,and $Z$ is the atomic number.
For a hydrogen atom with $n = 1$ and $Z = 1$,the radius is $r_H = a_0 \times \frac{1^2}{1} = a_0$.
Now,checking the options:
For $Be^{3+}$,$n = 2$ and $Z = 4$,so $r = a_0 \times \frac{2^2}{4} = a_0 \times \frac{4}{4} = a_0$.
Thus,the radius of $Be^{3+}$ with $n = 2$ is equal to the radius of a hydrogen atom with $n = 1$.

(D) According to Bohr's theory,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{Z^2 R_H}{n^2}$,which implies $E_n \propto -\frac{1}{n^2}$.
Given that the energy in the $3^{rd}$ orbit is $E_3 = -E$.
We know $E_n = \frac{k}{n^2}$ where $k$ is a constant.
For $n=3$,$E_3 = \frac{k}{3^2} = \frac{k}{9} = -E$,so $k = -9E$.
For the $1^{st}$ orbit $(n=1)$,$E_1 = \frac{k}{1^2} = k$.
Substituting the value of $k$,we get $E_1 = -9E$.
Therefore,the correct option is $D$.

(D) The wave number $\bar{\nu}$ for a hydrogen-like species 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 $He^{+}$ $(Z=2)$,the wave number is $y = R_H (2)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) ......(i)$.
For $Li^{2+}$ $(Z=3)$,the wave number is $\bar{\nu}' = R_H (3)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = 9 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) ......(ii)$.
Dividing equation $(ii)$ by equation $(i)$:
$\frac{\bar{\nu}'}{y} = \frac{9 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)}{4 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)} = \frac{9}{4}$.
Therefore,$\bar{\nu}' = \frac{9y}{4} \ cm^{-1}$.