Atomic models and Planck's quantum theory Questions in English

(C) The Rydberg formula for the hydrogen atom is given by: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$
Here,$n_1 = 1$ (stationary state) and $n_2 = \infty$ (infinity).
Substituting the values: $\frac{1}{\lambda} = 1.097 \times 10^7 \ m^{-1} \left[ \frac{1}{1^2} - \frac{1}{\infty^2} \right]$
Since $\frac{1}{\infty} = 0$,we have: $\frac{1}{\lambda} = 1.097 \times 10^7 \ m^{-1} \times 1$
$\lambda = \frac{1}{1.097 \times 10^7} \ m \approx 9.115 \times 10^{-8} \ m$
Converting to nanometers $(nm)$: $\lambda = 9.115 \times 10^{-8} \times 10^9 \ nm = 91.15 \ nm \approx 91 \ nm$.

(D) According to Bohr's model,the radius of the $n^{th}$ orbit is given by the relation $r_n \propto n^2$.
For the $1^{st}$ orbit $(n=1)$,the radius is given as $\gamma$.
For the $3^{rd}$ orbit $(n=3)$,the radius $r_3$ will be proportional to $3^2 = 9$.
Therefore,$r_3 = 9 \times r_1 = 9\gamma$.

(B) According to Bohr's model,the radius of the $n^{th}$ orbit is given by the formula:
$r_n = \frac{n^2 h^2}{4 \pi^2 m k Z e^2}$
In this expression,$h$,$\pi$,$m$,$k$,$Z$,and $e$ are constants for a given atom.
Therefore,the radius is directly proportional to the square of the principal quantum number:
$r_n \propto n^2$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula $E_n = \frac{-13.6 \ eV}{n^2}$.
For the energy level $n = 2$:
$E_2 = \frac{-13.6}{(2)^2} \ eV$
$E_2 = \frac{-13.6}{4} \ eV$
$E_2 = -3.4 \ eV$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by the formula: $E_n = \frac{-13.6}{n^2} \ eV$.
Given that the ground state energy $(n=1)$ is $-13.6 \ eV$.
For the energy level corresponding to the quantum number $n = 5$:
$E_5 = \frac{-13.6}{5^2} \ eV$
$E_5 = \frac{-13.6}{25} \ eV$
$E_5 = -0.544 \ eV \approx -0.54 \ eV$.

(A) The wavelength $(\lambda)$ of a transition is inversely proportional to the energy difference $(\Delta E)$ between the energy levels,given by the relation $\Delta E = \frac{hc}{\lambda}$.
For a hydrogen-like atom,the energy difference is $\Delta E = 13.6 Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$.
To have the minimum wavelength,the transition must have the maximum energy difference.
Comparing the transitions:
$(A)$ $n_4 \to n_1$: $\Delta E \propto (1 - \frac{1}{16}) = \frac{15}{16} = 0.9375$
$(B)$ $n_2 \to n_1$: $\Delta E \propto (1 - \frac{1}{4}) = \frac{3}{4} = 0.75$
$(C)$ $n_4 \to n_2$: $\Delta E \propto (\frac{1}{4} - \frac{1}{16}) = \frac{3}{16} = 0.1875$
$(D)$ $n_3 \to n_1$: $\Delta E \propto (1 - \frac{1}{9}) = \frac{8}{9} \approx 0.888$
Since the transition $n_4 \to n_1$ has the largest energy difference,it corresponds to the minimum wavelength.

(C) The correct pair is $(C)$.
Emission spectra consist of discrete lines at specific wavelengths,which directly provide evidence for the quantization of energy levels within an atom.
$X$-ray spectra relate to the atomic number $(Z)$,$\alpha$-particle scattering demonstrates the existence of a small,dense,positively charged nucleus,and the photoelectric effect demonstrates the particle nature of light (photons).

(D) As a result of electrostatic attraction between the nucleus and the electron,energy is released when an electron is brought from infinity to an orbital. Therefore,the energy of an electron is defined as $0$ at an infinite distance from the nucleus. As the electron approaches the nucleus,it loses energy,making its energy negative in any bound state. Thus,the maximum energy of an electron is at an infinite distance from the nucleus.

(B) The given statements describe the conditions for filling degenerate orbitals.
Statement $i$ refers to the preference for single occupancy before pairing.
Statement $ii$ refers to the stability gained by having parallel spins in degenerate orbitals.
These are the fundamental tenets of $Hund's$ rule of maximum multiplicity.

(D) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$.
This implies that $E \propto \frac{1}{\lambda}$.
Therefore,the ratio of the energies is $\frac{E_1}{E_2} = \frac{\lambda_2}{\lambda_1}$.
Given $\lambda_1 = 2000 \ \mathring{A}$ and $\lambda_2 = 4000 \ \mathring{A}$.
$\frac{E_1}{E_2} = \frac{4000}{2000} = 2$.

(B) According to Bohr's model,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{2.179 \times 10^{-11}}{n^2} \ erg$.
The energy change for a transition from $n_1 = 1$ to $n_2 = 3$ is $\Delta E = E_3 - E_1$.
$\Delta E = -\frac{2.179 \times 10^{-11}}{3^2} - (-\frac{2.179 \times 10^{-11}}{1^2})$.
$\Delta E = 2.179 \times 10^{-11} \times (1 - \frac{1}{9}) = 2.179 \times 10^{-11} \times \frac{8}{9}$.
$\Delta E = 1.9368 \times 10^{-11} \ erg = 0.19368 \times 10^{-10} \ erg$.
Comparing with the given options,the closest value is $0.1911 \times 10^{-10} \ erg$ (Option $B$).
Since the electron moves from a lower energy level to a higher energy level,energy is absorbed.

(A) 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 ground state,$n = 1$,so $E_1 = -13.6 \ eV$.
Excited states correspond to $n > 1$ (i.e.,$n = 2, 3, 4, \dots$).
For the first excited state,$n = 2$:
$E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \ eV$.
Thus,$-3.4 \ eV$ is a possible energy value for an excited state.

(C) 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 $E_n \propto Z^2$.
For the first orbit $(n=1)$,$E_1 \propto Z^2$.
Given for $He^{+}$ $(Z=2)$: $E_{1, He^{+}} = -871.6 \times 10^{-20} \ J$.
For Hydrogen $(Z=1)$: $E_{1, H} = E_{1, He^{+}} \times \frac{Z_H^2}{Z_{He^{+}}^2}$.
$E_{1, H} = -871.6 \times 10^{-20} \times \frac{1^2}{2^2} = -871.6 \times 10^{-20} \times \frac{1}{4} = -217.9 \times 10^{-20} \ J$.

(D) The radius of the $n^{th}$ Bohr orbit is given by the formula $r_n = r_0 \times n^2$,where $r_0 = 0.530 \ \mathring{A}$ is the radius of the ground state $(n = 1)$.
For the first excited state,$n = 2$.
Substituting the values: $r_2 = 0.530 \ \mathring{A} \times (2)^2$.
$r_2 = 0.530 \ \mathring{A} \times 4 = 2.12 \ \mathring{A}$.

(C) The series limit of a spectral series corresponds to the transition from $n_2 = \infty$ to $n_1 = 2$ for the Balmer series.
Using the Rydberg formula: $\frac{1}{\lambda} = R_H \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For $H$ atom,$Z = 1$,$n_1 = 2$,and $n_2 = \infty$.
$\frac{1}{\lambda} = R_H \times (\frac{1}{2^2} - \frac{1}{\infty^2}) = \frac{R_H}{4}$.
$\lambda = \frac{4}{R_H} = \frac{4}{109677 \ cm^{-1}} \approx 3.646 \times 10^{-5} \ cm = 3646 \ \mathring{A}$.

(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 required for excitation is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.
To convert this energy into Joules,we multiply by the conversion factor $1.602 \times 10^{-19} \ J/eV$:
$\Delta E = 10.2 \times 1.602 \times 10^{-19} \ J \approx 1.634 \times 10^{-18} \ J$.

(D) The $H^-$ ion consists of two electrons in the $1s$ orbital.
Due to the electron-electron repulsion,the electrons in $H^-$ are less tightly bound than the single electron in a neutral $H$ atom.
The first ionization energy of $H^-$ is significantly lower than the ionization energy of the $H$ atom $(13.6 \ eV)$.
Since the question specifies an 'excited' $H^-$ ion,the electron is in a higher energy level (e.g.,$n=2$ or higher).
The energy required to remove an electron from an excited state is even lower than that required for the ground state.
Therefore,the energy required is $\le 3.4 \ eV$ (the energy of the $n=2$ state of $H$ is $-3.4 \ eV$,and $H^-$ is even less stable).

(B) The Balmer series corresponds to electronic transitions where the final orbit is ${n_1} = 2$.
The lines in the series are defined by the initial orbit ${n_2}$:
- First line: ${n_2} = 3 \to {n_1} = 2$
- Second line: ${n_2} = 4 \to {n_1} = 2$
- Third line: ${n_2} = 5 \to {n_1} = 2$
Therefore,the third line corresponds to the transition from $n = 5$ to $n = 2$.

(C) For the Paschen series,the lower energy level is $n_1 = 3$.
The frequency $\nu$ is given by the Rydberg formula: $\nu = R_H \times c \times \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
Given $\nu = 2.340 \times 10^{11} \ Hz$,$R_H = 1.097 \times 10^7 \ m^{-1}$,and $c = 3 \times 10^8 \ m/s$.
$2.340 \times 10^{11} = (1.097 \times 10^7) \times (3 \times 10^8) \times \left( \frac{1}{3^2} - \frac{1}{n_2^2} \right)$.
$2.340 \times 10^{11} = 3.291 \times 10^{15} \times \left( \frac{1}{9} - \frac{1}{n_2^2} \right)$.
$\frac{2.340 \times 10^{11}}{3.291 \times 10^{15}} = \frac{1}{9} - \frac{1}{n_2^2}$.
$7.11 \times 10^{-5} = 0.1111 - \frac{1}{n_2^2}$.
$\frac{1}{n_2^2} = 0.1111 - 0.0000711 \approx 0.1110$.
$n_2^2 \approx \frac{1}{0.1110} \approx 9.009$.
$n_2 \approx 3$. However,the Paschen series requires $n_2 > n_1$. Given the options,if $n_2 = 4$,$\nu = 3.291 \times 10^{15} \times (1/9 - 1/16) = 1.599 \times 10^{14} \ Hz$. The provided frequency value in the question appears to be inconsistent with standard hydrogen transitions. Based on the options provided for a transition in the Paschen series,$n_2 = 4$ is the first line.

(C) The energy required for an electron transition in a hydrogen atom is given by the Rydberg formula: $\Delta E = 13.6 \times Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \text{ eV}$.
For a hydrogen atom,$Z = 1$.
$(A)$ $n = 1 \to n = 2$: $\Delta E = 13.6 \times (1 - 1/4) = 10.2 \text{ eV}$.
$(B)$ $n = 2 \to n = 3$: $\Delta E = 13.6 \times (1/4 - 1/9) \approx 1.89 \text{ eV}$.
$(C)$ $n = 1 \to n = \infty$: $\Delta E = 13.6 \times (1/1 - 1/\infty) = 13.6 \text{ eV}$.
$(D)$ $n = 3 \to n = 5$: $\Delta E = 13.6 \times (1/9 - 1/25) \approx 0.97 \text{ eV}$.
Comparing these values,the transition from $n = 1$ to $n = \infty$ requires the largest amount of energy.

(B) The Balmer series corresponds to electronic transitions ending at the $n = 2$ energy level.
Lines in the series are ordered by increasing energy and decreasing wavelength starting from the red end.
The first line (red end) corresponds to the transition $n = 3 \to n = 2$.
The second line corresponds to the transition $n = 4 \to n = 2$.
The third line corresponds to the transition $n = 5 \to n = 2$.

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

(A) The potential energy $(PE)$ of an electron in an atom is given by the formula $PE = -\frac{Ze^2}{4\pi\epsilon_0 r}$.
As the distance $(r)$ from the nucleus increases,the value of the negative term becomes smaller in magnitude,meaning the potential energy increases (becomes less negative).
Therefore,as an electron moves away from the nucleus,its potential energy increases.

(A) Vibrational energy is the sum of potential energy (due to the displacement of atoms from their equilibrium positions) and kinetic energy (due to the motion of atoms). Therefore,it is partly potential and partly kinetic.

(A) 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 $H$ atom: $n = 1, Z = 1$,so $r_H = a_0 \times \frac{1^2}{1} = a_0$.
For $He^{+}$ ion: $n = 1, Z = 2$,so $r_{He^{+}} = a_0 \times \frac{1^2}{2} = 0.5 \times a_0$.
The ratio $\frac{r_{He^{+}}}{r_H} = \frac{0.5 \times a_0}{a_0} = 0.5$.

(B) The energy of an orbit in a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \, eV$.
For $He^{+}$,the atomic number $Z = 2$.
For the first orbit $(n = 1)$: $E_1 = -13.6 \times \frac{2^2}{1^2} = -13.6 \times 4 = -54.4 \, eV$.
For the second orbit $(n = 2)$: $E_2 = -13.6 \times \frac{2^2}{2^2} = -13.6 \times \frac{4}{4} = -13.6 \, eV$.

(A) The electronic configuration of $Li$ is $1s^2 \, 2s^1$,so $Li^{2+}$ has the electronic configuration $1s^1$.
Bohr's model and the resulting spectra are applicable to hydrogen-like species,which contain only one electron.
Since $Li^{2+}$ has only one electron,its spectrum is similar to that of the hydrogen atom $(H)$,which also has the configuration $1s^1$.

(A) 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 ground state $(n=1)$,$E_1 = -\frac{13.6}{1^2} = -13.6 \ eV$.
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 the transition is $\Delta E = E_2 - E_1 = -3.4 - (-13.6) = 10.2 \ eV$.

(B) The ionisation 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 $E = 13.6 \ eV$.
For $He^+$ ion,$Z = 2$ and $n = 1$.
Therefore,$E = 13.6 \times \frac{2^2}{1^2} = 13.6 \times 4 = 54.4 \ eV$.

(A) The nuclear model of the atom was proposed by $Ernest \ Rutherford$ in $1911$ based on his famous $\alpha$-particle scattering experiment. He concluded that the positive charge and most of the mass of the atom are concentrated in a very small region called the nucleus.

(A) 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 $He^{+}$,the atomic number $Z = 2$ and for the ground state,$n = 1$.
Substituting these values: $E_1 = -13.6 \times \frac{2^2}{1^2} = -13.6 \times 4 = -54.4 \ eV$.
The energy required to remove the electron (ionization energy) is the negative of the ground state energy: $IE = -E_1 = -(-54.4) = +54.4 \ eV$.

(C) The energy of a transition in a hydrogen-like species is given by $\Delta E = 13.6 \ Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \ eV$.
For the first Lyman transition in $H$ $(Z=1)$: $\Delta E = 13.6 \times 1^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 13.6 \times \frac{3}{4} = 10.2 \ eV$.
For the second Balmer transition,$n_1 = 2$ and $n_2 = 4$ (since the first is $2 \to 3$ and the second is $2 \to 4$).
Setting the energy equal: $10.2 = 13.6 \times Z^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$.
$10.2 = 13.6 \times Z^2 \left( \frac{1}{4} - \frac{1}{16} \right) = 13.6 \times Z^2 \left( \frac{3}{16} \right)$.
$10.2 = 13.6 \times Z^2 \times 0.1875 = 2.55 \times Z^2$.
$Z^2 = \frac{10.2}{2.55} = 4$,so $Z = 2$.
The species with atomic number $Z = 2$ is $He^+$,which is a hydrogen-like ion.

(D) 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 $n_2 = 7$ and $n_1 = 1$:
$N = \frac{(7 - 1)(7 - 1 + 1)}{2}$
$N = \frac{6 \times 7}{2} = \frac{42}{2} = 21$
Thus,the total number of spectral lines is $21$.

(B) For the Balmer series,the transition occurs from $n_2 = \infty$ to $n_1 = 2$.
The frequency $\nu$ is given by the Rydberg formula: $\nu = R_\infty \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
For Hydrogen,$Z = 1$,$n_1 = 2$,and $n_2 = \infty$.
$\nu = 3.29 \times 10^{15} \times (\frac{1}{2^2} - \frac{1}{\infty^2}) \, s^{-1}$.
$\nu = 3.29 \times 10^{15} \times \frac{1}{4} \, s^{-1}$.
$\nu = 0.8225 \times 10^{15} \, s^{-1} = 8.225 \times 10^{14} \, s^{-1}$.

(B) The ionization energy of a hydrogen atom in its ground state is $13.6 \, eV$.
When a photon with energy $E = 14 \, eV$ interacts with the hydrogen atom,it provides more energy than the ionization energy.
The excess energy is converted into the kinetic energy of the ejected electron.
Kinetic Energy $(K.E.)$ = $E_{photon} - E_{ionization} = 14 \, eV - 13.6 \, eV = 0.4 \, eV$.
Therefore,the atom is ionized and the electron possesses a kinetic energy of $0.4 \, eV$.

(B) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -2.18 \times 10^{-18} \,J \times \frac{1}{n^2}$.
For $n = 1$,$E_1 = -2.18 \times 10^{-18} \,J$.
For $n = 4$,$E_4 = \frac{-2.18 \times 10^{-18}}{4^2} = -0.13625 \times 10^{-18} \,J$.
The energy difference is $\Delta E = E_4 - E_1 = (-0.13625 \times 10^{-18}) - (-2.18 \times 10^{-18}) = 2.04375 \times 10^{-18} \,J$.
Using the relation $\Delta E = h \nu$,the frequency $\nu$ is $\nu = \frac{\Delta E}{h} = \frac{2.04375 \times 10^{-18} \,J}{6.625 \times 10^{-34} \,J \cdot s} \approx 3.08 \times 10^{15} \,s^{-1}$.

(A) The radius of an orbit for a hydrogen-like species is given by the formula $r_n = 0.529 \times \frac{n^2}{Z} \ \mathring{A}$.
For the ground state,$n = 1$.
For a hydrogen atom,$Z = 1$,so $r_H = 0.529 \times \frac{1^2}{1} = 0.529 \ \mathring{A} \approx 0.53 \ \mathring{A}$.
For the $Li^{2+}$ ion,$Z = 3$,so $r_{Li^{2+}} = 0.529 \times \frac{1^2}{3} = \frac{0.529}{3} \ \mathring{A} \approx 0.176 \ \mathring{A}$.

(B) The hydrogen emission spectrum consists of various series based on the value of $n_1$ in the Rydberg formula.
For the Paschen series,the electron transitions to the third energy level,which means $n_1 = 3$.
The corresponding values for $n_2$ are $n_2 = 4, 5, 6, \dots$.
Therefore,the correct condition for the Paschen series is $n_1 = 3$ and $n_2 = 4, 5, 6, \dots$.

(A) The 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: $\text{Number of lines} = \frac{(n_2 - n_1)(n_2 - n_1 + 1)}{2}$.
Given $n_2 = 6$ and $n_1 = 2$.
Substituting the values: $\text{Number of lines} = \frac{(6 - 2)(6 - 2 + 1)}{2} = \frac{4 \times 5}{2} = \frac{20}{2} = 10$.
Therefore,the correct option is $A$.

(D) The frequency of an emitted photon is given by $\nu = \frac{\Delta E}{h}$. To maximize frequency,we must maximize the energy difference $\Delta E$ between the energy levels.
The energy of a level is given by $E_n = -13.6 \ \text{eV} \times \frac{Z^2}{n^2}$.
The Lyman series involves transitions to $n_1 = 1$. The first line of the Lyman series is $n_2 = 2 \to n_1 = 1$.
The energy difference is $\Delta E = 13.6 \times (1 - \frac{1}{4}) = 13.6 \times 0.75 = 10.2 \ \text{eV}$.
Comparing this with other series:
Balmer $(n_1 = 2)$: $\Delta E = 13.6 \times (\frac{1}{4} - \frac{1}{16}) = 13.6 \times 0.1875 = 2.55 \ \text{eV}$.
Paschen $(n_1 = 3)$: $\Delta E = 13.6 \times (\frac{1}{9} - \frac{1}{25}) \approx 0.96 \ \text{eV}$.
Humphrey $(n_1 = 6)$: $\Delta E$ is even smaller.
Thus,the first spectral line of the Lyman series has the largest energy gap and consequently the maximum frequency.

(C) For the $Balmer$ series,$n_1 = 2$ and for the limiting line,$n_2 = \infty$.
The Rydberg formula for frequency is $\nu = R_\infty \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Substituting $Z = 1$,$n_1 = 2$,and $n_2 = \infty$:
$\nu = 3.29 \times 10^{15} \times (\frac{1}{2^2} - \frac{1}{\infty^2}) \text{ s}^{-1}$.
$\nu = 3.29 \times 10^{15} \times \frac{1}{4} \text{ s}^{-1} = 8.225 \times 10^{14} \text{ s}^{-1}$.

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV$.
For $Z = 5$,the energy difference between the $4^{th}$ and $3^{rd}$ orbit is $\Delta E = E_4 - E_3$.
$\Delta E = -13.6 \times Z^2 \left( \frac{1}{4^2} - \frac{1}{3^2} \right) = -13.6 \times 25 \left( \frac{1}{16} - \frac{1}{9} \right)$.
$\Delta E = -13.6 \times 25 \left( \frac{9 - 16}{144} \right) = -13.6 \times 25 \left( \frac{-7}{144} \right)$.
$\Delta E = 13.6 \times 25 \times \frac{7}{144} \approx 16.53 \ eV$.

(C) For a hydrogen atom,the radius of the $n^{th}$ Bohr orbit is given by $r_n = n^2 a_0$,where $a_0$ is the Bohr radius.
For the second orbit $(n = 2)$,$r_2 = 2^2 a_0 = 4 a_0$.
The velocity of an electron in the $n^{th}$ orbit is $v_n = \frac{nh}{2\pi m r_n}$.
Substituting $r_n = n^2 a_0$,we get $v_n = \frac{nh}{2\pi m (n^2 a_0)} = \frac{h}{2\pi m n a_0}$.
For $n = 2$,$v_2 = \frac{h}{2\pi m (2) a_0} = \frac{h}{4\pi m a_0}$.
The kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v_2^2$.
$KE = \frac{1}{2} m \left( \frac{h}{4\pi m a_0} \right)^2 = \frac{1}{2} m \left( \frac{h^2}{16\pi^2 m^2 a_0^2} \right) = \frac{h^2}{32\pi^2 m a_0^2}$.

(A) The energy of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For a hydrogen atom $(H)$,$Z = 1$,so $E_n(H) = -13.6 \times \frac{1^2}{n^2} = E_n$.
For a singly ionized helium atom $(He^+)$,$Z = 2$,so $E_n(He^+) = -13.6 \times \frac{2^2}{n^2} = -13.6 \times \frac{4}{n^2}$.
Substituting $E_n$ into the expression for $He^+$,we get $E_n(He^+) = 4 \times (-13.6 \times \frac{1^2}{n^2}) = 4E_n$.

(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 $(n=1)$,$E_1 = -13.6 \, eV$.
For excited states,$n \geq 2$.
For the first excited state $(n=2)$,$E_2 = -\frac{13.6}{2^2} = -3.4 \, eV$.
Ionization energy is the energy required to remove the electron from its current state to $n = \infty$ (where $E = 0$).
For $n=2$,the ionization energy is $0 - (-3.4) = 3.4 \, eV$.
For higher excited states $(n > 2)$,the energy $E_n$ becomes closer to $0$,so the ionization energy $(0 - E_n)$ will be less than $3.4 \, eV$.
Therefore,the ionization energy for an excited hydrogen atom is $3.4 \, eV$ or less.

(A) The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \text{ eV}$.
For $He^{+}$,the atomic number $Z = 2$.
The first excited state corresponds to $n = 2$,and the second excited state corresponds to $n = 3$.
Energy in the first excited state $(E_1)$ = $-13.6 \times \frac{2^2}{2^2} = -13.6 \text{ eV}$.
Energy in the second excited state $(E_2)$ = $-13.6 \times \frac{2^2}{3^2} = -13.6 \times \frac{4}{9} \text{ eV}$.
The ratio of energy in the first excited state to the second excited state is $\frac{E_1}{E_2} = \frac{-13.6 \times (4/4)}{-13.6 \times (4/9)} = \frac{1}{4/9} = \frac{9}{4}$.
Thus,the ratio is $9 : 4$.

(C) 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 the ground state $(n_1 = 1)$,the energy is $E_1 = -13.6 \, eV$.
When the atom absorbs a photon of energy $12.75 \, eV$,the new energy $E_n$ becomes:
$E_n = E_1 + 12.75 \, eV = -13.6 + 12.75 = -0.85 \, eV$.
Using the formula $E_n = -13.6 / n^2$,we have:
$-0.85 = -13.6 / n^2$
$n^2 = 13.6 / 0.85 = 16$
$n = 4$.
Therefore,the principal quantum number of the excited state is $4$.

(D) The wavelength $\lambda$ is related to the energy transition by the Rydberg formula: $\frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For maximum wavelength,the energy difference must be minimum,which corresponds to the first line of the series.
For the Lyman series $(n_1 = 1, n_2 = 2)$: $\frac{1}{\lambda_L} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4}$.
For the Balmer series $(n_1 = 2, n_2 = 3)$: $\frac{1}{\lambda_B} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36}$.
The ratio of wavelengths is $\frac{\lambda_L}{\lambda_B} = \frac{1/\lambda_B}{1/\lambda_L} = \frac{5R/36}{3R/4} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27}$.

(A) According to the Bohr model for the hydrogen atom,the radius of the $n^{th}$ stationary orbit is given by the formula: $r_n = a_0 \times n^2$,where $a_0$ is the Bohr radius and $n$ is the principal quantum number.
Therefore,the radius of the stationary orbit is directly proportional to the square of the principal quantum number,$n^2$.

(B) The infrared region in the hydrogen spectrum includes the Paschen,Brackett,and Pfund series.
For transitions from $n_2 = 6$ to $n_1 = 3$:
$1$. Paschen series $(n_1 = 3)$: Transitions from $n_2 = 6, 5, 4$ to $n_1 = 3$. Number of lines = $6 - 3 = 3$.
$2$. Brackett series $(n_1 = 4)$: Transitions from $n_2 = 6, 5$ to $n_1 = 4$. Number of lines = $6 - 4 = 2$.
$3$. Pfund series $(n_1 = 5)$: Transition from $n_2 = 6$ to $n_1 = 5$. Number of lines = $6 - 5 = 1$.
Total number of lines in the infrared region = $3 + 2 + 1 = 6$.