Atomic models and Planck's quantum theory Questions in English

(A) The ionization energy of hydrogen is given by $(IE)_{H} = 13.6 \ eV = X$.
The energy of an electron in a hydrogen-like species is given by $E_n = -13.6 \times \frac{Z^2}{n^2} \ eV = -X \times \frac{Z^2}{n^2}$.
For $Li^{2+}$,the atomic number $Z = 3$.
The $2^{nd}$ excited state corresponds to $n_1 = 3$ and the $5^{th}$ excited state corresponds to $n_2 = 6$.
The energy required for excitation is $\Delta E = E_{n_2} - E_{n_1} = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.
Substituting the values: $\Delta E = X \times 3^2 \times (\frac{1}{3^2} - \frac{1}{6^2}) = 9X \times (\frac{1}{9} - \frac{1}{36})$.
$\Delta E = 9X \times (\frac{4-1}{36}) = 9X \times \frac{3}{36} = 9X \times \frac{1}{12} = \frac{3X}{4}$.

(A) The velocity of an electron in the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $v = v_0 \times \frac{Z}{n}$,where $v_0$ is the velocity in the $1^{st}$ orbit of the $H$ atom $(Z=1, n=1)$.
Given for $H$ atom: $v_1 = V = v_0 \times \frac{1}{1} = v_0$.
For $Be^{3+}$ ion,the atomic number $Z = 4$ and the orbit number $n = 4$.
Therefore,the velocity $v_2 = v_0 \times \frac{Z}{n} = V \times \frac{4}{4} = V$.
Thus,the velocity in the $4^{th}$ orbit of $Be^{3+}$ is $V$.

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = n$ and $n_2 = n+1$.
Substituting the values: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n^2} - \frac{1}{(n+1)^2} \right)$.
Simplifying the expression: $\frac{1}{\lambda} = R Z^2 \left( \frac{(n+1)^2 - n^2}{n^2(n+1)^2} \right) = R Z^2 \left( \frac{2n+1}{n^2(n+1)^2} \right)$.
Given the condition $n >> 1$,we can approximate $2n+1 \approx 2n$ and $(n+1)^2 \approx n^2$.
Thus,$\frac{1}{\lambda} \approx R Z^2 \left( \frac{2n}{n^2 \cdot n^2} \right) = \frac{2R Z^2}{n^3}$.
Since $v = \frac{c}{\lambda}$,we have $v = \frac{2cR Z^2}{n^3}$.

(C) The total energy $(E_n)$ of an electron in the $n^{th}$ shell is given by $E_n = \frac{E_1}{n^2}$,where $E_1$ is the energy of the ground state $(n=1)$.
Given $E_4 = -25 \, a.u.$,we have $-25 = \frac{E_1}{4^2}$.
Therefore,$E_1 = -25 \times 16 = -400 \, a.u.$
For a hydrogen-like atom,the potential energy $(PE)$ is related to the total energy $(E)$ by the relation $PE = 2 \times E$.
Thus,for the ground state $(n=1)$,$PE_1 = 2 \times E_1 = 2 \times (-400) = -800 \, a.u.$

(B) According to Bohr's theory,the time period $(T)$ of an electron in an orbit is given by the relation $T \propto \frac{n^3}{Z^2}$.
For a hydrogen atom ($H$-atom),the atomic number $Z = 1$,so $T \propto n^3$.
For the first orbit $(n_1 = 1)$,$T_1 \propto (1)^3 = 1$.
For the third orbit $(n_2 = 3)$,$T_2 \propto (3)^3 = 27$.
Therefore,the ratio of the time period is $\frac{T_1}{T_2} = \frac{1}{27}$,which is $1 : 27$.

(A) The minimum energy required to escape the atom (Ionization Energy) is $13.6 \ eV$.
Given that the electron absorbs $2$ times this energy,the total absorbed energy is $2 \times 13.6 \ eV = 27.2 \ eV$.
The kinetic energy of the emitted electron is calculated as: $\text{Kinetic Energy} = \text{Total Absorbed Energy} - \text{Ionization Energy}$.
$\text{Kinetic Energy} = 27.2 \ eV - 13.6 \ eV = 13.6 \ eV$.

(B) The velocity of an electron in the $n^{th}$ orbit is given by the relation $V \propto \frac{Z}{n}$.
For $He^{+}$,$Z = 2$ and $n = 2$. Thus,$V_1 \propto \frac{2}{2} = 1$.
For $B^{4+}$,$Z = 5$ and $n = 3$. Thus,$V_2 \propto \frac{5}{3}$.
The ratio is $\frac{V_1}{V_2} = \frac{1}{5/3} = \frac{3}{5}$.

(C) The energy of a photon is given by the equation $E = h \nu$,where $E$ is energy,$h$ is Planck's constant,and $\nu$ is frequency.
Given $E = 5.0 \times 10^{-5} \ erg$ and $h = 6.626 \times 10^{-27} \ erg \ sec$ (in $CGS$ units).
Rearranging for frequency: $\nu = \frac{E}{h}$.
$\nu = \frac{5.0 \times 10^{-5} \ erg}{6.626 \times 10^{-27} \ erg \ sec} \approx 7.54 \times 10^{21} \ sec^{-1}$.
Thus,the correct option is $C$.

(C) The series limit corresponds to the transition from $n_2 = \infty$ to $n_1$.
The Rydberg formula for the wavenumber is $\bar{v} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For the series limit,$n_2 = \infty$,so $\bar{v} = \frac{R}{n_1^2}$.
Given $\bar{v} = 12186.3 \ cm^{-1}$ and $R = 109677.76 \ cm^{-1}$,we have:
$12186.3 = \frac{109677.76}{n_1^2}$
$n_1^2 = \frac{109677.76}{12186.3} \approx 9$
$n_1 = 3$.
Since $n_1 = 3$,the series is the Paschen series.

(C) The energy of an electron at an infinite distance from the nucleus is defined as zero.
As an electron approaches the nucleus,the electrostatic attraction increases,causing the energy of the electron to decrease and become negative.
Therefore,as the value of $n$ decreases (i.e.,the orbit is closer to the nucleus),the energy of the electron becomes more negative,meaning it is more tightly bound,not loosely bound.
Thus,the statement in option $C$ is incorrect because it claims the electron is more loosely bound in the smallest orbit.

(D) The Rydberg formula for a hydrogen-like species is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.
For $He^{+}$ $(Z = 2)$,the transition from $n_2 = 4$ to $n_1 = 2$ gives $\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$.
For a hydrogen atom $(Z = 1)$,we need to find $n_1$ and $n_2$ such that $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$.
Comparing the two,we get $\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$.
This is satisfied when $n_1 = 1$ and $n_2 = 2$ because $\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$.

(A) The lines falling in the visible region of the hydrogen spectrum constitute the Balmer series,where the electron jumps to the $n_1 = 2$ orbit.
The lines are ordered by wavelength,with the red end representing the longest wavelength (lowest energy transition).
The first line is $3 \to 2$,the second line is $4 \to 2$,and the third line is $5 \to 2$.
Therefore,the third line from the red end corresponds to the transition $5 \to 2$.

(C) The Balmer series corresponds to transitions where $n_1 = 2$ and $n_2 = 3, 4, 5, ...$. Thus,the Assertion is correct.
For the highest wavelength,the energy difference between the levels must be the minimum.
Since $\Delta E = 13.6 \times Z^2 \times (\frac{1}{n_1^2} - \frac{1}{n_2^2}) \text{ eV}$ and $\lambda = \frac{hc}{\Delta E}$,the minimum energy transition for the Balmer series is from $n_2 = 3$ to $n_1 = 2$.
Therefore,the Reason is incorrect because the highest wavelength corresponds to the transition $n_2 = 3$ to $n_1 = 2$,not $n = 4$ and $6$.

(B) The assertion states that the angular momentum of an electron in an orbit is given by $mvr = \frac{nh}{2\pi}$,which is a fundamental postulate of the Bohr model of the atom. This statement is correct.
The reason states that the principal quantum number,$n$,can have any integral value $(n = 1, 2, 3, \dots)$. This is also a correct statement regarding the definition of the principal quantum number.
However,the fact that $n$ can take any integral value is not the reason why the angular momentum is quantized as $\frac{nh}{2\pi}$. The quantization of angular momentum is a separate postulate derived from the wave nature of the electron. Therefore,the reason is not the correct explanation of the assertion.

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula $r_n = 0.529 \ \mathring{A} \times \frac{n^2}{Z}$.
For the hydrogen atom,$Z = 1$.
For the first orbit,$n = 1$.
Substituting these values,$r_1 = 0.529 \ \mathring{A} \times \frac{1^2}{1} = 0.529 \ \mathring{A}$.
Thus,both the assertion and the reason are correct,and the reason provides the correct formula used to derive the assertion.

(B) In the spectrum of the hydrogen atom,the spectral lines of the $Balmer$ series lie in the visible region.
The $Lyman$ series lies in the ultraviolet region,while the $Paschen$,$Brackett$,and $Pfund$ series lie in the infrared region.

(D) The formula for the radius of the $n^{th}$ Bohr orbit is given by $r_{n} = \frac{n^{2} \times a_{0}}{Z}$.
For the $2^{nd}$ Bohr orbit of $Li^{2+}$:
$n = 2$
$Z = 3$ (atomic number of Lithium)
Substituting these values into the formula:
$r_{2} = \frac{2^{2} \times a_{0}}{3} = \frac{4 a_{0}}{3}$.

(B) For the Balmer series,$n_{1}=2$ and $n_{2}=3, 4, 5, \dots, \infty$.
$(I)$ As wavelength $\lambda$ decreases,the wave number $\bar{v}$ increases,and the lines converge towards the series limit,so this is correct.
$(II)$ By definition of the Balmer series,$n_{1}=2$,so this is correct.
$(III)$ The wavelength $\lambda$ is longest when the energy difference $\Delta E$ is minimum,which occurs for the transition $n_{2}=3$ to $n_{1}=2$. Thus,this is correct.
$(IV)$ The ionization energy corresponds to the transition from $n=1$ to $n=\infty$. While the Balmer series lines provide $R_{H}$,the ionization energy is specifically calculated using the Rydberg constant $R_{H}$ and the ground state energy,not directly from the wave numbers of the Balmer lines alone. Therefore,this statement is generally considered incorrect in this context.
Thus,statements $(I)$,$(II)$,and $(III)$ are correct.

Power of the bulb $P = 100 \, W = 100 \, J \, s^{-1}$.
Energy of one photon $E = \frac{hc}{\lambda}$.
$E = \frac{6.626 \times 10^{-34} \, J \, s \times 3 \times 10^{8} \, m \, s^{-1}}{400 \times 10^{-9} \, m} = 4.9695 \times 10^{-19} \, J$.
Number of photons emitted per second $n = \frac{P}{E}$.
$n = \frac{100 \, J \, s^{-1}}{4.9695 \times 10^{-19} \, J} \approx 2.012 \times 10^{20} \, s^{-1}$.

The energy $(E)$ of a $300 \, nm$ photon is given by $E = \frac{hc}{\lambda}$.
Energy per photon $= \frac{6.626 \times 10^{-34} \, J \, s \times 3.0 \times 10^{8} \, m \, s^{-1}}{300 \times 10^{-9} \, m} = 6.626 \times 10^{-19} \, J$.
The energy of one mole of photons $= 6.626 \times 10^{-19} \, J \times 6.022 \times 10^{23} \, mol^{-1} = 3.99 \times 10^{5} \, J \, mol^{-1}$.
The minimum energy (work function,$\Phi$) needed to remove one mole of electrons from sodium is the difference between the incident energy and the kinetic energy: $\Phi = (3.99 - 1.68) \times 10^{5} \, J \, mol^{-1} = 2.31 \times 10^{5} \, J \, mol^{-1}$.
The minimum energy for one electron $= \frac{2.31 \times 10^{5} \, J \, mol^{-1}}{6.022 \times 10^{23} \, mol^{-1}} = 3.84 \times 10^{-19} \, J$.
The maximum wavelength $(\lambda_{max})$ is given by $\lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \, J \, s \times 3.0 \times 10^{8} \, m \, s^{-1}}{3.84 \times 10^{-19} \, J} = 517 \, nm$.

According to Einstein's photoelectric equation:
Kinetic energy $(K.E.) = h(v - v_{0})$
Given:
$h = 6.626 \times 10^{-34} \,J \cdot s$
$v = 1.0 \times 10^{15} \,s^{-1} = 10.0 \times 10^{14} \,s^{-1}$
$v_{0} = 7.0 \times 10^{14} \,s^{-1}$
$K.E. = (6.626 \times 10^{-34} \,J \cdot s) \times (10.0 \times 10^{14} \,s^{-1} - 7.0 \times 10^{14} \,s^{-1})$
$K.E. = (6.626 \times 10^{-34} \,J \cdot s) \times (3.0 \times 10^{14} \,s^{-1})$
$K.E. = 1.9878 \times 10^{-19} \,J \approx 1.99 \times 10^{-19} \,J$

For a hydrogen atom,the energy change for a transition from $n_{i}=5$ to $n_{f}=2$ is given by the Rydberg formula:
$\Delta E = 2.18 \times 10^{-18} \, J \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$
$\Delta E = 2.18 \times 10^{-18} \, J \left( \frac{1}{2^{2}} - \frac{1}{5^{2}} \right) = 2.18 \times 10^{-18} \left( \frac{1}{4} - \frac{1}{25} \right) = 2.18 \times 10^{-18} \left( \frac{21}{100} \right) = 4.578 \times 10^{-19} \, J$
The frequency $\nu$ is calculated as $\nu = \frac{\Delta E}{h} = \frac{4.578 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, J \, s} = 6.91 \times 10^{14} \, Hz$
The wavelength $\lambda$ is calculated as $\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^{8} \, m \, s^{-1}}{6.91 \times 10^{14} \, Hz} = 4.34 \times 10^{-7} \, m = 434 \, nm$

(N/A) The energy of an electron in the $n^{th}$ orbit is given by the formula: $E_{n} = -\frac{(2.18 \times 10^{-18} \, J) Z^{2}}{n^{2}}$.
For $He^{+}$,the atomic number $Z = 2$ and for the first orbit $n = 1$.
Substituting these values: $E_{1} = -\frac{(2.18 \times 10^{-18} \, J) (2^{2})}{1^{2}} = -8.72 \times 10^{-18} \, J$.
The radius of the $n^{th}$ orbit is given by the formula: $r_{n} = \frac{(0.0529 \, nm) n^{2}}{Z}$.
Substituting $n = 1$ and $Z = 2$: $r_{1} = \frac{(0.0529 \, nm) (1^{2})}{2} = 0.02645 \, nm$.

$(i)$ The energy $(E)$ of a photon is given by the expression:
$E = h \nu$
Where $h = 6.626 \times 10^{-34} \, J \cdot s$ (Planck's constant) and $\nu = 3 \times 10^{15} \, Hz$.
Substituting the values:
$E = (6.626 \times 10^{-34}) \times (3 \times 10^{15}) = 1.988 \times 10^{-18} \, J$.
$(ii)$ The energy $(E)$ of a photon with wavelength $(\lambda)$ is given by:
$E = \frac{hc}{\lambda}$
Where $h = 6.626 \times 10^{-34} \, J \cdot s$,$c = 3 \times 10^{8} \, m/s$,and $\lambda = 0.50 \, \mathring{A} = 0.50 \times 10^{-10} \, m$.
Substituting the values:
$E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^{8})}{0.50 \times 10^{-10}} = 3.976 \times 10^{-15} \, J \approx 3.98 \times 10^{-15} \, J$.

(N/A) Energy $(E)$ of a single photon is given by $E = \frac{hc}{\lambda}$.
For $n$ photons, the total energy $(E_n)$ is $E_n = \frac{n \times hc}{\lambda}$.
Rearranging for $n$, we get $n = \frac{E_n \times \lambda}{hc}$.
Given values:
$E_n = 1 \ J$
$\lambda = 4000 \ pm = 4000 \times 10^{-12} \ m = 4 \times 10^{-9} \ m$
$h = 6.626 \times 10^{-34} \ J \cdot s$
$c = 3 \times 10^8 \ m/s$
Substituting these values:
$n = \frac{1 \times 4 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8}$
$n = \frac{4 \times 10^{-9}}{19.878 \times 10^{-26}}$
$n \approx 2.012 \times 10^{16}$.
Thus, the number of photons is $2.012 \times 10^{16}$.

$(i)$ Energy $(E)$ of a photon $= \frac{hc}{\lambda}$
$E = \frac{(6.626 \times 10^{-34} \, Js)(3 \times 10^{8} \, ms^{-1})}{4 \times 10^{-7} \, m} = 4.9695 \times 10^{-19} \, J$
Converting to $eV$: $E = \frac{4.9695 \times 10^{-19} \, J}{1.6020 \times 10^{-19} \, J/eV} = 3.1020 \, eV$
$(ii)$ Kinetic energy $(E_k) = E - \Phi = 3.1020 \, eV - 2.13 \, eV = 0.9720 \, eV$
$(iii)$ Velocity $(v)$ is given by $E_k = \frac{1}{2}mv^2$
$v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2 \times 0.9720 \times 1.6020 \times 10^{-19} \, J}{9.10939 \times 10^{-31} \, kg}}$
$v = \sqrt{0.3418 \times 10^{12} \, m^2s^{-2}} = 5.846 \times 10^{5} \, ms^{-1}$

Power of bulb,$P = 25 \ W = 25 \ J \ s^{-1}$.
Energy of one photon,$E = \frac{hc}{\lambda}$.
Substituting the values: $h = 6.626 \times 10^{-34} \ J \ s$,$c = 3 \times 10^8 \ m \ s^{-1}$,$\lambda = 0.57 \times 10^{-6} \ m$.
$E = \frac{(6.626 \times 10^{-34} \ J \ s)(3 \times 10^8 \ m \ s^{-1})}{0.57 \times 10^{-6} \ m} = 3.487 \times 10^{-19} \ J$.
Rate of emission of quanta per second = $\frac{P}{E} = \frac{25 \ J \ s^{-1}}{3.487 \times 10^{-19} \ J} = 7.17 \times 10^{19} \ s^{-1}$.

(A) The transition from $n_{i}=4$ to $n_{f}=2$ belongs to the Balmer series.
The Rydberg formula for the wavenumber $(\bar{\nu})$ is given by:
$\bar{\nu} = \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$
Where $R_H = 1.097 \times 10^7 \ m^{-1}$,$n_f = 2$,and $n_i = 4$.
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} \right)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{4} - \frac{1}{16} \right)$
$\frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{4-1}{16} \right) = 1.097 \times 10^7 \times \frac{3}{16}$
$\frac{1}{\lambda} = 2.056875 \times 10^6 \ m^{-1}$
$\lambda = \frac{1}{2.056875 \times 10^6} \approx 4.863 \times 10^{-7} \ m$
$\lambda = 486.3 \times 10^{-9} \ m = 486 \ nm$.

(N/A) The energy of an electron in the $n^{\text{th}}$ orbit is given by:
$E_{n} = \frac{-2.18 \times 10^{-18} \times Z^{2}}{n^{2}} \ J$
For a $H$ atom,$Z = 1$,so $E_{n} = \frac{-2.18 \times 10^{-18}}{n^{2}} \ J$.
For ionization from $n=5$ to $n=\infty$:
$\Delta E = E_{\infty} - E_{5} = 0 - \left( \frac{-2.18 \times 10^{-18}}{5^{2}} \right) \ J$
$\Delta E = \frac{2.18 \times 10^{-18}}{25} \ J = 8.72 \times 10^{-20} \ J$.
For ionization from $n=1$ to $n=\infty$ (Ionization Enthalpy):
$\Delta E' = E_{\infty} - E_{1} = 0 - \left( \frac{-2.18 \times 10^{-18}}{1^{2}} \right) \ J$
$\Delta E' = 2.18 \times 10^{-18} \ J$.
Conclusion: The energy required to ionize an electron from the $n=5$ orbit $(8.72 \times 10^{-20} \ J)$ is significantly less than the ionization enthalpy of the $H$ atom $(2.18 \times 10^{-18} \ J)$.

(B) When the excited electron of an $H$ atom in $n=6$ drops to the ground state,the possible transitions are from $n=6$ to $n=5, 4, 3, 2, 1$; $n=5$ to $n=4, 3, 2, 1$,and so on.
The total number of spectral lines produced when an electron in the $n^{th}$ level drops down to the ground state is given by the formula:
$\text{Number of spectral lines} = \frac{n(n-1)}{2}$
Given $n=6$:
$\text{Number of spectral lines} = \frac{6(6-1)}{2} = \frac{6 \times 5}{2} = 15$
Thus,a total of $15$ lines will be obtained in the emission spectrum.

$(i)$ The energy associated with the $n^{th}$ orbit of the hydrogen atom is given by $E_n = \frac{-2.18 \times 10^{-18}}{n^2} \, J \, atom^{-1}$.
For the fifth orbit $(n = 5)$:
$E_5 = \frac{-2.18 \times 10^{-18}}{5^2} = \frac{-2.18 \times 10^{-18}}{25} = -8.72 \times 10^{-20} \, J \, atom^{-1}$.
$(ii)$ The radius of Bohr's $n^{th}$ orbit for the hydrogen atom is given by $r_n = (0.0529 \, nm) \times n^2$.
For the fifth orbit $(n = 5)$:
$r_5 = 0.0529 \, nm \times 5^2 = 0.0529 \times 25 \, nm = 1.3225 \, nm$.

(N/A) For the Balmer series,the lower energy level is $n_{1} = 2$. The Rydberg formula for the wave number $(\bar{\nu})$ is given by:
$\bar{\nu} = R_{H} \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$
where $R_{H} = 1.097 \times 10^{7} \ m^{-1}$.
Since the wave number $(\bar{\nu})$ is inversely proportional to the wavelength $(\lambda)$,the longest wavelength corresponds to the smallest wave number.
For the Balmer series,the transition with the smallest wave number occurs from the nearest higher energy level,i.e.,$n_{2} = 3$ to $n_{1} = 2$.
Substituting the values:
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{1}{4} - \frac{1}{9} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{9 - 4}{36} \right)$
$\bar{\nu} = (1.097 \times 10^{7} \ m^{-1}) \left( \frac{5}{36} \right)$
$\bar{\nu} \approx 1.5236 \times 10^{6} \ m^{-1}$

The energy $(E)$ of the $n^{\text{th}}$ Bohr orbit of an atom is given by:
$E_{n} = \frac{-(2.18 \times 10^{-18} \ J) Z^{2}}{n^{2}}$
Given,ground state energy $= -2.18 \times 10^{-11} \ erg = -2.18 \times 10^{-18} \ J$.
Energy required to shift the electron from $n=1$ to $n=5$ is:
$\Delta E = E_{5} - E_{1} = -2.18 \times 10^{-18} [\frac{1}{5^{2}} - \frac{1}{1^{2}}]$
$\Delta E = -2.18 \times 10^{-18} [\frac{1}{25} - 1] = -2.18 \times 10^{-18} [-\frac{24}{25}]$
$\Delta E = 2.0928 \times 10^{-18} \ J$.
When the electron returns to the ground state,the energy emitted is equal to the energy absorbed,$\Delta E = 2.0928 \times 10^{-18} \ J$.
The wavelength $(\lambda)$ of the emitted light is given by:
$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \ J \cdot s) (3 \times 10^{8} \ m/s)}{2.0928 \times 10^{-18} \ J}$
$\lambda = 9.498 \times 10^{-8} \ m$.

(A) Given,$E_n = -\frac{2.18 \times 10^{-18}}{n^2} \ J$.
Energy required for ionization from $n = 2$ is $\Delta E = E_{\infty} - E_2$.
Since $E_{\infty} = 0$,$\Delta E = 0 - (\frac{-2.18 \times 10^{-18}}{2^2}) = \frac{2.18 \times 10^{-18}}{4} = 5.45 \times 10^{-19} \ J$.
Using $\lambda = \frac{hc}{\Delta E}$,where $h = 6.626 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$:
$\lambda = \frac{(6.626 \times 10^{-34} \ J \cdot s)(3 \times 10^8 \ m/s)}{5.45 \times 10^{-19} \ J} = 3.647 \times 10^{-7} \ m$.
Converting to $cm$: $\lambda = 3.647 \times 10^{-7} \ m \times 100 \ cm/m = 3.647 \times 10^{-5} \ cm$.

(A) For $He^{+}$ ion,the wave number $(\bar{v})$ associated with the transition $n_2=4$ to $n_1=2$ is given by the Rydberg formula:
$\bar{v} = \frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$
For $He^{+}$,$Z=2$,$n_1=2$,and $n_2=4$:
$\frac{1}{\lambda} = R(2)^2 \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = 4R \left( \frac{1}{4} - \frac{1}{16} \right) = 4R \left( \frac{3}{16} \right) = \frac{3R}{4}$
For the hydrogen atom,$Z=1$. Let the transition be from $n_2$ to $n_1$:
$\frac{1}{\lambda} = R(1)^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) = \frac{3R}{4}$
$\frac{1}{n_1^2} - \frac{1}{n_2^2} = \frac{3}{4}$
Comparing this with the Lyman series formula where $n_1=1$:
$\frac{1}{1^2} - \frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4}$
Thus,the transition $n=2$ to $n=1$ in the hydrogen spectrum has the same wavelength.

(D) The energy of an electron in a hydrogen-like species is given by the formula: $E_{n} = -2.18 \times 10^{-18} \left( \frac{Z^{2}}{n^{2}} \right) \, J \, atom^{-1}$.
For the process $He_{(g)}^{+} \rightarrow He_{(g)}^{2+} + e^{-}$,the electron is removed from the ground state $(n = 1)$ to infinity $(n = \infty)$.
The energy required is $\Delta E = E_{\infty} - E_{1}$.
Since $E_{\infty} = 0$,we have $\Delta E = -E_{1}$.
For $He^{+}$,the atomic number $Z = 2$. Substituting $Z = 2$ and $n = 1$ into the formula:
$E_{1} = -2.18 \times 10^{-18} \times \left( \frac{2^{2}}{1^{2}} \right) \, J = -2.18 \times 10^{-18} \times 4 \, J = -8.72 \times 10^{-18} \, J$.
Therefore,$\Delta E = 0 - (-8.72 \times 10^{-18} \, J) = 8.72 \times 10^{-18} \, J$.

(B) In Rutherford's experiment,the deflection of $\alpha$-particles is caused by the electrostatic repulsion between the positively charged nucleus and the $\alpha$-particles.
Heavier atoms (like gold,$Z = 79$) have a higher nuclear charge compared to lighter atoms (like Aluminium,$Z = 13$).
Since the repulsive force is directly proportional to the nuclear charge $(F \propto Z_1 Z_2)$,using a foil of lighter atoms results in a significantly weaker repulsive force.
Therefore,the deflection of $\alpha$-particles would be much less compared to the results obtained with heavy metal foils.

(D) Wavelength of radiation emitted $(\lambda) = 616 \, nm = 616 \times 10^{-9} \, m$ (Given).
$(a)$ Frequency of emission $(v) = \frac{c}{\lambda} = \frac{3.0 \times 10^{8} \, m/s}{616 \times 10^{-9} \, m} = 4.87 \times 10^{14} \, s^{-1}$.
$(b)$ Distance traveled by radiation in $30 \, s = \text{velocity} \times \text{time} = (3.0 \times 10^{8} \, m/s) \times (30 \, s) = 9.0 \times 10^{9} \, m$.
$(c)$ Energy of a quantum $(E) = hv = (6.626 \times 10^{-34} \, J \cdot s) \times (4.87 \times 10^{14} \, s^{-1}) = 3.227 \times 10^{-19} \, J$.
$(d)$ Number of quanta in $2 \, J$ of energy $= \frac{\text{Total Energy}}{\text{Energy of one quantum}} = \frac{2 \, J}{3.227 \times 10^{-19} \, J} = 6.197 \times 10^{18} \approx 6.2 \times 10^{18}$ quanta.

(N/A) The energy of a single photon $(E)$ is given by the formula:
$E = \frac{hc}{\lambda}$
Where:
$h = 6.626 \times 10^{-34} \, J \cdot s$ (Planck's constant)
$c = 3 \times 10^{8} \, m \cdot s^{-1}$ (Speed of light)
$\lambda = 600 \times 10^{-9} \, m$ (Wavelength)
Calculating the energy of one photon:
$E = \frac{(6.626 \times 10^{-34} \, J \cdot s)(3 \times 10^{8} \, m \cdot s^{-1})}{600 \times 10^{-9} \, m} = 3.313 \times 10^{-19} \, J$
The total number of photons $(n)$ is the total energy divided by the energy of one photon:
$n = \frac{\text{Total Energy}}{E} = \frac{3.15 \times 10^{-18} \, J}{3.313 \times 10^{-19} \, J} \approx 9.5$
Rounding to the nearest whole number,the detector receives approximately $10$ photons.

The energy of a photon is given by $E_{photon} = h\nu$. However,the frequency $\nu$ is related to the duration of the pulse $\Delta t$ as $\nu = \frac{1}{\Delta t}$.
Given:
Duration $\Delta t = 2 \, ns = 2.0 \times 10^{-9} \, s$
Number of photons $N = 2.5 \times 10^{15}$
Planck's constant $h = 6.626 \times 10^{-34} \, J \cdot s$
Frequency $\nu = \frac{1}{2.0 \times 10^{-9} \, s} = 5.0 \times 10^{8} \, s^{-1}$
Total energy $E = N \times h \times \nu$
$E = (2.5 \times 10^{15}) \times (6.626 \times 10^{-34} \, J \cdot s) \times (5.0 \times 10^{8} \, s^{-1})$
$E = 8.2825 \times 10^{-10} \, J$
Thus,the energy of the source is $8.2825 \times 10^{-10} \, J$.

(N/A) $(i)$ Given,
Wavelength associated with the first transition,$\lambda_{1} = 589 \, nm = 589 \times 10^{-9} \, m$
Wavelength associated with the second transition,$\lambda_{2} = 589.6 \, nm = 589.6 \times 10^{-9} \, m$
Frequency of the first wavelength is $\nu_{1} = \frac{c}{\lambda_{1}} = \frac{3 \times 10^{8} \, m s^{-1}}{589 \times 10^{-9} \, m} = 5.093 \times 10^{14} \, s^{-1}$
Frequency of the second wavelength is $\nu_{2} = \frac{c}{\lambda_{2}} = \frac{3 \times 10^{8} \, m s^{-1}}{589.6 \times 10^{-9} \, m} = 5.088 \times 10^{14} \, s^{-1}$
$(ii)$ Energy difference between the two excited states is given as:
$\Delta E = h(\nu_{1} - \nu_{2})$
$\Delta E = 6.626 \times 10^{-34} \, J s \times (5.093 \times 10^{14} - 5.088 \times 10^{14}) \, s^{-1}$
$\Delta E = 6.626 \times 10^{-34} \, J s \times 0.005 \times 10^{14} \, s^{-1}$
$\Delta E = 3.313 \times 10^{-22} \, J$

It is given that the work function $(W_{0})$ for a caesium atom is $1.9 \, eV$.
$(a)$ From the expression $W_{0} = \frac{hc}{\lambda_{0}}$,we get:
$\lambda_{0} = \frac{hc}{W_{0}}$
Where,$\lambda_{0} =$ threshold wavelength,$h =$ Planck's constant,$c =$ velocity of light.
Substituting the values:
$\lambda_{0} = \frac{(6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1})}{1.9 \times 1.602 \times 10^{-19} \, J} = 6.53 \times 10^{-7} \, m = 653 \, nm$.
$(b)$ From the expression $W_{0} = hv_{0}$,we get:
$v_{0} = \frac{W_{0}}{h} = \frac{1.9 \times 1.602 \times 10^{-19} \, J}{6.626 \times 10^{-34} \, Js} = 4.593 \times 10^{14} \, s^{-1}$.
$(c)$ For irradiation with $\lambda = 500 \, nm$:
Kinetic energy $(K.E.) = \frac{hc}{\lambda} - W_{0} = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}\right)$
$K.E. = (6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1}) \left(\frac{1}{500 \times 10^{-9} \, m} - \frac{1}{653 \times 10^{-9} \, m}\right) = 9.315 \times 10^{-20} \, J$.
Since $K.E. = \frac{1}{2} mv^{2}$,where $m = 9.109 \times 10^{-31} \, kg$:
$v = \sqrt{\frac{2 \times K.E.}{m}} = \sqrt{\frac{2 \times 9.315 \times 10^{-20} \, J}{9.109 \times 10^{-31} \, kg}} = 4.52 \times 10^{5} \, ms^{-1}$.

(A) The kinetic energy of the emitted electrons is given by the Einstein photoelectric equation: $h\nu - h\nu_{0} = \frac{1}{2}mv^{2}$,which can be written as $hc(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}) = \frac{1}{2}mv^{2}$.
Using the data provided for $\lambda = 500 \ nm$ $(v = 2.55 \times 10^{5} \ cm/s = 2.55 \times 10^{3} \ m/s)$ and $\lambda = 400 \ nm$ $(v = 5.35 \times 10^{3} \ m/s)$:
$hc(\frac{1}{500 \times 10^{-9}} - \frac{1}{\lambda_{0}}) = \frac{1}{2}m(2.55 \times 10^{3})^{2}$ $(I)$
$hc(\frac{1}{400 \times 10^{-9}} - \frac{1}{\lambda_{0}}) = \frac{1}{2}m(5.35 \times 10^{3})^{2}$ $(II)$
Dividing $(II)$ by $(I)$ and solving for $\lambda_{0}$ gives $\lambda_{0} \approx 540 \ nm$.
Substituting $\lambda_{0}$ back into equation $(I)$ allows for the calculation of Planck's constant $h$,yielding $h \approx 6.626 \times 10^{-34} \ J \ s$.

(N/A) The energy of the incident photon $(E)$ is given by $E = \frac{hc}{\lambda}$.
$E = \frac{(6.626 \times 10^{-34} \, Js)(3.0 \times 10^{8} \, ms^{-1})}{150 \times 10^{-12} \, m} = 1.3252 \times 10^{-15} \, J = 13.252 \times 10^{-16} \, J$.
The kinetic energy of the ejected electron $(K.E)$ is given by $K.E = \frac{1}{2} m_{e} v^{2}$.
$K.E = \frac{1}{2} (9.109 \times 10^{-31} \, kg)(1.5 \times 10^{7} \, ms^{-1})^{2} = 1.0248 \times 10^{-16} \, J$.
The binding energy $(B.E)$ is the difference between the incident photon energy and the kinetic energy of the ejected electron:
$B.E = E - K.E = 13.252 \times 10^{-16} \, J - 1.0248 \times 10^{-16} \, J = 12.2272 \times 10^{-16} \, J$.
Converting to $eV$:
$B.E = \frac{12.2272 \times 10^{-16} \, J}{1.602 \times 10^{-19} \, J/eV} \approx 7632 \, eV \approx 7.63 \times 10^{3} \, eV$.

(A) Given wavelength $\lambda = 1285 \, nm = 1285 \times 10^{-9} \, m$.
Frequency $\nu = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \, m/s}{1285 \times 10^{-9} \, m} \approx 2.334 \times 10^{14} \, Hz$.
Using the formula $\nu = 3.29 \times 10^{15} \left(\frac{1}{9} - \frac{1}{n^2}\right)$:
$2.334 \times 10^{14} = 3.29 \times 10^{15} \left(\frac{1}{9} - \frac{1}{n^2}\right)$.
$\frac{2.334 \times 10^{14}}{3.29 \times 10^{15}} = \frac{1}{9} - \frac{1}{n^2}$.
$0.07094 = 0.1111 - \frac{1}{n^2}$.
$\frac{1}{n^2} = 0.1111 - 0.07094 = 0.04016$.
$n^2 = \frac{1}{0.04016} \approx 24.9$.
$n = \sqrt{24.9} \approx 5$.
Since the Paschen series corresponds to transitions ending at $n=3$,and the wavelength $1285 \, nm$ is in the infrared range,the transition is from $n=5$ to $n=3$ and lies in the infrared region.

(D) The radius of the $n^{\text{th}}$ orbit of a hydrogen-like species is given by $r = 0.0529 \times \frac{n^2}{Z} \, nm = 52.9 \times \frac{n^2}{Z} \, pm$.
For the initial orbit $(r_1 = 1.3225 \, nm = 1322.5 \, pm)$:
$n_1^2 = \frac{1322.5 \times Z}{52.9} = 25Z$.
For the final orbit $(r_2 = 211.6 \, pm)$:
$n_2^2 = \frac{211.6 \times Z}{52.9} = 4Z$.
Taking the ratio: $\frac{n_1^2}{n_2^2} = \frac{25Z}{4Z} = 6.25$.
$\frac{n_1}{n_2} = \sqrt{6.25} = 2.5 = \frac{5}{2}$.
Thus, $n_1 = 5$ and $n_2 = 2$.
The transition is from $n = 5$ to $n = 2$, which belongs to the $\text{Balmer}$ series, and the spectral region is the visible region.
Using the Rydberg formula for wave number $(\bar{\nu})$:
$\bar{\nu} = R_H \left( \frac{1}{n_2^2} - \frac{1}{n_1^2} \right) = 1.097 \times 10^7 \, m^{-1} \left( \frac{1}{2^2} - \frac{1}{5^2} \right)$.
$\bar{\nu} = 1.097 \times 10^7 \times \left( \frac{1}{4} - \frac{1}{25} \right) = 1.097 \times 10^7 \times \frac{21}{100} = 2.3037 \times 10^6 \, m^{-1}$.
Wavelength $(\lambda)$ = $\frac{1}{\bar{\nu}} = \frac{1}{2.3037 \times 10^6} \, m \approx 4.34 \times 10^{-7} \, m = 434 \, nm$.

(N/A) The major problems faced by scientists after the discovery of sub-atomic particles were:
$1$. To account for the stability of the atom.
$2$. To compare the behavior of one element with another in terms of both physical and chemical properties.
$3$. To explain the formation of different kinds of molecules by the combination of different atoms.
$4$. To understand the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms.