Atoms and Nuclei

For an electron of mass $m$ and charge $e$ orbiting a stationary proton (charge $+e$ ) in a circular path of radius $r_k$ with speed $\mathrm{v}_{\mathrm{k}}$.

The electrostatic force provides the necessary centripetal acceleration,

$ \frac{\mathrm{m} \mathrm{v}_{\mathrm{k}}^2}{\mathrm{r}_{\mathrm{k}}}=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}^2} \Rightarrow \mathrm{~m} \mathrm{v}_{\mathrm{k}}^2=\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}} $

The kinetic energy of an electron in its orbit is,

$ \mathrm{K}_{\mathrm{k}}=\frac{1}{2} \mathrm{~m} \mathrm{v}_{\mathrm{k}}^2=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}} $

The potential energy of an electron in its orbit is,

$ \mathrm{U}_{\mathrm{k}}=-\frac{\mathrm{e}^2}{4 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}} $

So, the total energy of an electron in hydrogen atom in $\mathrm{k}^{\text {th }}$ orbit is,

$ \mathrm{E}_{\mathrm{k}}=\mathrm{K}_{\mathrm{k}}+\mathrm{U}_{\mathrm{k}}=-\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}}=-\mathrm{K}_{\mathrm{k}} $

According to Bohr's second postulate the orbital angular momentum is quantized as an integral multiple of $\frac{h}{2 \pi}$ :

$ \mathrm{L}_{\mathrm{k}}=\mathrm{m} \mathrm{v}_{\mathrm{k}} \mathrm{r}_{\mathrm{k}}=\frac{\mathrm{kh}}{2 \pi} \Rightarrow \mathrm{~m}=\frac{\mathrm{kh}}{2 \pi \mathrm{v}_{\mathrm{k}} \mathrm{r}_{\mathrm{k}}} $

For the ground state ( $\mathrm{k}=1$ )

$ \mathrm{m}=\frac{\mathrm{h}}{2 \pi \mathrm{v}_1 \mathrm{r}_1} $

For the $\mathrm{n}^{\text {th }}$ state ( $\mathrm{k}=\mathrm{n}$ )

$ \mathrm{m}=\frac{\mathrm{nh}}{2 \pi \mathrm{v}_{\mathrm{n}} \mathrm{r}_{\mathrm{n}}} $

When the electron undergoes a transition from a higher state n to the ground state 1 the magnitude of change in kinetic energy is:

$ |\Delta K|=\left|K_n-K_1\right| $

Since the mass of the electron is constant, $\mathrm{K}_{\mathrm{k}}=\frac{1}{2} \mathrm{~L}_{\mathrm{k}} \omega_{\mathrm{k}}$, where angular velocity $\omega_{\mathrm{k}}=\frac{\mathrm{v}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{k}}}$ :

$ \mathrm{K}_{\mathrm{k}}=\frac{1}{2}\left(\frac{\mathrm{k} \mathrm{~h}}{2 \pi}\right)\left(\frac{\mathrm{v}_{\mathrm{k}}}{\mathrm{r}_{\mathrm{k}}}\right)=\frac{\mathrm{k} \mathrm{hv}_{\mathrm{k}}}{4 \pi \mathrm{r}_{\mathrm{k}}} $

For $\mathrm{k}=\mathrm{n}$

$ \mathrm{K}_{\mathrm{n}}=\frac{\mathrm{nhv}_{\mathrm{n}}}{4 \pi \mathrm{r}_{\mathrm{n}}} $

And for $\mathrm{k}=1$ :

$ \mathrm{K}_1=\frac{\mathrm{hv}_1}{4 \pi \mathrm{r}_1} $

Therefore, the magnitude of the change in kinetic energy is:

$ |\Delta \mathrm{K}|=\left|\frac{\mathrm{nh}}{4 \pi} \frac{\mathrm{v}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}-\frac{\mathrm{h}}{4 \pi} \frac{\mathrm{v}_1}{\mathrm{r}_1}\right|=\frac{\mathrm{h}}{4 \pi}\left|\frac{\mathrm{n}_{\mathrm{n}}}{\mathrm{r}_{\mathrm{n}}}-\frac{\mathrm{v}_1}{\mathrm{r}_1}\right| $

Thus, option (A) is correct

The de Broglie wavelength $\lambda_{\mathrm{k}}$ of the electron is given by:

$ \lambda_{\mathrm{k}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{k}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}_{\mathrm{k}}}} $

From the kinetic energy equation

$ \mathrm{K}_{\mathrm{k}}=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{r}_{\mathrm{k}}} \Rightarrow \mathrm{r}_{\mathrm{k}}=\frac{\mathrm{e}^2}{8 \pi \epsilon_0 \mathrm{~K}_{\mathrm{k}}} $

Using Bohr's quantization condition $2 \pi \mathrm{r}_{\mathrm{k}}=\mathrm{k} \lambda_{\mathrm{k}}$ :

$ \lambda_k=\frac{2 \pi r_k}{k}=\frac{2 \pi}{k}\left(\frac{e^2}{8 \pi \epsilon_0 K_k}\right)=\frac{e^2}{4 \epsilon_0 k_k} $

The change in wavelength between the two levels is :

$ |\Delta \lambda|=\left|\lambda_n-\lambda_1\right|=\frac{\mathrm{e}^2}{4 \epsilon_0}\left|\frac{1}{\mathrm{nK}_{\mathrm{n}}}-\frac{1}{1 \cdot \mathrm{~K}_1}\right| $

$ |\Delta \lambda|=\frac{\mathrm{e}^2}{4 \epsilon_0}\left|\frac{1}{\mathrm{nK}_{\mathrm{n}}}-\frac{1}{\mathrm{~K}_1}\right| $

Thus, option (B) is incorrect.

According to Bohr's frequency condition, the energy of the photon emitted during a transition from level n to level 1 is equal to the change in total energy :

$ \mathrm{h} \nu=\mathrm{E}_{\mathrm{n}}-\mathrm{E}_1=\left(-\mathrm{K}_{\mathrm{n}}\right)-\left(-\mathrm{K}_1\right)=\mathrm{K}_1-\mathrm{K}_{\mathrm{n}} $

The kinetic energy in $k^{ ext{th}}$ energy level is $K_k = \frac{e^2}{8\pi \epsilon_0 r_k}$,

$h\nu = \frac{e^2}{8\pi \epsilon_0 r_1} - \frac{e^2}{8\pi \epsilon_0 r_n} = \frac{e^2}{8\pi \epsilon_0} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$

$\nu = \frac{e^2}{8\pi \epsilon_0 h} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$

Therefore, the frequency of radiation emitted can be expressed as $\frac{e^2}{8\pi \epsilon_0 h} \left( \frac{1}{r_1} - \frac{1}{r_n} \right)$. Thus, option (C) is correct.

The magnitude of the change in total energy is exactly equal to the magnitude of the change in kinetic energy because $|E_k| = K_k$.

$|\Delta E| = |\Delta K| = \frac{h}{4\pi} \left| \frac{n\,v_n}{r_n} - \frac{v_1}{r_1} \right|$

Hence, option (D) is incorrect.

Therefore, the correct options are (A) and (C).

The central force will provide necessary centripetal force

$\Rightarrow \mathrm{kr}=\frac{\mathrm{mv}^2}{\mathrm{r}}$

or, $\mathrm{kr}^2=\mathrm{mv}^2\quad \text{... (1)}$

By quantisation rule $\mathrm{n} \hbar=\mathrm{mvr}$

or, $\frac{\mathrm{n} \hbar}{\mathrm{r}}=\mathrm{mv}\quad \text{... (2)}$

$\begin{aligned} & \frac{(1)}{(2)^2} \Rightarrow \frac{\mathrm{kr}^2}{\frac{\mathrm{n}^2 \hbar^2}{\mathrm{r}^2}}=\frac{\mathrm{mv}^2}{\mathrm{~m}^2 \mathrm{v}^2} \\ & \Rightarrow \frac{\mathrm{k}}{\mathrm{n}^2 \hbar^2} \mathrm{r}^4=\frac{1}{\mathrm{~m}} \\ & \Rightarrow \mathrm{r}=\left(\frac{\mathrm{n}^2 \hbar^2}{\mathrm{~km}}\right)^{\frac{1}{4}} \Rightarrow \mathrm{r}^2=\frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}} \end{aligned}$

(B) Using (1), $\mathrm{K} \cdot \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}=\mathrm{mv}^2$

$\Rightarrow \mathrm{v}^2=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}^3}}$

(C) $\frac{\mathrm{L}}{\mathrm{mr}^2}=\frac{\mathrm{mvr}}{\mathrm{mr}^2}=\frac{\mathrm{v}}{\mathrm{r}}=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$ from (1)

(D) $\mathrm{E}=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{kr}^2=\frac{\mathrm{n} \hbar}{2} \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}+\frac{1}{2} \mathrm{k} \frac{\mathrm{n} \hbar}{\sqrt{\mathrm{mk}}}$

$\mathrm{E}=\mathrm{n} \hbar \sqrt{\frac{\mathrm{k}}{\mathrm{m}}}$

Total binding energy (without considering repulsions),

${E_b} = [Z{m_p} + (A - Z){m_n} - {m_x}]{c^2}$

Where, $_Z^AX$ is the nuclei under consideration.

Now, considering repulsion :

Number of proton pairs $ = {}^z{C_2}$

$\Rightarrow$ This repulsion energy $ \propto \,{{Z(Z - 1)} \over 2} \times {1 \over {4\pi { \in _0}}}{{{e^2}} \over R}$

Where R is the radius of the nucleus

$ \Rightarrow E_b^p - E_b^n \propto Z(Z - 1)$ $\therefore$ there will be no repulsion term for neutrons.

Also, since $R = {R_0}{A^{1/3}}$

$ \Rightarrow E_b^p - E_b^n \propto {A^{ - 1/3}}$

Because of repulsion among protons,

$E_b^p < E_b^n$

Since in $\beta^+$ decay, number of protons decrease $\Rightarrow$ repulsion would decrease

$ \Rightarrow E_b^p$ increases

(p_total) = 0 $\Rightarrow$ (p_total)_f = 0

So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.

Energy released = ($\Delta$m)c² = $\delta$c² = KE_P + KE_Q

$K{E_P} = {{{p^2}} \over {2{m_P}}},K{E_Q} = {{{p^2}} \over {2{m_Q}}}$

$ \Rightarrow K{E_P}:K{E_Q} = {1 \over {{m_P}}}:{1 \over {{m_Q}}} = {m_Q}:{m_P}$

$K{E_P} = \left( {{{{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_Q} = \left( {{{{m_P}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$

$K{E_P} + K{E_Q} = \delta {c^2}$

${{{p^2}} \over {2{m_P}}} + {{{p^2}} \over {2{m_Q}}} = \delta {c^2} \Rightarrow p = c\sqrt {2\left( {{{{m_P}{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta } $

For Balmer series

$ \frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{n^2}\right) n>2 $

$\lambda$ is minimum for transition from $n_1 \rightarrow \infty$ to $n_2=2$.

$ \frac{1}{\lambda_{\min }}=\frac{R}{4} $

$\lambda$ is maximum for transition from $n_1=3$ to $n_2=2$

$ \frac{1}{\lambda_{\max }}=\mathrm{R}\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5 \mathrm{R}}{36} $

Dividing the above two equations, we get

$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{R}{4} \times \frac{36}{5 R}=\frac{9}{5} $

Option (A) is correct.

For Lyman series,

$\lambda_{\min }=91.18 \mathrm{~nm}$

$\lambda_{\max }=121.57 \mathrm{~nm}$

For Balmer series,

$\lambda_{\min }=364.60 \mathrm{~nm}$

$\lambda_{\max }=656.30 \mathrm{~nm}$

For Pashen series,

$\lambda_{\min }=820.40 \mathrm{~nm}$

$\lambda_{\max }=1875 \mathrm{~nm}$

We can see from the above that there is no overlap between the wavelength ranges of Balmer and Paschen series. so, option (B) is incorrect.

Similarly wavelength ranges of Lyman and Balmer series also do not overlap. Hence, option (D) is correct.

Shortest wavelength of Lyman series

$ \begin{aligned} \frac{1}{\lambda_0} & =R\left(1-\frac{1}{\infty}\right) \\\\ \lambda_0 & =\frac{1}{R} \end{aligned} $

For Lyman series, $\frac{1}{\lambda}=\mathrm{R}\left(1-\frac{1}{n^2}\right)$

$ \begin{aligned} & =\frac{1}{\lambda_0}\left(1-\frac{1}{n^2}\right) \\\\ \lambda & =\frac{\lambda_0}{1-\frac{1}{n^2}}=\frac{\lambda_0 n^2}{n^2-1} \\\\ & =\frac{\lambda_0\left(n^2-1\right)+1}{n^2-1} \\\\ & =\lambda_0\left(1+\frac{1}{n^2-1}\right) \end{aligned} $

If $n^2-1=m^2$ ( $m$ is an integer)

then this is rarely true as $\left(n^2-1\right)$ cannot always be the square of an integer. Hence, option (C) is incorrect.

$\lambda \propto {1 \over V}$

So, when V double, then cut-off wavelength become half. Characteristic does not depend on voltage. When I is halved, then intensity will decrease.

$|Force| = \left| { - {{dV} \over {dr}}} \right| \Rightarrow F = {{m{v^2}} \over r}$ .....(i)

$mvr = {{nh} \over {2\pi }}$ .....(ii)

From Eqs. (i) and (ii), we get

$ \therefore $ $F = {m \over r}{\left( {{{nh} \over {2\pi mr}}} \right)^2}$ $ \because $ $\left( {v = {{nh} \over {2\pi mr}}} \right)$

or $F = {{m{n^2}{h^2}} \over {4{\pi ^2}{m^2}{r^3}}}$

or $F = {{{n^2}{h^2}} \over {4{\pi ^2}m{r^3}}}$

${r^3} \propto \,{n^2}$ (as F = constant)

$r \propto \,{n^{2/3}}$

$v = {{nh} \over {2\pi mr}}$

$v \propto {n \over {{n^{2/3}}}}$

or $v \propto {n^{1/3}}$

$ \therefore $ $KE = {1 \over 2}m{v^2} = {{Fr} \over 2}$

$ \because $ $TE = PE + KE = Fr + {{Fr} \over 2}$

$ \Rightarrow {{3Fr} \over 2} = TE$

or $TE = E = {{3F} \over 2} \times {\left( {{{{n^2}{h^2}} \over {4{\pi ^2}rnF}}} \right)^{1/3}}$

$ = {3 \over 2}{\left( {{{{n^2}{h^2}{F^2}} \over {4{\pi ^2}m}}} \right)^{1/3}}$

A hydrogen atom in its ground state (n = 1) absorbs a photon with wavelength $\lambda_a$ and gets excited to the n = 4 state. Subsequently, the electron transitions to the n = m state, emitting a photon with wavelength $\lambda_e$. Given that ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$, let's analyze the options systematically.

Calculation for the Wavelengths:

The energy difference between levels in a hydrogen atom can be given by the Rydberg formula for hydrogen:

$ E_n = - \frac{{13.6 \ eV}}{{n^2}} $

The energy of the absorbed photon that excites the electron from n = 1 to n = 4 could be expressed as:

$ E_a = E_4 - E_1 = - \frac{{13.6 \ eV}}{{4^2}} - (- \frac{{13.6 \ eV}}{{1^2}}) = - \frac{{13.6 \ eV}}{{16}} + 13.6 \ eV = 13.6 \ eV (1 - \frac{1}{16}) = 13.6 \ eV \cdot \frac{15}{16} = 12.75 \ eV$

Wavelength $\lambda_a$ associated with this energy is:

$ \lambda_a = \frac{{hc}}{{E_a}} = \frac{{1242 \ nm \cdot eV}}{{12.75 \ eV}} = 97.4 \ nm$

Given ${{{\lambda _a}} \over {{\lambda _e}}} = {1 \over 5}$, we get:

$ \lambda_e = 5 \lambda_a = 5 \cdot 97.4 \ nm = 487 \ nm$

Option A: The ratio of kinetic energy of the electron in the state n = m to the state n = 1 is ${1 \over 4}$

The kinetic energy of an electron in a hydrogen atom in the n^th state is proportional to $- \frac{{1}}{{n^2}}$. Therefore, the ratio of kinetic energy in state m to that in state n can be written as:

$ \frac{{E_m}}{{E_1}} = \frac{{- \frac{{13.6}}{{m^2}} eV}}{{- 13.6 \ eV}} = \frac{1}{{m^2}} $

According to the question, the ratio is $\frac{1}{4}$:

$ \frac{1}{{m^2}} = \frac{1}{4} \Rightarrow m = 2 $

Therefore, Option A is correct if m = 2.

Option B: m = 2

Based on the energy calculations above, we find m = 2. Therefore, Option B is correct.

Option C: ${{\Delta {p_a}} \over {\Delta {p_e}}} = {1 \over 2}$

The change in momentum for each photon is given by:

$ \Delta p = \frac{h}{\lambda} $

So, the ratio of the momentum changes is:

$ \frac{{\Delta p_a}}{{\Delta p_e}} = \frac{{\frac{h}{\lambda_a}}}{{\frac{h}{\lambda_e}}} = \frac{\lambda_e}{\lambda_a} = 5 $

This shows the given option C (i.e., $\frac{1}{2}$) is incorrect. The correct ratio is 5.

Option D: $\lambda_e = 418 \ nm$

From our previous calculations, we found $\lambda_e = 487 \ nm$, which does not match 418 nm. Option D is incorrect.

Hence, the correct options are A and B.

Given,

${}_{90}^{232}Th\buildrel {} \over \longrightarrow {}_{82}^{212}pb$

where N_$\alpha$ = number of $\alpha$ particles, N_$\beta$ = number of $\beta$ particles

We know that $\alpha$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z - 2}^{A - 4}Y + {\alpha ^ - }$ particle

and $\beta$-decay is given as

${}_Z^AX\buildrel {} \over \longrightarrow {}_{Z + 1}^AX + {\beta ^ - }$ particle

In above decay process change in mass number A is 20 and change in atomic number Z is 8.

Therefore, change in mass number A $\Rightarrow$ number of $\alpha$-particles emitted are

${N_\alpha } = {{20} \over 4} = 5$

Now, emission of 5 $\alpha$ particles implies atomic number reduces by 10 but change in atomic number is 8, therefore, number of $\beta$-particles emitted is 2.

Therefore, N_$\alpha$ = 5 and N_$\beta$ = 2

So, option (A) and option (C) are correct.

The radius of $n^{\text {th }}$ orbital for a hydrogen-like atom of atomic number $Z$ is given by $r_n={{{n^2}{a_0}} \over Z}$, where $a_0=0.53 $$\mathop A\limits^o $ is the Bohr's radius.

$ \therefore $ $ \text { Radius of orbit, } r \propto \frac{n^2}{Z} $

The relative change in the radii of two consecutive orbitals is

$ \frac{r_{n+1}-r_n}{r_n}=\frac{\frac{(n+1)^2}{Z}-\frac{n^2}{Z}}{\frac{n^2}{Z}}=\frac{2 n+1}{n^2} \approx \frac{2}{n} \quad(\because n>>1) $

The energy of the n^th orbital is given by

${E_n} = {{ - 13.6{Z^2}} \over {{n^2}}}eV$

$ \therefore $ $ E_n \propto \frac{-Z^2}{n^2} $

The relative change in the energy of two consecutive orbitals is

$ \begin{aligned} & \frac{E_{n+1}-E_n}{E_n}=\frac{\frac{-Z^2}{(n+1)^2}+\frac{Z^2}{n^2}}{\frac{-Z^2}{n^2}}=-\frac{(2 n+1)}{(n+1)^2} \\\\ & \approx-\frac{2 n}{n^2}=-\frac{2}{n} \quad(n>>1) \end{aligned} $

The angular momentum of the $n^{\text {th }}$ orbital is given by $L_n={{nh} \over {2\pi }}$. The relative change in the angular momentum of two consecutive orbitals is

$ \frac{L_{n+1}-L_n}{L_n}=\frac{(n+1) {{h} \over {2\pi }}-{{nh} \over {2\pi }}}{{nh} \over {2\pi }}=\frac{1}{n} $

The orbital angular momentum is

$L = {{nh} \over {2\pi }}$

${{3h} \over {2\pi }} = {{nh} \over {2\pi }} \Rightarrow n = 3$

The radius of the orbit is

$4.5({a_0}) = {a_0}\left( {{{{n^2}} \over Z}} \right)$

$Z = {{{n^2}} \over {4.5}} = {{{3^2}} \over {4.5}} = {9 \over {4.5}} = 2$

Thus, the possible transitions are 3$\to$2, 3$\to$1 and 2$\to$1. For the transition 3$\to$2, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {{1 \over 4} - {1 \over 9}} \right] = 4R\left[ {{{9 - 4} \over {36}}} \right] = {{5R} \over 9}$

$ \Rightarrow \lambda = {9 \over {5R}}$

For the transition 3$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 9}} \right] = 4R\left( {{8 \over 9}} \right) = {{32R} \over 9}$

$ \Rightarrow \lambda = {9 \over {32R}}$

For the transition 2$\to$1, the wavelength is

${1 \over \lambda } = R{(2)^2}\left[ {1 - {1 \over 4}} \right] = 4R\left( {{3 \over 4}} \right) = 3R$

$ \Rightarrow \lambda = {1 \over {3R}}$

When binding energy per nucleon increases for a nuclear process, energy is released.

(a) For 1 < A < 50, on fusion, mass number of the resulting nucleus will be less than 100. No energy will be released.

(b) For 51 < A < 100, on fusion, mass number of the resulting nucleus will be between 100 and 200. As BE increases, energy will be released.

(c) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 and 100. As BE decreases, no energy will be released.

(d) On fission for 200 < A < 260, the mass number for the fission nuclei will be between 100 and 130. Since BE will increase, energy will be released.