Atoms and Nuclei

Given, binding energy (BE) per

nucleon $=7.14 \mathrm{MeV}$

BE of element $=28.6 \mathrm{MeV}$

We know that,

$ \begin{aligned} \text { Number of nucleons } & =\frac{\text { BE of element }}{\text { BE per nucleon }} \\ & =\frac{28.6}{7.14}=4 \end{aligned} $

Wavelength for Lyman series is calculated as

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

For first line of Lyman series, $n=2$

$ \begin{array}{ll} \therefore & \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \frac{1}{\lambda}=\frac{3 R}{4} \\ \Rightarrow & \lambda=\frac{4}{3 R} \Rightarrow R=\frac{4}{3 \lambda} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{array} $

Wavelength of Balmer series is calculated as

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

For first line of Balmer series, $n=3$

$ \begin{array}{ll} \therefore & \frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{36} R \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} \times \frac{4}{3 \lambda}\,\,\,\,\,\text { [from Eq. (i)] } \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{27 \lambda} \Rightarrow \lambda^{\prime}=\frac{27 \lambda}{5} \end{array} $

Half-life of radioactive isotope,

$ T_{1 / 2}=30 \mathrm{~h} $

If $N_0$ be the initial amount of radioactive isotope, then remaining amount $(N)$ in time $t$ is given as

$ N=125 \% \text { of } N_0=\frac{125}{100} \times N_0 \Rightarrow N=\frac{N_0}{8} $

We know that,

$ \begin{aligned} \quad N & =N_0\left(\frac{1}{2}\right)^n \Rightarrow \frac{N_0}{8}=N_0\left(\frac{1}{2}\right)^n \\ \Rightarrow \quad\left(\frac{1}{2}\right)^3 & =\left(\frac{1}{2}\right)^n \Rightarrow n=3 \Rightarrow \frac{t}{T_{1 / 2}}=3 \\ \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad t & =3 \times T_{1 / 2}=3 \times 30=90 \mathrm{~h} \end{aligned} $

In atomic scale, the order of magnitude of forces in nature are related as,

Strong nuclear force> electromagnetic force > weak nuclear force > gravitational force

Hence, gravitational force is weakest force in nature in atomic scale.

In spectral line series, wavelength is given by

$ \frac{1}{\lambda}=R\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right) $

For Balmer transitions, $n_f=2$

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

$ \begin{aligned} & \text { Given, } \lambda=16 / 3 R \\ & \Rightarrow \frac{3 R}{16}=R\left(\frac{1}{4}-\frac{1}{n^2}\right) \Rightarrow \frac{3}{16}=\frac{1}{4}-\frac{1}{n^2} \\ & \Rightarrow \frac{1}{n^2}=\frac{1}{4}-\frac{3}{16}=\frac{1}{16} \Rightarrow n=4 \end{aligned} $

Decay scheme of given sample is as shown below

So, time required by sample to reduced to 7.5 g is

$ \Delta t=3 \times T_{1 / 2}=3 \times 5=15 \mathrm{~s} $

Nuclear forces are effective within the nucleus (radius $\approx 10^{-15} \mathrm{~m}$ or less) and these are responsible for keeping protons within this small volume. Hence, nuclear forces are short range attractive in nature.

Wavelength of spectral lines of Hydrogen like atom is given as

$ \frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right) $

For Balmer series, $n_1=2$ and $n_2=3,4,5,6, \ldots$.

$ \therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{2^2}-\frac{1}{n_2^2}\right) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

For minimum wavelength, $n_2=\infty$ From Eq. (i),

$ \begin{aligned} \frac{1}{\lambda_{\min }} & =R\left(\frac{1}{2^2}-\frac{1}{\infty^2}\right) \\ \frac{1}{\lambda_{\min }} & =\frac{R}{4} \\ \Rightarrow \quad \lambda_{\min } & =\frac{4}{R} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

For maximum wavelength, $n_2=3$

∴ From Eq. (i), we have

$ \begin{aligned} \frac{1}{\lambda_{\max }} & =R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=R\left(\frac{5}{36}\right) \\ \Rightarrow \lambda_{\max } & =\frac{36}{5 R}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

∴ From Eq. (ii) and (iii), we have

$ \frac{\lambda_{\max }}{\lambda_{\min }}=\frac{36 / 5 R}{4 / R}=\frac{36}{20}=\frac{9}{5} $

$\alpha$-rays are a bunch of particles consisting of 2 neutrons and 2 protons. Hence, a $\alpha$-particle is a doubly ionised helium-atom.

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.

(i) Radius of the orbit is given by

$r = {{{h^2}} \over {4\pi m{e^2}}} \times {{{n^2}} \over Z}$

Only, m_$\mu$ = 200 m_e rest are same. So, statement (A) is correct: The radius of muonic orbit is 200 times smaller than that of the electron.

(ii) Velocity of particle in an orbit is given by

$v = {{nh} \over {2\pi mr}} = {{nh \times 4\pi m{e^2} \times Z} \over {2\pi m \times {h^2} \times {n^2}}} = {{2{e^2}Z} \over {hn}}$

Therefore, speed does not change as it does not depend on mass. So, statement (B) is incorrect.

(iii) Ionisation energy is given by

$\Delta {E_n} = - {{2{\pi ^2}m{e^4}{Z^2}} \over {{h^2}{{(4\pi {\varepsilon _0})}^2}}} \times \left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$

Only, m_$\mu$ = 200 m_e rest are same. So, statement (C) is correct: The ionisation energy of muonic atom is 200 times more than that of an hydrogen atom.

(iv) Since Momentum $\propto$ Energy. So, statement (D) is correct: The momentum of the muon in he nth orbit is 200 times more than that of the electron.

The two nuclei have velocity in ratio 8 : 27. By conservation of momentum, we have

${m_1}{v_1} = {m_2}{v_2} \Rightarrow {{{v_1}} \over {{v_2}}} = {{{m_2}} \over {{m_1}}} \Rightarrow {{{m_2}} \over {{m_1}}} = {8 \over {27}}$

Now, since $m = \rho {4 \over 3}\pi {r^3}$

Therefore, ${{{m_2}} \over {{m_1}}} = {{\rho {4 \over 3}\pi r_2^3} \over {\rho {4 \over 3}\pi r_1^3}} \Rightarrow {{{m_2}} \over {{m_1}}} = {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} \Rightarrow {\left( {{{{r_2}} \over {{r_1}}}} \right)^3} = {8 \over {27}}$

$ \Rightarrow {{{r_2}} \over {{r_1}}} = {2 \over 3}$

Thus, ratio of radii of nuclei ${r_1}:{r_2} = 3:2$.

Given,

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

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 speedy of the particle in the orbit of an atom is $v = {{{I^2}} \over {2h{\varepsilon _0}}}$

We have the radius of the first orbit is

$r = {{{h^2}{\varepsilon _0}} \over {\pi m{e^2}}}$

$ \Rightarrow {\varepsilon _0} = {{r\pi m{e^2}} \over {{h^2}}}$

Therefore, the acceleration of an electron in the first orbit of the hydrogen atom (n = 1) is

${{{v^2}} \over r} = {{{I^6}\pi m} \over {4{h^2}\varepsilon _0^3}} = {{{h^2}} \over {4{r^3}{\pi ^2}{m^2}}}$

The power can be calculated by the relation

$P = {E \over {\Delta t}} = {{\Delta m{c^2}} \over {\Delta t}}$ ...... (1)

Therefore, from Eq. (1), the mass of the fuel consumed per hour in the reactor is

${{\Delta m} \over {\Delta t}} = {P \over {{c^2}}} = {{{{10}^9}} \over {{{(3 \times {{10}^8})}^2}}} = 4 \times {10^{ - 12}}$ g