Atoms and Nuclei

(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

To determine the minimum number of days after which the laboratory can be considered safe for use, we need to understand how the radioactive material decays over time.

The half-life of a radioactive material is the time it takes for half of the material to decay. For our given material, the half-life is 18 days. This means every 18 days, the amount of radioactive material is reduced to half of its previous amount.

Given that the initial radiation level is 64 times more than the permissible level, we can represent this situation mathematically. Let $ R $ be the initial radiation level and $ P $ be the permissible level of radiation. We are given:

$ R = 64P $

After each half-life period of 18 days, the amount of radioactive material will be halved. If $ n $ is the number of half-life periods required for the radiation to reduce to the permissible level, we can write:

$ \frac{R}{2^n} = P $

Using the given information ($ R = 64P $), we get:

$ \frac{64P}{2^n} = P $

Dividing both sides by $ P $, we get:

$ \frac{64}{2^n} = 1 $

This implies:

$ 2^n = 64 $

Knowing that $ 64 $ is a power of $ 2 $:

$ 64 = 2^6 $

Therefore, $ n = 6 $. This means it will take six half-life periods for the radiation to decay to a safe level.

Since each half-life period is 18 days, the total number of days required is:

$ 6 \times 18 = 108 $

Hence, the minimum number of days after which the laboratory can be considered safe for use is 108 days.

Answer: Option C (108)

The binding energies of $_7^{15}N$ and $_8^{15}O$ are given by

$B{E_N} = \Delta {m_N}{c^2} = (8{m_n} + 7{m_p} - {M_N}){c^2}$

$ = (8 \times 1.008665 + 7 \times 1.007825 - 15.000109 \times 931.5$

= 115.49 MeV,

$B{E_O} = \Delta {m_O}{c^2} = (7{m_n} + 8{m_p} - {M_O}){c^2}$

$ = (7 \times 1.008665 + 8 \times 1.007825 - 15.003065 \times 931.5$

= 111.95 MeV.

The difference in binding energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta BE = B{E_N} - B{E_O} = 115.49 - 111.95$

= 3.54 MeV. ........ (1)

The electrostatic energies of $_7^{15}N$ and $_8^{15}O$ are given by

${E_N} = {3 \over 5}{{Z(Z - 1){e^2}} \over {4\pi { \in _0}R}} = {3 \over 5}{{7(7 - 1)1.44} \over R}$

$ = {{36.288} \over R}$ MeV-fm

${E_O} = {3 \over 5}{{8(8 - 1)1.44} \over R} = {{48.384} \over R}$ MeV-fm.

The difference in electrostatic energies of $_7^{15}N$ and $_8^{15}O$ is

$\Delta E = {E_O} - {E_N} = (12.096/R)$ ...... (2)

Since, $\Delta E = \Delta BE$, equations (1) and (2) give $R = 12.096/3.54 = 3.42$ fm.

The nuclear fusion is responsible for energy production in stars via fusion of hydrogen nuclei into helium nuclei. In sun, the fusion takes place dominantly by proton-proton cycle, $4_1^1H \to _2^4He + 2{e^ + } + 2\nu + 2\gamma $. The neutrino ($\nu $) and $\gamma$-rays emissions are parts of this fusion reaction.

The uranium based fission reactions involve absorption of thermal neutrons by ${}_{92}^{235}$U nuclei to produce the highly fissionable ${}_{92}^{236}$U nuclei. This nuclei then fissions into two parts e.g., ${}_{92}^{236}$U $\to$ ${}_{63}^{137}$I + ${}_{39}^{97}$Y + 2n. The fission fragments are unstable and undergo $\beta$-decay to reduce their neutron to proton ratio. The fragments are generally formed in excited states and consequently emit $\gamma$-rays. The heavy water (D₂O) is used as a moderator to slow down the fast moving neutrons.

In $\beta$-decay, a neutron is converted into a proton. In this process, an electron and an antineutrino are created and emitted from the nucleus, n $\to$ p + e^$-$ + $\overline v $. The $\beta$-decay in ${}_{27}^{60}$Co $\to$ ${}_{28}^{60}$Ni + e^$-$ + $\overline v $. The daughter nuclei ${}_{28}^{60}$Ni is formed in excited state and comes to ground state by $\gamma$-ray emission.

The $\gamma$-rays are high energy electromagnetic rays. These rays are generally emitted when a nuclei in excited state (high energy) makes a transition to a lower state (low energy).

${}_{92}^{236}U \to {}_{54}^{140}Xe + {}_{38}^{94}Sr + x + y$

K_x = 2 MeV, K_y = 2 MeV, K_Xe = ?, K_Sr = ?

By conservation of charge number and mass number, x $\equiv$ y $\equiv$ n

B.E. per nucleon of ${}_{92}^{236}U = 7.5$ MeV

B.E. per nucleon of ${}_{54}^{140}Xe$ or ${}_{38}^{94}Sr = 8.5$ MeV

Q value of reaction,

Q = Net kinetic energy gained in the process

$ = {K_{Xe}} + {K_{Sr}} + 2 + 2 - 0 = {K_{Xe}} + {K_{Sr}} + 4$ ..... (1)

As number of nucleons is conserved in a reaction, so Q = Difference of binding energies of the nuclei

$ = 140 \times 8.5 + 94 \times 8.5 - 236 \times 7.5 = 219$ MeV ...... (2)

From eqns. (i) and (ii)

${K_{Xe}} + {K_{Sr}} = 219 - 4 = 215$ MeV ....... (3)

The linear momentum of a particle of mass m and kinetic energy K is given by $p = \sqrt {2mK} $. Since the masses and kinetic energies of x and y are very small in comparison to that of ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$, we can neglect the linear momentum of these particles. Initially, the linear momentum of ${}_{92}^{236}U$ is zero (at rest). Finally, the products ${}_{54}^{140}Xe$ and ${}_{38}^{94}Sr$ will move in opposite direction with equal linear momentum (by conservation of linear momentum). Thus,

$\sqrt {2{M_{Xe}}{K_{Xe}}} = \sqrt {2{M_{Sr}}{K_{Sr}}} $, i.e.,

$140{K_{Xe}} = 94{K_{Sr}}$. ...... (4)

Solve equations (3) and (4) to get K_Xe = 86 MeV and K_Sr = 129 MeV.

Let’s carefully match them:

• (a) Davisson and Germer Experiment → Demonstrated wave nature of electrons via electron diffraction. → (i)

• (b) Millikan’s Oil Drop Experiment → Measured charge of an electron. → (ii)

• (c) Rutherford Experiment → Showed the existence of nucleus in the atom through alpha-particle scattering. → (iv)

• (d) Franck–Hertz Experiment → Demonstrated quantisation of energy levels in atoms (mercury). → (iii)

So correct matching is:

(a)-(i), (b)-(ii), (c)-(iv), (d)-(iii)

✅ Answer: Option B

• Fresh wood: 20 decays/min (current activity if alive / just cut).

• Old museum sample: 2 decays/min.

• Half-life of C‑14 = 5730 years.

• Find the age of wooden piece in museum.

Step 1: Ratio of activities

$ \frac{A}{A_0} = \frac{2}{20} = 0.1 $

So, the museum sample’s activity is 10% of that in fresh wood.

Step 2: Radioactive decay law

$ A = A_0 e^{-\lambda t} $

$ \frac{A}{A_0} = e^{-\lambda t} $

So,

$ 0.1 = e^{-\lambda t} $

$ t = \frac{\ln(0.1)}{-\lambda} $

Step 3: Decay constant

$ \lambda = \frac{\ln 2}{T_{1/2}} = \frac{0.693}{5730 \,\text{yr}} \approx 1.2097 \times 10^{-4} \,\text{yr}^{-1} $

Step 4: Solve for $t$

$ t = \frac{\ln(0.1)}{-1.2097 \times 10^{-4}} $

$ \ln(0.1) = -2.3026 $

$ t = \frac{2.3026}{1.2097 \times 10^{-4}} $

$ t \approx 19039 \,\text{years} $

✅ Final Answer:

Option C: 19039 years

The wavelength of the K_$\alpha$ X-ray line is related to the atomic number of the element by Moseley's law, which can be expressed as:

$ \sqrt{\nu} = A(Z - B) $

Here, $\nu$ is the frequency of the emitted X-ray, $Z$ is the atomic number, and $A$ and $B$ are constants. Given that the wavelength $\lambda$ is inversely proportional to the frequency $\nu$ (since $\nu = \frac{c}{\lambda}$, where $c$ is the speed of light), we can rewrite this equation in terms of wavelength:

$ \lambda \propto \frac{1}{(Z - B)^2} $

Now, let's consider the ratio of the wavelengths for copper (Cu) and molybdenum (Mo):

$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{Z_{Mo} - B}{Z_{Cu} - B}\right)^2 $

For the K_$\alpha$ X-ray line, the constant $B$ can be approximated as 1. Using the atomic numbers $Z_{Cu} = 29$ and $Z_{Mo} = 42$, we can plug these values into the ratio:

$ \frac{\lambda_{Cu}}{\lambda_{Mo}} = \left(\frac{42 - 1}{29 - 1}\right)^2 = \left(\frac{41}{28}\right)^2 $

Calculating the above expression:

$ \frac{41}{28} \approx 1.464 $

$ \left(1.464\right)^2 \approx 2.14 $

Therefore, the ratio $\frac{\lambda_{Cu}}{\lambda_{Mo}}$ is close to 2.14. The correct option is:

Option B 2.14

We have

$m(_1^2H) + m(_2^4He) = 2.014102 + 4.002603 = 6.016705\,u$

$m(_3^6Li) = 6.015123\,u$

${m_1} + {m_2} > M$

Thus, option (A) is wrong.

$m(_1^1H) + m(_{83}^{209}Bi) = 1.007825\,u + 208.980388\,u = 209.988213\,u$

$m(_{84}^{210}Po) = 209.982876\,u$

${m_1} + {m_2} > M$

Thus, option (B) is wrong.

$m(_1^2H) + m(_2^4He) = 2.014102\,u + 4.002603\,u = 6.016705\,u$

$_3^6Li = 6.015123\,u$

$({m_3} + {m_4}) > M'$

Thus, option (C) is correct. Therefore, deuteron and alpha particle can go complete fusion.

$m(_{30}^{70}Zn) + _{34}^{82}Se = 69.925325\,u + 81.916709\,u = 151.842034\,u$

$_{64}^{152}Gd = 151.919803\,u$

${m_3} + {m_4} < M'$

Thus, option (D) is wrong.

The alpha decay of $_{84}^{210}Po$ is given by

$_{84}^{210}Po \to _{82}^{206}Pb + _2^4He$

The energy released during this process is

$Q = ({M_{Po}} - {M_{Pb}} - {M_{He}}){c^2}$

$ = (209.982876 - 205.974455 - 4.002603)\,u \times {c^2}$

$ = (0.005818\,u){c^2} = (0.005818\,u) \times 932\,MeV$

$ = 5.422\,MeV = 5422\,keV$

Kinetic energy of a particle, ${K_\alpha } = {{(A - 4)Q} \over A}$

${K_\alpha } = {{(210 - 4)} \over {210}} \times 5422\,keV = {{206} \over {210}} \times 5422\,keV = 5319\,keV$

In alpha decay, atomic number decreases by 2 and mass number decreases by 4.

P $\to$ 2

In $\beta$⁺ decay, atomic number decreases by 1 and mass number remains unchanged.

Q $\to$ 1

In fission, heavy nucleus breaks into two light nuclei.

R $\to$ 4

In proton emission, both atomic number and mass number decreases by 1.

S $\to$ 3

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}}$

${K_p} + {K_{\overline e }} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

When electron has zero kinetic energy is shared by antineutrino and proton.

Then, ${K_p} + {K_{\overline v }}$ = 0.8 $\times$ 10⁶ eV

As antineutrino is very light mass in comparison to proton so it will have almost contribution in total energy.

$\therefore$ Its energy is almost 0.8 $\times$ 10⁶ eV

Total energy remains conserved. Energy is shared by antineutrino, proton and electron. Kinetic energy of electron has continuous spectrum and it is maximum when antineutrino does not share any kinetic energy.

So total energy is shared with proton and electron only.

$\therefore$ K $\le$ 0.8 $\times$ 10⁶ eV

and kinetic energy of electron will be minimum or zero when total energy is shared by proton and antineutrino.

$\therefore$ 0 $\le$ K $\le$ 0.8 $\times$ 10⁶ eV

${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$

For singly-ionized helium atom, Z = 2. For hydrogen atom Z = 1.

For Balmer series n₁ = 2.

For first spectral line of hydrogen :

${1 \over {6561}} = {R_H} \times {(1)^2}\left[ {{1 \over {{2^2}}} - {1 \over {{3^2}}}} \right] = {{5R} \over {36}}$

$ \Rightarrow {R_H} = {{36} \over {5 \times 6561}}$

For second spectral line of helium,

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

$ = {3 \over 4} \times {{36} \over {5 \times 6561}}$

$ \Rightarrow \lambda = 1215\,\mathop A\limits^o $