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 ${r_1}:{r_2} = 3:2$.

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