Question

A nucleus of atomic mass $M$ emits a gamma ray photon of frequency $\nu $. Then
A) Kinetic energy acquired by the nucleus is $\dfrac{{{h^2}{\nu ^2}}}{{2M{c^2}}}$
B) Loss in the internal energy of the nucleus $h\nu \left( {1 + \dfrac{{h\nu }}{{2M{c^2}}}} \right)$
C) Momentum of the nucleus is equal and opposite to $\gamma $ ray photon
D) loss in internal energy of nucleus is $h\nu $

Answer 1

Hint: A nucleus of atomic mass $M$ emits gamma-ray photon $\gamma $, which is a penetrating form of EM radiation. This happens due to the radioactive decay of atomic nuclei. It consists of the photon of frequency $\nu $ and hence contains the internal energy of $h\nu $.

Complete step by step answer:
Step 1:
You can use the momentum conservation after the radiation of the gamma photon.
\[
  Mv = \dfrac{{{E_\gamma }}}{c} \\
   \Rightarrow Mv = \dfrac{{h\nu }}{c} \\
    \\
 \]
You can see that the nucleus recoils as the gamma photon emits. Hence, the direction of the momentum of the nucleus and the gamma photon is obviously the opposite.
The option (C) is correct.
So, the recoil velocity of the nucleus is
$v = \dfrac{{h\nu }}{{Mc}}$
Step 2:
Calculate the kinetic energy of the nucleus
 $
  K.E = \dfrac{1}{2}M{v^2} \\
   = \dfrac{1}{2}M{\left( {\dfrac{{h\nu }}{{Mc}}} \right)^2} \\
   = \dfrac{1}{2}M\dfrac{{{h^2}{\nu ^2}}}{{{M^2}{c^2}}} \\
   = \dfrac{1}{2}\dfrac{{{h^2}{\nu ^2}}}{{M{c^2}}} \\
 $
So, option (A) is the correct option.
Step 3:
Now you can calculate the change of Internal energy by using the conservation of energy after the gamma decay process.
Say, the current internal energy of the nucleus after the decay is ${E_N}$.
Hence the total energy of the system was
$
  {I_{Total}} = {E_N} + h\nu + \dfrac{1}{2}M{v^2} \\
   = {E_N} + h\nu + \dfrac{1}{2}\dfrac{{{h^2}{\nu ^2}}}{{M{c^2}}} \\
 $
Step 4:
Hence calculate the decrease in Internal energy
$
  \Delta I = {I_{Total}} - {E_N} \\
   = {E_N} + h\nu + \dfrac{1}{2}\dfrac{{{h^2}{\nu ^2}}}{{M{c^2}}} - {E_N} \\
   = h\nu + \dfrac{1}{2}\dfrac{{{h^2}{\nu ^2}}}{{M{c^2}}} \\
   = h\nu (1 + \dfrac{1}{2}\dfrac{{h\nu }}{{M{c^2}}}) \\
 $
Hence, the option (B) is correct but the option (D) is wrong.

So, the correct options are (A) Kinetic energy acquired by nucleus is $\dfrac{{{h^2}{\nu ^2}}}{{2M{c^2}}}$, (B) Loss in the internal energy of the nucleus $h\nu \left( {1 + \dfrac{{h\nu }}{{2M{c^2}}}} \right)$ and (C) Momentum of nucleus is equal and opposite to $\gamma $ ray photon.

Note:
The calculation should be taken care of keeping the relativistic nature of the gamma photon in mind. You should be careful in calculating the total internal energy before decay. There you should include the lost energies which are used in recoil and as gamma photon. Also, in place of calculation of the kinetic energy of the nucleus, the velocity of recoil is to be used only.