A nucleus of mass M emits $\mathrm{\gamma }$- ray photon of frequency ν. The loss of internal energy by the nucleus is

[Take, c is the speed of electromagnetic wave.]

Energy of $\mathrm{\gamma }\text{-ray, }{\mathrm{E}}_{\mathrm{\gamma }}=\mathrm{hv}$

and momentum of $\mathrm{\gamma }\text{-ray, }{\mathrm{p}}_{\mathrm{\gamma }}=\frac{\mathrm{h}}{\mathrm{\lambda }}$                 ...(i)

As,      ${\mathrm{p}}_{\mathrm{\gamma }}=\frac{{\mathrm{E}}_{\mathrm{\gamma }}}{\mathrm{C}}=\frac{\mathrm{hv}}{\mathrm{C}}$                                        ...(ii)

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

${\mathrm{p}}_{\mathrm{\gamma }}=\frac{\mathrm{hv}}{\mathrm{c}}=\frac{\mathrm{h}}{\mathrm{\lambda }}\left[\because \mathrm{\lambda }=\frac{\mathrm{c}}{\mathrm{v}}\right]$

Since, during the emission of $\mathrm{\gamma }$-ray photon, momentum is conserved.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{p}}_{\mathrm{\gamma }}+{\mathrm{p}}_{\text{decayed nuclei }}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{p}}_{\mathrm{\gamma }}={\mathrm{p}}_{\text{decayed nuclei }}\end{array}$

$\Rightarrow \frac{\mathrm{hv}}{\mathrm{C}}={\mathrm{p}}_{\text{decayed nuclei }}$                            ...(iii)

Kinetic energy of decayed nuclei,

$\mathrm{KE}=\frac{1}{2}{\mathrm{Mv}}^2=\frac{\left({\mathrm{p}}_{\text{decayed nuclei }}^2\right)}{2\mathrm{M}}$            ...(iv)

From Eqs. (iii) and (iv), we get

$\Rightarrow \mathrm{KE}=\frac{1}{2\mathrm{M}}{\left[\frac{\mathrm{hv}}{\mathrm{c}}\right]}^2$

$\begin{array}{c}\therefore \text{ Loss in internal energy }={\mathrm{E}}_{\mathrm{\gamma }}+{\mathrm{KE}}_{\text{decayed nuclei }}\\ =\mathrm{hv}+\frac{1}{2\mathrm{M}}{\left[\frac{\mathrm{hv}}{\mathrm{c}}\right]}^2=\mathrm{hv}\left[1+\frac{\mathrm{hv}}{2{\mathrm{Mc}}^2}\right]\end{array}$