A nucleus of mass M emits $\gamma$ ‐ray photon of frequency '$v$' . The loss of internal energy by the nucleus is : 

[Take 'c' as the speed of electromagnetic wave]

Energy of $\gamma$ ray $\left[E_{\gamma }\right]=hv$

Momentum of $\gamma$ ray $\left[P_{\gamma }\right]=\frac{h}{\lambda }=\frac{hv}{C}$

Total momentum is conserved. $\overrightarrow{P_{\gamma }}+{\overrightarrow{P}}_{Nu}=0$

Where  ${\overrightarrow{P}}_{Nu}=$Momentum of decayed nuclei

$\Rightarrow P_{\gamma }=P_{Nu}\Rightarrow \frac{hv}{C}=P_{Nu}\Rightarrow$K.E. of nuclei

$=\frac{1}{2}M{v}^2=\frac{{\left(P_{Nu}\right)}^2}{2M}=\frac{1}{2M}{\left[\frac{hv}{C}\right]}^2$

Loss in internal energy $=E_{\gamma }+K.E_{Nu}=hv+\frac{1}{2M}{\left[\frac{hv}{C}\right]}^2=hv[1+\frac{hv}{2M{C}^2}]$