A nucleus of mass M emits γ -ray photon of frequency v . The loss...

Energy of $\gamma$ ray \left[{E}_{\gamma}\right] = hv$</p><p>Momentum of$\gamma$ray \left[{P}_{\gamma}\right] = \dfrac{h}{\lambda} = \dfrac{hv}{c}$
Total momentum is conserved.

$\overset{\to}{P}\gamma + \overset{\to}{P}Nu = 0$
where $\overset{\to}{P}Nu =$ momentum of decayed nuclei

$⇒ {P}_{\gamma} = {P}_{Nu}$
$⇒ \dfrac{hv}{c} = {P}_{Nu}$
$⇒ KE$ of nuclei

$= \dfrac{1}{2}M{v}^{2} = \dfrac{{\left({P}_{Nu}\right)}^{2}}{2M} = \dfrac{1}{2M}{\left[\dfrac{hv}{c}\right]}^{2}$
Loss in internal energy $= {E}_{\gamma} + {KE}_{Nu}$
$= hv + \dfrac{1}{2M}{\left[\dfrac{hv}{c}\right]}^{2}$
$= hv\left[1 + \dfrac{hv}{2M{c}^{2}}\right]$