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, E_$\gamma$ = hv<br/><br/> and momentum of $\gamma$-ray, ${p_\gamma } = {h \over \lambda }$ .... (i)<br/><br/>As, ${p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}$ ..... (ii)<br/><br/>From Eqs. (i) and (ii), we get<br/><br/>${p_\gamma } = {{hv} \over c} = {h \over \lambda }$ [$\because$ $\lambda = {c \over v}$]<br/><br/>Since, during the emission of $\gamma$-ray photon, momentum is conserved.<br/><br/>$\therefore$ p_$\gamma$ + p_{decayed nuclei} = 0<br/><br/>$\Rightarrow$ p_$\gamma$ = p_{decayed nuclei}<br/><br/>$\Rightarrow {{hv} \over c}$ = p_{decayed nuclei} ..... (iii)<br/><br/>Kinetic energy of decayed nuclei,<br/><br/>$KE = {1 \over 2}M{v^2} = {{(p_{decayed\,nuclei}^2)} \over {2M}}$ ..... (iv)<br/><br/>From Eqs. (iii) and (iv), we get<br/><br/>$\Rightarrow KE = {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2}$<br/><br/>$\therefore$ Loss in internal energy = E_$\gamma$ + KE_{decayed nuclei}<br/><br/>$$ = hv + {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2} = hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]$$