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Q.
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 $\left[E_{\gamma}\right]=h v$
Momentum of $\gamma$ ray
$\left[P_{\gamma}\right]=\frac{h}{\lambda}=\frac{h v}{C}$
Total momentum is conserved.
$\vec{P}_{\gamma}+\vec{P}_{N u}=0$
Where $\vec{P}_{N u}=$ Momentum of decayed nuclei
$\Rightarrow P_{\gamma}=P_{N_{u}} $
$\Rightarrow \frac{h v}{C}=P_{N u} $
$\Rightarrow K . E .$ of nuclei
$=\frac{1}{2} M v^{2}=\frac{\left(P_{N u}\right)^{2}}{2 M}=\frac{1}{2 M}\left[\frac{h v}{C}\right]^{2} $
Loss in internal energy $=E_{\gamma}+K \cdot E_{N u}$
$=h v+\frac{1}{2 M}\left[\frac{h v}{C}\right]^{2} $
$=h v\left[1+\frac{h v}{2 M C^{2}}\right]$