Energy of $$\gamma$$-ray, E_$$\gamma$$ = hv

and momentum of $$\gamma$$-ray, $${p_\gamma } = {h \over \lambda }$$ .... (i)

As, $${p_\gamma } = {{{E_\gamma }} \over c} = {{hv} \over c}$$ ..... (ii)

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

$${p_\gamma } = {{hv} \over c} = {h \over \lambda }$$ [$$\because$$ $$\lambda = {c \over v}$$]

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

$$\therefore$$ p_$$\gamma$$ + p_{decayed nuclei} = 0

$$\Rightarrow$$ p_$$\gamma$$ = p_{decayed nuclei}

$$ \Rightarrow {{hv} \over c}$$ = p_{decayed nuclei} ..... (iii)

Kinetic energy of decayed nuclei,

$$KE = {1 \over 2}M{v^2} = {{(p_{decayed\,nuclei}^2)} \over {2M}}$$ ..... (iv)

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

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

$$\therefore$$ Loss in internal energy = E_$$\gamma$$ + KE_{decayed nuclei}

$$ = hv + {1 \over {2M}}{\left[ {{{hv} \over c}} \right]^2} = hv\left[ {1 + {{hv} \over {2M{c^2}}}} \right]$$