Question

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]

• A. $hv$
• B. $0$
• C. $hv\left[1-\frac{hv}{2M{c}^2}\right]$
• D. $hv\left[1+\frac{hv}{2M{c}^2}\right]$

Answer 1

Answer: (D)

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.
${\vec{P}}_{\gamma }+{\vec{P}}_{Nu}=0$
Where ${\vec{P}}_{Nu}=$ Momentum of decayed nuclei
$\Rightarrow P_{\gamma }=P_{N_u}$
$\Rightarrow \frac{hv}{C}=P_{Nu}$
$\Rightarrow K.E.$ 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 $=E_{\gamma }+K\cdot E_{Nu}$
$=hv+\frac{1}{2M}{\left[\frac{hv}{C}\right]}^2$
$=hv\left[1+\frac{hv}{2M{C}^2}\right]$