Question

A nucleus of mass $M$ emits $\gamma$-ray photon of frequency $\nu$. The loss of internal energy by the nucleus is :

Take $c$ as the speed of electromagnetic wave.

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

Answer 1

The correct option is D $h\nu [1+\frac{h\nu }{2M{c}^2}]$

Energy of $\gamma$-ray, $E_{\gamma }=h\nu$
Momentum of $\gamma$-ray, $p_{\gamma }=\frac{h}{\lambda }=\frac{h\nu }{c}$

As, the total momentum is conserved.

$\overrightarrow{p_{\gamma }}+\overrightarrow{p_n}=0$

$\Rightarrow \left|\overrightarrow{p_{\gamma }}\right|=\left|\overrightarrow{p_n}\right|$

$\Rightarrow \frac{h\nu }{c}=p_n$

Further, KE of nuclei,

$K{E}_n=\frac{1}{2}M{v}^2=\frac{p_n^2}{2M}=\frac{1}{2M}{\left[\frac{h\nu }{c}\right]}^2$

Now, loss in internal energy,

$\Delta E=E_{\gamma }+K{E}_n$

$\Rightarrow \Delta E=h\nu +\frac{1}{2M}{\left[\frac{h\nu }{c}\right]}^2$

$\Rightarrow \Delta E=h\nu [1+\frac{h\nu }{2M{c}^2}]$

Hence, option $\left(D\right)$ is the correct answer.