A nucleus of mass M emits

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]

1. hv

2. 0

3. $\mathrm{h} v\left[1-\frac{\mathrm{hv}}{2 \mathrm{Mc}^{2}}\right]$

4. $\mathrm{h} v\left[1+\frac{\mathrm{hv}}{2 \mathrm{Mc}^{2}}\right]$

Correct Option: , 2

Solution:

Energy of $\gamma$ ray $\left[\mathrm{E}_{\gamma}\right]=\mathrm{h} v$

Momentum of $\gamma$ ray $\left[\mathrm{P}_{\gamma}\right]=\frac{\mathrm{h}}{\lambda}=\frac{\mathrm{hv}}{\mathrm{C}}$

Total momentum is conserved.

$\overrightarrow{\mathrm{P}}_{\gamma}+\overrightarrow{\mathrm{P}}_{\mathrm{Nu}}=0$

Where $\overrightarrow{\mathrm{P}}_{\mathrm{Nu}}=$ Momentum of decayed nuclei

$\Rightarrow P_{\gamma}=P_{\mathrm{Nu}}$

$\Rightarrow \frac{\mathrm{hV}}{\mathrm{C}}=\mathrm{P}_{\mathrm{Nu}}$

$\Rightarrow \mathrm{K} . \mathrm{E}$. of nuclei

$=\frac{1}{2} \mathrm{Mv}^{2}=\frac{\left(\mathrm{P}_{\mathrm{Nu}}\right)^{2}}{2 \mathrm{M}}=\frac{1}{2 \mathrm{M}}\left[\frac{\mathrm{hv}}{\mathrm{C}}\right]^{2}$

Loss in internal energy $=\mathrm{E}_{\gamma}+\mathrm{K} \cdot \mathrm{E}_{\mathrm{Nu}}$

$=h v+\frac{1}{2 \mathrm{M}}\left[\frac{\mathrm{hv}}{\mathrm{C}}\right]^{2}$

$=\mathrm{hv}\left[1+\frac{\mathrm{hv}}{2 \mathrm{MC}^{2}}\right]$

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