QJEEM21Ph2f3d: JEE Main 2021 Physics - Class 12 JEEM21Ph2f3d Solution: A nucleus of mass M emits -ray photon of frequency 'v'. The…

Question

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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]

Accepted Answer

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]

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