A nucleus of mass M emits a gamma-ray photon | vedclass

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Question

(B) The energy of the $\gamma$-ray photon is $E_{\gamma} = h
u$.The momentum of the $\gamma$-ray photon is $p_{\gamma} = \frac{h
u}{c}$.By the law o

Vedclass · Accepted Answer

$h
u \left[1 + \frac{h
u}{2Mc^2}ight]$

Vedclass · Answer

(B) The energy of the $\gamma$-ray photon is $E_{\gamma} = h
u$.The momentum of the $\gamma$-ray photon is $p_{\gamma} = \frac{h
u}{c}$.By the law of conservation of momentum,the nucleus must recoil with an equal and opposite momentum $p_N = p_{\gamma} = \frac{h
u}{c}$.The kinetic energy of the recoiling nucleus is $K_N = \frac{p_N^2}{2M} = \frac{(h
u/c)^2}{2M} = \frac{(h
u)^2}{2Mc^2}$.The total loss in internal energy of the nucleus is the sum of the energy of the emitted photon and the kinetic energy of the recoiling nucleus:$\Delta E = E_{\gamma} + K_N = h
u + \frac{(h
u)^2}{2Mc^2} = h
u \left[1 + \frac{h
u}{2Mc^2}ight]$.

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

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