Q 12-13-103JEE MainJEE Main 2021 (20 Jul, Shift 1)Medium
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]
Answer: (D) $h\nu\left[1 + \frac{h\nu}{2Mc^2}\right]$
The photon carries momentum $\dfrac{h\nu}{c}$, so the nucleus recoils with the same momentum and kinetic energy $\dfrac{p^2}{2M} = \dfrac{(h\nu)^2}{2Mc^2}$.
The internal energy lost supplies both:
$$\Delta E = h\nu + \frac{(h\nu)^2}{2Mc^2} = h\nu\left[1 + \frac{h\nu}{2Mc^2}\right]$$
Solution by Sreeraj P, M.Sc Physics