JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Modern Physics

A nucleus of mass M emits

γ

-ray photon of frequency 'v'. The loss of internal energy by the nucleus is: [Take 'c' as the speed of electromagnetic wave]

(A) $h v [1 + \frac{h v}{2 Mc ^{2}}]$ ✓ Correct
(B)0
(C) $h v [1 - \frac{h v}{2 M c ^{2}}]$
(D) $h v$

Answer: (A)

Solution

The nucleus is initially at rest, so momentum conservation makes it recoil with a momentum equal in magnitude to the photon's,

p = \frac{h ν}{c}

. Its recoil kinetic energy is

K = \frac{p ^{2}}{2 M} = \frac{( h ν ) ^{2}}{2 M c ^{2}}

The internal energy given up has to supply both the photon and this recoil, so

Δ E = h ν + \frac{( h ν ) ^{2}}{2 M c ^{2}} = h ν [1 + \frac{h ν}{2 M c ^{2}}]

The internal energy lost is

h ν [1 + \frac{h ν}{2 M c ^{2}}]

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Single correct option

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