Question

A radioactive nucleus of mass M emits a photon of frequency f and the nucleus recoils. The recoil energy will be:

$(a)hf \\ (b)M {c}^{2} - hf \\ (c) \frac{ {h}^{2} {f}^{2} }{2M {c}^{2} } \\ (d) \: zero$

Answer is (c).
⏩Explanation required!!!⏪

Answer 1

Formula based question

Answer 2

Answer:option c

Explanation

Use mv = (h×frequency)/c....(from lamda=h/p)

and then,

E = 1/2mv^2 and find the answer.