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

Options

A. `Mc^(2)-hv`
B. `h^(2)v^(2)//2Mc^(2)`
C. zero
D. hv

Answer

B

Explanation

Recoil energy of nucleus `=1/2Mv^(2) =1/2`
`=1/2p^(2)/M`
Momentum of nucleus is same as momentum of photon.
`p=E/C =(hv)/C`
Substituting in equation (i)
Recoil energy `=1/(2M)((hv)/C)^(2)=(h^(2)v^(2))/(2MC^(2))`