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

Momentum of emitted photon, `p_("photon")=(hv)/c`.
Let v be the speed of recoil nucleus, then
`p_("nucleus")=p_("photon")`
`:. Mv=(hv)/c or v=(hv)/(Mc)`
The recoil energy of the nucleus
`=1/2Mv^(2)=1/2M((hv)/(Mc)) =(h^(2)v^(2))/(2Mc^(2))`