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

Answer

B

Explanation

Momentum of emitted photon can be written as
`=p_("photon")`
`=(hv)/(c)`
From applying the law of conservation of linear momentum, momentum of recoil nucleus
`impliesn_("nucleus")=p_("photon")`
`thereforeMv=(hv)/(c)`
Where, v is the recoil speed of the nucleus
or `v=(hv)/(Mc)` . . . (i)
The recoil energy of the nucleus is ltBrgt `=(1)/(2)Mv^(2)=(1)/(2)((hv)/(Mc))^(2)`
`=(h^(2)v^(2))/(2Mc^(2))` [using eq. (i)]