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`

Explanation

Momentum of emitted photon `p_("photon")=(hupsilon)/(c )`
From the law of conservation of linear momentum,
Momentum of recoil nucleus `=p_("nucleus")=p_("photon")`
`therefore" "Mv=(h upsilon)/(c) or v=(hupsilon)/(Mc)" ….(i)"`
where v is the recoil speed of the nucleus
The recoil energy of the nucleus
`=(1)/(2)Mv^(2)=(1)/(2)M((hupsilon)/(Mc))^(2)=(h^(2)upsilon^(2))/(2Mc^(2))" (Using (i))"`