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) - hu`
B. `(h ^(2) u ^(3))/( 2 M c ^(2))`
C. zero
D. `hu`

Answer

B

Explanation

Momentum of emitted photon
`= p _("photon") = (hv)/(c )`
From the law of conservtion of linear momentum, Momentum of recoil nucleus
`= p _("nucleus") = P _("photon")`
`therefore Mv = (hv)/(c )`
where v is the recoil speed of the nucleus
`or v = (hv)/(Mc)`
The recoil energy of the nucleus
` 1/2 Mv ^(2) =1/2 M ((hv )/(Mc)) = (hv)/(2Mc ^(2))`
Using (i)