A free nucleus of mass `24` amu emits a gamma photon (when initially at rest). The energy of the photon is `7 MeV`. The recoil energy of the nucleus in `KeV` is (Assuming `1 "amu"=931 MeV`).

Options

A. 2.2
B. 1.1
C. 3.1
D. 22

Answer

B

Explanation

`(hc)/(lambda)=7MeV`
`p_(N)=p_(Ph)`
`sqrt(2MKE)=(h)/(lambda)`
`2M(K.E.)=((E)/(c))^(2)`
`K.E.=((E)/(c))^(2)xx(1)/(2M)`
Substituting value
`"KE = 1.0965 keV"`