A nucleus of amss 20 u emits a `gamma` photon of energy 6 MeV . If the emission assume to occur when nuclues is free and at rest then the nulceus will have kintetic energy nearest to
(take `1u=1.6xx10^(-27)Kg`)

Options

A. 10 KeV
B. 1 KeV
C. 0.1 KeV
D. 100 KeV

Answer

B

Explanation

By conservation of linear momentum
Momentum of photon = momentum of nucleus
`(E)/(c )=sqrt(2m(KE))`
`(6xx10^(6)xx1.6xx10^(-19))/(3xx10^(8))=sqrt(2xx20xx1.6xx10^(-27)xxKE)`
`3.2xx10^(-21)=sqrt(64xx10^(-27)xxKE)`
`((3.2)^(2)xx10^(-42))/(64xx10^(-27))=KE`
`KE=1.6xx10^(-16)J`
`KE=(1.6xx10^(-16))/(1.6xx10^(-19))eV=10^(3)eV=1keV`