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

Answer

b

Explanation

Mass of nulceus M=20u
`E_lamda=6MeV=6xx1.6xx10^(-13)j`
Using momentum conservation priciple ,
`O=MV+(E_(lamda))/(C )`
`therefore v=-(E_(lamda))/(Mc)=-6xx(1.6xx10^(-13))/(20xx1.6xx10^(-27)xx3xx10^(8))`
`=-10^(5)ms^(-1)`

K.E. of nucleus , `K=(1)/(2)Mv^(2)`
`= (1)/(2)xx20xx1.6xx10^(-27)xx(10^(5))^(2)=16xx10^(-17)j`
`=(16xx10^(-17))/(1.6xx10^(-16))KeV=1keV`