The iron `Fe^(27)` nucleus emits a y-axis of energy 14.4KeV. If mass of nucleus is 56.935u, calculate the recoil energy of the nucleus. Take `1u=1.66xx10^(-27)kg.`

Explanation

Here, energy of y-ray
`E=14.4keV=14.4xx1.6xx10^(-16)J`
Mass of nucleus `=56.935u`
`=56.935xx1.66xx10^(-27)kg`
Recoil energy fo nucleus, `E^(')=?`
Momentum of y-ray photon
`mxxc=(mc^(2))/(c)=(E)/(c)=(14.4xx1.6xx10^(-16))/(3xx10^(8))`
`=7.68xx10^(-24)kg ms^(-1)`
According to the principle of conservation of linear momentum,
momentum fo residual nucleus, `p^(')=p=7.68xx10^(-24)kgm//s`
Recoil energy of residual nucleus
`E^(')=(p^(').^(2))/(2m)=((7.68xx10^(-24))^(2))/(2xx56.935xx1.66xx10^(-27))`