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

The nuclear decay is represented as
Accordinig to be Broglie hypothesis, linear momentum of photon,
`p=(E)/(c)=(14.4xx1.6xx10^(-16)J)/(3xx10^(8)m//s)`
`=7.68xx10^(-24)kgms^(-1)`
From conservation of linear momentum,
momentum of daughter nucleus `=` momentum of photon
`p=7.68xx10^(-24)ms^(-1)`
`:.` Recoil energy of nucleus
`K=(p^(2))/(2m)=((7.68xx10^(-24))^(2))/(2xx56.935xx1.66xx10^(-27))`
`=0.312xx10^(-21)J`
`K=(0.312xx10^(-21))/(1.6xx10^(-16))keV`
`=1.95xx10^(-6)keV`