A nucleus of radium ` (._(88)Ra^(226))`
decays to `._(86)Rn^(222)`
by emisson of `alpha-` particle `(._(2)He^(4))` of energy `4.8MeV`. If meass of `._(86)Rn^(222)=222.0 a.m.u` mass of `._(2)He^(4)` is `4.003 ` a.m.u. and mass of `._(88)Ra^(226)` is `226.00826` a.m.u., then calculate the recoil energy of the daughter nucleus. Take ` 1 a.m.u. =931MeV`

Explanation

Mass defect,
`Deltam=226.00826-(222.000+4.003)`
`=0.00526 am u`
`:.` Total energy released
`=0.00526xx931MeV`
`=4.897MeV`
Recoil energy of daughter nucleus
`=4.897-4.8`
`=0.097MeV`