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

The nuclear decay is represented as
`._(88)Ra^(226)rarr ._(86)Rn^(222)+ ._(2)H^(4)`
The KE of a particle is given by `K=(p^(2))/(2m)`
` :. p=sqrt(2mK)`
As momentum is conserved in nuclear decy.
`:. mK=` constant
`m_(Rn)K_(Rn)=m_(alpha)K_(alpha)`
or `K_(Rn)=(m_(alpha) K_(alpha))/(m_(Rn))`
`=(4.003xx4.8)/(222)=0.08666MeV`