A nucleus of mass `218` amu in Free State decays to emit an `alpha`-particle. Kinetic energy of the `beta-`particle emitted is `6.7 MeV`. The recoil energy (in MeV) of the daughter nucleus is

Options

A. `1.0`
B. `0.5`
C. `0.25`
D. `0.125`

Answer

D

Explanation

Kinetic energy of `alpha` - particle `= ((A- 4)/(A)) Q`
`Q rarr` energy release in `alpha`- particle emission
`rArr E_(alpha)= ((214 -4)/(214)) xx 6.7` MeV
`rArr E_(alpha)= (0.9813 xx 6.7) MeV` = 6.575MeV
Energy of daughter nucleus
`= (6.7 - 6.575) MeV`= 0.125 MeV