A nucleus with mass number 220 initially at rest emits an `alpha` particle. If the Q value of the reaction is 5.5 MeV , the kinetic energy of the `alpha` particle is

Options

A. `4.4 Me V`
B. `5.4 MeV`
C. `5.6 MeV`
D. `6.5 MeV`

Answer

B

Explanation

The magnitude of momentum of the daughter nucleus and `alpha`-particles will be equal `Q=KE` of daughter nucleus `+ KE` of `alpha`-particles
`=(p^(2))/(2m_(d))+(p^(2))/(2m_(alpha))`
`KE` of `alpha`-particle =`(p^(2))/(2m_(alpha))=(1)/(m_(alpha))xx(m_(alpha).m_(d))/(m_(alpha)+m_(d)).Q`
`=(216)/(220)xx5.5 MeV.=5.4 MeV`.