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 MeV`
B. `5.4 MeV`
C. `5.6 MeV`
D. `6.5 MeV`

Answer

B

Explanation

Given that `K_(1)+K_(2)=5.5 MeV` …(i)
From conservation of linear momentum. `p_(1)=p_(2)`
`implies sqrt(2K_(1)(216 m))=sqrt(2K_(2)(4m))` as `P=sqrt(2Km)`
`:. K_(2)=54 K_(1)` …(ii)
solving equation (i) and (ii).
We get `K_(2)=KE` of `alpha`-particle `=5.4 MeV`