A nucleus with mass number `220` initially at rest emits an `alpha`-particle. If the Q-value of the reaction is `5.5 MeV`, calculate the kinetic energy of the `alpha`-particle.
(a) 4.4 MeV (b) 5.4 MeV (c) 5.6 MeV (d) 6.5 MeV

Options

A. `4.8` Me V
B. `5.4` Me V
C. `5.5 `Me V
D. `6.8` Me V

Answer

B

Explanation

m=220 `m_(1)=216 m_(2)4`
Q-value of the reaction is 5.5 MeV
i.e., `K_(1)+K_(2)=5.5MeV.....(i)`
By conservation of linear momentum,
`p_(1)=p_(2)impliessqrt(2(216)K_(1))=sqrt(2(4)K_(2))`
`K_(2)=54K_(1).....(ii)`
On solving Eqs. (i) and (ii), we have
The energy of emitted `alpha-"particle`, `K_(2)=5.4MeV`