Question

A nucleus with mass number $224$ initially at rest emits an $\alpha$ particle. If the $Q$ value of the reaction is $8MeV$ and only one photon of energy $2.00MeV$ is subsequently emitted after the emission of the $\alpha$ particle, the kinetic energy (in $MeV$ ) of the $\alpha$ particle is : (Approximate the answer to the nearest integer.)

Answer 1

Answer: 6

$T_{\alpha }=\frac{\left(Q-\Delta E\right)Y}{Y+4}=\frac{\left(8-2\right)220}{224}=5.89MeV$