Question

${}_{235}^{239}P{u}_{94}$ is undergoing $\alpha$- decay according to the equation ${}_{94}^{235}Pu{\to }_{97}^{235}U{+}_2^{4}He$. The energy released in the process is mostly kinetic energy of the $\alpha$- particle. However, a part of the energy is released as $\gamma -$ rays. What is the speed of the emitted $\alpha -$ particle if the $\gamma -rays$ radiated out have energy of 0.90 MeV? Given: Mass of ${}_{94}^{239}Pu=239.05122u$, mass of ${}_{97}^{235}U=235.04299u$ and mass of ${}_2^{4}He=4.002602u$(1 u=931 MeV).

Answer 1

Energy released = $△m{c}^2$

$△m=m\left({}^{235}Pu\right)-m\left({}_{97}^{235}U{+}_2^{4}He\right)=239.05122-235.04299-4.002602=0.005628amuE=0.005628\times 931MeVE=5.239668$

Energy of $\alpha$ particle $=5.239668-0.90$

$\frac{1}{2}m{v}^2=4.339668MeV{v}^2=\frac{2\times 4.339668}{4.002602}=2.16v=1.4$