Question

Don't hate me for asking this question, as it will need you to work your calculator! So, in the radioactive decay -
${}^{238}Pu{\to }^{234}U+\alpha ,$
What will be the kinetic energy of the outgoing α-particle?
Things you will need:

• A. 5.58 MeV
• B. 4.72 MeV
• C. 5.03 MeV
• D. 3.34 MeV

Answer 1

The correct option is A

5.58 MeV

This apparent loss in mass of the system of decay products will manifest itself as kinetic energy of the alpha particle. This energy is called the Q value of the decay. Therefore,
$Kineticenergyof\alpha =Qvalue=\left[m{(}^{238}Pu\right)-\left\{m{(}^{234}U\right)+m{(}^{4}He\left)\right\}]\times c^2$
$=[238.04955u-(234.04095u+4.002603u\left)\right]\times 931\frac{MeV}{u}$
$=\begin{array}{c}5.58MeV\end{array}.$