Question

In the radioactive decay - ${}^{238}Pu{\to }^{234}U+\alpha$,
What will be the kinetic energy of the outgoing α-particle?
Given:-

-Atomic masses:

$m{(}^{238}Pu)=238.04955u;m{(}^{234}U)=234.04095u;$

$m{(}^{4}He)=4.002603u$

-Constant: $c^2=931\frac{MeV}{u}$

• 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

The 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,
$Kineticenergyof\alpha =Qvalueis$

$\Rightarrow \left[m{(}^{238}Pu\right)-\left\{m{(}^{234}U\right)+m{(}^{4}He\left)\right\}]\times c^2$
$=[238.04955u-(234.04095u+4.002603u\left)\right]\times 931\frac{MeV}{u}$
$=\begin{array}{c}5.58MeV\end{array}$