Question

What is the energy (in MeV) released in the alpha decay of${}^{239}\mathrm{Pu}\text{?}$

Short answer

Therefore, the energy is$5.2452765\mathrm{MeV}$

Step-by-step solution

Given information

$$\begin{gathered} m\left({}^{239}P_u\right)=239.052157\mathrm{u} \\ \mathrm{m}\left({}^{235}\mathrm{U}\right)=235.043924\mathrm{u} \\ \mathrm{m}\left({}^4\mathrm{H}_{\mathrm{e}}\right)=4.002602\mathrm{u} \\ 1\mathrm{u}=931.5\frac{\mathrm{MeV}}{{\mathrm{c}}^2} \end{gathered}$$

Explanation

The decay is given as:

${}^{239}P_u\to {}^{235}U+{}^{4}H_e$

We have the formula:

$E=\Delta m{c}^2\to \left(\text{Equation 1.}\right)$

Calculation

Now, finding the value of

$$\begin{gathered} \Delta m=m\left({}^{239}P_u\right)m\left(-{}^{235}U\right)-m\left({}^{4}H_e\right) \\ \Delta m=(239.052157\mathrm{u})-(235.043924\mathrm{u})-(4.002602\mathrm{u}) \\ \Delta m=0.005631\mathrm{u} \end{gathered}$$

We will now find the energy:

$$\begin{gathered} E=\Delta m{c}^2 \\ E=(0.005631\mathrm{u})c^2\times \frac{931.5\frac{\mathrm{MeV}}{{\mathrm{c}}^2}}{1\mathrm{u}} \\ E=5.2452765\mathrm{MeV} \end{gathered}$$