Answer
251 MeV
Work Step by Step
The electric potential energy associated with the two circular nuclei touching each other is given by
$$
U=\frac{1}{4 \pi \varepsilon_{0}} \frac{Z_{\mathrm{Xe}} Z_{\mathrm{Sr}} e^{2}}{r_{\mathrm{Xe}}+r_{\mathrm{Sr}}}
$$
where $Z_{\mathrm{Xe}}$ is the atomic number of xenon, $Z_{\mathrm{Sr}}$ is the atomic number of strontium, $r_{\mathrm{Xe}}$ is the radius of a xenon nucleus, and $r_{\mathrm{Sr}}$ is the radius of a strontium nucleus.
The radii are given by $r=$ $(1.2 \mathrm{fm}) A^{1 / 3},$ where $A$ is the mass number. Where fm is fermi meter which is $10^{-15}$m
The mass number of Xe = 140 and for Sr is 96
Using the above information, the following calculations can be done to calculate radius of Xe and Sr
Thus,
$$
r_{\mathrm{Xe}}=(1.2 \mathrm{fm})(140)^{1 / 3}=6.23 \mathrm{fm}=6.23 \times 10^{-15} \mathrm{m}
$$
and
$$
r_{\mathrm{Sr}}=(1.2 \mathrm{fm})(96)^{1 / 3}=5.49 \mathrm{fm}=5.49 \times 10^{-15} \mathrm{m}
$$
Hence, the potential energy is
$$
\begin{aligned}
U &=\left(8.99 \times 10^{9} \mathrm{V} \cdot \mathrm{m} / \mathrm{C}\right) \frac{(54)(38)\left(1.60 \times 10^{-19} \mathrm{C}\right)^{2}}{6.23 \times 10^{-15} \mathrm{m}+5.49 \times 10^{-15} \mathrm{m}}=4.08 \times 10^{-11} \mathrm{J} \\
&=251 \mathrm{MeV}
\end{aligned}
$$
