Answer
260 MeV of potential energy, which is about 30 percent larger than the fission energy released.
Work Step by Step
We are told that the uranium-236 splits into two roughly equal fragments. The atomic mass number of each piece will be 118 (half of 236). The atomic number of each piece will be 46 (half of 92).
The two daughter nuclei start at a distance equal to the sum of their radii; use this to calculate the initial electrical potential energy. Use equation 30–1 to find the radii.
$$PE=k\frac{q_{1}q_{2}}{ r_{1}+r_{2}}$$
$$=(8.99\times10^9 Nm^2/C^2)\frac{ (46)^2(1.60\times10^{-19}C)^2}{(1.2\times10^{-15}m)(118^{1/3}+ 118^{1/3})}$$
$$=4.14\times10^{-11}J\approx 260MeV$$
This electric potential energy is about 30 percent larger than 200 MeV, the fission energy that is released.
