A uranium nucleus fissions in two spherical fragments of charge 38e and 54e and radii 5.5 * 10^-15m and 6.2 *10^-15m, respectively. Assume that the charge is distributed uniformly throughout the volumes, that just before separation each fragment is at rest, and that fragments surfaces are in contact. The electric potential energy of two spherical fragments in MeV (1 MeV = 1.6 *10^-13J) is

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The electric potential energy (U) between two charges can be calculated using the formula $U = \frac{kq_1q_2}{r}$, where $k$ is Coulomb's constant ($8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2$), $q_1$ and $q_2$ are the charges, and $r$ is the distance between the centers of the charges. Given that the charges are $38e$ and $54e$ (where $e = 1.6 \times 10^{-19} \, \text{C}$), and the radii of the spheres are $5.5 \times 10^{-15} \, \text{m}$ and $6.2 \times 10^{-15} \, \text{m}$, the distance $r$ is the sum of the radii, $5.5 \times 10^{-15} \, \text{m} + 6.2 \times 10^{-15} \, \text{m} = 11.7 \times 10^{-15} \, \text{m}$.Substituting the values into the formula:$$U = \frac{(8.99 \times 10^9) \times (38 \times 1.6 \times 10^{-19}) \times (54 \times 1.6 \times 10^{-19})}{11.7 \times 10^{-15}}$$$$U = \frac{(8.99 \times 10^9) \times (60.8 \times 10^{-19}) \times (86.4 \times 10^{-19})}{11.7 \times 10^{-15}}$$$$U = \frac{(8.99 \times 10^9) \times (5251.52 \times 10^{-38})}{11.7 \times 10^{-15}}$$$$U = \frac{47.2 \times 10^{-19}}{11.7 \times 10^{-15}}$$$$U = 4.03 \times 10^{-14} \, \text{J}$$To convert joules to MeV, use the conversion factor $1 \, \text{MeV} = 1.6 \times 10^{-13} \, \text{J}$:$$U = \frac{4.03 \times 10^{-14}}{1.6 \times 10^{-13}} \, \text{MeV}$$$$U \approx 250 \, \text{MeV}$$Therefore, the correct answer is 250 MeV.

A Uranium Nucleus Fissions in Two Spherical Fragments of Charge