Answer
29 MeV.
Work Step by Step
Energy is conserved. In the beginning, when the particles are very far apart, there is KE but no PE. Assume that the uranium nucleus is at rest, and all of the initial KE belongs to the alpha particle.
At the end, there is PE but no kinetic energy (we ignore the negligible recoil of the uranium nucleus).
$$KE_i=PE_f$$
The two particles end up at a distance equal to the sum of their radii; use this to calculate the final electrical potential energy. Use equation 30–1 to find the radii.
$$KE_i=PE_f=k\frac{q_{\alpha}q_{U}}{ r_{\alpha}+r_{U}}$$
$$=(8.99\times10^9 Nm^2/C^2)\frac{(2)(92)(1.60\times10^{-19}C)^2}{(1.2\times10^{-15}m)(4^{1/3}+ 232^{1/3})}$$
$$=4.56\times10^{-12}J$$
This is about 29 MeV.