Answer
$4.8\times 10^{-12}J$, which is about 30 MeV.
Work Step by Step
Energy is conserved. At the beginning there is PE but no kinetic energy. At the end, when the particles are very far apart, there is KE but no PE.
$$PE_i=KE_f$$
The two particles 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.
$$PE_i=k\frac{q_{\alpha}q_{Fm}}{ r_{\alpha}+r_{Fm}}$$
$$=(8.99\times10^9 Nm^2/C^2)\frac{(2)(100)(1.60\times10^{-19}C)^2}{(1.2\times10^{-15}m)(4^{1/3}+ 257^{1/3})}=4.8\times10^{-12}J$$
This is the system's final KE. Because the nucleus is so much more massive than the alpha particle, the nucleus is hardly moving, and the alpha particle has almost all of the final KE.
