"If an alpha particle is fired straight at the nucleus of an uranium atom (q = 92e), at a velocity of 5 x 10^5 m/s, how close will it get to the uranium nucleus?
Ek = (mv^2)/2
Ee = (kq1q2)/d
The Attempt at a Solution
Since the alpha particle and uranium atom are both positively charged, they will naturally repel each other. Firing the alpha particle at the uranium atom will therefore store electrical energy. At the closest point the alpha particle gets to the nucleus, all of its kinetic energy has been converted into electrical potential energy. Therefore, I equated Ee and Ek, and solved for d, taking the mass of an alpha particle to be quadruple that of a proton. However, the answer I got was 1.02 x 10^-10 m, but the correct answer is 5.1 x 10^-11 as per the provided solution.