An alpha particle of kinetic energy $K$ is fired head-on at a fixed gold nucleus and momentarily stops at a distance $d_{0}$ from it. A proton of the same kinetic energy $K$ is fired head-on at the same nucleus. Its distance of closest approach is

Answer: B

At closest approach all kinetic energy becomes electrostatic potential energy: K = k Z1 Z2

e^{2} / d

, so d is proportional to the projectile charge Z1 at fixed K. The proton (Z1 = 1) has half the charge of the alpha particle (Z1 = 2), so d = d0/2.