Question

Problem 8 In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles (having charges of +2e and masses of 6.64 X 10-27 kg) were fired toward a gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.02 X 107 m/s directly toward the nucleus, as in the figure below. How close does the alpha particle get to the gold nucleus before turning around? Assume the gold nucleus remains stationary 79 2e v=0

Answer 1

Charge of alpha particle, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

Charge of gold nucleus, Q = 79e = 79 x 1.6 x 10^-19 C = 126.4 x 10^-19 C

Initial speed of the alpha particle, v = 2.02 x 10⁷ m/s

Mass of alpha particle, m = 6.64 x 10^-27 kg

Kinetic energy of alpha particle, K = ½ mv² = ½ x 6.64 x 10^-27 x (2.02 x 10⁷ )² = 13.55 x 10^-13 J

The alpha particle turns back when its kinetic energy completely converts into electric potential energy due to repulsion of gold nucleus. If d is the distance at which the alpha particle turns back then,

K = U = 1/4πε₀ qQ/d

The distance, d = 1/4πε₀ qQ/K

d= 9 x 10⁹ x 3.2 x 10^-19 x 126.4 x 10^-19 /13.55 x 10^-13 = 2.68 x 10^-14 m

the distance, d = 2.68 x 10^-14 m