Question

In 1911, Ernest Rutherford discovered the nucleus of the atom by observing the scattering of helium nuclei from gold nuclei. If a helium nucleus with a mass of 6.68 10-27 kg, a charge of +2e, and an initial velocity of 1.18 107 m/s is projected head-on toward a gold nucleus with a charge of +79e, how close will the helium atom come to the gold nucleus before it stops and turns around? (Assume the gold nucleus is held in place by other gold atoms and does not move.)

Answer 1

Given :
mass of He nucleus = 6.68 * 10^-27 kg
charge of He nucleus = +2e = 2(1.60217*10^-19) Coulombs = 3.20434*10^-19 C
initial velocity = u = 1.18 * 10^7 m/s
charge of Au nucleus = +79e = 79(1.60217 * 10^-19) Coulombs = 1.2657143 * 10^-17 C

potential energy between the ii nuclei of the He nucleus = U = kQq / r

where
k = Coulomb's constant = 9 * 10^9 N.m²/Coulomb²
r is the distance of separation

when the kinetic energy has been converted entirely to potential energy the He nucleus will reverse direction

(1/2)[6.68 * 10^-27][1.18 * 10^7]^2 = [9 * 10^9][79(1.60217 * 10^-19)][2(1.60217*10^-19)] / r

r = 7.84885×10^-14 m