Question

Rutherford fired a beam of alpha particles (helium nuclei) at a thin sheet of gold. An alpha particle was observed to be deflected by 90.0°; its speed was unchanged. The alpha particles used in the experiment had an initial speed of 2.2 ✕ 10^7 m/s and a mass of
6.7 ✕ 10^−27 kg.
Assume the alpha particle collided with a gold nucleus that was initially at rest. Find the speed of the nucleus after the collision.

Answer 1

well we need mass of a gold atom
gold atomic mass = 197 g/mol
so 197/6*10^23 = 32.7 * 10^-23 g/atom = .0327 *10^-23 kg
original x momentum = 6.7 ✕ 10^−27 * 2.2*10^7 = 14.74*10^-20 kg m/s
final x momentum = .0327*10^-23 u
u = 14.74/.0327 * 10^3

do the same thing for y but original = final = 0
m alpha * v alpha = m atom * v atom