posted by Hryn on .
Ernest Rutherford fired alpha particles (helium nuclei) at a sheet of gold foil rolled as thin as he could get it. Some of these particles passed straight through the foil, some bounced off at various angles, and some rebounded directly backwards. From this experiment Rutherford induced the atomic nucleus.
If the initial speed of the alpha particles was 2.00*10^7 m/s, how close to an Au nucleus could a particle get before it bounced straight back?
Assume that, in a head-on collision with the nucleus, the He nucleus interacts only with the nucleus of gold. (Ignore the electron cloud around the gold ataom). At closest approach, the alpha particle's kinetic energy is converted to Coulomb potential energy of
(1/2) M Vo^2 = k Z1 Z2 / R
M is the He mass
V is the initial velocity
R is the closest approach to the nucleus
Z1 = 2 e (the charge of helium nucleus)
Z2 = 79 e
e = electron charge
k = Coulomb's-Law constant
Solve for R