Question

In Rutherford's scattering experiment, alpha particles (m = 6.68 x 10-27 kg, q = +2e) of
7.5 MeV kinetic energy were aimed at a thin gold foil (Z = 79). Some of the alpha
particles scattered straight back: 180 degrees. From this data, what is his estimate for the
size of the gold nucleus?

Answer 1

e = magnitude of charge on electron = 1.6 x 10-19 C

q = charge on alpha particle = 2e = 2 x 1.6 x 10-19 C

Q = charge on gold foil = Ze = 79 x 1.6 x 10-19 C

r = radius

Kinetic energy of alpha particle = 7.5 MeV = 7.5 x 1.6 x 10-13 J

Using conservation of energy

Electric potential energy = Kinetic energy of alpha particle

k Q q/r = 7.5 x 1.6 x 10-13

(9 x 109)(79 x 1.6 x 10-19 )(2 x 1.6 x 10-19)/r = 7.5 x 1.6 x 10-13

r = 3.03 x 10-14 m

Size = diameter = 2r = 2 x 3.03 x 10-14 = 6.06 x 10-14 m