Question

What fraction of the α particles in Rutherford's gold foil experiment are scattered at large angles? Assume the gold foil is two layers thick, as shown in the figure, and that the approximate diameters of a gold atom and its nucleus are 2.6 Å and 1.3×10−4Å, respectively. Hint: Calculate the cross-sectional area occupied by the nucleus as a fraction of that occupied by the atom. Assume that the gold nuclei in each layer are offset from each other.
Express your answer to two significant figures.

Answer 1

The fraction of incident particles scattered through an angle greater than θ is given by f = σnt, where σ is the cross-sectional area, n is the number of atoms per volume and t is the thickness of the foil. Taking that the large angles are angles more than 90 degrees, θ = 90°. The differential cross-section in Rutherford scattering model is given by

where KE is the kinetic energy (one can be derived from the energy of alpha-particles).

Therefore, in order to calculate the fraction of the α particles in Rutherford's gold foil experiment that are scattered at large angles, the energy of beam required.