Question

A narrow beam of alpha particles with kinetic energy T = 600 keV falls normally on a golden foil incorporating n= 1.1x10¹⁹ nuclei/cm². Find the fraction of alpha particles scattered through the angles θ < θ₀ = 20°.

Answer 1

Because of the cosec⁴θ/2 dependence of the scattering, the number of particles (or fraction) scattered through θ < θ₀ cannot be calculated directly. But we can write this fraction as 

P(θ₀)=1-Q(θ₀)

where Q (θ₀) is the fraction of particles scattered through θ ≥ θ₀. This fraction has been calculated before and is 

where n here is a number of nuclei/cm². Using the data we get 

Q = 0.4 

Thus

P(θ₀)=0.6