Question

A narrow beam of alpha particles with kinetic energy T = 500 keV falls normally on a golden foil incorporating 1.0 × 10¹⁹ nuclei cm⁻³. Calculate the fraction of alpha particles scattered through the angles θ < θ₀ = 30^◦

Answer 1

Given nt = 1.0 × 10¹⁹ nuclei/cm² 

The fraction ΔN/N = 1−(πR₀² cot²(θ/2).nt)/4 = 1−(π/4)(1.44zZ/T)² 

cot²(θ/2).nt = 1 − π(1.44 × 2 × 79/0.5)² cot²(30/2) × (1.0 × 10¹⁹ × 10⁻²⁶)/4 = 0.82 

The factor 10⁻²⁶ has been introduced to convert fm into cm².