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◦
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.
