Question

A narrow beam of protons with kinetic energy T = 1.5MeV falls normally on a brass foil whose mass thickness ρt = 2.0mg cm−2. The weight ratio of copper and zinc in the foil is equal to 7:3. Find the fraction of the protons scattered through the angles exceeding θ = 45◦. For copper, Z = 29 and A = 63.55 and for zinc Z = 30 and A = 65.38

Answer 1

The fraction of particles scattered in brass at angles exceeding θ is given by ΔN/N = (π/4)(1.442/T2) (0.7Z1 2 /A1 + 0.3Z22/A2) ρtN0 cot2θ/2 × 10−26 where Z1 = 29 for copper and Z2 = 30 for Zinc, A1 = 63.55 and A2 = 65.38 are the atomic masses, respectively, and N0 = 6.02 × 1023 is the avagadro’* number, T = 1.5 MeV, ρt = 2 × 10−3 g cm−2 and θ = 450. The factor 10−26 is introduced to convert fm2 into cm2. Using these values in the above equation ΔN/N = 6.78 × 10−4