Nuclear & Particle Physics - Scattering cross section

A proton beam is incident on a scatterer 0.1 centimeter thick. The scatterer contains 1020 target nuclei per cubic centimeter. In passing through the scatterer, one proton per incident million is scattered. The scattering cross section is 

A. 10−29 cm²
B. 10−27 cm²
C. 10−25 cm²
D. 10−23 cm²
E. 10−21 cm²
(GR8677 #42 )
Solution:

Given:
Thickness: T = 10−1 cm
Density: ρ = N/V = 1020 nuclei/cm³
Number of proton per incident million: N = 10−6

The scattering cross section → Area A = Volume/Thickness
A = V/T =  (V/N) × (N/T) = (1/ρ) × (N/T) = N/ρT
A = 10−6 / (1020 × 10−1) = 10−6−20+1 = 10−25 cm²

Answer: C

No comments :