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

Nuclear & Particle Physics - Photon Scattering

Photons of wavelength λ scatter elastically on free protons initially at rest. The wavelength of the photons scattered at 90^o is increased by

1. λ ⁄ 137
2. λ ⁄ 1836
3. ħ ⁄ (m_ec) where ħ Planck's constant, m_e the rest mass of an electron, and c the speed of light
4. ħ ⁄ (m_pc) where ħ Planck's constant, m_p the rest mass of a proton, and c the speed of light
5. zero

(GR8677 #45)

Solution:

Compton Scattering: the inelastic scattering of photons from charged particles.

θ = 90^o → cos θ = 0
For proton scattered at 90^o, Δλ = ħ ⁄ (m_pc)

Answer: D