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 – Scattering cross section

A beam of 10¹² protons per second is incident on a target containing 10²⁰ nuclei per square cm. At an angle of 10 degrees, there are 10² protons per second elastically scattered into a detector that subtends a solid angle of 10⁻⁴ steradians. What is the differential elastic scattering cross section, in units of sq cm per steradian?

A. 10²⁴
B. 10²⁵
C. 10²⁶
D. 10²⁷
E. 10²⁸

(GR9277 #95)

Solution:

Using dimension analysis:

the fractional scattering rate:



In units of sq cm per steradian:



Answer: C