Physics Problems & Solutions: Nuclear & Particle Physics

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:

$\frac{\text{Rate}_\text{{scattered}}}{\text{Rate}_\text{{incident}}}=\frac{R_s}{R_i}=\frac{10^2}{10^{12}} \frac{\text{proton}}{\text{second}}\frac{\text{second}}{\text{proton}}=10^{-10}$

In units of sq cm per steradian:

$10^{-10} \left( \frac{1}{10^4} \frac{1}{\text{steradians}} \frac{1}{10^{20}} \frac{\text{cm}^2}{\text{nuclei}}\right ) =10^{-10+4-20}=10^{-26}\frac{\text{cm}^2}{\text{steradians}}$

Answer: C

Physics Problems & Solutions

Nuclear & Particle Physics – Scattering cross section

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