Answer to Question #121960 in Physics for Aqela

Question

a) A beam of 7.8-MeV  particles scatters from a gold foil of thickness 0.30 m.
i) What fraction of the  particles is scattered between 1.5° and 2.5°?
ii) What is the ratio of  particles scattered through angles greater than 2° to the number
scattered through angles greater than 10°?

Answer 1

Given:

$Z_1=2,\\ Z_2=79, \\t=30\cdot10^{-6}\text{ m},\\ \rho=19300\text{ km/m}^3,\\ \Mu=197\cdot10^{-3}\text{ kg/mol},\\ \theta_1=1.5°,\\ \theta_2=2.5°,\\ \theta_3=2°,\\ \theta_4=10°.$

Solution

The fraction of particles scattered at angle θ is

$$f(\theta)=\pi nt\bigg(\frac{Z_1Z_2e^2}{8\pi\epsilon_0K}\bigg)^2\text{cot}^2\bigg(\frac{\theta}{2}\bigg).$$

In this equation, t - thickness, K - kinetic energy, n is the number of atoms per unit volume of material (gold):

$$n=\frac{\rho N_A}{\Mu}=5.9\cdot10^{28}\text{ atoms/m}^3,$$

where $\Mu$ is the molar mass of Gold and $\rho$ is its density.

i) Between 1.5° and 2.5° the fraction will be

$$f(\theta_1)-f(\theta_2)=\pi nt\bigg(\frac{Z_1Z_2e^2}{8\pi\epsilon_0K}\bigg)^2\bigg[\text{cot}^2\bigg(\frac{\theta_1}{2}\bigg)-\text{cot}^2\bigg(\frac{\theta_2}{2}\bigg)\bigg]=\\\space\\ =0.0442.$$

ii) Now it is much easier to find the ratio of particles scattered through angles greater than 2° to the number scattered through angles greater than 10°:

$$\frac{f(\theta_3)}{f(\theta_4)}=\frac{\text{cot}^2\frac{\theta_3}{2}}{\text{cot}^2\frac{\theta_4}{2}}=25.$$