Answer to Question #121960 in Physics for Aqela

Question #121960
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°?
1
Expert's answer
2020-06-15T10:21:43-0400

Given:

Z1=2,Z2=79,t=30106 m,ρ=19300 km/m3,M=197103 kg/mol,θ1=1.5°,θ2=2.5°,θ3=2°,θ4=10°.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(θ)=πnt(Z1Z2e28πϵ0K)2cot2(θ2).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=ρNAM=5.91028 atoms/m3,n=\frac{\rho N_A}{\Mu}=5.9\cdot10^{28}\text{ atoms/m}^3,

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


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


f(θ1)f(θ2)=πnt(Z1Z2e28πϵ0K)2[cot2(θ12)cot2(θ22)]= =0.0442.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°:


f(θ3)f(θ4)=cot2θ32cot2θ42=25.\frac{f(\theta_3)}{f(\theta_4)}=\frac{\text{cot}^2\frac{\theta_3}{2}}{\text{cot}^2\frac{\theta_4}{2}}=25.

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