Answer to Question #121938 in Quantum Mechanics for Moel Tariburu

Question

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

Answer 1

I)

$$f(1.5)-f(2.5)=\pi nt\left(\frac{Z_1Z_2e^2}{8\pi\epsilon_0K}\right)^2\\(\cot^2{0.5(1.5)}-\cot^2{0.5(2.5)})$$

$$f(1.5)-f(2.5)=\pi (5.9\cdot10^{-6})(0.3\cdot10^{28})\\\left(\frac{(2)(79)(1.6\cdot10^{-19})^2}{8\pi(8.85\cdot10^{-12})(7.8\cdot10^{6})(1.6\cdot10^{-19})}\right)^2\\(\cot^2{0.5(1.5)}-\cot^2{0.5(2.5)})$$

$$f(1.5)-f(2.5)=0.044$$

II)

$$\frac{f(2)}{f(10)}=\frac{\cot^2(0.5(2))}{\cot^2(0.5(10))}=25.1$$