Answer to Question #121946 in Quantum Mechanics for Ranil

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

Given, energy of alpha particle $E=7.8\:MeV=12.5\times 10^{-13}J$ ,Thickness $t=0.30\mu m=3\times 10^{-7}$ .

Let, $Z_1,Z_2$ be atomic number of Gold and alpha particle respectively.

$n=\rho_{gold}N_{A}=5.9\times 10^{28} \:atoms/m^3$ .

To calculate the number of fractions of alpha particle scattered by angle grater than or equal to $\phi$

is given by,

$$F(\phi)=n\pi t\bigg(\frac{Z_1Z_2e^2}{8\pi\epsilon_0E}\bigg)^2\cot^2(\phi/2)$$

i).

The fraction of alpha particle scattered between $1.5^{\circ}\leq\phi\leq2.5^{\circ}$ is

$$F(1.5^{\circ})-F(2.5^{\circ})=0.1176\times10^{-3}\bigg(\cot^2(\frac{1.5^{\circ}}{2})-\cot^2(\frac{2.5^{\circ}}{2})\bigg)\\ F(1.5^{\circ})-F(2.5^{\circ})=0.1176\times10^{-3}\times3735.10\\ F(1.5^{\circ})-F(2.5^{\circ})=0.439$$

ii).

Again, from above main formula we get,

$$\frac{F(2^{\circ})}{F(10^{\circ})}=\frac{\cot^2(\frac{2^{\circ}}{2})}{\cot^2(\frac{10^{\circ}}{2})}\\ \frac{F(2^{\circ})}{F(10^{\circ})}=\frac{\cot^2(1^{\circ})}{\cot^2(5^{\circ})}\\ \frac{F(2^{\circ})}{F(10^{\circ})}=25.12$$