Question

A beam of 3 MeV α particles scatters from a thin gold foil. What is the ratio of the number of α particles scattered to angles greater than 4° to the number scattered to angles greater than 16°?

Answer 1

Answer:

Ratio of number of alpha particles is 16.20

Explanation:

In Rutherford Scattering, the fraction of incident beam particles scattered with angle θ and grater than θ is given by the relation :

$f(\theta)=\pi nt(\frac{Z_{1}Z_{2}e^{2} }{8\pi\epsilon_{0}K } )^{2} \cot^{2}\frac{\theta}{2}$

Here Z₁ is the atomic number of target, Z₂ is atomic number of incident particles, e is electronic charge, K is kinetic energy of incident particles, n is target particle density and t is thickness of the target foil.

According to the problem, number of alpha scattered to angles greater than 4° is given by the relation :

$f(4)=\pi nt(\frac{Z_{1}Z_{2}e^{2} }{8\pi\epsilon_{0}K } )^{2} \cot^{2}\frac{4}{2}$      ....(1)

Number of alpha scattered to angles greater than 16° is given by the relation :

$f(16)=\pi nt(\frac{Z_{1}Z_{2}e^{2} }{8\pi\epsilon_{0}K } )^{2} \cot^{2}\frac{16}{2}$    .....(2)

Ratio of equation (1) and (2) gives :

$\frac{f(4)}{f(16)} =\frac{cot^{2}\frac{4}{2}}{cot^{2}\frac{16}{2}}$

$\frac{f(4)}{f(16)} =16.20$