A narrow beam of alpha particles with kinetic energy $K$ is directed at a stationary heavy target nu

Question

A narrow beam of alpha particles with kinetic energy $K$ is directed at a stationary heavy target nucleus of charge $Ze$. Which of the following statements is/are correct regarding the elastic scattering process? (Neglect the recoil of the heavy target nucleus).

Professor PC Thomas Classes · Answer

•	(A) Distance of closest approach $r_0$ is found by equating kinetic and potential energy: $K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0} \implies r_0 \propto 1/K$.
•	(B) Rutherford's scattering formula shows $dN \propto \frac{1}{\sin^4(	heta/2)} = \operatorname{cosec}^4(	heta/2)$.
•	(C) The impact parameter is given by $b = \frac{z Z e^2}{4 \pi \epsilon_0 (2K)} \cot(	heta/2)$. Notice that the term $\frac{z Z e^2}{4 \pi \epsilon_0 K}$ is exactly $r_0$. Thus, $b = \frac{r_0}{2} \cot(	heta/2)$. For $	heta = 90^{\circ}$, $\cot(45^{\circ}) = 1$, so $b = r_0/2$.
•	(D) Momentum transferred to the nucleus is $\Delta \vec{p} = \vec{p}_f - \vec{p}_i$. Since the collision is elastic and the nucleus is very heavy, $|\vec{p}_f| = |\vec{p}_i| = p$. The magnitude $|\Delta \vec{p}| = \sqrt{p^2 + p^2 - 2p^2 \cos	heta} = 2p \sin(	heta/2)$. This is maximum at $	heta = 180^{\circ}$.