The Rutherford scattering formula
The Rutherford scattering formula describes the final direction $\theta$ that a particle will scatter in the presence of a repulsive inverse square force between it and a stationary point mass (see figure). This is precisely the situation encountered in Rutherford's famous gold foil experiment when a beam of alpha particles approach heavy nuclei. One version of the formula relates the deflection angle ($\theta$) to the particle's initial speed ($v_0$) and the so-called scattering parameter $b$, which is the shortest distance between the two particles that would occur if the particle were to continue undeflected.
Derivation
We begin with the impulse momentum relation in the $y$-direction:
\begin{equation} \Delta p_y = \int F_y \diff t \label{eq:impulse-momentum} \end{equation}Considering a particle approaching from the left ($\psi = \pi$), the inital $y$-momentum is zero. Since the scattering is elastic, the particle's final speed will be equal to its initial speed. Therfore,
\begin{equation} \Delta p_y = m v_0 \sin \theta. \label{eq:momentum} \end{equation}Meanwhile, the component of the force in the $y$-direction is given by
\begin{equation} F_y = \frac{k q_1 q_2}{r^2} \sin \psi \label{eq:force} \end{equation}Substitution of \eqref{eq:momentum} and \eqref{eq:force} into \eqref{eq:impulse-momentum} yields
\begin{equation} mv_0 \sin\theta = \int_{-\infty}^\infty \frac{kq_1q_2}{r^2} \sin\psi \diff t \label{eq:subs impulse-momentum} \end{equation}At this point, we perform a change of variables $t \to \psi$
\begin{equation} mv_0 \sin\theta = \int_\pi^\theta \frac{kq_1q_2}{r^2} \sin\psi\frac{\diff t}{\diff\psi}\diff\psi \label{eq:changed vars} \end{equation}We can rewrite this equation in terms of $b$ and $v_0$ by applying the conservation of angular momentum ($r^2 \dot \psi = b v_0$)
\begin{equation} mv_0 \sin\theta = \int_\pi^\theta \frac{kq_1q_2}{bv_0}\sin \psi \diff \psi \end{equation} and integrating: \begin{equation} mv_0 \sin \theta = \left.-\frac{kq_1q_2}{bv_0} \cos\psi\right]^\theta_\pi = \frac{kq_1q_2}{bv_0} (1+\cos\theta) \end{equation}After noting that $E = \frac{1}{2}mv_0^2$ and applying the trig identity
$$\tan\left(\frac{x}{2}\right) = \frac{\sin x}{1+\cos x} $$ We finally have \begin{equation} \tan\left(\frac{\theta}{2}\right) = \frac{kq_1q_2}{2bE} \label{eq:rutherfordscattering1} \end{equation} or \begin{equation} \boxed{b = \frac{k q_1 q_2}{2 E}\cot\left(\frac{\theta}{2}\right)} \label{eq:rutherfordscattering} \end{equation}This formula can be also be derived from the equation of motion.
Differential scattering cross section
It is common practice to express the Rutherford scattering formula in terms the differential scattering cross section, $D$. For an beam of particles scattered over a small solid angle $\diff \Omega$, $D$ is defined as the rate at which the beam's cross-sectional area, $\sigma$ changes in response to a change in $\Omega$. \begin{equation} D \equiv \frac{\diff\sigma}{\diff\Omega} \label{eq:DSCSdef} \end{equation} It is convenient to define (see figure) \begin{equation} \diff\sigma = |b \diff b \diff \phi|, \hspace{35pt} \diff\Omega = |\sin\theta \diff\theta \diff\phi| \end{equation}