Impact Parameter and Distance of Closest Approach

Impact parameter. For any single alpha particle in the beam, its impact parameter $b$ is defined as the perpendicular distance between the nucleus and the line along which the alpha particle would have travelled had there been no force of deflection at all -- in other words, how far off-centre, aimed at the nucleus, the incoming particle's undeflected path would have passed. An alpha particle with a LARGE impact parameter passes far from the nucleus and feels only a weak Coulomb repulsion, so it is deflected through only a small angle $\theta$; an alpha particle with a SMALL impact parameter passes close to the nucleus, feels a much stronger repulsion, and is deflected through a large angle. In the limiting case $b=0$ (a perfectly head-on encounter, aimed directly at the nucleus), the alpha particle is repelled straight back along its own incoming path, giving the maximum possible scattering angle, $\theta=180^\circ$.

Working out the trajectory of a positive charge under a repulsive inverse-square (Coulomb) force gives the exact relation between the impact parameter and the resulting scattering angle:

where $K=\frac12 m v^2$ is the alpha particle's kinetic energy far from the foil and $Ze$ is the nuclear charge. This relation (which Exercise 2 asks for only in statement form, not derived in full) is exactly what Rutherford used to predict how many alpha particles should scatter into any given angular range for a nucleus of a given size and charge -- and the excellent agreement between this prediction and Geiger and Marsden's measured counts, across a wide range of angles, was itself powerful confirmation that the nuclear model (Section 1.3) was correct.

Distance of closest approach. For the special case of a perfectly head-on collision ($b=0$), the alpha particle travels straight toward the nucleus, is continuously slowed by the repulsive Coulomb force, and comes MOMENTARILY to rest at some minimum separation $r_0$ from the nucleus before being repelled straight back. At this instant, all of the alpha particle's initial kinetic energy $K$ has been converted entirely into electrostatic potential energy, since its speed (and hence its kinetic energy) is zero:

using the alpha particle's charge $+2e$ and the nucleus's charge $+Ze$. Solving for $r_0$:

$\boxed{r_0 = \frac{1}{4\pi\epsilon_0}\cdot\frac{2Ze^2}{K}}$ …