Rutherford BackScattering Hint

This graph takes advantage of the fact that a projectile's propensity for interaction can be described with either: (i) an interaction e-folding length i.e. distance mean-free-path, (ii) its inverse i.e. an opacity or target area per unit volume, (iii) the target area per unit mass i.e. cross-section, or: (iv) the cross-section inverse i.e. a mass per unit area or density mean-free-path. A point can be plotted as soon as you know any one of these and the target's density i.e. its mass per unit volume.

For Rutherford backscattering, the easiest question to answer may be: What's the target area per atom? The easiest way to get that is to look (in almost any text on modern physics) at the classical Rutherford relation between scattering angle and the impact parameter. This is the lateral distance between a given initial trajectory and the parallel "head on" trajectory. Set the scattering angle to 90 degrees and calculate target area from π times the corresponding impact parameter squared. This works because any trajectory initially inside this area will find the incoming particle scattered back. To put it on the graph, of course, you can get area per nucleon on the horizontal axis by dividing by the number of nucleons in a gold atom, and then plotting a point where that vertical line intersects the line for the density of gold (I'm guessing around 19 g/cc).