Rutherford-scattering can be described as:
$$ \frac{\mathrm d\sigma}{\mathrm d\Omega} \propto \frac{1}{\sin(\theta/2)^4} $$
As we can see, the mathematical function is undefined at $\theta = 0^\circ$, but I am curious what happens physically at this singularity.
My guess would be that, the case of $\theta = 0^\circ$ is impossible, because the particle had to go through the target's nuclei. On the other hand, couldn't it tunnel through the nuclei?
And another question: shouldn't the cross section be infinite for $\theta = 180^\circ$ since the particle "bounces back" ?
Best Answer
In a classical treatment of electromagnetic scattering (which is what Rutherford used, since the relevant parts of quantum mechanics hadn't been invented yet), there's a one-to-one relationship between the final scattering angle $\theta$ and the impact parameter, usually called $b$:
[By Tonatsu, Public Domain, source]
The differential cross section for forward scattering (that is, $\theta=0^\circ$) diverges, because you get zero-angle scattering in the limit of large $b$, where the alpha particle and the nucleus don't interact very much at all. There's just an awful lot of ways to miss the nucleus and go straight.
Likewise, you only get $\theta \approx 180^\circ$ for a very small range of $b$ near zero, which is why the cross section has a minimum (but doesn't vanish) in the backwards limit.