Is Rutherford scattering formula inconsistent with reality? On our way to deriving the famous Rutherford scattering formula, we get a formula for the fraction ($f$) of incident alpha particles scattered by $\theta$ or more and this formula has the form
$$f=\pi n t\left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2)$$
where $n$ is the number of atoms per unit volume, $t$ is the thickness of the foil etc. My issue with this is that the right handed limit of $\cot^2(\theta/2)$ as $\theta \to 0+$ is infinity. But this leads to a contradiction because we expect that the fraction of particles scattered by an angle of $0$ degrees or more should be exactly one. This formula makes the claim that the fraction is in fact infinite. So what is going on here? Is it that the formula breaks down for all angles smaller than that particular angle $\theta_0$ for which $f(\theta_0)=1$?
Any help on this issue would be greatly appreciated!
 A: Let's take a step back.
We begin with the impact parameter $b(\theta)$
and the total cross section $\sigma(\theta)$
for an alpha particle being scattered by an angle of $\theta$ or more
by a single atomic nucleus.
$$b(\theta) = \frac{Ze^2}{4\pi \epsilon_0 K_E}\cot(\theta/2)$$
$$\sigma(\theta) = \pi b^2(\theta) = \pi\left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2) \tag{1}$$
The important thing is: This formula was derived from analyzing a two-body-problem
(an alpha particle and a single atomic nucleus).
I.e. the alpha particle suffered only one single scattering event.
This approximation is fine when you consider only sufficiently
large scattering angles $\theta$, or equivalently, sufficiently
small cross section areas $\sigma(\theta)$.
Then there is only one atomic nucleus inside a cylindrical tube
of cross section area $\sigma(\theta)$.
And you can apply statistical reasoning to find the fraction
$f(\theta)$ of alpha particles being scattered by the many
atomic nuclei of a gold foil.
$$\begin{align}
f(\theta) &= n t \sigma(\theta) \\
 &= \pi n t \left(\frac{Ze^2}{4\pi \epsilon_0 K_E}\right)^2\cot^2(\theta/2)
\end{align} \tag{2}$$
The situation becomes more complicated when you consider smaller
scattering angles $\theta$, or equivalently, bigger cross section
areas $\sigma(\theta)$.
Then there will be several atomic nuclei within the cylindrical
tube of area $\sigma$.
And hence the alpha particle will be scattered by more than one atomic nucleus.
That means, the requirement (single scattering event) used to derive
formulas (1) and (2) is no longer valid
and the formulas can no longer be applied.
So formula (2) is only applicable if $f(\theta)\ll 1$.
