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!

  • 1
    $\begingroup$ Yes, the cross section for scattering by an infinite range potential is, well, infinite. Stars in the Andromeda galaxy are ‘scattering’ off our Sun as we speak. $\endgroup$
    – Jon Custer
    Apr 3, 2021 at 13:49

1 Answer 1


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 $f(\theta)\ll 1$.

  • $\begingroup$ Won't we still have the same problem if we consider the fraction of particles scattered by a single nucleus? In this case the single scattering event requirement would be fullfilled, but the fraction would still become $>1$ for small enough angles. $\endgroup$
    – Ruslan
    Apr 4, 2021 at 10:18

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