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$$N(\theta)=\frac{N_i nLZ^2e^4}{64\pi^2\varepsilon_0^2 R^2E_\alpha^2}\cdot\frac1{\sin^4(\theta/2)}$$ Where:
$N_i$ - Total amount of incident particles
$n$ - amount of gold atoms per unit volume
$L$- thickness of gold foil
$Z$ - atomic number of gold
$e$ - electron charge
$E_0$- absolute permittivity
$R$ - distance between foil and detector
$E_a$ - initial energy of alpha particles
$\theta$ - scattering angle

Everything apart from $N_i$ and theta is a constant for some particular aperture. If I understand the derivation of this equation correctly I should get a number of particles $N(\theta)$ that is a fraction of $N_i$. Or in other words, a number between $0$ and $N_i$.

However at small angles ($\theta \in(0,20)$) my $N(\theta)$ numbers above $10^3$.
Actually I'll never get the expected behaviour from this equation, because $\sin^{-4}$ function reaches infinity at $0$ and never drops below $1$.

Hence the question. Am I misunderstanding and misusing this equation? Or maybe it's not an equation I'm looking for?

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$N(\theta)/N_i$ is a probability density not a probability. To get a probability you have to integrate it over $\theta$ or maybe multiply it by $d\theta$. That makes the numbers reasonable again.

Usually the difference is clear because of dimensions, if you're dealing with distributions in things like time or position or energy. But angles are dimensionless that doesn't help here.

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  • $\begingroup$ What I understand from your comment is area below said function, from -180 to 180 is...1. In other words there would be 100% probability of finding all incident particles in that range of angles. Then integrate function f.e. from -1 to 1 to find probability of finding particles in that range of angles. Then divide probability by probability to find the exact fraction of total incident particles I'd find in that range of angles. $\endgroup$
    – Gabriel Rej
    May 16, 2018 at 20:23
  • $\begingroup$ Almost. $\theta$ goes from 0 to 180. Integrate over a range of $\theta$ to find the number in that range. $\endgroup$
    – RogerJBarlow
    May 16, 2018 at 20:33
  • $\begingroup$ In practice from almost 0 to 90, then double the result because in 0 it reaches + $\infty$ and in 180 - $\infty$ . Thank you! $\endgroup$
    – Gabriel Rej
    May 16, 2018 at 21:37
  • $\begingroup$ No you do have to consider $\theta>90$ - that's what surprised Rutherford, after all, that the alpha particles occasionally bounce backwards! $\endgroup$
    – RogerJBarlow
    May 17, 2018 at 9:26
  • $\begingroup$ The problem of using N/Ni as density probability is that you can't normalize it. The area under this curve must be one. You can't do this because the area diverges when you consider theta from 0 to 180 degrees. $\endgroup$
    – Guilherme Correa Teixeira
    Oct 8, 2021 at 12:38

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