# What is the actual formula for fraction of particles scattered at certain angle in Rutherford scattering?

$$N(\theta)=\frac{N_i nLZ^2e^4}{64\pi^2\varepsilon_0^2 R^2E_\alpha^2}\cdot\frac1{\sin^4(\theta/2)}$$ 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?

$N(\theta)/N_i$ 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.
• Almost. $\theta$ goes from 0 to 180. Integrate over a range of $\theta$ to find the number in that range. Commented May 16, 2018 at 20:33
• In practice from almost 0 to 90, then double the result because in 0 it reaches + $\infty$ and in 180 - $\infty$ . Thank you! Commented May 16, 2018 at 21:37
• No you do have to consider $\theta>90$ - that's what surprised Rutherford, after all, that the alpha particles occasionally bounce backwards! Commented May 17, 2018 at 9:26