I have been trying to look for a derivation of the Rutherford scattering formula from Hyperphysics, but cannot find one. It doesn't show up in the original paper How is this equation derived?

$$N(\theta) = \frac{N_i nLZ^2k^2e^4}{4r^2 KE^2\sin^4(\theta/2)}$$ where

  • $N_i$ is the number of incident alpha particles
  • $n$ is the number of atoms per unit volume in the target
  • $L$ is the thickness of the target
  • $Z$ is the atomic number of the target
  • $e$ is the electron charge
  • $k$ is Coulomb's constant
  • $r$ is the target-to-detector distance
  • $KE$ is the kinetic energy of the alpha particle
  • $\theta$ is the scattering angle
  • $\begingroup$ Check Modern Physics: Arthur Beiser chapter 4: appendix here $\endgroup$
    – user36790
    Jan 22, 2016 at 7:37
  • $\begingroup$ Do you need an explanation of the derivation of the formula from scratch or you got stuck during the studying of the article? $\endgroup$ Nov 11, 2017 at 11:55

1 Answer 1


In the Wikipedia article about rutherford scattering the derivation of the scattering cross section $$ \frac{d\sigma}{d\Omega} =\left(\frac{ Z_1 Z_2 e^2}{8\pi\epsilon_0 m v_0^2}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)}$$ is given. Let's rewrite that in your notation: $Z_1 = Z$, $Z_2 = 4$, $k = \frac{1}{4\pi\varepsilon_0}$ and $KE = \frac{1}{2}m v^2$: $$ \frac{d\sigma}{d\Omega} =\left(\frac{ Z e^2 k}{KE}\right)^2 \csc^4{\left(\frac{\Theta}{2}\right)} = \frac{Z^2 k^2 e^4}{ KE^2\sin^4(\theta/2)}$$ The relation between the cross section and the number of detected particles is $$\frac{N(\theta)}{N_i} = \frac{nL}{4r^2} \frac{d\sigma}{d\Omega}$$ $nL$ gives the number of targets to scatter from and $1/(4r^2)$ is because at a bigger distance the intensity goes down by this factor (I'm sure about the $r^2$ but not exactly about the factor. I thought it would be $2\pi$ so maybe I did an error there.)

I hope this helps.


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