Derivation of the Rutherford scattering formula

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
• Check Modern Physics: Arthur Beiser chapter 4: appendix here
– user36790
Jan 22, 2016 at 7:37
• Do you need an explanation of the derivation of the formula from scratch or you got stuck during the studying of the article? Nov 11, 2017 at 11:55

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.)