Derivation
The differential cross section can be derived from the equations of motion for a particle interacting with a central potential. In general, the equations of motion describing two particles interacting under a central force can be decoupled into the motion of the center of mass and the motion of the particles relative to one another. For the case of light alpha particles scattering off heavy nuclei, as in the experiment performed by Rutherford, the reduced mass is essentially the mass of the alpha particle and the nucleus off of which it scatters is essentially stationary in the lab frame.
Substituting into the Binet equation yields the equation of trajectory
where, is the speed at infinity, and is the impact parameter.
The general solution of the above differential equation is
and the boundary condition is
If we choose
then the deflection angle Θ can be seen from solving as
b can be solved to give
To find the scattering cross section from this result consider its definition
Since the scattering angle is uniquely determined for a given and, the number of particles scattered into an angle between and must be the same as the number of particles with associated impact parameters between and . For an incident intensity, this implies the following equality
For a radially symmetric scattering potential, as in the case of the Coulombic potential, yielding the expression for the scattering cross section
Finally, plugging in the previously derived expression for the impact parameter we find the Rutherford scattering cross section
Read more about this topic: Rutherford Scattering