the scattered amplitude is:

$f(\theta) = \frac {m } { 2 \pi \hbar^2 } \int { V(r') Exp ( i q \cdot r' )dr'^3}$

where $q = k - k'$ . using this:

$Exp( i q \cdot r ) dr^3 = \frac { sin(q r) } {q r } 4 \pi r^2 dr$

then the amplitude reduce to

$f(\theta ) = \frac { 2 m } { \hbar^2 q } \int { r V(r) sin(q r ) dr }$

For Coulomb force:

$V(r) = Z e^2 / r$

$f(\theta ) = 2 m Z e^2 / q^2$

which is Rutherford Scattering equation!

For electron scattering, the potential from finite size of the nucleus is

$\Delta V(r) =\frac { Ze^2 } { |r- r'| } \rho(r') dr'$

$f(\theta) = \frac { m Z e^2 } { 2 \pi \hbar ^2 } \int \int \frac {\rho(r') } {|r-r'| } Exp( i q \cdot ( r' + |r - r'| ) ) d r' d |r-r'|$

rearrange

$f(\theta) = \int \rho(r') Exp( i q \cdot r' ) dr' \times f_{Ruth} (\theta) = f_{Ruth}(\theta ) \times F(\theta)$

the $F(\theta)$ is called the Form factor, which is due to the distribution of the charge.

$F(\theta ) = \int \rho(r) Exp( i q \cdot r) dr$

the Form-factor is Fourier transform of the charge distribution alone. If reader familiarized with optics, the Form-factor is also the diffraction pattern. so, we can see, when the incoming particle wave has wavelength similar to the size of the charge distribution, then a diffraction occurred. here are some form of the Form factor.

for point charge:

$\rho(r) = \delta(r)/4 \pi$                  $F(\theta) = 1$

for exponential :

$\rho(r) = (a^3/8 \pi) Exp( - a r)$          $F(\theta ) = ( 1+ q^2/a^2 \hbar^2)^{-2}$

Solid sphere :

$\rho(r) = 3/4 \pi R^3 , r < R$

$F(\theta) = 3 \alpha^{-3} ( sin ( \alpha) - \alpha cos(\alpha) ) , \alpha = |q| R / \hbar$