My question relates to Rutherford Scattering of particles. When we calculate the "differential cross-section" expression for a nucleus with finite size, it is said that the expression is almost the same as that if the nucleus was a point only a "Form Factor" multiplies,which is nothing but the fourier transform of the charge density, I want to know the derivation of this expression with the form factor. I cant find it anywhere.

  • $\begingroup$ I would like to add, please do not refer any Quantum Treatment of the problem, I just want to know even classically does it show up. $\endgroup$
    – Utkarsh
    Commented Jun 18, 2014 at 9:52
  • $\begingroup$ Most practicable answers to your question involve quantum mechanics. But, of course, classical Rutherford scattering is modified by a classical charge distribution form factor. (Compare to center-of-mass motion in mechanics.) You should then insert your restriction "classical" in your title and in several points in the body of the question, to specify what you don't understand. $\endgroup$ Commented Jan 16, 2018 at 15:13

1 Answer 1


My discussion essentially follows this article.

Form Factors are an intuitive and simple tool used to describe the scattering particles from extended targets. Here I'm going to show how the Form Factor comes about in the context of the scattering of spinless electrons

As with many scattering experiments, the quantity we are interested in is the differential cross section $\frac{d\sigma}{d\Omega}$ of our scattered electrons off our target. The differential cross section is related to the scattering amplitudes through the relation: $$\frac{d\sigma}{d\Omega}=\frac{k}{k_i}|f(\theta,\phi)|^2$$ where $\theta$ is the scattered angle.

The scattering amplitudes $f(\theta,\phi)$ can be obtained in approximate form using the Born Approximation. To first order (and up to a normalization) the Born Approximation can be written as: $$f_{B1}=\int\phi_{k_f}^*(\vec{r})V(\vec{r})\phi_{k_i}(\vec{r})d^3\vec{r}$$

In the first Born Approximation the initial incoming wave and the outgoing waves are assumed to be plane waves of the form: \begin{align} \phi_{k_i}(\vec{r})=e^{ik_i\cdot r}\\ \phi_{k_f}(\vec{r})=e^{ik_f\cdot r} \end{align}

We can describe an extended charge distribution by $Ze\rho(\vec{r})$ with $$\int\rho(\vec{r})d^3\vec{r}=1$$

In this case, the potential experienced by an electron located at $\vec{r}$ is given by the Coloumb potential: $$V(\vec{r})=-\frac{Ze^2}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{|r-r'|}d^3\vec{r}'$$

And substituting this potential into the general expression for the first Born Approximation to the scattering amplitudes $f(\theta,\phi)$ gives $$f_{B1}=-\frac{Ze^2}{4\pi\epsilon_0}\int e^{\frac{iq\cdot r}{h}}\int\frac{\rho(\vec{r}')}{|r-r'|}d^3\vec{r}d^3\vec{r}'$$

Making the substitution $\vec{R}=\vec{r}-\vec{r}'$ and noticing that $d^3\vec{R}=d^3\vec{r}$ $$f_{B1}=-\frac{Ze^2}{4\pi\epsilon_0}\int \frac{e^{\frac{iq\cdot \vec{R}}{h}}}{\vec{R}}d^3\vec{R} \left[\int e^{\frac{iq\cdot \vec{r}'}{h}}\rho(\vec{r}')d^3\vec{r}'\right]$$

This bracked factor is known as the Form Factor, $F(q)$.

$$F(q)=\int e^{\frac{iq\cdot \vec{r}'}{h}}\rho(\vec{r}')d^3\vec{r}' $$

It can be shown that when the expression for $f_{B1}$ is used to determine $\frac{d\sigma}{d\Omega}$, that:


This expression can be interpreted intuitively as Rutherford scattering modulated by the square of the Form Factor. In other words, electron scattering off an extended source is equal to scattering off a point source modulated by the form factor.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.