We know that, $\sigma (\theta)=\sigma _M (\theta){F(q)}^2$, where $\sigma _M (\theta)$ is the scattering cross section for a point nucleus while $\sigma (\theta)$ is the observed scattering cross section from a nucleus of a finite size.Here, F(q) is the form factor.

$\sigma _M (\theta)=\frac{Ze^2}{8\pi \epsilon _0 E}\frac{cos^2 \theta /2}{sin^4 \theta/2}$ In a book, it's given that the form factor F(q) is lower than one as the scattered electron waves from different parts of the nucleus interfere destructively. If you could help me to understand why they interfere destructively, I'd appreciate that.


1 Answer 1


One must be clear when one is talking of wave solutions whether it is classical sound or electromagnetic or water waves, or whether one is talking about a quantum mechanical wave equation.

Interference effects always appear in wave solutions of wave equations but the context is very important.

As you are talking about a nucleus, you are talking about quantum mechanical wave equations and their wave function solutions. By axiomatic definition the wave effects are measurable only as probability distributions, in the case you are talking these will be the interference effects seen in the cross section. The cross section is in one to one correspondence with the probability distribution given by the $Ψ^*Ψ$ , where $Ψ$ is the wavefunction solution of the scattering for the nucleus you are talking about.

It is simple to understand this with the double slit experiment one electron at a time : "electrons scattering off two slits given distance apart , given width" is the experiment.


Single electrons seem random, the interference shows the wave nature of the underlying equation.

The experiment "electrons impinging on a nucleus" , if set up for one electron at a time, again would look random until a statistical accumulation would show the interference.

So the destructive interference the book is talking about is in the probability distribution, due to the wave nature of all quantum mechanical equations. It is more probable to scatter in specific directions than at others, and the interference is in the probability, not on the electrons themselves. The above double slit is given as an intuitive example of how shapes can appear in a quantum mechanical level.

Now the mathematical definition of a form factor is such as to describe the spatial extent of a composite object In contrast to the simple point scattering of Rutherford. In experiments with nuclei, interference means that a shape appears when a large accumulation of interactions is obtained. This means that there will be enhancements and depletions in the spatial distribution of scattering, depending on how much the nucleus departs from a point scattering formula. The form factor is a function, of spacetime, and this gives it a variable value dependent on the four vectors describing the experiment; usually $Q^2$ is used as the relevant variable, as far as I remember.

  • $\begingroup$ But that doesn't explain why the form factor's value is 1 everywhere. If it's similar to a double-slit experiment, then at some point electron waves should interfere constructively which should result in F(q)>1 at such points. $\endgroup$ Aug 11, 2019 at 16:39
  • $\begingroup$ @DebasishDhal I edited $\endgroup$
    – anna v
    Aug 12, 2019 at 3:31
  • $\begingroup$ @DebasishDhal For interference in the quantum world I think of allowed paths for the particles, for photons in the double slit the path allowed is based on integer multiples of the wavelength, so bright areas contain all the photons as the path length there is n times wavelength. For electrons the same should be true, they have a wavelength and maybe they only interact with matter at certain path lengths.? $\endgroup$ Aug 12, 2019 at 17:05
  • $\begingroup$ The interaction is given by the mathematics of the wavefunction a probability to happen. i.e. it is a statistical effect of many measurements. individual events have continuous variables in the integral. $\endgroup$
    – anna v
    Aug 12, 2019 at 18:01

Your Answer

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

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