Why are diffraction rings closer together when electrons travel at a greater speed in electron diffraction?

I know that at higher speeds the de Broglie wavelength decreases so the electrons diffract less, but does the fact electrons repel affect it in any other way?

What I was thinking was that since electrons now reach the screen in less time because of the greater speed, they have less time to repel and thus repel less so rings are closer together. Is this a correct reason and does this relate in any way to the wavelength reason?

These images are from Electron diffraction.

A beam of electrons is accelerated in an electron gun to a potential of between 3500 V and 5000 V and then allowed to fall on a very thin sheet of graphite (see diagram above). The electrons diffract from the carbon atoms and the resulting circular pattern on the screen (see diagrams below) is very good evidence for the wave nature of the electrons.

The diffraction pattern observed on the screen is a series of concentric rings. This is due to the regular spacing of the carbon atoms in different layers in the graphite. However since the graphite layers overlay each other in an irregular way the resulting diffraction pattern is circular. It is an example of Bragg scattering.

What I was thinking was that since electrons now reach the screen in less time because of the greater speed, they have less time to repel and thus repel less so rings are closer together. Is this a correct reason and does this relate in any way to the wavelength reason?

Short Answer is no. You would get the exact same pattern if you shot the electrons one by one and not in a beam where they could theoretically interact with each other via coulomb repulsion. Note that the pattern tells you something about the distribution of the "impact positions of the particles".

Stating in the graph here it is purely particle-like and here it is purely wave-like is oversimplifying. You should think of the particles as something different, which happens to behave like waves in one limit and like particles in the other, and how it behaves in between is more complex and described by shrödinger equation.

• I understand that electrons would form the same pattern if they were fired one by one, but why would the electrostatic repulsion reason be wrong? It sounds logical, and there must be some repulsion at least?
– XXb8
Mar 30, 2020 at 9:48
• I would assume it is not significant here. I'm a bit lazy, but you can calculate the deviation due to classical coulomb during the flight time. You can assume it is a particle, calculate how long it stays in flight, and just assume there is a particle in another trajectory leading to the other diffraction ring. It should be dramatically less than the diffraction length scale. My physical intuition speaking here, though. Mar 30, 2020 at 9:57

Assuming that we are dealing with parallel electron beams, this can be explained through Bragg's Law, $$n\lambda = 2d sin\theta$$, with $$2\theta$$ the angle between the incident and diffracted ray. As $$\lambda$$ is decreased, $$sin\theta$$ should decrease for the same value of d, hence the rings are now closer to the central spot.

• I started to draw this but then remembered drawing Bragg diffraction takes longer in PowerPoint than I wanted to invest and it never looks convincing when you do manage to do it right :-) commons.wikimedia.org/wiki/File:Bragg_legea.jpg and youtube.com/watch?v=Cjce4QumZNk
– uhoh
Mar 30, 2020 at 4:01
• @uhoh These answers make sense, thank you. I was looking for a more intuitive explanation, such as the fact electrons repel or the de Broglie wavelength explanation. How would you go about this explanation?
– XXb8
Mar 30, 2020 at 8:30
• @XXb8 well $\]ambda$ in Bragg's law is the de Broglie wavelength of the electron. If this were Bragg scattering in X-ray diffraction then we'd just call it the wavelength. The article mentions that the energy is 3500 to 5000 eV, which makes the wavelength between 0.21 and 0.17 Angstroms and the distance between planes of graphene atoms is about 3.35 Angstroms.
– uhoh
Mar 30, 2020 at 9:25
• @XXb8 The electrons really scatter off of the atoms which have both repulsive electrons and attractive nuclei, so you can think of each atom as a source of spherically shaped but forward peaked waves just like the diagram in your question shows.
– uhoh
Mar 30, 2020 at 9:25

The momentum of an electron $$p$$ increases with energy.

Let's say the lattice spacing of a material is $$a$$. Then we can assign a de Broglie momentum to the lattice (of sorts anyways) that is $$q=\hbar \frac{2\pi}{a}$$.

We can think of the crystal/material as giving a kick off momentum to the electron that is of order $$q$$, so that the electron exits with momentum $$\mathbf{p}+\mathbf{q}$$. If $$q$$ is perpendicular to $$p$$, then we have the exiting angle is approximately $$\theta\sim q/p$$. Thus, for larger electron energy $$p$$ goes down and so does the scattering angle $$\theta$$.

This behavior is ompletely identical to light passing through a diffraction grating. As you decrease the wavelength (larger momentum), the angular spacing between the diffracted beams becomes smaller.