In the 2 slit experiment with electrons, is the distance between the slits related to the individual electron's de broglie wavelength?

In other words, if the slits are too far apart which would prevent the electron's matter wave from passing through both slits, does the interference pattern then fail?

A broader question is what is the relationship between the size of the slits, the distance between the slits, and the observed interference pattern?


2 Answers 2


You ask:

A broader question is what is the relationship between the size of the slits, the distance between the slits, and the observed interference pattern?

The answer here covers your question.

The de Broglie wavelength describes the effective wavelength that a particle would have when it was behaving as a wave.

to decide what wavelength an electron should have so as to be able to see the interference pattern one has to see separation and and distance to the screen that the slits should have.

The interference pattern is dictated by the distance from one bright line (coherence) to the next:

$$\Delta y = \frac{\lambda D}{d}$$

where D is the distance from the slit to the screen (or detector), little d is the spacing between the slits, and λ is going to be our de Broglie wavelength.

Let's assume we want to use electrons for our experiment. We build a setup with the screen placed 1 meter from the slits, and the two slits 1 millimeter apart (maybe we found this equipment in a storage closet in the physics department...). This setup will make the distance between the bright spots on our screen 1000 times what the de Broglie wavelength of our incoming electron is. We want to be able to actually see the interference pattern in our detectors, so perhaps we should request that the spacing of the bright spots be about 1 millimeter (this would depend on the detectors, of course). This means the de Broglie wavelength of our electron has to be about one meter. Now we go back to the equation for de Broglie wavelength, and see that we know h and we now know λ, so we can calculate what p should be. Since we know the mass of the electron, calculating the momentum is essentially the same as calculating the speed; for our experiment, we find the electron needs to be going about 0.0007 m/s! That's a tiny speed... about 2 inches a minute (kind of like pouring ketchup)!

So experiments are not easy with electrons.

For the buckyball experiment , the researchers used slits about 100 nanometers apart (a nanometer is one millionth of a millimeter), and shot the buckyballs through the slits at about 200 meters per second (roughly 500 mph), much slower than the speed of light.


First experiments to get intensity distribution of electrons behind an edge was done by Möllensted and Düker with a thin wire and a biprisma (they are mentioned here). There was no slit at all. And electron diffraction an a single edge was explored in 1940 by H.Boersch, but there was mentioned Arkadiew, who has worked on this in 1913. The link to my elaboration is in German only.

Using a biprisma it was shown that an electrical potential lead to variing deflection. So electrons with the same speed and hence the same de brogli wavelength get deflected different in different electric potential.

  • $\begingroup$ I appreciate the additional examples. But I am looking for the set up where individual electrons one by one were "shot" at 2 slits/holes in a double slit experiment. The question is whether one can hypothesize whether a single electron's wave packet would interfere with itself if it in effect was diffracted through the 2 slits. $\endgroup$
    – Greg
    Nov 5, 2015 at 6:39
  • $\begingroup$ Hi @Greg. Read my last sentence? Perhaps the intensity pattern is the result not of some waves but is the result of interaction between electron and electric potential of the edge. Does not play role, with single electron or a bunch of electrons $\endgroup$ Nov 5, 2015 at 6:46
  • $\begingroup$ This electrical potential is a quantised of course. $\endgroup$ Nov 5, 2015 at 6:51
  • $\begingroup$ Thanks for reply. I don't understand that claim. Are you referencing the Bohm-Aharanov effect? Getting an intensity distribution behind an edge is different than generating an interference pattern one electron at a time. They measure different phenomena I think. $\endgroup$
    – Greg
    Nov 5, 2015 at 6:52
  • $\begingroup$ physics.stackexchange.com/questions/158105/… $\endgroup$ Nov 5, 2015 at 6:57

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