Suppose I have a 50-50 mix of two different kinds of atoms in a crystal. The crystal is BCC and the atoms have different scattering factors $f$.

I know that if the crystal is organized in the regular BCC way (say, like CsCl) then I can easily calculate the structure factor. This page shows what it would be for CsCl, for example. I understand this part.

But what would happen if the crystal were still BCC but the atom locations were picked randomly? If you performed neutron diffraction on this crystal, would you be able to see any (h,k,l) planes?

My thinking is either: a) you see no (h,k,l) planes (because the disorder means that you don't know the atom locations so the calculation for the structure factor is impossible) or b) you see all planes (because there are no interference effects so all planes are equally "visible")?

I am trying to wrap my head around crystal diffraction and I'm finding this one tough to think about. This is partially a continuation of this question.

  • $\begingroup$ Well, we certainly get good patterns from $NaCl$ ; my suspicion (being too foggy to think this AM) is that you'd have to have awfully different atomic number species to see any effect from nucleus size. $\endgroup$ Mar 4, 2016 at 15:07
  • $\begingroup$ Of course you will still see diffraction. There are many binary bcc metals with mutual solubility, and many have both ordered and disordered regions on a phase diagram. In randomly ordered bcc alloys the spot intensity will be that of an appropriately averaged scattering intensity. Once ordered, you would of course see different intensities. $\endgroup$
    – Jon Custer
    Mar 4, 2016 at 15:29

1 Answer 1


Yes you will see the usual BCC pattern and I think it's fairly easy to see why.

If you take some plane then this creates a bright reflection at a certain angle because along a line at that angle all the lattice points radiate in phase, so we get constructive interference. At other angles the lattice points don't radiate in phase so there is destructive interference and no reflection.

Now imagine randomly removing some of the lattice points i.e. making the biggest decrease possible in the structure factor. The remaining lattice points still radiate in phase, so there's still a strong reflection at the same angle, but since there's fewer points radiating the intensity of the reflection is decreased. At the same time the destructive interference in other directions is less complete, so the background reflection goes up. But you'd still see a strong reflection at the same angle.

As you removed more and more of the lattice points the contrast would decrease, and eventually you'd no longer see a pattern. However even removing half the lattice points wouldn't make much difference because the in phase reflection is so much stronger than the background.

  • $\begingroup$ Great answer, thanks. So you would expect to see a different diffraction pattern for the randomly-distributed bcc crystal and the "regular" bcc crystal? $\endgroup$
    – Wise Owl
    Mar 4, 2016 at 17:18
  • $\begingroup$ @WiseOwl: The spots/lines will be in the same place but their relative intensities might change. Bear in mind that the structure factors aren't likely to be very difference if the two atoms are interchangeable. In most of the alloys Jon Custer mentions in a comment the atoms are similar atomic numbers. $\endgroup$ Mar 4, 2016 at 17:24
  • $\begingroup$ @JohnRennie - A quick glance does raise interesting counter examples. The largest differences that I found quickly in atomic number are the ordered bcc alloys of Mg with rare earths including Dy, Gd, Ho, Tb, and Y. By the actinides the tendency to want to be bcc in the first place is much weaker (hence the weird allotropes of, e.g., U and Pu). $\endgroup$
    – Jon Custer
    Mar 4, 2016 at 19:24

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