I thought that I understand the "Bragg's Law" understanding of crystal diffraction, but recently I read something that made me confused. I understand that if the planes in the crystal have sufficiently small spacing relative to wavelength (more specifically, such that $sin \theta$ in Bragg's Law >1 even for lowest order), then there would be no angles for which constructive interference occurs. Now, I read that if I shoot neutrons at the crystal of sufficiently large corresponding wavelength, then the neutrons will make it through. Why did I get confused by this statement?

Because, suppose we are shooting waves. I would think that scattering still occur, just that at every angle, the waves "cancel" each other out so nothing is observed. In particular, there is still nothing that makes it to the end of the crystal because the wave is scattered as it moves through the crystal. But apparently this is not the case for neutrons. So my question is:

  1. Is my understanding with shooting of waves correct? [That the waves are still scattered in all directions, but not observed due to destructive interference.]
  2. Am I right to conclude that the wave model is therefore not an adequate description of neutrons? [I think I can see why neutrons make it to the end if I use probability amplitude analysis instead of thinking of the neutrons as waves.]
  3. What about something like light which can be understood as a wave and as a stream of photons?

The X-Rays make it through as well, or some of them do. For anything travelling through the crystal there will be a scattering probability, so some of the wave is scattered and some makes it through. You'll get a roughly exponential fall-off of the intensity with distance into the crystal.

I can't see how neutrons are behaving differently from X-rays, except that the scattering probabilities will be different - X-Rays scatter off electrons while neutrons scatter off nuclei.

| cite | improve this answer | |
  • $\begingroup$ But for neutrons, if you study the probability amplitude of being scattered to a fixed angle $\theta \neq 0$, it would be sum of things that cancel out to yeild 0. Most of the probability amplitude would concentrate on the direction that made it through thee crystal, no? $\endgroup$ – suncup224 May 24 '14 at 1:12
  • $\begingroup$ Anyway, this was the source where I read about the neutron bits. Assume we are firing long-enough wavelength neutrons only. feynmanlectures.caltech.edu/III_02.html#Ch2-F7 $\endgroup$ – suncup224 May 24 '14 at 1:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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