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Crystal x-ray diffraction

  1. Nov 7, 2005 #1
    I was given the following question: A crystal of hydrogen atoms and a crystal of lead atoms have exactly the same crystal structure, including lattice constant. How can they be distinguished by x-ray diffraction? The solutions say that the beams diffracted off the lead crystal would be much brighter, but those diffracted off hydrogen would scatter to larger angles. Can anyone explain why this is?
  2. jcsd
  3. Nov 7, 2005 #2
    The atomic scattering factors are different. The scattered intensity depends on the scattering factor squared at a given magnitude of the scattering vector (or angle). I didn't check but if you check the tabulated values for the scattering factor for H and Pb I'm pretty sure you'll see behaviour that explains this in them.
  4. Nov 7, 2005 #3
    Is there a more qualitative way to explain it? This was a question on a test, so we wouldn't have had access to any tables.
  5. Nov 8, 2005 #4
    Tell me if this sounds right. The atomic form factor, which is basically the Fourier transform of the charge density of a single atom, determines the magnitude of the refractions. If the incoming wave vector is k and the refracted wave vector is k', then the magnitude squared of the form factor evaluated at k-k' gives the intensity. So for lead, the charge distribution is more spread out, but larger in magnitude, so the form factor will be more localized and larger in magnitude than for hydrogen. So small k-k', corresponding to small Bragg angles, will scatter with more intensity off lead, but since the form factor is more localized, the refractions will not extend to as large a Bragg angle as those for hydrogen.
    Last edited: Nov 8, 2005
  6. Nov 8, 2005 #5
    That seems ok to me. Magnitude of charge distribution could use rephrasing though. I know what you mean by it but I don't think the magnitude of a distribution really means anything.
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