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Bragg Reflection = Standing Waves

  1. Oct 22, 2008 #1
    1. The problem statement, all variables and given/known data
    I've got my exam in solid state physics tomorrow, and although I've understood most of the most complex subjects now, I feel I'm missing the understanding on one fundamental phenomena; Bragg reflection..

    Throughout Kittel it's mentioned, that when we're at the first brillouin zone with k=½G, then we have standing waves. I don't think it's mentioned anywhere why though?

    Is there some equation from which I can easily realize, that when k=½G then the wavefunctions are made up of equal parts of waves traveling to the right and left?
     
  2. jcsd
  3. Oct 22, 2008 #2
    If you look in Kittel at Bragg diffraction you will see that Bragg's condition for diffraction can be also put in the same form: k=1/2 G where k is the wave-vector of the X-ray.
    For electrons, k is the wave-vector of the electron wave.
     
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