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Solid State: Diffraction Conditions

  1. Sep 11, 2009 #1
    In my Solid State course we've reached the topic of crystal diffraction and reciprocal lattices. I haven't had any problems so far, but I've hit a little snag in understanding how the diffraction condition [tex]2\vec{k} \cdot \vec{G}=G^2[/tex] is equivalent to the Bragg law [tex]2d\sin{\theta}=n\lambda[/tex].

    In the textbook (Introduction to Solid State Physics - Kittel 8th Ed.), the math is as follows:

    1. The reciprocal lattice vector [tex]\vec{G}=h\vec{b}_1+k\vec{b}_2+l\vec{b}_3[/tex] is normal to the lattice plane given by the indices [tex](hkl)[/tex].
    2. The spacing between these planes is [tex]d_{hkl}=\tfrac{1}{h}\hat{n}\cdot\vec{a}_1=\tfrac{1}{Gh}\vec{G}\cdot\vec{a}_1=\tfrac{2\pi}{G}[/tex] where [tex]\vec{a}_1[/tex] is a primitive axis for the direct lattice.
    3. The diffraction condition [tex]2\vec{k} \cdot \vec{G}=G^2[/tex] can then be written as [tex]2k\sin{\theta}=2(\tfrac{2\pi}{\lambda})\sin{\theta}=\tfrac{2\pi}{d_{hkl}} \Rightarrow 2d_{hkl}\sin{\theta}=\lambda[/tex]. Here, [tex]\theta[/tex] is the angle between [tex]\vec{k}[/tex] and the plane [tex](hkl)[/tex], not between [tex]\vec{k}[/tex] and [tex]\vec{G}[/tex].
    4. Somehow [tex]d_{hkl}=\tfrac{d}{n}[/tex], so this becomes [tex]2d\sin{\theta}=n\lambda[/tex] which is just the Bragg law.

    Now, most of this I understand. The only things I don't get are why the lattice plane spacing [tex]d_{hkl}[/tex] is equal to [tex]\tfrac{1}{h}\hat{n}\cdot\vec{a}_1[/tex] and what the relationship [tex]d_{hkl}=\tfrac{d}{n}[/tex] means. What's the difference in physical meaning between [tex]d_{hkl}[/tex] and [tex]d[/tex]?
     
  2. jcsd
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