Miller Indices of Simple Cubic Lattice

  • Thread starter mkphysics
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  • #1
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Main Question or Discussion Point

I'm trying to get my head around indexing x-ray diffraction patterns and this thought experiment stumps me. Lets say I have a crystal with a simple cubic lattice structure:

http://ecee.colorado.edu/~bart/book/sc.gif

I align a narrow x-ray beam incident on the crystal so that the x-ray beam direction vector is normal to the crystal 100 plane. I then observe the pattern of spots created on an observation screen downstream of the crystal. I would expect to see a spots with index [100]. Would I expect to see spots with indexes 200, 300, etc? That is would I expect to see spots with indexes [X00] where X is larger than 1? If so why? My understanding is that the planes associated with Miller Indices are determined by planes passing through atoms within the crystal lattice and that higher index values indicate closer plane spacing. In a simple cubic lattice [100] would be defined by adjacent planes of atoms within the lattice but [200] would require a plane of atoms with a spacing half the minimum distance possible between atoms. Therefore I would NOT expect to see higher order spots in the diffraction pattern.

Am I thinking about this correctly or not?
 

Answers and Replies

  • #2
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OK. I think I worked it out. The key is to remember that Braggs law is nl = 2d sin(T) NOT l = 2d sin(T) where n is diffraction order, l is wavelength, d is plane spacing and T is angle of diffraction. If d = a/sqrt(h^2 + k^2 + l^2) then Braggs law is

nl = 2*a/sqrt(h^2 + k^2 + l^2)*sin(T)

After rearranging the equation

l = 2*a/sqrt(n^2(h^2 + k^2 + l^2))*sin(T)
l = 2*a/sqrt(nh^2 + nk^2 + nl^2))*sin(T)

Photons which are second order diffracted (n=2) from the [hkl] = [100] plane occur at equivalent locations to x-rays which are first order diffracted (n=1) from the [hkl] = [200] plane (if a [hkl] = [200] plane existed).

Therefore I WOULD expect to see HIGHER INDEX, X00 (X>1), spots due to higher order diffraction NOT the presence of tighter packed planes.
 
  • #3
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I was bothered with the same "controversy".
It didn't take you long to solve your own question.

Thanks.
 

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