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Calculating the interplanar distance d111 for an FCC lattice

  1. Feb 23, 2010 #1
    1. The problem statement, all variables and given/known data

    As a part of a lab report, I need to calculate the distance of the (111) planes of an FCC lattice made out of spheres with diameter D.

    2. Relevant equations



    3. The attempt at a solution
    The course assistant has given me the value of [LATEX]\frac{\sqrt{6}}{3}D[/LATEX]. I can understand where the [LATEX]\sqrt{6}D[/LATEX] comes from; it's the space diagonal of the cubic unit cell of the FCC lattice. But why is it divided by three? That means that there are three (111)-planes in one unit cell, but I have no idea why is that.
     
  2. jcsd
  3. Feb 23, 2010 #2

    Gokul43201

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    Your interpretation is more or less correct. The body diagonal of a single FCC unit cell intersects (or terminates at) 4 successive (111) planes, with 3 inter-planar regions between them.

    Perhaps this figure might help...

    [​IMG]

    However, to calculate the interplanar spacing for a set of planes in a cubic lattice there is a pretty straightforward formula based on the Miller Indices of the plane. If you know the definition of the Miller Indices in terms of intercepts along the crystal axes, you can derive this formula using simple geometry.
     
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