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Nearest Neighbor lattice question

  1. Dec 19, 2009 #1


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    1. The problem statement, all variables and given/known data

    An element crystalises in a face-centred cubic lattice with a basis group of two atoms at 000 and 1/4 1/4 1/4. The lattice constant is 3.55Angstroms.

    i) what is the separation of nearest neighbor atoms
    ii) how many nearest and second nearest neighbors does each atom have.


    i) The nearest neighbor distance is just going to be from the 000 atom to the one 1/4 1/4 1/4 away from it? Which gives
    [tex]\sqrt{3(\frac{1}{4}a)^3} = 1.54Angstroms[/tex]

    ii) If the guess for part i is right then each atom will have one nearest neighbor. I can't get my head around the second nearest neighbors.

    Any help appreciated, thanks.
  2. jcsd
  3. Dec 20, 2009 #2
    see attachment...

    1) distance = 1/2*(2)^(1/2) * lattice constant
    2) 6 next-nearest neighbors at distance of lattice constant

    Attached Files:

    • fcc.jpg
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