Finding the lattice constant for a zincblende (modified fcc) crystal ...

Posted on 2006-03-31
Last Modified: 2008-03-17

I want to know how you would find the lattice constant (edge length) for a crystal unit cell if it is a zincblende, in effect, composed of atoms of two different radii. In this case (a homework problem I'm stuck on) the atoms are In and SB, but generally speaking ... isn't zincblende essentially the same as a diamond structure? How do the differing radii change the standard formulas, ie, the ones I actually know?
Question by:princess_of_reason
    LVL 11

    Accepted Solution

    What is the "touching direction" of Zinc Blend, i.e. the direction of the atoms where the most "touching" occurs.

    For instance in FCC the touching direction would be accross the face diagional where a=root(2) x a and in BCC the touching direction is the body diagional where a = root(3) x a. Here a is the edge length.

    And yes, ZnS does have 2 different sized ions ( 14 Zinc and 4 Sulfur ) but the sulfur are arranged in a specific tetrahedral pattern inside the FCC structure. And yes, it is exactly like the diamond cubic structure, just with sulfur instead of zinc in the center of the cube.

    So, I would use a corner atom and one at the (1/4,1/4,1/4) position (starting up the body diagional) to get a length = ( body diagional ) / 4. Then go from there.


    Author Comment

    With your guidance, I've worked it out. Thank you! (And I apologise for taking so long to respond.)
    LVL 11

    Expert Comment

    No problem!

    Glad i could help.


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