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Lattice constant

  1. Mar 3, 2009 #1
    Calculate the lattice constant of caesium chloride with ionic radii Cs+ 0.167nm and cl- is 0.181 nm

    I dont really get it,
    from all the structures i have seen the distance between one atom and the next is (a) or (a/2)
    but i dont think im right.Any ideas. What is the difference between the radii (what do they correspond to.

    Wouldnt you just add them together.
    I think Iam supposed to do some sort of pythagoras but i dont know why.
  2. jcsd
  3. Mar 3, 2009 #2
    I found an equation that might apply but dont know if its right or why its right

    2(r + R)/√3

    Where the r's correspond to the different radii
  4. Apr 15, 2009 #3
    Hi Fabsuk,

    I know this answer is a little late in coming, but I only came across your post now. The reason I only came across it now is because I'm having problems with a similar problem.

    The CsCl lattice constant is just the edge length of its conventional unit cell (a BCC structure), that is, the distance between the centers of the Cl atoms on two adjacent corners.

    The two radii add up to the distance between the centers of a Cl atom at the corner and the Cs atom in the center of the BCC structure. It is the therefore the distance between a corner and the center of the BCC structure.

    You could use either Pythagoras or trigonometry to relate the length of a body diagonal (2 x sum of the two radii given) to the edge length, i.e., the lattice constant, in order to calculate it.

    Hope this helps.

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