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Lattice constant problem

  1. Sep 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A material contains two types of atoms (A and B) in a zincblende structure. Atom A is group V with a hard sphere radius of 2.5 Å. Atom B is group III with a hard sphere radius of 1.6 Å.
    (a) What is the lattice constant of this material assuming that nearest neighbour atoms touch? (working on this one right now)
    (b) What is the volume density of valence electrons in this material?
    (c) What is the surface density of B atoms on the (110) plane in this material?

    2. Relevant equations


    3. The attempt at a solution

    a)
    Since it's a face centered structure it should have 8X 1/8 + 6* 1/2 atoms, so 4
    Is it correct to say ( r1+r2 ) *4 = my lattice constant for A?
     
  2. jcsd
  3. Sep 23, 2015 #2

    ehild

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    No it is not right. If there are 4 units in a cube, how can the side of the cube 4 times the interatomic distance???
    this helps http://www.ehow.com/how_8721263_determine-lattice-parameter-zincblende.html
     
  4. Sep 23, 2015 #3
    a) (4/3)^1/2*(1.6+2.5) = 4.73 A

    b) Volume density of valence electrons is 8/((4.73*10^-8)^3) = 7.56*10^22 cm^3

    c) 4/(sqrt2(4.73*10^-8)^2) = 1.26*10^15 atoms/cm^3
     
  5. Sep 23, 2015 #4

    ehild

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    How many valence electrons are there in one AB unit? How many units are there in the unit cell?
    The surface density is asked, it has atoms/cm^2 unit.
    How did you get that formula?
     
  6. Sep 23, 2015 #5
    4 *(1/4) + 2*(1/2) + 2*1=4?
    I confused it with a volume density formula.

    4/(4.73*10^-8)^3) = 3.78*10^22 cm^3 = volume density

    sqrt2/((4.73*10^-8)^2)=6.32*10^14 surface density
     
    Last edited: Sep 23, 2015
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