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Number of atoms in FCC gold cube

  1. Dec 25, 2011 #1
    Hello i need help in this question

    it is given that the radius of a gold atom sphere is 144.2 pm, gold is

    packed in an FCC manner

    Give the ratio Ns/Nv as a function of L (gold cube side) and a parameter

    of FCC latice, knowing that Ns is the number of surface atoms and Nv is

    the number of atoms per volume.

    thank you
     
  2. jcsd
  3. Dec 25, 2011 #2
    what i am thinking is:
    in a unit cell we have 1/8 x 8 corner atoms = 1 atom and in the center of each phase we have 1/2 x 6 = 3 then the total number of atoms will be 4
    atoms per unit = 4
    volume of the cube is L^3 and volume of the unit is a^3 so if we want to know how many units are there in this cube we can divide the volume of the cube over the volume of the unit cell and we will have L^3/a^3
    and if we multiply them by 4 which is the number of atoms per unit we will have the number of atoms in a cubic volume Nv= 4L^3/a^3

    now talking about the surface is the thing confusing me
    counting the atoms in unit cell is it 2 coming from 4 corners shared by 4 cells which is 1/4 x 4 and 1 center atom and by this we will have 2 atoms or my thinking is not reasonable?
     
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