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Homework Help: Packing of Particles

  1. Jul 18, 2006 #1
    I am doing a lab on packing methods. Hexagonal closest, face-centred and body-centred.

    I had no trouble drawing the diagrams for these and and determining cordination numbers and seeing which is more efficient. However I am having trouble explaining the relationship between coordination number and density. It gives mass/volume as the definition of density.

    What I did was place one unit in an imaginary box - assign each atom a value of 1 mass and divide by volume. This gives me a higher density for body-centred which I know is wrong.

    Can anyone help explain to me what I am doing wrong?

  2. jcsd
  3. Jul 18, 2006 #2


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    Last edited: Jul 18, 2006
  4. Jul 19, 2006 #3
    Ok this seems to help but I still dont quite have it.

    Lets do a test on say cubic and give an imagingary radius of 1 for each of the 13 units
    V = (4/3)Pi r^3
    = 1.33 x 3.14 x 1
    = 4.2 x 13 units = 54.6 units of volume in 13 spheres

    Sphere volume (radius would be 3)
    V= (4/3)pi r^3
    = 1.33 x 3.14 x 27
    = 113 units

    54.6 / 113 = 48.3?? (I have read that it is 74% so I did something wrong but can't put my finger on it)
  5. Jul 20, 2006 #4


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    Where are you getting 13 and 3 from? Your approach - which I can't say I understand - appears to be wrong. Here's how you do it.

    You have an FCC lattice. Look at the unit cell - it has 8 corner atoms and 6 face centers. Each corner atom is shared by 8 neighboring unit cells and each face center atom is shared by two. So the total number of atoms per unit cell in the FCC lattice = (8*1/8) + (6*1/2) = 1+3 = 4

    Nearest neighbor atoms ("touching" each other) may be found along a face diagonal. If the side of the unit cell is 'a', the length of the face diagonal is a*sqrt(2) = about 1.4a. Going from one corner to another, along the face diagonal, you encounter the radius of the first corner atom, then the diameter of the face center atom, and finally the radius of the second corner atom. In terms of the atomic radius, the length of the face diagonal is then r + 2r + r. So, we have 1.4a = 4r or roughly a = 2.83r

    That establishes a relationship between the sphere radius and the unit cell size. Next you say that the unit cell volume is a^3, and the volume occupied by spheres is 4*(4/3)*pi*r^3. The 4 comes from the number of atoms per unit cell, calculated abouve. The packing density is then the ratio of the second volume to the first.

    PD = [(16/3)*pi*r^3]/[a^3] = 16.75*r^3/a^3 = [16.75]/[2.83^3] = 16.75/22.62 = 0.74
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