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Hexagonal Closest Packed Unit Cell Height

  1. Nov 29, 2006 #1
    My attempts to derive the height as a function of radius of the spheres packing the unit cell have failed. The best attempt so far was to create a diagonal line from the top atom to one of the middle ones. With that distance I thought I could use trig to find the component parallel to the height and multiply by two. Of course that didn't work. Any hints or suggestions? Even good 3d visualizations would help.
     
  2. jcsd
  3. Dec 1, 2006 #2

    Gokul43201

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    The trick is in realizing that the middle plane of atoms occupy positions directly above the centroids of the triangles in the base plane. This follows directly from a symmetry argument.

    #1. Each basal plane has nearest neighbor atoms making equilateral triangles. So, a=2R (where R is the sphere radius).

    #2. Each atom at height c/2 above the basal plane is positioned directly above the centroid of the triangles in the base plane. For an equilatreral triangle, the distance from a vertex to the centroid is two-thirds the length of the median.

    #3. Each atom in the base plane has a nearest neighbor in this middle plane. So, the distance from the corner atom in the base plane to the nearby atom in the mid-plane is 2R.

    #4. Finally, use Pythagoras to write (2R)^2 = (c/2)^2 + (dist. calculated in #2)^2. That should take you to the required ratio.
     
  4. Jun 18, 2011 #3
    I tried to find it with the help of packing efficiency (i.e. 0.74). the number of contributing spheres inside one unit cell is 6. therefore volume of 6 spheres/volume of hexagonal unit cell=0.74
    volume of hexagonal unit cell is 6*sqrt(3)*r*r*h
    then u will get answer of "h".
     
    Last edited: Jun 18, 2011
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