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Proof that c/a=RootOf(8/3) for Hexigonal Close Packed (HCP)

  1. Oct 23, 2012 #1
    This is a typical problem in an introductory course to condensed matter physics. I recently solved this problem and had some trouble conceptualizing it, so I thought I would make a post on how to solve the problem.

    In general the problem will be stated like this: "Prove that for an ideal HCP [itex]\frac{c}{a}[/itex]=[itex]\sqrt{\frac{8}{3}}[/itex]."

    First of all "Ideal" means that the atoms in the basis are solid spheres which are touching one another. Hexigonal close packed HCP looks something like this: hcp.png . Where 'a' is the distance between the center of the atoms (spheres), and 'c' is the distance between A-Layers (height). The radius of the atoms will be 'a/2' because there are two radii per length 'a'.

    The three atoms in the center of the unit cell form a tetrahedron with the base being defined by the three atoms in the B-Layer, and the top vertex is defined by the center atom in the top A-Layer. The height of the tetrahedron is [itex]\frac{c}{2}[/itex] and the length of its base side is 'a'.

    In the limiting case: we define 'x' to be the distance between the center of spheres in the A-Layer and B-Layer.

    We have two formulas for 'x':

    a2-x2=([itex]\frac{c}{2}[/itex])2 (1)

    x=[itex]\frac{a/2}{cos(30)}[/itex] =[itex]\frac{a}{\sqrt{3}}[/itex] (2)


    Next substitute (2) into (1) and after some algebra you should get the result of

    [itex]\frac{c}{a}[/itex]=[itex]\sqrt{\frac{8}{3}}[/itex]
     
  2. jcsd
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