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Calculating ratio c/a as 1.633 in HCP?

  1. Feb 19, 2010 #1
    Hi All,

    This is a similar post to one before, but that didn't offer much more than i know already.

    a = width of cell - depth of cell
    c= height of cell
    c/2 = height of central atom

    So i have various formula for a few isosceles lengths, namely a/2*cos30 for the base line of the vertical right-angle triangle connecting to the 'central' atom. and c/2 for the line connecting the base to the atom its self.

    I am simply having trouble relating the two unknowns from here. i have tired pythagoras but cant yield anything suitable. Any hints?

  2. jcsd
  3. Feb 19, 2010 #2
    Im currently working on the same problem.

    Apparently there is a slightly older thread but it is full of confusion, namely poorly written equations and explanations, probably due to a language barrier.

    I found a ratio of c/a = 7/6 = 1.166, so I must have did something incorrect. For a geometry problem this question appears to be surprisingly tricky.

    First I found the length l to the midpoint of the isoceles triangle in the plane, cos30 = (a/2)*l,
    ie l = 0.577a.

    Next I used Pythagora's Theorem to set an equation containing c and a:
    (c/2)^2 = [(a/3)+(a/3)+(c/2)]^2 + (0.577a)^2

    Then I solved for c/a. I got 7/6. I would appreciate any help
  4. Feb 19, 2010 #3
    A couple things I did wrong... I meant to say equilateral triangle in the plane.

    Also, Pythagora's Theorm should be:
    [(a/3)^2 + (a/3)^2 + (c/2)^2)]^2 = l^2 + (c/2)^2

    ...but it still doesnt work out. Is there something I am completely missing in this proof? Ive tried plugging in the lattice constants for the primitive lattice vectors a1, a2, and a3 into the vector a1/3 + a2/3 + a3/2 and then took the absolute value of that vector to find its magnitude, but I still dont end up with anything sensible.
  5. Feb 19, 2010 #4
    Well, first problem: c/2 is not the height of the central atom. Otherwise the problem would be trivial, c = 2a. (Remember, every atom in hcp is equivalent to every other one.) Unless you mean the position of the atom, not its height.

    Look at how the layers stack. In one layer, you have a close packed spheres forming a triangular lattice. The atoms in the next layer sit in the "holes" above the first layer. That is, if you look at a single equilateral triangle in one layer, then in the next layer there is an atom sitting above the center of that equilateral triangle. The position of this atom is [tex](a/2, 1/(2\sqrt{3}) a, z)[/tex], if you have an atom at the origin. (Center of equilateral triangle = average of the location of points. You have one at (0,0,0) and one at (a,0,0) and one more that is easy to find geometrically.) The distance between the atom at [tex](a/2, 1/(2\sqrt{3}) a, z)[/tex] and any of the others should be a. Then you know that z = c/2.
  6. Feb 20, 2010 #5
    I worked it out and found my magic ratio.

    I understand that c != 2a, because the stacking constant is independent of the spacing of the equilateral triangle in the plane.

    If we define each point in the lattice to be a sphere of radius r, such that the each sphere touches its nearest neighbour, then a = 2r.

    And we know the primitive vectors of a hcp structure is:
    a1 = a xhat = 2r xhat (1);
    a3 = c zhat (2);
    and define the coplanar vector as h = a1/3 + a2/3 + a3/2,

    so that h^2 = |1/3(a1 + a2 + (a3/2)|^2 = (2r)^2 = 4r^2
    and from (1) and (2) above: h^2 = |1/3(a1 + a2 + c/2)|^2 = a^2.

    Work out |a1 + a2|, then solve for c/a.

    So there wasnt any need for Pythagora this time
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