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Homework Help: Volume of HCP unit cell from radius + c/a ratio

  1. Nov 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Cobalt has an HCP (hexagonal close packed) crystal structure, an atomic radius of 0.1253 nm and a c/a ratio of 1.623. Compute the volume of the unit cell for cobalt.


    2. Relevant equations
    area of hexagon = apothem*perimeter/2


    3. The attempt at a solution
    calculate 1 of the 6 sides (2 atomic radii per side):
    2*.1253 = 2.506e-1

    calculate perimeter:
    2.506e-1*6 = 1.5036e+0

    calculate height:
    2.506e-1*1.623 = 4.06724e-1

    calculate hexagon area (apothem = 2 radii):
    1.5036e+0*2.506e-1 / 2 = 1.88401e-1

    calculate volume:
    1.88401e-1*4.06724e-1 = 7.66272e-2

    my answer:
    7.66272e-2

    correct answer:
    6.64e-2 nm3
     
  2. jcsd
  3. Nov 29, 2008 #2
    mistook the apothem for the hexagon radius.
     
  4. Sep 11, 2011 #3
    Hi... I'm currently studying for my first test on Solid State Physics and looking for some info I found your question... I agree that considering the right expression for the apothem one gets to the answer proposed... Nevertheless, I think there is a consideration to make in the problem...

    I think they are asking for six times the volume of the parallelipiped formed by the vectors

    a1 = (a,0,0)
    a2 = (a/2,a*sqr(3)/2,0)
    b3 = a1/3 + a2/3 + (0,0,c/2)

    . In other words, just the volume proposed by the previous answers, 3*sqr(3)*a*a*c/2.

    But, due to the geometry of the hcp the distance to the nearest neighbors depends on c/a (It is not always a). One finds that this distance is the minimum between a and sqr(a*a/3 +c*c/4) (This last value is just the norm of the vector b3 I used before). Then if c/a > sqr(8/3) the closest neighbors are just the 6 ones in the xy planes (the hexagon around the point). But if c/a < sqr(8/3) (around 1.63299) then the closest neighbors are 6 different points (3 in the upper plane and 3 in the lower plane). So, for this case the radius given for the "sphere-atom" wouldn't be a/2 but 0.5*sqr(a*a/3 +c*c/4) (in order to avoid overlapping). Using this fact and the value of c/a you find a and then replacing in the volume you'll get a slightly different answer (6.7179e-2 nm^3). Am I making a mistake? If I am, I apologize for any slip either in the solution or in my English... it's my first post. Good luck.
     
    Last edited: Sep 11, 2011
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