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Hexagonal Close Packing (Why do I get c/a = 3^1/2 ?)

  1. Apr 1, 2007 #1
    Hexagonal Close Packing (Why do I get c/a = 3^1/2 ??)

    1. The problem statement, all variables and given/known data

    For the hexagonal close packed unit cell, show that c/a = 1.6333. Where c is the height of the cell, and a is the length of the sides of the base. (See picture below)

    2. Relevant equations

    http://img254.imageshack.us/img254/5609/hcpqr6.th.jpg [Broken]

    Note: There's a typo in the picture above: [itex]c = h^2 - a^2[/itex] should be [itex] c^2 = h^2 - a^2[/itex].

    3. The attempt at a solution
    h = 4r , \
    \\ a = 2r , \
    a^2 + c^2 = h^2 \Rightarrow c^2 = h^2 - a^2 = 16r^2 - 4r^2 = 12r^2
    \Rightarrow c = 2 \sqrt{3}r \Rightarrow
    \frac{c}{a} = \frac{2 \sqrt{3}r}{2r} = \sqrt{3} \approx 1.732 \neq 1.633

    Can someone at least give me a hint at what I did wrong? It just seems like I am certainly correct, unless the angle between a and c is not [itex]90^o[/itex] (which I really doubt)


    Tangential Latex Question: Does anyone know why \\ is not producing a new line for me? Thanks.
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 1, 2007 #2
    It is hard to tell from the picture but side labeled h is not 4r. The third layer is in the same spot except for that is translated vertically. The second layer fits in the valleys of the first layer. The points of contact of the sphere make the radii line not travel along one of the faces but into the hexagon. You can still have an h that is 4r but then you have to figure out what the bottum side of the triangle is and it can't just be 'a'. Three dimension stuff is hard to draw, describe and think about sometimes. Is my communication make sense to you?
    Last edited: Apr 1, 2007
  4. Apr 1, 2007 #3
    I think I get the idea of what you're saying. I'll try it out and let you know if I get 1.633. Thanks.

    The thing I hate about these crystal problems is: How in the world could I ever figure this out for myself? All we are given is pictures of the unit cell, and if they are misleading, we're screwed. It's really frustrating. Never mind my rant. :)
  5. Apr 1, 2007 #4
    You probably won't want to set h = 4r. I mean that isn't the easiest way. I was just trying to get you to realize that path of radii entered into the crystal unit. The way I solved was by visualizing four spheres making a pyramid. Then find the height of that period as function of the radius of the spheres. From there finding the height of the unit is easier.
  6. Apr 1, 2007 #5
    I don't know what's going on. I tried to do a 4 sphere pyramid, and I still get [itex]c/a = \sqrt{3}[/itex]. I'm about to give up. This is ridiculous.
    Last edited: Apr 2, 2007
  7. Apr 1, 2007 #6
    the following link shows a better picture in my opinion

    http://www.uncp.edu/home/mcclurem/lattice/hcp.htm [Broken]

    the first picture shows the just two layers. The third layer should stack right on top of the other layer. The conceptual point is that a pyramid is formed from three spheres on the bottum layer and one in the middle. The height of that pyramid is half the height of c. However just finding the height of that pyramid is hard. Another word for that type of pyramid is tetrahedron. If you have trouble solving the math for the height of tetrahedron I can assure there are derivations on the web. Keywords: tetrahedron height
    It is a pretty difficult problem. Material science can be difficult. I was flipping through a book the other day on it and it had tensor calculus on it which I have know idea how to do. Some parts are fun and sometimes the math requires more time than it worth depending on the discipline type.
    Last edited by a moderator: May 2, 2017
  8. Apr 1, 2007 #7
  9. Jan 19, 2008 #8
    Sorry, to bring this up again, but I also get c/a = 3^1/2. When I draw it out I get a rhombus with side-lengths 2 and a diagonal of length 2. I get that c/a should be the length of the long diagonal divided by 2.

    I realized that I was calculating the wrong thing because the primitive cell I used has its origin in the A layer, has two vectors the go to the same A layer, and then a third that goes to the next A layer. So, it basically skips the B layer. Can someone give me a description of the primitive cell that I should be using (I know this is hard!)?
  10. Jan 19, 2008 #9
    I just solved it. One thing I was missing was that THE PRIMITIVE CELL ACTUALLY CONTAINS 2 SPHERE CENTERS. All of the lattice points will be on one of the two plane layers either A or B .

    Also, the problem really does reduce to finding the height of a tetrahedron as NotMrX said. The MOST ESSENTIAL POINT IS THAT THE CONFIGURATION WE ARE INTERESTED IN IS NOT THE SAME AS http://www.soyouwanna.com/site/syws/shootpool/img/SHOOTPOOL_4-1.gif
    That is where the sqrt(3) comes from and it is downright wrong.
    Last edited by a moderator: Apr 23, 2017
  11. Feb 3, 2009 #10
    Re: Hexagonal Close Packing (Why do I get c/a = 3^1/2 ??)

    its an entire year that no one has replied..
    anyways, i had the same problem too...
    but now its crystal clear for me ...

    instead of solving with the diagram(of the HP cell) given by others, try this one,

    on the site they have proved it by using volume of a tetrahedron(scalar triple product).
    but i proved it in a simpler way.

    choose any 3 adjacent atoms from the middle plane. they form a equilateral triangle of side = 2r (r=radius of atom).
    Any atom on the top(or bottom) plane is positioned directly above the centroid of the equilateral triangle formed by the atoms of the middle plane, it touches.

    then consider the right triangle formed by one atom(say atom A) of the top plane, the centroid below it (in the middle plane) and one of the atoms of the middle plane touching atom A.
    the distance between the atom A and the centroid = (height of hexagonal cell)/2
    use pythagoras theorem in this triangle to find the height in terms of radius r.
  12. May 12, 2011 #11
    Re: Hexagonal Close Packing (Why do I get c/a = 3^1/2 ??)

    I know its 2 years later, but i had the same problem as the OP. i just wanted you to know that even if noone else commented, your pun did brighten my revision session, thanks.

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