Solve HCP Packing C/A Ratio: Geometry & 1.67 Explained

[benndamann33]

I'm trying to solve for the ideal c/a ratio of hcp packing. why doesn't c/2 =a/2, looks like they pass through the same amoutn of radius's so I'm not sure the geometry to use. Used this pic http://www.engr.ku.edu/~rhale/ae510/lecture2/sld013.htm but it is more confusing to me how 1.67 comes out. Help please?

 

[Astronuc]

Look at the middle picture in the link you provided.

In the c-direction, there is an extra layer of atoms (triangle) between the top and bottom layers (hexagonal). The a-length is the side of the hexagon, but also the distance among adjacent atoms in the 'basal' plane. The dimension 'a' is in the closest-packing direction. The packing is less efficient in the c-direction.

 

[benndamann33]

I understand what you're saying about the picutre now, but I don't understand what you're using to compare C and A. I know where a is and i know where c is but i don't know the other length of triangle built between them, so i'ts unclear to me as to what the link is between their length
Ben

 

[Gokul43201]

It's not obvious.

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, and is hence $(2/3)*(a\sqrt{3}/2) = a/\sqrt{3}$.

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. This distance can also be calculated from Pythagoras, giving:
$$4R^2 = a^2 = (a/\sqrt{3})^2 + (c/2)^2$$

That should get you home.

 

[benndamann33]

thank you I was feeling like an idiot that I couldn't girue this out, I'm glad it doesn't appear to really be a trivial process. I appreciate the help,
Ben