# Solve for c/a ratio

1. Sep 24, 2006

### 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 [Broken] but it is more confusing to me how 1.67 comes out. Help please?

Last edited by a moderator: May 2, 2017
2. Sep 24, 2006

### Astronuc

Staff Emeritus
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.

3. Sep 24, 2006

### 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 thier length
Ben

4. Sep 24, 2006

### Gokul43201

Staff Emeritus
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.

5. Sep 24, 2006

### 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