# Homework Help: Packing fraction of spheres in a HCC structure

1. Sep 19, 2014

### CAF123

1. The problem statement, all variables and given/known data
Show that the ratio of atomic sphere to unit cell volume in HCP (hexagonal close packing) is 0.74.

2. Relevant equations
volumes of spheres, geometry

3. The attempt at a solution
I did the same problem for FCC and BCC and it was fine.
My unit cell structure is that shown below. I labeled the height between the two hexagonal planes by $h$ and the length of the equilateral triangles comprising the hexagon by $2r$. If we then consider a 1/6 of this structure in the obvious way and orient it suitably so that one of the sides of the lower triangles coincides with the x axis say, then the volume of 1/6 of this structure is $$V = 2 \int_0^{r} \int_0^{\sqrt{3}x} \int_0^h dx dy dz = \sqrt{3}hr^2.$$ Multiply this by 6 to get the whole volume of the unit cell structure shown below.

I would like to try to relate the height of this structure to the radius $r$ of the spheres so that in the ratio, I get cancellation. I am assuming that the three spheres on the middle layer of the structure (labeled B in the sketch) are wholly contained within the structure?

#### Attached Files:

• ###### HPC.jpg
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2. Sep 19, 2014

### nasu

Yes, for a closed packed structure there is a specific value for the ration between the height and the side of the hexagon. Imagine that you put some hard spheres in each node so that each one is in contact with nearest neighbors.

3. Sep 19, 2014

### ZetaOfThree

There is a definite relationship between $h$ and $r$ that you could either look up in your textbook, or derive it yourself (it's pretty easy).

4. Sep 21, 2014

### CAF123

Yes, if we look at just the hexagon shape and put six spheres at each of the nodes then each side of the hexagon has a length two times that of the radius of one of the spheres, which was my motivation for calling the length $2r$. In the HCP structure, as seen from the diagram I posted, there are three atoms in the middle as well. It seems to me that these three atoms are wholly contained within the structure.
So would it be right that $h = 2 (2r \cos 30^o) = 2 \sqrt{3} r$ and so $V = 6r^3$. As mentioned in the OP, this corresponds to 1/6 of the structure so the total volume is $36r^3$.

So then I need to find the volumes of all the spheres inside the structure. Let $v$ be the volume of a sphere of radius $r$. The three middle ones contribute $3v$. The ones in the middle of the planar hexagons contribute $(1/2)v$. And I figured that each of the spheres at the 12 nodes (6 per planar hexagon at the sides) contribute $(1/6)v$. So the total volume I get is $V' = 12((1/6)v) + 2((1/2)v) + 3v$ which seems to give the wrong answer.

My reasoning for the nodal spheres was that we can neglect 1/2 of the volume each time (since exactly 1/2 of the sphere will be above the structure). Each node is in contact with two triangles so this is 1/3 of the amount making a full hexagon. Therefore the volume of a nodal sphere inside the structure is $(1/2) (1/3) v$.

Adding up and taking the ratio I get about 0.70 which is close, but not quite right. Thanks.

Last edited: Sep 21, 2014
5. Sep 22, 2014

### CAF123

I am thinking that I got my expression for $h$ incorrect, could you give me some pointers on that?

6. Sep 22, 2014

### ehild

I can not quite follow your derivation. You are right that there are 6 whole spheres inside the unite cell.
The height of the cell is twice the height of a tetrahedron made of the centres of four atoms, so the sides of the tetrahedron is twice the radius. What is the height of a tetrahedron?
The volume ratio is the volume of six atoms divided by the volume of the hexagon.

ehild

7. Sep 23, 2014

### CAF123

Thanks ehild, I got the right answer ;) Just out of interest, I know the choice of unit cell is not unique so if I tried the question using a different unit cell (e.g perhaps just a sixth of the structure considered before) then I should get the same answer?

8. Sep 23, 2014

### ehild

If it is really a unit cell, so as the whole crystal can be covered by its translations and you count the number of atoms in it correctly you have to get the same answer. We use a bigger and symmetrical unit cell instead of the smallest one (the primitive cell) because it is easier to see the symmetry and to count the number of atoms.

ehild

9. Sep 23, 2014

### CAF123

Thank you ehild.