# Volume of Diamond Unit Cell

roam
I have some difficulty understanding how the volume per unit cell for the diamond structure is calculated.

I've seen in various websites that this volume is:

##v=a^3 = \left( \frac{8r}{\sqrt{3}} \right)^3##

Here ##r## is the radius of an atom. But how did they work out ##a## (the edge of the cube)?

Here is a picture of the structure.

The unit cell has 8 atoms as:

##8 \times \frac{1}{8} + 6 \times \frac{1}{2} + 4 = 8##

But to work out ##a##, I've simply considered one of the faces, in this case the top face of the diagram which looks like a fcc structure.

The diagonal would then be 4r, hence using Pythagoras

##4r = \sqrt{2}a \implies a = 2 \sqrt{2} r##

But my answer is not correct. Any help with this problem is greatly appreciated.

daveyrocket
The two closest atoms in the diamond structure aren't the two on the face of the fcc cell, it's the one in the corner and the one it's nearest inside the cell. See the purple bond in your picture. That distance is 2r.

roam
The two closest atoms in the diamond structure aren't the two on the face of the fcc cell, it's the one in the corner and the one it's nearest inside the cell. See the purple bond in your picture. That distance is 2r.

Thank you for your input. But in that case I will get

##d=8r##

##8r=\sqrt{a^2+a^2} = \sqrt{2} a##

##\therefore a= \frac{8r}{\sqrt{2}} = 4 \sqrt{2} r##

But this is wrong as the answer must be 8r/√3. Where did the ##\sqrt{3}## come from?

daveyrocket
I'm not sure what d is. Is that your diagonal in the cube?
You're using the wrong triangle for the Pythagorean theorem. The diagonal of a cube is not the same length as the diagonal of a square.

roam
Oops, I was using the wrong formula. I used the pythagorean theorem in three dimensions and got the correct solution. Thank you so much for your help! :)