- #1

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## Main Question or Discussion Point

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.

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.