# Determining lattice points per cell in FCC

I'm having hard times understanding the concept of lattice point and unit cell.

For example, let's say we need to determine the number of atoms per unit cell in SC BCC and FCC.

For SC, in my understanding each corner is like the center of the atom. So there are 4 quarters of the atom which makes one whole atom.

In BCC, there is one whole atom in the center and the rest is the same as in SC, so therefore there are 2 atoms.

However, when it comes to FCC, I don't understand why there are 4 atoms per unit cell. Can someone explain why there are 4 atoms per unit cell ?

Also, my logic might be wrong, so if it is wrong, can someone explain the right way of calculating the atoms per unit cell and the relationship between unit cell and lattice point?

## 1 Answer

Since you're talking about quarters of atoms, I assume you consider the two-dimensional case. To answer your question, you have to go the 3D-case.

As you can see in the picture below, the FCC unit cell in three dimensions contains

$$\left( 8 \cdot \frac{1}{8} + 6 \cdot \frac{1}{2} \right) \text{Atoms} = 4 \,\text{Atoms}. \\[1cm]$$

• Please copy the image into the answer (with attribution). Too many links go dead rendering the answers useless. – garyp Oct 1 '16 at 23:20
• Thanks for adding the figure. The reason I'm aware of it is that I've been frustrated by broken links on stack exchange sites many times. – garyp Oct 2 '16 at 13:56