# Number of atoms per unit cell

I don't understand the concept of unit cell and the number of atoms per unit cell in a cubic lattice also the calculations for the number of atoms. For example in the $\ce{fcc}$ lattice, number of atoms per unit cell is:

$$8\cdot\frac{1}{8} + 6\cdot\frac{1}{2}=4$$

• what does the 2 and 8 in the denominator stand for?
• also 4?

The denominator signifies the number of cubes that are needed to completely encompass the whole point. For example, a corner point can be thought of as a center of 8 whole cubes, while a face centre is enclosed by 2 cubes and an edge center by 4. Hence, only 1/8 of a corner atom is in a specific unit cell and so on and so forth.

Consequently, the total number of atoms in a unit cell (say a FCC) would be equal to -

(no of corners)(fraction of corner in the unit cell) = 8(1/8)

plus

(no of face centers)(fraction of face center in the unit cell) = 6(1/2)

which equals to 4

• It should be noted that this answer (and the question!) only make sense for a unit cell which contains an atom in the corner and one on each symmetry-equivalent face. You might then translate the unit cell (or the atoms) to show how one can arrive at the end result of four in a different way. – Jan Oct 8 '17 at 15:59