How do I calculate lattice points per unit volume? [duplicate]

This question already has an answer here:

My professor was explaining simple cubic, body centered cubic, and face centered cubic lattices and I didn't quite understand the calculations below:

• Simple Cubic: $8 \cdot 1/8 = 1$ lattice point per unit volume

• Body Centered Cubic: $(8 \cdot 1/8) + 1 = 2$ lattice points per unit volume

• Face Centered Cubic: $(8 \cdot 1/8) + (6 \cdot 1/2) = 4$ lattice points per unit volume

Can someone explain the reasoning behind these values in the equations?

marked as duplicate by Mithoron, andselisk♦, Jon Custer, airhuff, Melanie Shebel♦Oct 11 '17 at 23:54

You have to imagine a cube with spheres (the atoms) on the corresponding locations. Now you have to look at how much of each of these spheres is inside the cube (please see Patrick Moloneys answer for very descriptive images).

For Simple Cubic: You have spheres on all 8 corners. Because the spheres are on the corners of the cube, only $1/8$ of each sphere is inside the cube, while $7/8$ of the sphere are on the outside. Now, you have 8 spheres on the corners, giving you 8 times $1/8$ of a sphere. So there is $8 \cdot 1/8 = 1$ sphere inside the cube.

For Body Centered Cubic you have a sphere completely inside the cube additionally to the 8 on the corners ($8 \cdot 1/8$). Now we add the $1$ on the inside, giving you a total of ($8 \cdot 1/8) + 1 = 2$ spheres inside the cube.

Face Centered Cubic has the same 8 spheres on the corners ($8 \cdot 1/8$) and an additional sphere on all the faces of the cube. For each of these spheres, only half is inside the cube, while the other half is not. There are 6 faces, giving you $6 \cdot 1/2$ spheres for the faces. Add these to the corner spheres and you get $(8 \cdot 1/8) + (6 \cdot 1/2) = 4$ spheres inside the cube.

• I felt as though this answer was easier to understand, so I marked it as correct. Thanks!! – whatwhatwhat Oct 12 '17 at 20:44

Atoms on the corners, edges, and faces of a unit cell are shared by more than one unit cell.

An atom on a face is shared by two unit cells $\implies$ half an atom belongs to each unit cell. An atom on the edge is shared by four unit cells and an atom on the corner is shared by eight unit cells What does this have to do with the equations?

Well lets say we have an sodium atom that is crystallized in a simple cubic unit cell, then there would be an sodium atom on each of the eight corners of the cell. However, only $\frac{1}{8}$ of these atoms can be assigned to a given cell. Hence the simple cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8}=1 \, \text{atom}$$

if sodium formed a body centered cubic structure, there would be two atoms per unit cell (as seen in the top diagram). It isn't shared with an other cells. Therefore the body-centered cubic structure is given by

$$8 \, \text{(corners)}\, \times \frac{1}{8} + 1 \, \text{(body)}= 2 \, \text{atoms}$$

now finally if sodium was a face-centered cubic structure the six atoms on the faces of the unit cell would contribute three net sodium atoms, for a total of four atoms per unit cell. Hence the face-centered cubic structure is

$$8 \, \text{(corners)} \times \frac{1}{8} + \Big{(}6 \, \text{(faces)} \times \frac{1}{2}\Big{)}= 4 \, \text{atoms}$$

Hope this helps