The Cubic Unit Cells & Cubic Lattices

We now look at cubic unit cells.  First, let us do a little math.  Consider a cube.  It has 6 faces, 12 edges, and 8 vertices.  Each edge has a length which we shall call a.  Again, consider the three cubic unit cells, now shown.

The problem we now consider is how to relate the cell volume to the radii of the atoms in the cell; here, we shall consider all atoms to be identical.  This is true for metals and there are metals (pure elements, that is) which exist for all three possible structures.  Furthermore, for all of these, the volume of the cell is just

V = a3.

The next, rather primitive, drawing is of the face of a simple cubic unit cell.

The volume of this cell is easily related to the size of the component (r is the atomic radius) atoms; you can simply note that a = 2r and go from there.  In short, the volume of this cell is just

V = a3 = (2r)3 = 8r3.

Let us now look at a cube in general.  The following picture is as good as any.

The side of the cube is a.  The diagonal along a face (in blue) is b.  The inner diagonal (in green) is c.  Using the Pythagorean theorem, it is easily seen that

and, in turn, it can be shown that the inner diagonal is just

.

Now, let us look at the face of a face-centered-cubic unit cell.  The following plain picture shows all we need to know.

It should be clear from this picture that the length of this diagonal is just 4r.  We can then relate the size of the unit cell to that of the individual atoms by

Similarly, with a body-centered cubic lattice with the three touching atoms on the inner diagonal, we see immediately that

This relationship is shown clearly in the next figure.


 


We can use these various expressions to relate the density of a metal to its atomic size, to calculate the size of a unit cell, or even to determine the type of crystal structure a given metal has.  Examples of some of these things are shown with the problems and some discussion is given in the next section.