Past exam of the materials science course of the University of Cambridge / 2015 / Paper 1 / 1 / a / Solution

Use fractional coordinates in a conventional cubic cell with an atom at the origin. The diamond cubic structure consists of an FCC Bravais lattice with the two-atom basisThe FCC translations within the conventional cell are $(0,0,0)$, $(0,\frac{1}{2},\frac{1}{2})$, $(\frac{1}{2},0,\frac{1}{2})$ and $(\frac{1}{2},\frac{1}{2},0)$. Adding the basis gives eight atoms per conventional cell. Equivalently, two identical FCC sublattices are displaced by $a[111]/4$. It is not a monatomic close-packed arrangement: every atom has four tetrahedrally arranged nearest neighbors rather than twelve.

In the projection, label each atom by its fractional height. The $(0,0)$ and $(1/2,1/2)$ columns have height0; $(0,1/2)$ and $(1/2,0)$ have height1/2. The additional four columns are $(1/4,1/4)$ and $(3/4,3/4)$ at height1/4, and $(1/4,3/4)$ and $(3/4,1/4)$ at height3/4. Boundary columns repeat periodically; they must not be counted as full extra atoms.