= 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 basis
$$
(0,0,0),\qquad(\tfrac14,\tfrac14,\tfrac14).
$$
The FCC translations within the conventional cell are $(0,0,0)$, $(0,\tfrac12,\tfrac12)$, $(\tfrac12,0,\tfrac12)$ and $(\tfrac12,\tfrac12,0)$. Adding the basis gives eight atoms per conventional cell. Equivalently, \b[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.
\Image[/past-exam-of-the-materials-science-course-of-the-university-of-cambridge/2015/paper-1-diamond-projection.png]
{title=Original fractional-coordinate projection of diamond silicon, with atom heights, vertical mirror traces and a bond-centred inversion point}
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