 Solution

The nearest-neighbor displacement is $a[111]/4$, of length $a\sqrt{3}/4$. Touching hard spheres therefore obeyThere are eight atoms per conventional cubic cell, givingFor monatomic FCC close packing, four atoms per cell touch along a face diagonal, so $r_{\mathrm{atom}}=a/(2\sqrt{2})$ andThus the diamond packing fraction is much smaller. The tetrahedral network leaves substantial open volume. Although its Bravais lattice is FCC, the two-atom motif and the shorter hard-sphere contact distance prevent it from inheriting the FCC close-packing fraction.