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

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.

The vertical mirrors of the diamond cubic structure have the required orientation: their normals lie in the $(001)$ plane. In the chosen atomic-origin convention, their traces satisfywith $x,y$ expressed as fractions of $a$. These are the two orientations of vertical $\{110\}$ mirrors and their translated copies. For example, the plane $x=y$ interchanges the first two coordinates and leaves every diamond site in an equivalent site. Reflection in $x+y=1/2$ sends $(x,y,z)$ to $(1/2-y,1/2-x,z)$, which also preserves the full three-dimensional motif.

The projected traces alone do not test a mirror: the atom heights must also agree under reflection. In particular, a vertical $\{100\}$ trace can appear plausible in an unlabelled projection but does not preserve all heights as a pure mirror; glide operations are different symmetry elements.

Bond-centred inversion in diamond cubic gives one inversion centre at the midpoint of a nearest-neighbor bond:Mark its projected point $(1/8,1/8)$ and label its height $z/a=1/8$. Inversion about this point sends $\mathbf{r}$ to $(1/4,1/4,1/4)-\mathbf{r}$. It exchanges the two identical FCC sublattices and therefore preserves the crystal. An individual atom at the chosen origin is not an inversion centre.

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.

The four FCC translations give a lattice structure factor that is nonzero only when $h,k,l$ are all even or all odd. The identical two-atom motif then multiplies it byFor $(200)$ the FCC part is allowed, but the motif factor is $1+e^{i\pi }=0$. HenceIn the projection, equal numbers of atoms occur at $x/a=0,1/4,1/2,3/4$. Their $(200)$ phases are respectively $+1,-1,+1,-1$. The paired layers cancel exactly for equal silicon scattering amplitudes. This diamond-basis diffraction extinction is additional to the FCC centring rule; it is not evidence that no atoms lie on the nominal reflecting planes.

For indexing a diamond-cubic powder pattern, $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$ and Bragg's law gives ${\sin }^2\theta ={\lambda }^2(h^2+k^2+l^2)/(4a^2)$. Using half of each printed scattering angle, the ratios to the first peak are approximatelyMultiplying by3 gives the allowed sequence $3,8,11,16,19$. ThusThe numerical calculation, expressed without assuming a historical Data Book wavelength, isTherefore $a\simeq 3.51537\lambda$. As a separately identified reference convention, choosing Cu K${\alpha }_1$ with $\lambda \simeq 0.15406\mathrm{nm}$ givesThat wavelength follows from the copper $K{L}_3$ transition near $8047.8227\mathrm{eV}$ in the NIST transition database, using $\lambda =hc/E$.^[ref]

The paper says Cu K$\alpha$ but does not give the line-selection convention or an explicit wavelength. A rounded unresolved-doublet wavelength, for example $0.15418\mathrm{nm}$, would instead give about $0.54200\mathrm{nm}$. These are conditional wavelength evaluations, not claims about the unavailable historical Data Book. The angles are internally consistent with one cubic parameter; they should not be replaced by literature peak positions simply to obtain a familiar silicon value.