Past exam of the condensed matter physics course of the University of Oxford / 2018 / 1

A Bravais lattice is a set of points generated by integer translations along three independent primitive lattice translation vectors:Every lattice point has the same environment under lattice translations. A crystal structure consists of this Bravais lattice with an atomic basis of a crystal repeated at every lattice point; the set of atomic positions need not itself be a Bravais lattice.

The two positions in the conventional cubic cell of alpha-tungsten give a body-centred cubic lattice, with one tungsten atom per primitive cell and two per conventional cell. One possible set of primitive lattice translation vectors isTheir primitive-cell volume is $a^3/2$.

For the conventional body-centred cubic lattice, the crystal basis structure factor of identical atoms isIt vanishes for odd $h+k+l$ and equals $2f_{\mathrm{W}}$ for even $h+k+l$. Thus the first powder diffraction ring is the $\{110\}$ family, with interplanar spacing $d_{110}=a/\sqrt{2}$.

The measured deflection is the angle between incident and scattered rays, $2\theta =40.26^{\circ }$, so the Bragg angle is $\theta =20.13^{\circ }$. Using first-order Bragg diffraction, $2d\sin \theta =\lambda$, givesThe mass density follows from two atoms per conventional unit cell, with molar mass $M=0.1838\mathrm{kg}\mathrm{mo}{\mathrm{l}}^{-1}$ and Avogadro constant $N_A$:Using $40.26^{\circ }$ as the Bragg angle would confuse $\theta$ with $2\theta$ and give an incorrect lattice constant and mass density.

The beta phase has a primitive cubic lattice with an eight-atom atomic basis of a crystal, rather than a body-centred cubic lattice. A lattice translation must map the entire atomic arrangement onto itself. For example, the putative body-centering vector $(1/2,1/2,1/2)$ maps $(1/4,1/2,0)$ to $(3/4,0,1/2)$ modulo integer coordinates, where the given atomic basis of a crystal has no atom. None of the seven nonzero basis positions generates a translation that preserves all eight sites. ThereforeThe presence of atoms at both the origin and the body centre alone is insufficient to establish body-centering of the full crystal structure.

Initially approximate every tungsten atom by the same atomic scattering factor $f$. Pairing the quarter-coordinate sites in the crystal basis structure factor givesFor cubic powder diffraction, increasing deflection corresponds to increasing $h^2+k^2+l^2$, since $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. The first possibilities are

Hence the first two strong rings have Miller indicesThe powder diffraction multiplicity is six for $\{200\}$, from three choices of nonzero axis and two signs. For $\{210\}$ it is $3!2^2=24$, from six permutations and four sign choices. If the atomic scattering factors and angular corrections are taken as approximately equal for these nearby rings, their integrated diffraction intensity obeyThis is a multiplicity-based estimate. An exact measured ratio also depends on the momentum-transfer dependence of the atomic scattering factor, the Lorentz-polarization factor, the Debye–Waller factor, absorption and instrument geometry. In particular, the source does not provide the beta-phase lattice constant or instrument corrections needed to calculate those angular factors. Peak heights additionally depend on peak widths; the estimate refers to integrated ring intensities.

The two body-centre/origin sites and the other six sites have different local environments. Their effective atomic scattering factors therefore need not be identical even in an elemental, chemically pure crystal. Write them as $f_A$ and $f_B$, respectively. The crystal basis structure factor becomesFor the low-angle families this givesThus $\{110\}$ can produce a weak extra powder diffraction ring below $\{200\}$. This is an accidental diffraction extinction from equal atomic scattering factors: its cancellation in part d is exact only in the identical-atom-scattering approximation. Small differences in the bonding electron density, or site-dependent Debye–Waller factors, can leave a small nonzero amplitude. For instance, taking $f_A=f+\delta f$ and $f_B=f$ gives $F_{110}=2\delta f$ while $F_{200}=4f+2\delta f$; the corresponding intensity suppression is quadratic in $|\delta f/f|$. No impurity is required. The $\{100\}$ and $\{111\}$ cancellations remain for arbitrary scalar $f_A,f_B$ in this two-site-class model.