Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B7 Structure of Materials, National Examination
May 2013 — a closed-book examination (Casio or Sharp approved calculators only; all
necessary equations, constants and diagrams supplied in the paper's own appendix). Candidates
attempt any five of the eight questions, each worth 20 marks; every question is nonetheless
answered in full below so the paper remains a complete study resource.
Reference texts. W. D. Callister Jr. & D. G. Rethwisch, Materials
Science and Engineering: An Introduction, 10th ed. (atomic bonding, crystal structure and
packing, point defects, diffusion, dislocations and slip, mechanical properties, phase diagrams
and the lever rule, precipitation hardening, X-ray diffraction); D. J. Griffiths, Introduction
to Quantum Mechanics, 3rd ed. (de Broglie wavelength, Heisenberg uncertainty).
Part 1 — Given. Cubic planes (111) and (210); hexagonal (Miller–Bravais)
plane $(1\bar{1}00)$ and direction $[11\bar{2}0]$.
Find. A drawing of each plane/direction on its unit cell.
(a) The (111) plane cuts the $a$, $b$, $c$ axes each at 1 lattice
parameter, forming the shaded triangle through three cube corners.
(a) The (210) plane has intercepts $a/2,\,b,\,\infty$: it crosses the $a$
edge at the half-way point, the $b$ edge at a full lattice parameter, and never meets $c$
(parallel to it), giving the shaded quadrilateral spanning the full cell height.
(b) $(1\bar{1}00)$ is one of the six vertical prism faces (shaded); $[11\bar
{2}0]$ is an $a$-type direction lying in the basal plane, from the cell centre to the marked
corner — the Miller–Bravais third index is redundant, $i=-(h+k)$.
Part 2 — Given. MgO adopts the rock-salt (NaCl) structure: an FCC anion
sublattice with cations filling every octahedral hole, so it is really two interpenetrating FCC
lattices (4 formula units per cell). $r_{\text{Mg}^{2+}}=0.078$ nm, $r_{\text{O}^{2-}}
=0.132$ nm, $M_{\text{Mg}}=24.31$ g/mol, $M_{\text{O}}=16.00$ g/mol.
Find. The ionic packing factor (IPF) and the mass density.
Approach. Get the lattice parameter from cation–anion contact along the
cube edge, then divide the volume of the 4 cation + 4 anion spheres per cell by the cell volume;
get the density from the same 4 formula units' mass over the same cell volume.
Lattice parameter. In the rock-salt structure, cations and anions touch along
the cell edge: $$a=2\left(r_{\text{Mg}^{2+}}+r_{\text{O}^{2-}}\right)
=2(0.078+0.132)=\boxed{0.420\ \text{nm}}.$$
Ionic packing factor. Each cell holds 4 $\text{Mg}^{2+}$ and 4
$\text{O}^{2-}$ ions:
$$V_{\text{ions}}=4\left(\tfrac{4}{3}\pi r_{\text{Mg}}^3\right)+4\left(\tfrac{4}{3}\pi
r_{\text{O}}^3\right)=4(0.001988)+4(0.009633)=0.04648\ \text{nm}^3,$$
$$\text{IPF}=\frac{V_{\text{ions}}}{a^3}=\frac{0.04648}{0.420^3}=\frac{0.04648}{0.07409}
=\boxed{0.627\ (62.7\%)}.$$
Mass density. Mass of the 4 formula units per cell,
$$m_{\text{cell}}=\frac{4(24.31)+4(16.00)}{N_A}=\frac{161.24}{6.023\times10^{23}}
=2.677\times10^{-22}\ \text{g},$$
over the cell volume $a^3=7.409\times10^{-23}\ \text{cm}^3$:
$$\rho=\frac{m_{\text{cell}}}{a^3}=\frac{2.677\times10^{-22}}{7.409\times10^{-23}}
=\boxed{3.61\ \text{g/cm}^3}.$$ (Real periclase MgO is tabulated at $\approx3.58$ g/cm$^3$
— a useful independent check that the hard-sphere touching assumption is reasonable.)
