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17-Phys-B7 Structure of Materials · December 2018

Question 3 of 7: Question III: Crystal Structure I

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. 17-Phys-B7 Structure of Materials, National Examination December 2018 — 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 seven 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).

Check — two points on the printed paper. (1) Question I.2 prints the ion as "Cl- (Z = 16)"; Z = 16 is sulfur, not chlorine (Cl is Z = 17) — a printed typo. The electronic structure below uses the correct Z = 17 for chlorine. (2) Question III.1(d) prints the hexagonal plane as $(2\bar{2}10)$, i.e. Miller–Bravais indices $h=2,\,k=-2,\,i=1$; the third index of a valid Miller–Bravais symbol is never independent — it is fixed by $i=-(h+k)$, here $i=-(2-2)=0$, not 1. Reducing the self-consistent index $(2\bar200)$ by its common factor of 2 gives $(1\bar100)$, a standard prism-type plane. The drawing below uses the symmetry-equivalent, non-degenerate face $(10\bar10)$ of the same $\{1\bar100\}$ family (chosen because it renders visibly in the cell projection used here) and states this substitution explicitly.

Question III: Crystal Structure I (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two cubic indices, [111] (direction) and (132) (plane); two hexagonal Miller–Bravais indices, $[1\bar2 10]$ (direction) and $(2\bar210)$ (plane, corrected to $(1\bar100)$ — see the check note at the top of this solution).

Find. A drawing of each direction/plane on its unit cell.

Approach. For a direction $[uvw]$, draw the vector from the origin to the lattice point $u\,\vec a+v\,\vec b+w\,\vec c$; for a plane $(hkl)$, take axis intercepts $a/h,\,b/k,\,c/l$ (a zero index means "parallel to that axis," i.e. no finite intercept) and join them.

[111] direction [111]
(a) [111] runs along the cube's main body diagonal, from the origin (0,0,0) to the opposite corner (1,1,1) — equal steps along all three axes.
(132) plane intercepts: a, b/3, c/2
(c) $(132)$: $h=1,k=3,l=2$ give intercepts $a/1=a$, $b/3$, $c/2$; joining the three intercept points on the $a$-, $b$- and $c$-edges gives the shaded triangular plane.
Hexagonal unit cell (1 0 -1 0) prism face* [1 -2 1 0]
(b, d) $[1\bar210]$: writing the vector $R=U\vec a_1+V\vec a_2+T\vec a_3$ with $\vec a_3=-(\vec a_1+\vec a_2)$ gives $R=\vec a_1-2\vec a_2+(-(\vec a_1+\vec a_2))=-3\vec a_2$, i.e. simply the $-\vec a_2$ basal direction (drawn centre-to-vertex, red arrow) — a member of the close-packed $\langle11\bar20\rangle$ family. $(1\bar100)^*$: the corrected, self-consistent index from the check note; the equivalent, clearly-visible face $(10\bar10)$ of the same $\{1\bar100\}$ family is shaded (orange) — it is parallel to the $c$-axis and to $\vec a_2$, crossing $\vec a_1$ at $+1$ and $\vec a_3$ at $-1$.

Part 2 — FCC vs. HCP stacking. Both are close-packed structures built from identical close-packed atomic planes (74% atomic packing factor), but they differ in how those planes STACK. FCC stacks the planes in the sequence $\ldots\text{ABCABCABC}\ldots$ — the third layer sits over neither the first nor the second layer's atom positions, so the pattern only repeats every THIRD layer; this is the $\{111\}$-family stacking of the face-centred-cubic lattice. HCP stacks $\ldots\text{ABABAB}\ldots$ — the third layer sits directly over the first, so the pattern repeats every SECOND layer. Both give the same packing density and the same coordination number (12), but the different long-range stacking symmetry is what makes FCC cubic and HCP hexagonal, and is also why they have different numbers of independent close-packed slip planes (4 in FCC $\{111\}$, versus 1 in HCP basal $(0001)$) — a key reason FCC metals are generally more ductile than HCP metals.

Given. GaAs, zinc blende structure, lattice parameter $a=0.565$ nm; $M_{\text{Ga}}=69.72$ g/mol, $M_{\text{As}}=74.91$ g/mol.

Find. Theoretical density $\rho$ (g/cm³).

Approach. The zinc blende structure is two interpenetrating FCC sublattices (Ga on one, As on the other), giving 4 GaAs formula units per conventional cubic cell; divide their mass by the cell volume.

  1. Cell volume. $$V_c = a^3 = (0.565\times10^{-7}\ \text{cm})^3 = 1.8036\times10^{-22}\ \text{cm}^3.$$
  2. Theoretical density. With $n=4$ formula units per cell, $$\rho = \frac{n\,(M_{\text{Ga}}+M_{\text{As}})}{V_c\,N_A} = \frac{4(69.72+74.91)}{(1.8036\times10^{-22})(6.023\times10^{23})} = \frac{577.32}{108.63} = \boxed{5.33\ \text{g/cm}^3}.$$ (The accepted handbook density of GaAs is $5.32$ g/cm³ — a close independent check on the hard-sphere zinc-blende model.)
Final results — Question III
QuantityValue
[111] directioncube body diagonal, origin to (1,1,1)
(132) planeintercepts $a,\,b/3,\,c/2$
$[1\bar210]$ direction$-3\vec a_2$, i.e. the $-\vec a_2$ basal direction
$(2\bar210)\to(1\bar100)$ planeprism face, corrected index (see check note)
GaAs theoretical density5.33 g/cm³