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17-Phys-B7 Structure of Materials · December 2018

Question 4 of 7: Point Defects in Crystalline Solids

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 IV: Point Defects in Crystalline Solids (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. Copper at $T=1000^\circ\text{C}$; vacancy formation energy $Q_v=0.9$ eV/atom; atomic weight $A_w=63.5$ g/mol; density $\rho=8.4$ g/cm³ (at this elevated temperature).

Find. Equilibrium vacancy concentration $N_v$ (per m³).

Approach. Get the total lattice-site density $N$ from the given density and atomic weight, then apply the Arrhenius vacancy-equilibrium relation $N_v=N\exp(-Q_v/kT)$.

  1. Absolute temperature. $T = 1000+273 = 1273\ \text{K}$.
  2. Lattice-site density. $$N = \frac{\rho N_A}{A_w} = \frac{(8.4)(6.023\times10^{23})}{63.5} = 7.97\times10^{22}\ \text{atoms/cm}^3 = \boxed{7.97\times10^{28}\ \text{atoms/m}^3}.$$
  3. Boltzmann factor. With $k=8.62\times10^{-5}$ eV/atom-K, $$\frac{N_v}{N} = \exp\!\left(-\frac{0.9}{(8.62\times10^{-5})(1273)}\right) = \exp(-8.202) = \boxed{2.74\times10^{-4}}.$$
  4. Equilibrium vacancy concentration. $$N_v = N\exp(-Q_v/kT) = (7.97\times10^{28})(2.74\times10^{-4}) = \boxed{2.18\times10^{25}\ \text{vacancies/m}^3}.$$

Part 2 — Frenkel vs. Schottky defects. Both are STOICHIOMETRIC point defects that occur in pairs in ionic solids, but they differ in WHERE the displaced ion ends up. A Frenkel defect is a cation that leaves its normal lattice site and lodges in a nearby INTERSTITIAL site, leaving a vacancy behind — the cation is not lost from the crystal, only displaced, so this is a "vacancy + interstitial" pair on the SAME sublattice; it is favoured when the cation is much smaller than the anion (so it fits into an interstitial void), e.g. Ag⁺ interstitials in AgCl (Ag⁺ is small enough to squeeze between the Cl⁻ ions). A Schottky defect is instead a CATION vacancy paired with an ANION vacancy in the correct stoichiometric ratio (e.g. one cation + one anion vacancy in NaCl, or one Mg²⁺ + two O²⁻ vacancies in MgO) — both ions are effectively removed to the crystal surface, so, unlike a Frenkel defect, a Schottky defect measurably lowers the crystal's density. NaCl is the textbook Schottky example.

Given. Nb–W BCC substitutional alloy; lattice parameter $a=0.32554$ nm; alloy density $\rho=11.95$ g/cm³. Pure-element data: $A_{\text{Nb}}=92.91$ g/mol, $\rho_{\text{Nb}}=8.57$ g/cm³; $A_{\text{W}}=183.84$ g/mol, $\rho_{\text{W}}=19.35$ g/cm³.

Find. Atomic (and weight) fraction of tungsten, $x_{\text{W}}$.

Approach. A BCC cell holds 2 atoms regardless of composition; write the alloy density as a function of the unknown atomic fraction $x_{\text{W}}$ using a composition-weighted average atomic mass, and solve for the $x_{\text{W}}$ that reproduces the given alloy density at the given (measured) lattice parameter.

  1. Cell volume from the measured lattice parameter. $$V_c = a^3 = (0.32554\times10^{-7}\ \text{cm})^3 = 3.4501\times10^{-23}\ \text{cm}^3.$$
  2. Density as a function of composition. With 2 atoms/cell (BCC) and an average atomic weight $\bar A = (1-x_{\text{W}})A_{\text{Nb}} + x_{\text{W}}A_{\text{W}}$, $$\rho(x_{\text{W}}) = \frac{2\,\bar A}{V_c\,N_A}.$$
  3. Solve $\rho(x_{\text{W}})=11.95$ for $x_{\text{W}}$. Substituting and solving the resulting linear equation in $x_{\text{W}}$, $$x_{\text{W}} = \boxed{0.344\ (34.4\ \text{at\%})}.$$
  4. Weight-fraction cross-check. $$w_{\text{W}} = \frac{x_{\text{W}}A_{\text{W}}}{(1-x_{\text{W}})A_{\text{Nb}}+x_{\text{W}}A_{\text{W}}} = \boxed{0.509\ (50.9\ \text{wt\%})}.$$
Final results — Question IV
QuantityValue
Total lattice-site density $N$ (Cu, 1273 K)7.97 × 10²⁸ atoms/m³
Equilibrium vacancy concentration $N_v$2.18 × 10²⁵ vacancies/m³
Tungsten atomic fraction34.4 at%
Tungsten weight fraction50.9 wt%