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21-Mat-A4 Deformation Behaviour and Properties 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. National Exams, December 2018 — 10-Met-A4, Structure of Materials. Three hours, closed book, one approved calculator (Casio or Sharp). Seven questions of 20 marks each (Roman numerals I–VII); the rubric asks for any five, with only the first five in the answer book marked. All seven are solved here, because this set is a study resource rather than an exam script. All necessary equations, constants and an error-function table are provided in the exam's own appendix (reproduced where used below).

Note on the exam title. The printed exam header reads 10-Met-A4, Structure of Materials. Only Question VI.2(c) (grain-size strengthening) touches mechanical/deformation properties directly; the paper as a whole is a broad introductory materials-science survey — atomic structure, bonding, crystal structure/directions/planes, point defects, XRD, dislocations and phase diagrams — and is answered as such below.

Reference texts. The answers below are keyed to the works normally recommended for this syllabus code:

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.

IV.1 — Equilibrium Vacancy Concentration in Cu at 1000°C

Given. $T=1000\,{}^{\circ}\text{C}=1273$ K; $Q_v=0.9$ eV/atom; $A_{wt}=63.5$ g/mol; $\rho=8.4$ g/cm³ (at this temperature); $k=8.62\times10^{-5}$ eV/atom·K (appendix).

Find. $N_v$, the equilibrium number of vacancies per m³.

Approach. First get the total lattice-site density $N=\rho N_A/A_{wt}$ (appendix), then apply the Arrhenius vacancy relation $N_v=N\exp(-Q_v/kT)$ (appendix).

  1. Total lattice-site density. $$N=\frac{\rho N_A}{A_{wt}}=\frac{(8.4)(6.023\times10^{23})}{63.5}=7.967\times10^{22}\ \text{atoms/cm}^3=7.967\times10^{28}\ \text{atoms/m}^3$$
  2. Boltzmann factor. $$kT=(8.62\times10^{-5})(1273)=0.1097\ \text{eV}, \qquad \frac{Q_v}{kT}=\frac{0.9}{0.1097}=8.20$$
  3. Vacancy density. $$N_v=N\exp\!\left(-\frac{Q_v}{kT}\right)=(7.967\times10^{28})(2.74\times10^{-4})=\boxed{2.18\times10^{25}\ \text{vacancies/m}^3}$$
Question IV.1 — summary
QuantityValue
Total lattice-site density $N$$7.97\times10^{28}$ m⁻³
Equilibrium vacancy density $N_v$$2.18\times10^{25}$ m⁻³ ($2.18\times10^{19}$ cm⁻³)
Vacancy fraction $N_v/N$$2.74\times10^{-4}$ (0.027%)

IV.2 — Frenkel vs. Schottky Defects

Both are equilibrium point-defect pairs found predominantly in ionic solids. A Frenkel defect is a cation that leaves its regular lattice site and lodges in a nearby interstitial site, leaving behind a vacancy — a vacancy–interstitial pair of the same ion, with no change in overall stoichiometry or charge neutrality (e.g. Ag⁺ displaced into an interstitial site in AgCl, or Zn⁺⁺ in ZnO). A Schottky defect is instead a matched pair of vacancies — one cation vacancy and one anion vacancy, in the stoichiometric ratio required to preserve overall charge neutrality — with both ions effectively removed to the crystal surface rather than relocated internally (e.g. a paired Na⁺/Cl⁻ vacancy pair in NaCl). The key structural distinction is that a Frenkel defect keeps the displaced ion within the crystal (as an interstitial), while a Schottky defect removes both ions from the bulk entirely.

IV.3 — Tungsten Fraction in the Nb–W BCC Alloy

Given. BCC structure ($n=2$ atoms/cell); $a=0.32554$ nm; $\rho=11.95$ g/cm³; $M_{Nb}=92.91$ g/mol; $M_W=183.84$ g/mol.

Find. The atomic (mole) fraction of tungsten, $x_W$.

Approach. Invert $\rho=\dfrac{n\,\bar{A}_{wt}}{V_c N_A}$ to get the alloy's mean atomic weight $\bar A_{wt}$, then solve $\bar A_{wt}=x_W A_W+(1-x_W)A_{Nb}$ for $x_W$.

  1. Unit cell volume. $$V_c=a^3=(3.2554\times10^{-8}\,\text{cm})^3=3.4500\times10^{-23}\ \text{cm}^3$$
  2. Mean atomic weight from the measured density. $$\bar A_{wt}=\frac{\rho V_c N_A}{n}=\frac{(11.95)(3.4500\times10^{-23})(6.023\times10^{23})}{2}=124.15\ \text{g/mol}$$
  3. Solve for the tungsten atomic fraction. $$\bar A_{wt}=A_{Nb}+x_W(A_W-A_{Nb}) \ \Rightarrow\ x_W=\frac{\bar A_{wt}-A_{Nb}}{A_W-A_{Nb}}=\frac{124.15-92.91}{183.84-92.91}=\frac{31.24}{90.93}$$ $$\boxed{x_W \approx 0.344\ (34.4\ \text{at\%})}$$ (equivalently $w_W=x_W A_W/\bar A_{wt}=50.9$ wt% W).
Question IV.3 — summary
QuantityValue
Alloy mean atomic weight $\bar A_{wt}$124.15 g/mol
Tungsten atomic fraction $x_W$34.4 at%
Tungsten weight fraction $w_W$50.9 wt%
Check
The individual densities of pure Nb (8.57 g/cm³) and pure W (19.35 g/cm³) given in the question are not needed for this calculation — the alloy's own measured lattice parameter and density are sufficient, via the unit-cell mass-balance route above, to back out the mean atomic weight and hence composition directly. The pure-metal densities would only be needed for a (less direct, rule-of-mixtures) volume-based cross-check, which is not required here.