Question 4 of 7: Crystal Structure II (20 marks)

Question IV — Crystal Structure II (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 — Vanadium: FCC or BCC?

Given. $\rho=5.8$ g/cm³ (measured); $a=0.303$ nm; $A_w=50.94$ g/mol; $Z=23$.

Find. Which cubic structure (FCC, $n=4$, or BCC, $n=2$) is consistent with the measured density.

Approach. Compute $\rho=nA_w/(a^3N_A)$ for both $n=2$ and $n=4$ and compare against the measured 5.8 g/cm³.

1. Unit-cell volume. $$a^3=(0.303\times10^{-7}\ \text{cm})^3=2.782\times10^{-23}\ \text{cm}^3$$
2. Trial densities. $$\rho_{BCC}=\frac{2(50.94)}{(2.782\times10^{-23})(6.023\times10^{23})}=6.08\ \text{g/cm}^3, \qquad \rho_{FCC}=\frac{4(50.94)}{(2.782\times10^{-23})(6.023\times10^{23})}=12.16\ \text{g/cm}^3$$

The measured 5.8 g/cm³ sits close to the BCC prediction (6.08 g/cm³, 4.6% high) and far below the FCC prediction (12.16 g/cm³, more than double), so vanadium is body-centered cubic at this lattice parameter — consistent with real vanadium, which is BCC with a measured density of about 6.0–6.1 g/cm³.

IV.2 — Solubility factors; Zn and Pb in Cu

Factors governing substitutional solid solubility (Hume-Rothery rules). Extensive solid solubility of one element in another requires all four of: (1) atomic size factor — atomic radii within about 15% of each other, or the solvent lattice distorts too much to accept the solute; (2) crystal structure — solute and solvent should share the same crystal structure, for complete solubility; (3) electronegativity — a small electronegativity difference favours solid-solution formation, while a large difference favours a stable intermetallic compound instead; (4) valence — other factors equal, a metal of higher valence is more soluble in a lower-valence solvent than the reverse (the relative valence effect).

1. Zinc in copper. Size: $\Delta r=(0.133-0.128)/0.128=3.9\%$ (well within 15%, favourable). Structure: Zn is HCP vs. Cu's FCC (unfavourable). Electronegativity: $|1.8-1.7|=0.1$ (small, favourable). Valence: both +2 (favourable). Three of four factors favourable, one (structure) unfavourable.
2. Lead in copper. Size: $\Delta r=(0.175-0.128)/0.128=36.7\%$ (far above 15%, strongly unfavourable). Structure: Pb is FCC, matching Cu (favourable). Electronegativity: $|1.8-1.6|=0.2$ (small, favourable). Valence: Pb offers +2 or +4 vs. Cu's +2 (roughly comparable). One dominant factor (size) is strongly unfavourable.

Prediction. Zinc, satisfying three of the four Hume-Rothery factors (only the crystal-structure mismatch counts against it), should show substantially greater solid solubility in copper than lead, whose 36.7% size mismatch dominates despite its matching FCC structure. This matches the real Cu-Zn and Cu-Pb systems: Cu-Zn forms the extensive α-brass solid solution (up to about 35 wt% Zn), while Cu-Pb shows essentially negligible mutual solid solubility — molten Pb solidifies as discrete, nearly pure particles within the Cu matrix (exploited deliberately in leaded bronzes for machinability).

IV.3 — Equilibrium vacancy concentration in Mg at 700°C

Given.

Find. Equilibrium vacancy concentration $N_v$ per cubic metre.

Approach. First get the atomic site density $N_0=\rho N_A/A_w$; then apply the appendix's Arrhenius vacancy relation $N_v=N_0\exp(-Q_v/kT)$.

1. Atomic site density. $$N_0=\frac{\rho N_A}{A_w}=\frac{(1.74)(6.023\times10^{23})}{24.304}=4.312\times10^{22}\ \text{atoms/cm}^3=4.312\times10^{28}\ \text{atoms/m}^3$$
2. Boltzmann factor. $$\frac{Q_v}{kT}=\frac{0.8}{(8.62\times10^{-5})(973)}=9.538 \quad\Rightarrow\quad \exp(-9.538)=7.20\times10^{-5}$$
3. Vacancy concentration. $$N_v=N_0\exp\left(-\frac{Q_v}{kT}\right)=(4.312\times10^{28})(7.20\times10^{-5})=\boxed{3.11\times10^{24}\ \text{vacancies/m}^3}$$