Question 4 of 8: Crystalline Imperfections (20 marks)

Question IV — Crystalline Imperfections (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 — Hume-Rothery rules and Cu–Ni solubility

The four Hume-Rothery rules for extensive (ideally complete) substitutional solid solubility are: (1) atomic size factor — atomic radii must differ by less than about 15%; (2) crystal structure — both metals must share the same crystal structure; (3) electronegativity — the two elements should have similar electronegativity (a large difference favours compound formation over solid solution); (4) valence — the metals should have the same valence (a metal of lower valence dissolves a higher-valence solute more readily than the reverse).

All four Hume-Rothery criteria are satisfied, so $\boxed{\text{Cu and Ni are completely (isomorphously) miscible}}$ — they form a continuous substitutional solid solution across the full 0–100 at% composition range at all temperatures below their solidus, exactly as observed in the real Cu–Ni binary phase diagram (the textbook example of a simple isomorphous system).

IV.2 — BCC iron: theoretical density and vacancies

Given. BCC Fe, $a=0.2866$ nm, $A_w=55.85$ g/mol, $n_{BCC}=2$ atoms/cell.

Find. (a) $\rho_{theoretical}$; (b) vacancy concentration needed to reduce this to $\rho_{actual}=7.874\ \text{g/cm}^3$.

Approach. (a) direct $\rho=nA_w/(a^3N_A)$. (b) A vacancy removes one atom's mass from the same unit-cell volume (vacancies do not measurably change the lattice parameter), so the actual/theoretical density ratio equals the fraction of lattice sites that remain occupied.

1. (a) Theoretical density. $$\rho_{th}=\frac{nA_w}{a^3N_A}=\frac{2(55.85)}{(0.2866\times10^{-7}\text{cm})^3(6.023\times10^{23})}=\boxed{7.878\ \text{g/cm}^3}$$
2. (b) Fraction of vacant sites. $$\frac{N_v}{N}=1-\frac{\rho_{actual}}{\rho_{th}}=1-\frac{7.874}{7.878}=4.98\times10^{-4}\ (0.0498\%)$$
3. Vacancies per unit cell / per unit volume. $$N_v\text{ per cell}=\left(\frac{N_v}{N}\right)n_{BCC}=(4.98\times10^{-4})(2)=\boxed{9.96\times10^{-4}\text{ vacancies/cell}}$$ i.e. about 1 vacant site for every 1004 unit cells, or, per unit volume, $$\frac{N_v}{V}=\frac{N_v}{N}\cdot\frac{n_{BCC}}{a^3}=\boxed{4.23\times10^{19}\ \text{vacancies/cm}^3}$$

A vacancy fraction of order $5\times10^{-4}$ is a physically reasonable equilibrium value for iron somewhat below its melting point (equilibrium vacancy concentrations are typically $10^{-4}$–$10^{-3}$ near $0.7$–$0.9\,T_m$); it is not, however, the room-temperature equilibrium concentration (which is many orders of magnitude smaller) — the question is testing the density-defect relationship itself, not proposing this as room-temperature Fe's true vacancy population.

IV.3 — Slip systems and Burgers vectors, FCC and BCC

FCC: slip occurs on the close-packed $\{111\}$ octahedral planes along the close-packed $\langle110\rangle$ directions. There are 4 distinct $\{111\}$ plane families $\times$ 3 $\langle110\rangle$ directions per plane $=\boxed{12\ \text{slip systems}}$. The Burgers vector of a perfect (edge or screw) dislocation is $\vec b=\tfrac{a}{2}\langle110\rangle$, magnitude $$|\vec b|_{FCC}=\frac{a}{2}\sqrt{1^2+1^2+0^2}=\boxed{\frac{a\sqrt2}{2}=0.707\,a}$$

BCC: slip occurs on the most densely-packed $\{110\}$ planes (6 of them) along the close-packed $\langle111\rangle$ body-diagonal directions (2 independent directions per plane) $=\boxed{12\ \text{slip systems}}$ (BCC additionally cross-slips on $\{112\}$ and $\{123\}$ at elevated temperature/stress, but $\{110\}\langle111\rangle$ is the primary system asked for here). The Burgers vector is $\vec b=\tfrac{a}{2}\langle111\rangle$, magnitude $$|\vec b|_{BCC}=\frac{a}{2}\sqrt{1^2+1^2+1^2}=\boxed{\frac{a\sqrt3}{2}=0.866\,a}$$