Question 3 of 8: Question III — Point Defects in Crystals (20 marks)

Question III — Point Defects in Crystals (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.

III.1 — Interstitial vs. substitutional solid solutions

A substitutional solid solution forms when solute atoms directly replace solvent atoms on the parent lattice sites; an interstitial solid solution forms when (typically small) solute atoms instead squeeze into the voids between the solvent atoms on their normal sites, without displacing them. Interstitial solutes are almost always much smaller than the solvent atom (carbon or nitrogen in iron is the classic example); substitutional solutes are comparable in size to the solvent.

The Hume-Rothery rules identify the parameters that favour extensive (ideally complete) substitutional solid solubility: (i) an atomic size difference below about 15% — larger mismatches build up too much lattice strain energy; (ii) similar crystal structures for the two pure elements; (iii) similar electronegativity — a large difference favours compound formation (ionic/intermetallic) over solid solution; (iv) similar valence, or a solute valence no higher than the solvent's (a higher-valence solute donates electrons the solvent lattice can only accommodate in limited amounts before a different phase becomes favourable).

III.2 — Largest FCC interstitial void

Given. FCC γ-iron, atomic radius $R = 0.129\ \text{nm}$; the largest voids are the octahedral sites at $(\tfrac12,0,0)$-type positions.

Find. The radius of the largest interstitial (octahedral) void.

Approach. The octahedral site in FCC is surrounded by six touching host atoms along the cell-edge directions; the geometric ratio of void radius to host radius for this coordination is a standard result, $r_{\text{void}}/R = \sqrt{2}-1$.

1. Geometry of the octahedral hole. The site at $(\tfrac12,0,0)$ sits at the midpoint of a cube edge, midway between two corner atoms and equidistant from four face-centred atoms; along that edge, $a = 2R\sqrt{2}$ (FCC face-diagonal contact) and the void radius satisfies $R + r_{\text{void}} = a/2$, giving $r_{\text{void}} = R(\sqrt2-1)$.
2. Evaluate. $$r_{\text{void}} = (\sqrt{2}-1)(0.129) = \boxed{0.0534\ \text{nm} = 53.4\ \text{pm}}$$ This is the maximum sphere that fits in the octahedral hole without displacing the surrounding Fe atoms — carbon (atomic radius ≈0.077 nm) is substantially larger than this void, which is exactly why interstitial carbon locally strains the γ-iron lattice and why carbon solubility in γ-Fe (up to ≈2 wt%) is far higher than in the smaller octahedral/tetrahedral voids of α-Fe (BCC).

III.3 — Tin-bronze: atomic concentration of Sn

Given. Sn substitutionally dissolved in Cu (FCC), lattice parameter $a = 0.376\ \text{nm}$, density $\rho = 8.772\ \text{g/cm}^3$; $M_{Sn}=118.71\ \text{g/mol}$, $M_{Cu}=63.546\ \text{g/mol}$; FCC has $n=4$ atoms/cell.

Find. The atomic fraction (at%) of tin in the alloy.

Approach. Write the cell mass as a composition-weighted average atomic mass and solve the density formula for the tin atomic fraction $x$.

1. Density formula with a mixed atomic mass. With $x$ the atomic fraction Sn (so $1-x$ is Cu), the average atomic mass is $\bar M = xM_{Sn}+(1-x)M_{Cu}$ and $$\rho = \frac{n\bar M}{N_A a^3}$$
2. Solve for $x$. With $a = 3.76\times10^{-8}\ \text{cm}$, $a^3 = 5.316\times10^{-23}\ \text{cm}^3$: $$\bar M = \frac{\rho N_A a^3}{n} = \frac{(8.772)(6.023\times10^{23})(5.316\times10^{-23})}{4} = 70.09\ \text{g/mol}$$ $$x = \frac{\bar M - M_{Cu}}{M_{Sn}-M_{Cu}} = \frac{70.09-63.546}{118.71-63.546} = \boxed{0.1209 = 12.1\ \text{at\% Sn}}$$