Question 3 of 8: Question III: Point Defects in Crystals

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.

Part 1. In an interstitial solid solution, solute atoms are small enough to squeeze into the void spaces between the host (solvent) atoms without displacing them from their lattice sites — carbon in $\gamma$-iron is the classic example. In a substitutional solid solution, solute atoms instead directly replace solvent atoms on the parent lattice sites, so the solute is comparable in size to the solvent. The Hume-Rothery rules govern how readily an appreciable substitutional solid solution forms: (i) an atomic-radius difference below about 15% between solute and solvent (too large a mismatch generates excessive lattice strain); (ii) similar crystal structures for the two pure elements; (iii) similar electronegativities, so the solute does not preferentially form an intermetallic compound instead; and (iv) similar valence, or a solute valence equal to or greater than the solvent's (Hume-Rothery valence rule), which favours mutual solubility. Interstitial solubility is instead governed almost entirely by the size of the available void relative to the interstitial atom, since interstitials are always much smaller than the host atom.

Part 2 — Given. FCC $\gamma$-iron, atomic radius $R=0.129$ nm; the largest interstitial voids sit at the octahedral positions $(\tfrac12,0,0)$ etc.

Find. The radius of the largest interstitial void.

1. Void radius from the octahedral-site geometry. An FCC octahedral void at an edge-centre position is bounded by two atoms touching across the cell edge; standard geometry for this site gives the universal ratio $$\frac{r_{\text{void}}}{R}=\sqrt2-1=0.4142,$$ independent of which specific FCC metal, so $$r_{\text{void}}=(\sqrt2-1)(0.129)=\boxed{0.0534\ \text{nm}\ (53.4\ \text{pm})}.$$

Part 3 — Given. Sn-in-Cu substitutional solid solution (FCC), $a=0.376$  nm, $\rho=8.772$ g/cm$^3$, $M_{\text{Sn}}=118.71$ g/mol, $M_{\text{Cu}}=63.546$  g/mol.

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

Approach. An FCC cell always holds 4 atoms regardless of species mix; write the cell mass as 4 atoms of a composition-weighted average molar mass, and solve the density equation for the tin atomic fraction $x$.

1. Set up the density equation. With $x$ the atomic fraction of Sn among the 4 atoms per cell, $$\rho=\frac{4\left[xM_{\text{Sn}}+(1-x)M_{\text{Cu}}\right]}{N_Aa^3}.$$ The cell volume is $a^3=(0.376\times10^{-7}\,\text{cm})^3=5.316\times10^{-23}\ \text{cm}^3$.
2. Solve for x. Rearranging for the average atomic mass, $$\bar M=\frac{\rho N_Aa^3}{4}=\frac{8.772\times6.023\times10^{23}\times5.316\times10^{-23}}{4} =70.21\ \text{g/mol},$$ $$x=\frac{\bar M-M_{\text{Cu}}}{M_{\text{Sn}}-M_{\text{Cu}}}=\frac{70.21-63.546}{118.71-63.546} =\boxed{0.1209\ \ (12.1\ \text{at.\% Sn})}.$$