Question 3 of 7: Question III: Crystal Structure I

Question III: Crystal Structure I (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 — Miller Indices of Planes A and B

Given. The unit cell figure shows Plane B through the unit intercepts $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, and Plane A through the three points $(0,1,0)$, $(1,\tfrac{2}{3},0)$ and $(1,1,\tfrac{1}{2})$ (all in units of the cell edge $a$).

[Figure not reproduced: Fig. III.1 — the two shaded planes reproduced from the exam figure, with the vertices used to derive each plane's equation. See the official exam paper.]

Find. The Miller indices $(hkl)$ of Plane A and Plane B.

Approach. Fit each set of points to the general plane equation $ax+by+cz=d$, read the axis intercepts from that equation, take reciprocals, and clear to the smallest integers.

1. Plane B. Its three points are already the unit intercepts on the three axes, $(1,0,0)$, $(0,1,0)$, $(0,0,1)$, so the reciprocal intercepts are $(1,1,1)$ directly: $$\boxed{\text{Plane B} = (111)}$$
2. Plane A — fit the plane equation. Substituting the three points $(0,1,0)$, $(1,\tfrac23,0)$, $(1,1,\tfrac12)$ into $ax+by+cz=d$ gives three equations: $b=d$; $\ a+\tfrac23 b=d$; $\ a+b+\tfrac12 c=d$. Solving (take $d=3$ for integer results): $b=3$, then $a+2=3\Rightarrow a=1$, then $1+3+\tfrac12 c=3\Rightarrow c=-2$. The plane is $x+3y-2z=3$.
3. Intercepts and Miller indices. The axis intercepts of $x+3y-2z=3$ are $x_0=3$, $y_0=1$, $z_0=-3/2$. Reciprocals: $1/3,\ 1,\ -2/3$; clearing to the smallest integers (×3): $$\boxed{\text{Plane A} = (1\ 3\ \bar2)}$$ the coefficients of the fitted plane equation ($a{,}b{,}c$) are already proportional to $(hkl)$, confirming the result directly.

III.2 — Theoretical Volume Change, BCC → FCC

Given. Hard-sphere atomic model; atomic radius $R$ unchanged by the transformation; BCC: $a=4R/\sqrt3$, 2 atoms/cell; FCC: $a=2\sqrt2\,R$, 4 atoms/cell (appendix relations).

Find. The percentage volume change on transforming from BCC to FCC.

Approach. Compute the volume per atom (unit-cell volume divided by atoms per cell) for each structure in terms of the common radius $R$, then compare.

1. Volume per atom, BCC. $$V_{BCC}=\frac{a_{BCC}^3}{2}=\frac{1}{2}\left(\frac{4R}{\sqrt3}\right)^3=\frac{32}{3\sqrt3}R^3=\boxed{6.158\,R^3}$$
2. Volume per atom, FCC. $$V_{FCC}=\frac{a_{FCC}^3}{4}=\frac{1}{4}\left(2\sqrt2\,R\right)^3=4\sqrt2\,R^3=\boxed{5.657\,R^3}$$
3. Percentage volume change. $$\frac{\Delta V}{V_{BCC}}=\frac{V_{FCC}-V_{BCC}}{V_{BCC}}\times100\%=\frac{5.657-6.158}{6.158}\times100\%=\boxed{-8.14\%}$$ — the metal contracts by about 8.1% in volume per atom on transforming from BCC to FCC, consistent with FCC's higher atomic packing factor (0.74 vs. 0.68 for BCC) at the same atomic radius.

III.3 — Theoretical Density of CdTe (Zinc Blende)

Given.

Find. The theoretical density $\rho$ of CdTe, g/cm³.

Approach. Use $\rho=n\cdot A_{wt}/(V_c N_A)$ (appendix), with $n$ the number of CdTe formula units per unit cell and $A_{wt}$ their combined molar mass. The zinc blende structure is an FCC sublattice of one species (4 atoms/cell) with the other species filling all 4 tetrahedral interstitial sites, so there are 4 CdTe formula units per cell.

1. Formula units per cell and combined molar mass. $n=4$ CdTe/cell; $A_{wt}=112.4+127.6=\boxed{240.0\ \text{g/mol}}$.
2. Unit-cell volume. Converting $a=0.648\ \text{nm}=6.48\times10^{-8}\ \text{cm}$: $$V_c=a^3=(6.48\times10^{-8})^3=\boxed{2.721\times10^{-22}\ \text{cm}^3}$$
3. Theoretical density. $$\rho=\frac{n\,A_{wt}}{V_c N_A}=\frac{4(240.0)}{(2.721\times10^{-22})(6.023\times10^{23})}=\frac{960.0}{163.9}=\boxed{5.858\ \text{g/cm}^3}$$