Question 3 of 8: Question III — Crystal Structure (20 marks)

Question III — Crystal Structure (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 — Four definitions

(a) Atomic packing factor (APF) is the fraction of a unit cell's volume actually occupied by atoms, modelled as hard, touching spheres: $\text{APF} = V_s/V_c$ (appendix), where $V_s$ is the total sphere volume of the atoms belonging to the cell and $V_c$ is the cell volume. It ranges from about 0.68 (BCC) to 0.74 (FCC/HCP, the densest possible packing of equal spheres).

(b) Electronegativity is a dimensionless measure (Pauling scale, roughly 0.7–4.0) of an atom's tendency to attract shared bonding electrons toward itself in a compound. A large electronegativity difference between two bonded atoms (e.g. Na, 0.9, and Cl, 3.0) favours ionic bonding (electron transfer); a small difference (e.g. C and H, both ≈2.1–2.5) favours covalent bonding (electron sharing).

(c) Hexagonal close-packed (HCP) structure is one of the two ways to stack close-packed atomic planes with maximum density: successive close-packed layers follow an ABABAB… stacking sequence (each atom sits directly above one two layers below), giving a hexagonal unit cell with $c/a \approx 1.633$ for ideal hard-sphere packing, 6 atoms per (hexagonal-prism) cell, coordination number 12 and $\text{APF}=0.74$ — identical packing density to FCC (ABCABC… stacking) but different long-range symmetry. Mg, Zn, Ti and the Co examined in Question IV.3 below are common HCP metals.

(d) Miller indices $(hkl)$ are the standard notation for a crystallographic plane: take the plane's intercepts on the three crystal axes (in units of the lattice parameters), invert them, and clear fractions to the smallest integer set — e.g. a plane cutting the axes at $(1,1,1)$ is $(111)$; one parallel to an axis has intercept $\infty$, reciprocal 0. A bar over an index, e.g. $(1\bar{1}0)$, denotes a negative intercept. Miller indices identify planes for X-ray diffraction, slip-plane and cleavage-plane analysis, exactly as used in III.3 and V.2 below.

III.2 — APF derivation for FCC

Given. FCC unit cell: atoms touch along the face diagonal, $n=4$ atoms/cell (8 corners × 1/8 + 6 faces × 1/2), each atom of radius $R$.

Find. $\text{APF}=V_s/V_c$.

Approach. Relate the lattice parameter $a$ to $R$ from the face-diagonal touching condition, then take the ratio of sphere volume to cell volume.

1. Relate $a$ to $R$. The face diagonal has length $4R$ (four atomic radii: corner–centre–centre–corner along the diagonal) and, geometrically, length $a\sqrt2$: $$a\sqrt2 = 4R \ \Rightarrow\ a = \frac{4R}{\sqrt2} = 2\sqrt2\,R$$ (appendix relation $a=2\sqrt2R$, confirmed by this derivation.)
2. Sphere volume per cell. $n=4$ atoms, each of volume $\tfrac{4}{3}\pi R^3$: $$V_s = 4\times\frac{4}{3}\pi R^3 = \frac{16}{3}\pi R^3$$
3. Cell volume. $$V_c = a^3 = (2\sqrt2R)^3 = 16\sqrt2\,R^3$$
4. Ratio. $$\text{APF} = \frac{V_s}{V_c} = \frac{\tfrac{16}{3}\pi R^3}{16\sqrt2\,R^3} = \frac{\pi}{3\sqrt2} = \boxed{0.740}$$

III.3 — Miller indices of planes A and B

Given. Plane A intercepts the axes at $x=1$, $y=2/3$, $z=1/2$ (lattice-parameter units). Plane B passes through the origin $O$, $(0,0,1)$ and $(1,1,0)$.

Find. $(hkl)$ for both planes.

Approach. For plane A, invert the intercepts and clear fractions. Plane B passes through the origin, so intercepts are undefined there; instead take its normal vector from two in-plane vectors, which is directly proportional to $(hkl)$ for a cubic system.

1. Plane A — invert and clear. Intercepts $(1,\ 2/3,\ 1/2)$ invert to $(1,\ 3/2,\ 2)$; multiplying by 2 to clear the fraction gives $$\boxed{(234)}$$
2. Plane B — normal-vector method. Two in-plane vectors from the origin: $\mathbf{v_1}=(0,0,1)-(0,0,0)=(0,0,1)$ and $\mathbf{v_2}=(1,1,0)-(0,0,0)=(1,1,0)$. The plane normal (proportional to $(hkl)$ for a cubic cell) is $$\mathbf{n}=\mathbf{v_1}\times\mathbf{v_2} = \big((0)(0)-(1)(1),\ (1)(1)-(0)(0),\ (0)(1)-(0)(1)\big) = (-1,\ 1,\ 0)$$ $$\boxed{(1\bar{1}0)}$$
3. Cross-check plane B by the origin-shift method. Shift the origin to $(1,0,0)$, an adjacent cell corner not on the plane: in shifted coordinates the plane equation $-x+y=0$ becomes $-x'+y=1$ (with $x'=x-1$), giving intercepts $x'=-1$, $y=1$, $z=\infty$. Reciprocals $(-1,\ 1,\ 0)$ — the same $(1\bar{1}0)$, confirming step 2.