Question 1 of 7: FCC Lattice — Packing Fraction, Miller Indices, Reciprocal Lattice

Question 1: FCC Lattice — Packing Fraction, Miller Indices, Reciprocal Lattice (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.

Given. A conventional fcc unit cell, edge length $a$: one atom at each of the 8 corners plus one at each of the 6 face centres (4 lattice points per conventional cell). Figure P1b shows a lattice plane cutting the three axes at $x_0=2a$, $y_0=4a$, $z_0=a$. Figure P1a shows the primitive rhombohedral cell built from the three primitive vectors $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ drawn from one corner atom to the centres of the three cube faces that meet at that corner.

Find. (a) the atomic packing fraction of the fcc lattice; (b) the Miller indices $(hkl)$ of the plane in Fig. P1b; (c) the reciprocal-lattice primitive vectors $\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3$.

Approach. (a) impose the hard-sphere touching condition along the face diagonal; (b) invert the axis intercepts and clear to the smallest integers; (c) build $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ from Fig. P1a and apply the exam's own reciprocal-vector formula, eq. (6): $\mathbf{b}_1=2\pi\dfrac{\mathbf{a}_2\times\mathbf{a}_3}{\mathbf{a}_1\cdot\mathbf{a}_2\times\mathbf{a}_3}$ (cyclic).

1. Part (a) — hard-sphere touching condition. Nearest neighbours in an fcc lattice touch along the face diagonal of length $a\sqrt{2}$, which spans 4 atomic radii: $4r=a\sqrt{2}\Rightarrow r=\dfrac{\sqrt{2}}{4}a$. The conventional cell holds $8\times\tfrac18+6\times\tfrac12=4$ atoms, so $$\text{packing fraction}=\frac{4\cdot\frac{4}{3}\pi r^3}{a^3}=\frac{16\pi}{3}\left(\frac{\sqrt2}{4}\right)^3=\frac{\pi}{3\sqrt2}$$ Evaluating, $\boxed{\text{packing fraction}=\dfrac{\pi}{3\sqrt2}=0.7405\ (74.05\%)}$ — the same close-packing fraction as hcp, the densest possible packing of identical spheres.
2. Part (b) — Miller indices from the intercepts. Figure P1b is read directly off its own axis tick marks: the plane cuts $x$ at $2a$, $y$ at $4a$, $z$ at $1a$. Taking reciprocals and clearing to the smallest integer triple (multiply by the LCD, 4): $$\left(\frac1{2},\ \frac1{4},\ \frac1{1}\right)\times4=(2,\ 1,\ 4)$$ $\boxed{(hkl)=(2\,1\,4)}$.
3. Part (c) — reciprocal vectors via eq. (6). From Fig. P1a, $\mathbf{a}_1=\tfrac{a}{2}(\hat y+\hat z)$, $\mathbf{a}_2=\tfrac{a}{2}(\hat z+\hat x)$, $\mathbf{a}_3=\tfrac{a}{2}(\hat x+\hat y)$. The primitive-cell volume is $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)=a^3/4$ (checks: $4$ primitive cells per conventional cell of volume $a^3$). Substituting into eq. (6): $$\begin{aligned} \mathbf{b}_1&=2\pi\frac{\mathbf{a}_2\times\mathbf{a}_3}{a^3/4}=\frac{2\pi}{a}(-\hat x+\hat y+\hat z)\\ \mathbf{b}_2&=\frac{2\pi}{a}(\hat x-\hat y+\hat z)\\ \mathbf{b}_3&=\frac{2\pi}{a}(\hat x+\hat y-\hat z) \end{aligned}$$ each of magnitude $|\mathbf{b}_i|=\sqrt3\,(2\pi/a)$. $\boxed{\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3\text{ as above}}$ — this is exactly the primitive-vector set of a bcc lattice with conventional cube edge $4\pi/a$, the standard result that the reciprocal of fcc is bcc.