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

Question 1: FCC Lattice — Primitive Vectors, Packing Fraction, Reciprocal Lattice, Miller Indices (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. Conventional fcc cube of edge $a$; the primitive vectors $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ are drawn from a corner atom to the three adjacent face-centre atoms (Figure P1, whose origin is the hidden back corner where the dashed $x$, $y$, $z$ axes meet: $\mathbf{a}_1$ points to the centre of the $y$–$z$ face ($x=0$), $\mathbf{a}_2$ to the centre of the $x$–$z$ face ($y=0$), $\mathbf{a}_3$ to the centre of the $x$–$y$ face ($z=0$)). Close-packed atomic volume $V_{atom}=\dfrac{\pi\sqrt2}{24}a^3$ (standard hard-sphere volume for the fcc touching radius).

Find. (a) unit cell vs. primitive cell; (b) $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$; (c) packing fraction; (d) $\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3$; (e) Miller indices $(hkl)$ of the shaded plane.

[Figure not reproduced: Fig. 1 — Primitive rhombohedral cell of the fcc lattice, built from a corner atom (blue) to its three nearest face-centre neighbours (red), redrawn in the same projection as Figure P1: $\mathbf{a}_1=\tfrac{a}{2}(\hat y+\hat z)$, $\mathbf{a}_2=\tfrac{a}{2}(\hat x+\hat z)$, $\mathbf{a}_3=\tfrac{. See the official exam paper.]

Approach. (a) recall the defining property of each cell type; (b) read the vectors directly off Figure P1; (c) compute $V_{cell}=|\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)|$ and divide the given $V_{atom}$ into it; (d) apply the paper's own reciprocal-vector formula, eq. (5); (e) identify the lattice points at the corners of the grey area, find the plane through them, and take the reciprocals of its axis intercepts.

1. Part (a) — unit cell vs. primitive cell. A unit cell is any volume that, repeated by lattice translations, fills all space and reproduces the crystal; it may contain more than one lattice point (the conventional fcc cube has 4). A primitive cell is the smallest such volume — it contains exactly one lattice point and is spanned directly by a set of primitive translation vectors $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ (eq. 1, $\mathbf{T}=u_1\mathbf{a}_1+u_2\mathbf{a}_2+u_3\mathbf{a}_3$, generates every lattice point from these).
2. Part (b) — primitive vectors from Figure P1. Each vector runs from the corner atom to the centre of the adjacent face: $$\mathbf{a}_1=\frac{a}{2}(\hat y+\hat z),\qquad \mathbf{a}_2=\frac{a}{2}(\hat x+\hat z),\qquad \mathbf{a}_3=\frac{a}{2}(\hat x+\hat y)$$ $\boxed{\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3\text{ as above}}$. (Reading tip: in Figure P1 the $x$ axis runs toward the lower left, $y$ to the right and $z$ up from the hidden back corner, so $\mathbf{a}_1$, drawn up and to the right, lies in the $x=0$ plane, and $\mathbf{a}_2$, drawn up and to the left, lies in the $y=0$ plane.) These form a right-handed set: $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)=+a^3/4$.
3. Part (c) — packing fraction. The primitive-cell volume from eq. (4): $$V_{cell}=|\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)|=\frac{a^3}{4}$$ (consistency check: 4 primitive cells per conventional cube of volume $a^3$, matching fcc's 4 lattice points per conventional cell). The packing fraction is the given atomic volume divided by the cell volume: $$\text{PF}=\frac{V_{atom}}{V_{cell}}=\frac{\pi\sqrt2/24\ a^3}{a^3/4}=\frac{\pi\sqrt2}{6}$$ $\boxed{\text{PF}=\dfrac{\pi\sqrt2}{6}=0.7405\ (74.05\%)}$ — exactly the textbook fcc close-packing fraction, which confirms $V_{atom}=\tfrac{\pi\sqrt2}{24}a^3$ is the correct (not the misprinted $\tfrac{\sqrt2}{12}a^3$) hard-sphere volume.
4. Part (d) — reciprocal vectors via eq. (5). Using $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)=a^3/4$ from Part (c) and $\mathbf{b}_i=2\pi\,(\mathbf{a}_j\times\mathbf{a}_k)/V_{cell}$ (cyclic): $$\begin{aligned} \mathbf{b}_1&=\frac{2\pi}{a}(-\hat x+\hat y+\hat z)\\ \mathbf{b}_2&=\frac{2\pi}{a}(\hat x-\hat y+\hat z)\\ \mathbf{b}_3&=\frac{2\pi}{a}(\hat x+\hat y-\hat z) \end{aligned}$$ $\boxed{\mathbf{b}_1,\mathbf{b}_2,\mathbf{b}_3\text{ as above}}$ — this is exactly the primitive-vector set of a bcc lattice of cube edge $4\pi/a$, the standard result that the reciprocal of fcc is bcc.
5. Part (e) — Miller indices of the shaded plane. The grey area in Figure P1 is the rhombus-shaped face of the primitive cell whose corners are the lattice points $\mathbf{a}_3=(\tfrac a2,\tfrac a2,0)$ (bottom-face centre), $\mathbf{a}_1+\mathbf{a}_3=(\tfrac a2,a,\tfrac a2)$ (centre of the $y=a$ face), $\mathbf{a}_1+\mathbf{a}_2+\mathbf{a}_3=(a,a,a)$ (the front top corner of the cube) and $\mathbf{a}_2+\mathbf{a}_3=(a,\tfrac a2,\tfrac a2)$ (centre of the $x=a$ face). It is the face parallel to $\mathbf{a}_1$ and $\mathbf{a}_2$, so its normal is $\mathbf{a}_1\times\mathbf{a}_2\propto(1,1,-1)$, and all four corners satisfy $$x+y-z=a$$ The plane therefore cuts the axes at $x_0=a$, $y_0=a$, $z_0=-a$; the reciprocals of the intercepts (in units of $a$) are $(1,1,-1)$, already the smallest integer triple. (The plane does not pass through the tips of $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$: those three points lie on $x+y+z=a$, a different plane, so reading the grey area as that triangle would give $(111)$.) $\boxed{(hkl)=(11\bar1)}$ — one of the four close-packed $\{111\}$ planes of the fcc structure, consistent with it being a face of the fcc rhombohedral primitive cell.