Question 1 of 7: Bravais Lattices — Naming, Packing Fraction, Reciprocal Lattice, Miller Indices

Question 1: Bravais Lattices — Naming, 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. Three cubic Bravais lattices of identical edge length $a$: Figure P1a is a cube with atoms at the 8 corners only; Figure P1b is the same cube plus one atom at the body centre, with a grey diagonal plane through the corner atoms at $(0,0,0),(0,a,0),(a,a,a),(a,0,a)$ (and through the body-centre atom), read off the printed drawing with the origin where the three axes meet; Figure P1c is the same cube plus one atom at each of the 6 face centres.

Find. (a) the name of each lattice; (b) the packing fraction of P1c; (c) the primitive direct and reciprocal translation vectors of P1b; (d) the Miller indices of the plane family parallel to the grey area in P1b.

[Figure not reproduced: Fig. P1a Fig. P1b Fig. P1c Fig. 1 — the three cubic Bravais lattices at the same edge length $a$: simple cubic (corner atoms only; face $x=a$ shaded as printed), body-centred cubic (corners + 1 body-centre atom, grey plane $x=z$ shown), face-centred cubic (corners + 6 face-centre atoms). See the official exam paper.]

Approach. Identify each lattice by its atom decoration; apply the hard-sphere touching condition for the packing fraction; build the primitive vectors geometrically from the BCC cell and invert them with the paper's own eq. (6) for the reciprocal lattice; read the grey plane's intercepts for the Miller indices (parallel planes share the same $(hkl)$).

1. Part (a) — identify the three lattices. Figure P1a decorates only the 8 cube corners: this is Simple Cubic (SC). Figure P1b adds one atom at the cube's geometric centre $(a/2,a/2,a/2)$: this is Body-Centred Cubic (BCC). Figure P1c adds one atom at the centre of each of the 6 faces instead: this is Face-Centred Cubic (FCC).
2. Part (b) — FCC packing fraction. In FCC the nearest-neighbour contact is along the face diagonal (length $a\sqrt2$), which spans 4 atomic radii: $4r=a\sqrt2\Rightarrow r=\dfrac{\sqrt2}{4}a$. The conventional cell holds $8\times\tfrac18+6\times\tfrac12=4$ atoms, so $$\text{packing fraction}=\frac{4\cdot\frac{4}{3}\pi r^3}{a^3}=\frac{16\pi}{3}\left(\frac{\sqrt2}{4}\right)^3=\frac{\pi}{3\sqrt2}$$ $$\boxed{\text{packing fraction}=\dfrac{\pi}{3\sqrt2}=0.7405\ (74.05\%)}$$ the densest possible packing of identical spheres (tied with HCP).
3. Part (c) — BCC primitive vectors. Take the origin at a corner atom. Reaching the body-centre atom of the origin cell and of the three adjacent cells along $-x,-y,-z$ respectively gives a standard, symmetric primitive set $$\mathbf{a}_1=\frac{a}{2}(\hat x+\hat y-\hat z),\quad \mathbf{a}_2=\frac{a}{2}(-\hat x+\hat y+\hat z),\quad \mathbf{a}_3=\frac{a}{2}(\hat x-\hat y+\hat z)$$ Their cell volume is $\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)=a^3/2$ (checks: BCC has 2 atoms per conventional cell of volume $a^3$, so 2 primitive cells per conventional cell). Substituting into the paper's eq. (6), $\mathbf{b}_1=2\pi\dfrac{\mathbf{a}_2\times\mathbf{a}_3}{\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)}$ (cyclically for $\mathbf{b}_2,\mathbf{b}_3$): $$\boxed{\mathbf{b}_1=\frac{2\pi}{a}(\hat x+\hat y),\quad \mathbf{b}_2=\frac{2\pi}{a}(\hat y+\hat z),\quad \mathbf{b}_3=\frac{2\pi}{a}(\hat x+\hat z)}$$ — the reciprocal lattice of BCC is itself FCC (with cubic cell edge $4\pi/a$), the standard textbook duality between the two.
4. Part (d) — Miller indices of the plane parallel to the grey area. Reading Figure P1b with the origin at the corner where the $x$, $y$ and $z$ axes meet ($x$ to the right, $z$ up, $y$ drawn receding up-left), the grey area is bounded by the corner atoms $(0,0,0)$ and $(0,a,0)$ (the bottom-left edge running along $y$) and $(a,0,a)$ and $(a,a,a)$ (the top-right edge running along $y$), and it contains the body-centre atom $(a/2,a/2,a/2)$. All five points satisfy $x=z$ ($0=0$, $0=0$, $a=a$, $a=a$, $a/2=a/2$), so the grey area lies in the plane $x-z=0$. That plane passes through the origin, so its intercepts cannot be read directly — which is exactly why the question asks for the parallel plane through the two bottom right-hand atoms, $(a,0,0)$ and $(a,a,0)$. Both satisfy $x-z=a$, and the edge joining them runs along $y$, which lies in the plane. The plane $x-z=a$ cuts the axes at $x_0=a$, $y_0=\infty$ (parallel to $y$), $z_0=-a$. Taking reciprocals in units of $a$ and clearing to the smallest integers, $$\left(\frac1{1},\ \frac1{\infty},\ \frac1{-1}\right)=(1,\ 0,\ -1)$$ $$\boxed{(hkl)=(1\,0\,\bar1)}$$ (equivalently $(\bar1\,0\,1)$, the same plane described with its normal reversed). Miller indices depend only on orientation, so the same label also describes the grey area itself and every plane parallel to it.