Question 1 of 7: Bravais Lattices, BCC Packing, Primitive Vectors, and Miller Indices

Question 1: Bravais Lattices, BCC Packing, Primitive Vectors, and 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. The body-centred-cubic (BCC) lattice of Figure P1: cubes of edge $a$ with one atom at each of the 8 corners plus one atom at the body centre. The figure draws three adjoining cubes (one below the $xy$-plane, two above it) that share a single lattice point at the origin, and the primitive vectors $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ run from that shared point to the body-centre atoms (open circles) of the three cubes; eight corner atoms lie in the top layer at height $z=a$.

Find. (a) the count of 3-D Bravais lattices; (b) the BCC packing fraction; (c) a primitive vector set $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$; (d) the primitive cell volume; (e) the Miller indices of the plane family containing the "top eight" corner atoms.

[Figure not reproduced: Fig. 1 — the BCC conventional cell (blue: corner atoms; red: body-centre atom), redrawn as a single cube. Here $\mathbf{a}_1,\mathbf{a}_2,\mathbf{a}_3$ are drawn from the body-centre atom to three of its corner neighbours; they are the same three displacement vectors that Figure P1 draws from . See the official exam paper.]

Approach. (a)–(b) are direct recall/geometry; (c)–(d) build the primitive vectors from the body-centre-to-corner geometry and take their scalar triple product; (e) reads the plane orientation from the atoms' stacking direction.

1. Part (a) — number of 3-D Bravais lattices. Crystallography admits exactly $\boxed{14}$ distinct three-dimensional Bravais lattices (7 crystal systems, some split into primitive/body-centred/face-centred/base-centred variants) — a standard classification result, not a derived number.
2. Part (b) — BCC packing fraction. In BCC the nearest-neighbour contact is along the cube's body diagonal (length $a\sqrt3$), which spans 4 atomic radii (corner–centre–corner, each contact $2r$ twice): $4r=a\sqrt3\Rightarrow r=\dfrac{\sqrt3}{4}a$. The conventional cell holds $8\times\tfrac18+1=2$ atoms (8 corners shared 8 ways, plus 1 full body-centre atom), so $$\text{packing fraction}=\frac{2\cdot\frac43\pi r^3}{a^3}=\frac{8\pi}{3}\left(\frac{\sqrt3}{4}\right)^3=\frac{\pi\sqrt3}{8}$$ $$\boxed{\text{packing fraction}=\dfrac{\pi\sqrt3}{8}=0.6802\ (68.02\%)}$$ noticeably more open than the FCC/HCP close-packed value of 74.05%, consistent with BCC not being a close-packed structure.
3. Part (c) — primitive translation vectors. Figure P1 takes the origin at the lattice point shared by its three cubes and draws each vector to the body centre of one cube: $\mathbf{a}_1$ down into the lower cube (which extends toward $+x,+y,-z$), $\mathbf{a}_2$ into the upper cube extending toward $-x,+y$, and $\mathbf{a}_3$ into the upper cube extending toward $+x,-y$. Each body centre sits half a cube edge along each axis from the origin, so $$\boxed{\mathbf{a}_1=\frac a2(\hat x+\hat y-\hat z),\quad \mathbf{a}_2=\frac a2(-\hat x+\hat y+\hat z),\quad \mathbf{a}_3=\frac a2(\hat x-\hat y+\hat z)}$$ each of length $\tfrac{\sqrt3}2a$ (half the body-diagonal), matching the three arrows of Figure P1 ($\mathbf{a}_1$ heading toward $-\hat z$, $\mathbf{a}_2$ and $\mathbf{a}_3$ toward $+\hat z$). This is the standard symmetric BCC primitive set (Kittel, Ch. 1).
4. Part (d) — primitive cell volume. The primitive cell volume is the scalar triple product $$V_{\text{prim}}=\mathbf{a}_1\cdot(\mathbf{a}_2\times\mathbf{a}_3)$$ Evaluating with the vectors above gives $V_{\text{prim}}=a^3/2$ — consistent with the cross-check that BCC has 2 atoms per conventional cell of volume $a^3$, i.e. exactly 1 atom (hence 1 lattice point) per primitive cell: $$\boxed{V_{\text{prim}}=\dfrac{a^3}2}$$
5. Part (e) — Miller indices of the "top eight atoms" plane family. The eight black corner atoms at the top of Figure P1 all sit at the same height, $z=a$, one cube edge above the origin (they are the top corners of the two upper cubes plus the corner directly above the lower cube, i.e. every lattice point $(x,y,a)$ of the $3\times3$ top grid except the hidden back one). A plane containing all of them is therefore the horizontal plane $z=a$, normal to $\hat z$, with intercepts $(\infty,\infty,1)$ in units of $a$. Taking reciprocals gives $$\boxed{(hkl)=(0\,0\,1)}$$