= Solution
Let the <graphene> <lattice constant> be $a=\sqrt3b$, and choose direct primitive vectors $\mathbf a_1=a(1/2,\sqrt3/2)$ and $\mathbf a_2=a(-1/2,\sqrt3/2)$. Reciprocal primitive vectors satisfying $\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}$ are
$$
\mathbf b_1=\frac{2\pi}{a}(1,1/\sqrt3),\qquad\mathbf b_2=\frac{2\pi}{a}(-1,1/\sqrt3).
$$
The first <Brillouin zone> is a hexagon. Its corners K and K' are inequivalent valleys; M denotes an edge midpoint. In the nearest-neighbour model, $E_\pm(\mathbf k)=\pm t|f(\mathbf k)|$, where $f(\mathbf k)=\sum_{j=1}^3e^{i\mathbf k\cdot\boldsymbol\delta_j}$ and $\boldsymbol\delta_j$ are the three bond vectors. The band separation is $2t|f|$: it \b[vanishes at K/K' and is $2t$ at M], varying continuously along each edge.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2019/paper-2-graphene-zone.png]
{title=Graphene reciprocal primitive vectors and the orientation-dependent band separation on its zone boundary}
{source=paper-2-graphene-zone.py}
This <graphene zone-boundary band separation> is a local conduction-to-valence separation. Ideal <graphene> itself remains globally gapless because the bands touch at the corners. Circumferential quantization in a nanotube selects lines through this two-dimensional band structure; whether they pass through a valley determines the ideal metallic versus semiconducting distinction.
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