Discussion
Graphene is a two dimensional material, in which carbon atoms are arranged in a honeycomb lattice with lattice constant \(a\). As shown in the figure, \(\vec{a}_{1}\) and \(\vec{a}_{2}\) are two lattice vectors. Which one of the following is the area of the first Brillouin zone for this lattice?
Answer: A
The area is \(\frac{8\pi^{2}}{3\sqrt{3}a^{2}}\), with \(a\) read, as in the figure, as the carbon-carbon spacing.
The arrows \(\vec{a}_{1}\) and \(\vec{a}_{2}\) in the figure each join one carbon atom to a neighbouring atom, so the length \(a\) here is the C-C distance. The honeycomb's true Bravais lattice is triangular, with primitive vectors of length \(\sqrt{3}a\) at 60° to each other, so the real-space unit cell has area
\[ \begin{aligned} A_{cell} &= (\sqrt{3}a)^{2}\sin 60^{\circ} \\ &= \frac{3\sqrt{3}}{2}a^{2} \end{aligned} \]
The first Brillouin zone has area \((2\pi)^{2}/A_{cell}\):
\[ \begin{aligned} A_{BZ} &= \frac{4\pi^{2} \times 2}{3\sqrt{3}a^{2}} \\ &= \frac{8\pi^{2}}{3\sqrt{3}a^{2}} \end{aligned} \]
If \(a\) were instead the length of the primitive vectors, the same steps would give \(8\pi^{2}/\sqrt{3}a^{2}\); the key follows the figure's labelling. The options with \(4\pi^{2}\) drop the factor 2 that comes from \(\sin 60^{\circ} = \sqrt{3}/2\).
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