Consider a two-dimensional honeycomb lattice with nearest-neighbour (bond) distance \(a\), as shown in the figure below.
\(\dfrac{8\pi^2}{\sqrt{3}\,a^2}\)
Step 1 - Area of the real-space primitive cell. The primitive vectors of the triangular Bravais lattice have magnitudes \(|\vec a_1|=|\vec a_2|=a\sqrt{3}\) and the angle between them is \(60^{\circ}\). Hence the area of the real-space primitive cell is
\[A_{\text{cell}}=|\vec a_1\times\vec a_2|=|\vec a_1|\,|\vec a_2|\sin 60^{\circ}=(a\sqrt{3})(a\sqrt{3})\cdot\tfrac{\sqrt{3}}{2}=\tfrac{3\sqrt{3}}{2}\,a^{2}.\]Step 2 - Area of the first Brillouin zone. For any two-dimensional Bravais lattice the area of the first Brillouin zone is
\[A_{\text{BZ}}=\frac{(2\pi)^{2}}{A_{\text{cell}}}=\frac{4\pi^{2}}{\tfrac{3\sqrt{3}}{2}a^{2}}=\frac{8\pi^{2}}{3\sqrt{3}\,a^{2}}.\]Step 3 - Match with options. The value \(\dfrac{8\pi^{2}}{3\sqrt{3}\,a^{2}}\) corresponds to option (C).
Answer: (C) \(\dfrac{8\pi^{2}}{3\sqrt{3}\,a^{2}}\).
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