Discussion
For a two-dimensional hexagonal lattice with lattice constant \(a\), the atomic density is
Answer: C
The key's answer is \(\frac{4}{3\sqrt{3}a^{2}}\), which treats the hexagonal lattice as a honeycomb of regular hexagons of side \(a\).
In that reading, atoms sit at the corners of hexagons of side \(a\). Each corner is shared by three hexagons, so one hexagon holds \(6 \times \frac{1}{3} = 2\) atoms, and its area is \(\frac{3\sqrt{3}}{2}a^{2}\):
\[ \begin{aligned} n &= \frac{2}{\frac{3\sqrt{3}}{2}a^{2}} \\ &= \frac{4}{3\sqrt{3}a^{2}} \approx \frac{0.77}{a^{2}} \end{aligned} \]
Strictly, the honeycomb is not a Bravais lattice; the triangular Bravais lattice with spacing \(a\) has density \(\frac{2}{\sqrt{3}a^{2}}\), which is not among the options, so the paper intends the honeycomb arrangement. \(\frac{1}{3\sqrt{3}a^{2}}\) counts only one-third of an atom per hexagon, and the other options use wrong areas.
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