Discussion
The Fermi-momentum and dimension of a mono-atomic 2D square crystal are given by kF and L. If each atom is contributing one electron to the Fermi gas, the size of the primitive cell is :
Answer: A
The primitive cell has area \( \frac{2\pi}{k_{F}^{2}} \).
In two dimensions each \( k \)-state occupies an area \( (2\pi/L)^{2} \), and each holds two electrons. Filling a Fermi circle of radius \( k_{F} \):
\[ \begin{aligned} N &= 2 \times \frac{\pi k_{F}^{2}}{(2\pi/L)^{2}} = \frac{k_{F}^{2}L^{2}}{2\pi} \\ \frac{L^{2}}{N} &= \frac{2\pi}{k_{F}^{2}} \end{aligned} \]
With one electron per atom, N is also the number of atoms, so \( L^{2}/N \) is the area of one primitive cell. \( \pi/k_{F}^{2} \) forgets the spin factor of 2, and the options containing L grow with the size of the crystal, whereas the area of one cell cannot depend on L.
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