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The fluorine sites form a simple cubic sublattice of spacing $\ell =a/2=3.10\mathring{\mathrm{A}}$. Adopt uncorrelated nearest-neighbour hopping, identify the field-correlation time with the total jump waiting time, and use homonuclear dipolar relaxation: $f_D=c=1$ and $x_{\ast }=0.616$. ThenIf the simplified single-frequency condition ${\omega }_0{\tau }_c=1$ is used instead, the corresponding estimates are $1.01\times 10^{-12}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $1.9\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. These are explicit model-dependent estimates, not a uniquely measured tracer diffusivity: a different correlation-time-to-jump-time conversion or defect correlation factor multiplies them by $f_{D}c$. The lattice constant supplies a jump distance, not proof that all possible neighbouring sites are vacant.

Let the graphene lattice constant be $a=\sqrt{3}b$, and choose direct primitive vectors ${\mathbf{a}}_1=a(1/2,\sqrt{3}/2)$ and ${\mathbf{a}}_2=a(-1/2,\sqrt{3}/2)$. Reciprocal primitive vectors satisfying ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$ areThe 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 $f(\mathbf{k})={\sum }_{j=1}^3{e}^{i\mathbf{k}\cdot {\mathbf{\delta }}_j}$ and ${\mathbf{\delta }}_j$ are the three bond vectors. The band separation is $2t|f|$: it vanishes at K/K' and is $2t$ at M, varying continuously along each edge.

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.