# How are the q-states in graphene spaced?

So I've (relatively) recently been studying the electronic states in solids, where I learned that the spacing of states in the energy dispersion relation in periodic boundary conditions is $\Delta k=\frac{2\pi}{L}$, where $L$ is the length of the dimension of that solid in question.

I am now studying graphene, where we derived the following energy dispersion relation:

I also know that the energy is is given by $E_{\vec{q}}=\hbar v_F|q|$, where $\hbar v_F = 3Aa/2$, where $A=2.8eV$ and $a$ is the lattice spacing in graphene. (Note that this is the dispersion relation around the Dirac points, where it is appropriate to approximate it as a linear function.)

I am trying to find an expression for the number of q-states. Considering a Fermi circle of states (since we are in 2D), I have $$\frac{N_q}{2}=\frac{\pi q^2}{(\Delta q)^2},$$ where I divided by two to account for electron spin. This is analogous to doing $$\frac{N_k}{2}=\frac{\pi k^2}{(\Delta k)^2}=\frac{\pi k^2}{(2\pi /L)^2}$$ for a 2D square arrangement of atoms.

Can anyone help me in determining $\Delta q$?

EDIT:

They are spaced the same as in normal materials. I.e. in period boundary conditions, $$\Delta q = \frac{2\pi}{L}$$