# Number of electrons within Fermi Surface

This question is regarding problem 9.5 in Ashcroft and Mermin where we have to calculate the radius of the Fermi circle in a 2D square lattice with lattice constant $$a$$ and $$m$$ electrons per primitive cell. This is how I suppose the solution must go:

The area of the Fermi circle is $$\pi k_F^2$$ and the volume of a unit k-space square is $$\frac{(2\pi)^2}{A}$$ where A is the physical area of the square lattice (From the Born-Von Karman boundary conditions). Thus the total number of electrons is $$Nm = 2\frac{\pi k_F^2}{\frac{(2\pi)^2}{A}}= 2\frac{\pi k_F^2A}{(2\pi)^2}=2\frac{\pi k_F^2 Na^2}{(2\pi)^2}$$ $$\Longrightarrow k_F =\sqrt{\frac{m}{2\pi}}$$ Where I expresses $$k_F$$ in units of $$\frac{2\pi}{a}$$ and $$N$$ is the total number of lattice points.

The Question:
In the above solution I assumed that the total area within the Fermi surface was $$\pi k_F^2$$. But since there's a Fermi surface associated with every lattice point, shouldn't the total Fermi area be $$N\pi k_F^2$$? This would lead to a factor of $$N$$ in the denominator in my expression for $$k_F$$ which would make it really small with respect to the reciprocal lattice vector. Is there some point that I am missing?

There is not a Fermi surface associated with every lattice point. The Fermi circle as a concept exists in k-space, and has a fixed center, which is the origin of k-space. The point is that you are filling the circle up to a surface of fixed energy.

No, your original aproach is the correct one. The Fermi surface is in the reciprocal space. I don't understand what you mean by Fermy surface associated with every lattice point. If you care to explain I will add more to the answer.