I want to draw the energy ($E$) diagrams for a simple cubic cell of parameter $a$, where each atom provides two electrons for the almost free electron levels for planes [100], [110] and [111].
I calculate $k$ for the first Brillouin zone:
$k_{\text{ first Brillouin zone}}=\frac{\pi}{a}=\frac{3,14}{a}$
I calculate $k_{\text{Fermi}}$:
$n=\frac{2}{a^3}$
$k_{F}=(3\pi^2n)^{1/3}=\big(3\pi^2\frac{2}{a^3}\big)=\frac{(6\pi^2)^{1/3}}{a}\approx \frac{3,90}{a}$
I calculate $k$ for the plane [100]:
$k_{x}=\frac{\pi}{a}\hat{x}=\frac{3,14}{a}$
$k_{[100]}=\frac{\pi}{a}=\frac{3,14}{a} = k_{\text{ first Brillouin zone}}$
I calculate $k$ for the plane [110]:
$\vec{k}=\frac{\pi}{a}\hat{x}+\frac{\pi}{a}\hat{y}$
$k_{[110]}=\frac{\sqrt{2}\pi}{a}$
I calculate $k$ for the plane [111]:
$\vec{k}=\frac{\pi}{a}\hat{x}+\frac{\pi}{a}\hat{y}+\frac{\pi}{a}\hat{z}$
$k_{[111]}=\frac{\sqrt{3}\pi}{a}$
Finally I made the drawings of the energy bands for each plane:
I understand why in the first zone of Brillouin there is a discontinuity as in the case of plane [100], where the material is conductor, but I do not understand for the planes [110] and [111].
As for planes [110] and [111], $k$ is greater than $k$ of the first Brillouin zone, should there also be in the drawing a discontinuity in the first Brillouin zone?
Why is it just a single semi-filled band?
Are they insulators or conductors in this case (planes [110] and [111])?
