# How can I draw the energy bands for the first and second zones of Brillouin? Is it conductor or insulator? [closed]

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 ,  and .

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 :

$$k_{x}=\frac{\pi}{a}\hat{x}=\frac{3,14}{a}$$

$$k_{}=\frac{\pi}{a}=\frac{3,14}{a} = k_{\text{ first Brillouin zone}}$$

I calculate $$k$$ for the plane :

$$\vec{k}=\frac{\pi}{a}\hat{x}+\frac{\pi}{a}\hat{y}$$

$$k_{}=\frac{\sqrt{2}\pi}{a}$$

I calculate $$k$$ for the plane :

$$\vec{k}=\frac{\pi}{a}\hat{x}+\frac{\pi}{a}\hat{y}+\frac{\pi}{a}\hat{z}$$

$$k_{}=\frac{\sqrt{3}\pi}{a}$$

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 , where the material is conductor, but I do not understand for the planes  and .

• As for planes  and , $$k$$ 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  and )?

• sorry that I cannot answer to your question. In any case it is not clear to me how you calculated the energy gap, that is the energy at which the second band for  starts. Can you kindly explain? Jan 3, 2021 at 14:54