# 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 [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])?

• 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 [100] starts. Can you kindly explain? – gryphys Jan 3 at 14:54