[Physics] Derivation of existence of energy band gap in semiconductor (solid State)

electronic-band-theoryquantum mechanicsquasiparticlessemiconductor-physicssolid-state-physics

I am looking for both a mathematical and a physical reason for energy band gap in metals. For Physical reason, I was told that at each reciprocal lattice, you could have Bragg scattering, that would cause the band gap, but this does not really make full sense to me.

For a mathematical reason, do we have to solve the Schrodinger equation? Before we can see why?

Similarly these kind of bang gap have been found in a diatomic chain and I have seen the derivation which makes perfect sense.

Best Answer

Consider the fermi-dirac state density distribution:

$ f(E) = \frac{1}{1 + \exp^{\frac{E}{k_{B}T}}} $

Fermi-Dirac Distribution

(Derived from the Boltzmann canonical formulae)

This shows that as the temperature of a system of spin 1/2 particles increases, the density of states lowers itself over more states.

To model the solid-state of a metal, we take the k-space of dimension:

$ V = (\frac{\pi}{L})^3 $

and the fermi-sphere, of radius equal to the fermi wave-vector:

$V = \frac{4}{24} \pi r^{3} $

because $ r = k_{F} $ and add factor of two for each spin +-1/2 of N, number of electrons in volume, then you can get:

$ k_{F} = (\frac{2\pi^{2}V}{N})^{1/3} $

Where N is the number of electrons, generally $ N_{A} \times N_{e} $, V is the volume.

This enables us to find the fermi-energy:

$ E_{F} = \frac{\hbar^{2}k_{F}^{2}}{2m_{e}} $

In the first formula and graph, this energy defines what energy the states reach at 0K.

This distribution of states relates to electricity due to the Band Structure of metals, of the valency and conduction bands. As you can see in the image the fermi-energy is plotted on the axes for different types of metal and DOS along the bottom stands for density of states which is the first formula.