Filling of energy bands In the context of electron energy bands for crystals, I have heard the phrase 

There are $x$ electrons per band, therefore the electrons fill $y$
  bands.

I am trying to understand the 'therefore' in that sentence. I believe it could be related to the number of atoms per unit cell, and the number of valence electrons per atom.


*

*What is the rationale behind the statement? How can electrons fill a band? I thought the bands were the eigenvalues of the Bloch wave?

*Can you relate the number of valence electrons of Silicon to the valence bands? Here is a plot of the bands for fcc silicon in case it is helpful:


 
 A: Bloch's theorem states that the eigenvalues and eigenvectors of a periodic one-electron Hamiltonian can be labeled by Bloch's wavevectors $\bf k$ lying in the first Brillouin zone. For each $\bf k$ there is an infinite set of solutions of the eigenvalue problem, labeled by another quantum number, $n$, the band index. The eigenvalues $E_n({\bf k})$ depend on both $n$ and  $\bf k$, being continuous functions of  $\bf k$. Each function  $E_n({\bf k})$ , at fixed $n$, and for $\bf k$ in the first Brillouin zone, is an energy band.
The one-electron eigenstates $\psi^{(n)}_{\bf k}({\bf r})$ can be used to build a totally antisymmetric $N_e$-electron ground state wavefunction by using the lowest lying eigenvectors in a Slater determinant where each spatial state enters twice, once  multiplied by a spin up and then multiplied by a spin down wavefunction.
Filling the n-th band is a pictorial expression equivalent to say that all the $\bf k$ states of the $n$-th band appear in the Slater determinant with double occupation (to take into account the spin)
In a finite periodic crystal (it can be obtained by using 3D periodic boundary conditions (pbc)), there is a finite number of ${\bf k}$ vectors in the first Brilluoin zone, exactly equal to the total number $N$ of cells in the pbc crystal. So, all the $N$ orbital ${\bf k}$ states of a band can accomodate  a maximum of $N_e=2N$ electrons.  
In order to have the lowest possible energy (the ground state), one has to "fill", i.e. to use, the one-electron states, sorting the values of  $E_n({\bf k})$, starting from the lowest values   and using all the states in increasing order of energy until the number of used values of $n$ and  ${\bf k}$ is exactly equal to $\frac{N_e}{2}$ (remember that there are two spin values for each $n$ and  ${\bf k}$).
For insulators and semiconductors, such an ordered filling of one-electron states stops when all the ${\bf k}$ values of a finite number of bands have been used. The first empty state is separated from the highest filled state by a finite gap of energy. Things are more complex in the case of metals because in that case there is no gap neatly separating filled and empty bands and partially filled bands appear.
About question n.2, Silicon, in its equilibrium crystalline structure at normal conditions  is a semiconductor with the diamond structure. The diamond structure is not a pure Bravais lattice but it can described as an fcc lattice with a  2 atoms base. If the pbc finite size crystal contains $N$ cells, there will be $2N$ atoms. Each atom will contribute to the valence bands with $4$ electrons ( $3s^23p^2$ electronic configuration of the atom). In total, the finite size crystal will have $N_e=8N$ valence electrons. Consequently, the number of filled bands will be $4$.
