# Density of states of electrons

The problem is as follows:

Let the density of states of the electrons in some sample be assumed to be a constant D for $\epsilon > 0$ ($D=0$ for $\epsilon<0$) and the total number of electrons be equal to N. Calculate the Fermi energy $\epsilon_0$ at $T=0$ K.

I'm not sure what expression to use for density of states, is it $dn = \frac{dNV}{f} = D$ where $f=\frac{1}{e^{(\epsilon-\mu)/kT} + 1}$ and $dN = \frac{4\pi p^2 dp}{h^3}$?

Just remember how you calculate the total number of electrons, and work backwards from there. Number of electrons, $N$, equals the number of states for a given energy (D) times the probability that an electron will occupy a state at that energy ($f$), integrated over energy. That is, $N=\int D(\epsilon)f(\epsilon)d\epsilon$. In this case, D is just a constant and $f$ is a step function since $T=0$ (but, ask yourself, at what energy does $f$ step?). See where you get from there.
• Ok, I find $\epsilon_0 = N/D$, is that correct? – user2132672 Aug 30 '17 at 1:43