# Confusion about average KE in free electron model via Density of States (DOS)

Assume a low temperature regime in which levels up to the Fermi Level, $$E_F$$, are populated.

I have evaluated the density of states in energy space as $$D(E)=\frac{L^3}{\pi^2\hbar^3}(2m_e^2E)^{1/2},$$

and the Fermi Energy as $$E_F=\frac{(3\pi^2n_e)^{2/3}}{2m_e}\hbar^2,$$

where symbols have usual meanings and $$n_e=\frac{N}{L^3}$$. So far, so good. Now, I wish to evaluate the average kinetic energy of the electrons. Intuitively, I would integrate $$\int_0^{E_F}E\times D(E) dE$$ which would give me the expectation value of kinetic energy $$\equiv \left< E_K\right>$$ - or so I thought. Apparently, I need to divide by the number of molecules $$N$$ once more, such that $$\left< E_K\right> = \frac{\int_0^{E_F}E\times D(E) dE}{\int_0^{E_F} D(E) dE}.$$ I can understand that to find the average energy of an electron, we need to divide through by $$N$$. But then physically, what is the integral in the numerator above? As far as I remember, for other distribution functions, the expectation value or average of a quantity was only this integral in the numerator. Have I misunderstood something basic? Would be grateful if someone could quickly and logically clear this up for me.

Did some more digging on this and answered myself. The integral in the numerator in fact determines the average energy for the entire system, in other words, the internal energy $$U$$.
By analogy relating to thermodynamics and kinetic theory for a monatomic ideal gas, each molecule is said to have a kinetic energy $$E_K=\frac{3}{2}k_BT$$ and internal energy $$U$$ is defined as $$U=\frac{3}{2}Nk_BT$$.