# How do I calculate the surface energy density from Fermi's energy?

I have the following question:

Consider the free eletron model in 2D to answer:

Calculate an expression for the fermi energy and for the surface energy density at T= 0 K, expressed in function of the fermi energy, supposing we have 1 eletron per Angstrom squared $$\require{mediawiki-texvc}$$ ($$1 \AA {}^{-2}$$)

What i did was :

$$E_f=\frac{\hbar^2k_F^2}{2m}$$

we know $$n=\frac{1}{(10^{-10})^2}m^{-2}$$ so we want to express $$k_f$$ in function of $$n$$ in 2D.

Total area of states is the circumference $$\pi k_f^2$$. The area of one state is $$\left(\frac{2\pi}{L}\right)^2$$, so number of allowed states is $$N=\frac{\pi k_f^2}{\left(\frac{2\pi}{L}\right)^2}=\frac{k_f^2}{4\pi}L^2$$

Accounting for the spin degeneracy we multiply by $$2$$ and have $$\frac{N}{L^2}=\frac{k_f^2}{2\pi}\leftrightarrow n=\frac{k_f^2}{2\pi}\leftrightarrow k_f^2=2\pi n$$

So that $$E_f=\frac{\hbar^2\pi n}{m}$$

At $$T=0$$ the energy of the system should be $$E_f$$ right? but how do I calculate the surface energy density. Do I divide by the total area? I don't understand what to do next.

It's more understandable if you use standard names. The first step that you did was to calculate the density of states of 2D free electrons defined by: $$n = \int_0^{E_f} D(E)dE$$ Translating your approach, it is more intuitive to use the density of states in wave number (taking into account the spin 1/2 degeneracy): \begin{align} D(k)dk &= \frac{kdk}{\pi} \\ D(E)dE &= \frac{2m}{\hbar^2}\frac{1}{2\pi}dE \\ &= \frac{m}{\pi\hbar^2}dE \\ E_F &= \frac{\pi\hbar^2n}{m} \end{align}
Remember that $$E_F$$ is not the energy of every electron. It is rather the maximal possible energy of the electrons. All the energies from $$0$$ to $$E_F$$ are populated according to the pdf: $$p(E)dE = \frac{D(E)}{n}dE$$ In particular, the average energy per area is: $$\bar E = \int_0^{E_F}D(E)EdE$$ and will necessarily be less than $$E_F$$. In your case, it is: $$\bar E = n\frac{E_F}{2}$$ i.e. on average, an electron has energy $$E_F/2$$. Note that this is less than $$E_F$$, but not too far off since the density increases with energy, which is why higher energies are more heavily weighted in the average.