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

Question

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.

Answer 1

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.

Hope this helps.