Question

Calculate the Fermi energy for noninteracting electrons in a two-dimensional infinite square well. Let σ be the number of free electrons per unit area.

Short answer

The Fermi energy for electrons in a two-dimensional infinite square well is

$\therefore E_F=\frac{\pi {\sout{h}}^2\sigma }{m}$

Step-by-step solution

Definition of Fermi energy of electron

The greatest energy that an electron may hold at 0K is known as the Fermi energy.

Equation 5.50

$$\begin{gathered} E_{n_x{n}_y}=\frac{{\pi }^2{\sout{h}}^2}{2m}\left(\frac{n_x^2}{l_x^2}+\frac{n_y^2}{l_y^2}\right) \\ =\frac{{\sout{h}}^2{k}^2}{2m},\mathrm{with} \\ k=\left(\frac{\pi n_{\mathit{x}}}{l_{\mathit{x}}},\frac{\pi n_{\mathit{y}}}{l_{\mathit{y}}}\right) \end{gathered}$$

Calculating the Fermi energy for electrons in a two-dimensional infinite square well

Each state is represented by an intersection on a grid in k-space”-this time a plane-and each state occupies an area ${\pi }^2/l_x{l}_y={\mathrm{\pi }}^2/A$( whereA$\equiv l_x{l}_y$ is the area of the well). Two electrons per state means

$$\begin{gathered} E_{n_x{n}_y{n}_z}=\frac{{\sout{h}}^2}{2m}\left(\frac{n_x^2}{l_x^2}+\frac{n_y^2}{l_y^2}+\frac{n_z^2}{l_z^2}\right) \\ =\frac{{\sout{h}}^2{k}^2}{2m} \end{gathered}$$ …(5.50).

$$\begin{gathered} \frac{1}{4}\pi k^2=\frac{Nq}{2}\left(\frac{{\pi }^2}{A}\right),or \\ {k}_F={\left(2\pi \frac{Nq}{A}\right)}^{1/2} \\ ={\left(2\pi \sigma \right)}^{1/2} \end{gathered}$$

where $\sigma \equiv Nq/A$is the number of free electrons per unit area.

$$\begin{gathered} E_F=\frac{{\sout{h}}^2{k}_F^2}{2m} \\ =\frac{{\sout{h}}^2}{2m}2\pi \sigma \\ =\frac{\pi {\sout{h}}^2\sigma }{m} \end{gathered}$$

Thus the Fermi energy for electrons in a two-dimensional infinite square well is

$E_F=\frac{\pi {\sout{h}}^2\sigma }{m}$