Question

Calculate the electron and hole density in a 2D quantum well with a bandgap of 1 . The effective mass of the conduction and valence bands is the free electron mass mo. By assuming that there is only a single subband in the quantum well’s conduction and valence bands,

(a) write down the expression for calculating the electron and hole densities and simplify it as much as possible

Answer 1

The electron and hole densities in a 2D quantum well can be calculated using the constant density of states for a two-dimensional system and considering the Fermi-Dirac distribution. The densities are equal and given by n = p = (mo EF)/(piℏ2), where EF is the Fermi energy.

Step-by-step explanation:

To calculate the electron and hole densities in a 2D quantum well with a bandgap of 1 eV and effective masses equal to the free electron mass mo, we can use the density of states function for a two-dimensional system and integrate it over the energy with the Fermi-Dirac distribution. Since there is only a single subband, the densities are greatly simplified.

The density of states g(E) in a two-dimensional system is constant within the subbands, and is given by:

g(E) = mo / (πℏ2)

where ℏ is the reduced Planck's constant. Now considering the Fermi-Dirac distribution at absolute zero temperature for simplicity, all states below the Fermi energy are filled and all states above are empty. Each electron that transitions into the conduction band leaves behind a hole; therefore, if the conduction band is originally empty, the electron density n is equal to the hole density p.

n = p = (mo EF) / (πℏ2)

where EF is the Fermi energy. This relationship becomes more complex at non-zero temperatures or with additional subbands.