Engineering What Is the Bandgap of a Semiconductor with Light Emission at 326.3 nm?

[jisbon]

Homework Statement

A semiconductor material is characterized by photoluminescence (PL)
measurement, light emission at the wavelength of 326.3 nm due to band-to-
band recombination is detected. In addition, radiation due to the transition
from the conduction band to an impurity level at the wavelength of 885.7 nm is
also observed. Calculate the bandgap (in eV) of the semiconductor, and
determine the energy (in eV) of the impurity level with respect to the valence
band. The wavelength of light for PL excitation is 280 nm.

Relevant Equations

Eg=1.24/wavelength

Just wanted to check my understanding as well as my attempted answers here:

Since light emission at the wavelength of 326.3 nm due to band-to-band recombination is detected, this means that the electron falls down to the valence band to recombine with a hole (to occupy an empty state close to Ev) and emit a photon with energy close to Eg, hence 1.24/0.3263 = 3.8eV is the bandgap of the semiconductor?

For part 2, since the electron is captured by an energy level induced by impurities such as the energy level from a donor (ED), energy = 1.24/0.8857 - 1.4eV?

Thanks.

 

[trurle]

Just wanted to check my understanding as well as my attempted answers here:

Since light emission at the wavelength of 326.3 nm due to band-to-band recombination is detected, this means that the electron falls down to the valence band to recombine with a hole (to occupy an empty state close to Ev) and emit a photon with energy close to Eg, hence 1.24/0.3263 = 3.8eV is the bandgap of the semiconductor?

For part 2, since the electron is captured by an energy level induced by impurities such as the energy level from a donor (ED), energy = 1.24/0.8857 - 1.4eV?

Thanks.

Correct.