This experiment focused at the luminescence from three semiconductors and their heterostructures. The experiments used Ar+ lasers as the source of power to excite the luminescence, and a double spectrometer to determine the spectral shape of the luminescence.
For each spectrum, the energy of the emitted photon vs. The intensity of the photons was plotted with the peak being the largest areas of photon emission. This peak of the spectrum is also considered as the transition energy (CB=VB)
The photon also has a cosecant drive thereby creating a spectrum with a linear energy. Therefore since α sin θ=mλ
Therefore any instrument that can scan linearly when in sin θ will only be linear when λ. additionally, any instruments that can linearly in cosec θ=1/sin θ will also be linear in its energy. This is why the unit of energy in the instrument with sin θ is 1/λ, whereby λ is in cm.
A better example is of E4880=20492cm-1
It is important to note that the software was tweaked in such a way that it runs with the energy axis can read the energy separations under the laser energy. In this case therefore, changes in the peak energy causes changes in the absolute transition energy. For exempla, when the weak energy is 700cm-1
The absolute energy si as indicatedn in the calculation bellow;
from the above, it is qute clear that the intensiyt scale will only be read in uin terms iof photons per second. Therefore, for 20kscale spectra reading of 1000 on the vertical axis directly cotrepspjnds to 20,000 photons persecond
Figure 1: schematic diagram of the energy profile of conduction and valence bands in GaAs/AIGaAs heterostructures. The horizontal axis is the growth direction z.
From the experiment, the lowest energy level is found in the GaAs layer for both the electrons irrespective of the sources, as they will finally merge in the GaAas layer. It is also important to note the recombinations and the energies of emission will be characteristically GaAs type.
It is also important to note that in case the deBrogolie wavelength of the electron is longer that the width of the GaAs layer, the potential barriers of the AIGaAs layer will be detected especially when: LZ=<500λ in GaAs
Because of the layers are very wide in both the x, and y-axis, it is important to note that the energy levels in that direction will be huge. This cases the electron to have discrete quantized energies noted with the quantum number n.
In many solids, the effective potential detected by electrons can inherently affect the motion as they change the mass of the electrons. On the other hand, Holes tend to have different effective masses because of the varying number of valence bands each with its own hole mass.
Of the hole masses, the heavy hole mass is estimated at = 0.80me. the quantized energies Qn of each mass is different for all the electron and the heavy metals denoted as Qn,e, and Qn,e respectively, as well as the light holes denoted by Qn,l these variations are caused by varying m*. Therefore, the emission energy in luminescence required to transits from the nth quantized electron state to the nth quantized heavy whole state can be determined using the formula below:
E (n, h0=E (GaAs gap) +Qn, electron+Qn, heavy hole
Therefore, at 300k, the heavy hole will be equal to=
The key peak seen directly corresponds to the n=one electron to heavy whole transition E (1h), the smallest peak is the (electron to light whole transition, while the small peak on the low energy side can be attributed to the impurity and thus is negligible.
By scaling the energies of E (1h), and E (1l) and the equation above, determine the effective mass of the light hole
Multiwall GaAs/ AIGaAs sample
For this sample, an isolated quantum well of five different widths grown directly above the other are used tom determine the luminescence spectrum. Out of this will arise four different E (1h) peaks and possible other four E (1l) peaks. Based on the luminance spectrum, you are required to determine the energies of all peaks then plot the energies of the transition as a function of the temperature below: 50, 70, 90, and 110 for all the transitions. Then plot the diagram and obtain the curve that best describe the energy of transition as a major function of the formula E (T) given by E (T) =E (T=0)-
For all values, E in eV, and T in K. α=5.4×10-4ev/k, and β=204k. Calculate E (t=0) for all peaks, then plot the mission energies obtained for all the energy peaks as a function of the formula 1/L2z