Hello,
If I get UV-vis absorption peat at 340 nm for a semiconductor and if I excite the material with 290 nm wavelength, where shall I expect the PL peak in emission spectrum assuming no other defect state present in sample?
What is the concept behind that ?
If there is no other defects in the crystal (ideal crystal) then for a direct band gap semiconductor only conduction band to valence band transition related peak will be observed in the PL spectrum. But for that excitation energy should be higher than the band gap which you have mentioned in question. For a semiconductor with defects you also can find the band gap energy (the highest energy peak) form PL. It will depends on the material you are using
if there is any peak at all, which is not guaranteed, it should correspond to the (direct) bandgap of your material. Please read more books
Excitation with higher energy will take the electron to higher excited state energy level. As per Kasha's rule, excess energy will be quickly dissipated (internal conversion) and electron will return to lowest vibrational level of excited state S1 and will then relax to the ground ground state. So there will be no change in the position of emission peak. Only thing expected is the change in intensity of the emission peak.
If there is no other defects in the crystal (ideal crystal) then for a direct band gap semiconductor only conduction band to valence band transition related peak will be observed in the PL spectrum. But for that excitation energy should be higher than the band gap which you have mentioned in question. For a semiconductor with defects you also can find the band gap energy (the highest energy peak) form PL. It will depends on the material you are using
With the details provided, here is an explanation
There are two competing mechanisms that decide the fate of carriers in direct band gap semiconductors. One is the fast carrier thermalization where carriers lose energy to the surrounding
matrix (non-radiative) and come to the bottom of conduction band and the second is energy loss due to emission of a photon (radiative). The coupling of the
matrix element between the levels decides how fast each of these processes occur. Since the value corresponding to former process is high, this process occurs very
rapidly and the carrier comes down rapidly and accumulates at the bottom of the conduction band and later losing a photon the carrier jumps back to the valence band.
This is the reason for observing emission at 340 nm even when the sample was excited at 290 nm. Some energy is lost to the surroundings hence the energy shift or wavelength shift.
Thermalization occurs in indirect band gap semiconductors also but there will not be any luminescence there.
This happens in organic molecules also, it is referred to as Stokes' shift.
The feature you are seeing may be due to excitonic absorption and such features are more prominent in organic semiconductors rather than inorganic semiconductors due to differences in excitonic binding energy (1 eV vs 10 meV).
If your material is an organic semiconductor material, you will see emission at 340 nm itself corresponding to excitonic band gap.
If your material is an inorganic semiconductor then the band gap corresponds to the energy at which the onset of absorption occurs and this can be obtained easily from Tauc's plot. So emission in an inorganic semiconductor will occur at >340 nm becoz the band onset will be at longer wavelengths (lower in energy) to the excitonic peak.
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