is interesting, I think that relation is the nature radiation ionizing

The optically detrapped electrons go to conduction band or to higher excited states of carbon defects? Which level is emitting?

Dear  Roman, The optically detrapped electrons go to conduction band and then recombine at the luminescence centers (F/F+ in case of Al2O3:C). Though the exact origin of traps in Al2O3:C are still not known but their thermal trap depth are well tabulated. 

One possible answer of the above query is that the transfer of energy optically to the defect is not  as straight forward as in case  by thermal way. The term "photoionization cross-section"  which is the basic thing for any trap to accept optical energy, doesn't depend on optical energy only, it also depends on others factors such as effective mass of electron and optical ionization threshold. 

You should ask Regina DeWitt. She´s a specialist for Al2O3:C and OSL traps. Other authors linking the OSL and Thermal traps (especially in Quartz) are e.g. Richard Bailey or Andrew Murray. Some traps are directly linked, but not all of them, which is that not for every thermal trap you find a corresponding optical one. Especially towards higher temperatures, if I remember it correctly.

It seems to me that there are two questions here. In answer to the top question, the trap depth is the same whether sensed optically or thermally. The stimulation method will not cause the trap depth to change. The answer to the sub-question is rather complicated but Anuj’s suggestion is on the button. The probability of optical ionisation is related to the photoionisation cross-section. This parameter is not continuous with photon energy but has a threshold. Thus to ionize a trap optically, one needs to input energy equal to or better still, above the above the photoionisation threshold.

Dear Chithambo Sir, 

                                 You are right, trap depth will remain same whether we sense it optically or thermally. As far as the second question is concerned, we have enough examples to prove what you and me suggested about the optical ionization threshold. For example of Feldspar, it can be optically stimulated by Iinfrared (~1.5 eV) though the trap depth is also ~ 1eV. So it is a function of threshold energy. Further, in case of CaSO4:Dy, it is also having trap depth ~ 1.2 eV, but it doesnt get stimulated even with the optical energy of 2.6 eV, suggesting the threshold value is too hight for it or in otherwords, the photoionization cross-section for it is very poor for visible wavelength.

I was not aware of the case of CaSO4:Dy that you mention.  This is interesting.  In a related case, it was thought for a long time that you couldn't optically stimulate CaCO3 but this was achieved with careful filtering of the stimulating light.  I wonder if a similar method would work for CaSO4:Dy as well.

Anuj, you have to think in terms of configuration coordinate diagrams. Thermally one can acccess the difference between the two potential minima, i.e. the ground state and the excited state (or conduction band in this case). However, since they are often displaced spatially, one needs higher optical energy to reach the excited state potential. The latter is therefore the 'true' trap depth, and E(thermal) typically less than E(optical). It becomes confusing because one calls both of these as trap depth; in my view one should use 'thermal activation energy' for the thermal process and 'trap depth' for the optical..

Dear Dr. Soni talking of thermal trap depth and the optical trap depth as same can create a lot of confusion. Because in an optical excitation method we are measuring the optical energy needed to excite the electron from the trap to the conduction band and in a thermal excitation we are measuring the thermal energy required to excite the electron from the same trap to the conduction band. Thus we are measuring the same trap by two different units i.e optical energy and thermal energy. We can quantify them only if we find a relation between the two. It is same as measuring a length by a meter scale and a centimeter scale.

Thanks a lot McKeever Sir for recommending my point of view.