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Re^6: How to check for acceptance of Kasha's rule from TDDFT data...?

Siddheshwar Chopra
sidhusai@gmail.com

Dear Pavlo,
Pardon me if I am not clear in my question. Perhaps I am really in deep need of understanding. By de-excitation I mean a vertical transition from S1 to S0 state. For more details please have a look at following absorption and emission data:

EMISSION CALCULATION:: ISTATE=1, NSTATE=3

EXCITED STATE   1  ENERGY=     -1355.7217994424  S = 0.0  SPACE SYM = A

EXCITED STATE   2  ENERGY=     -1355.6780193460  S = 0.0  SPACE SYM = A

EXCITED STATE   3  ENERGY=     -1355.6756729663  S = 0.0  SPACE SYM = A

UV CALCULATION:::: ISTATE=1, NSTATE=3

EXCITED STATE   1  ENERGY=     -1355.7080939866  S = 0.0  SPACE SYM = A

EXCITED STATE   2  ENERGY=     -1355.6840777343  S = 0.0  SPACE SYM = A

EXCITED STATE   3  ENERGY=     -1355.6782292662  S = 0.0  SPACE SYM = A

Can you explain the absorption process and emission process in terms of energy levels? I would be glad if you could just draw the energy transitions for both processes. Also when I am obtaining the emission spectra, the TERM "EXCITED STATE" stands for what? Are these the SO vibrational levels now? If yes, then IS THE TRANSITION FROM S1 TO "n" S0 VIBRATIONAL LEVELS IS STUDIED OR OBTAINED HERE?
I hope I am clear this time..

Kind Regards,

On Fri Sep 26 '14 0:27am, Pavlo Solntsev wrote
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>To be honest, i have no idea what you are trying to ask. What do you call de-excitation?

>I really recommend you to play with small molecule first to understand how it works.
>Do a search on this forum. People already discussed a calculation of luminescence.

>-Pavlo.
>
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>On Thu Sep 25 '14 2:18pm, Siddheshwar Chopra wrote
>--------------------------------------------------
>>Dear Pavlo,

>>Please have a look at the energies for 3 excited states from absorption and emission runs:
>>EMISSION CALCULATION:: ISTATE=1, NSTATE=3

>> EXCITED STATE   1  ENERGY=     -1355.7217994424  S = 0.0  SPACE SYM = A

>> EXCITED STATE   2  ENERGY=     -1355.6780193460  S = 0.0  SPACE SYM = A

>> EXCITED STATE   3  ENERGY=     -1355.6756729663  S = 0.0  SPACE SYM = A
>>
>>
>>
>>UV CALCULATION:::: ISTATE=1, NSTATE=3

>> EXCITED STATE   1  ENERGY=     -1355.7080939866  S = 0.0  SPACE SYM = A

>> EXCITED STATE   2  ENERGY=     -1355.6840777343  S = 0.0  SPACE SYM = A

>> EXCITED STATE   3  ENERGY=     -1355.6782292662  S = 0.0  SPACE SYM = A

>>
>> IN ALL THE TRANSITIONS... EXCITED STATE 1 HAS THE LOWEST ENERGY...BE IT ABSORPTION OR EMISSION...So for emission, if the initial state is S1,
>>then what are these excited states? Ideally speaking in this run, S1 is the ground state. The excited state 1 has the lowest energy (-1355.7217994424). Then it should be the lowest vibrational level of S0.

>>Kind Regards,

>>On Thu Sep 25 '14 6:47am, Siddheshwar Chopra wrote
>>--------------------------------------------------
>>>Dear Pavlo,
>>>That is some real relief... Actually the same under study is a doped one. Sir please read my last message and answer them. The picture of Emission process is unclear to me, with regard to the firefly calculations. The ISSTATE=1 is ENSURING that we are finding a transitions from S1 only..Is it correct?
>>>So now SPECIFICALLY about the emission calculations::: What are these EXCITED STATE 1, 2 and so on (depending on NSTATE vaue)? I mean are these the S0 vibrational levels now (because now we are studying de-excitation)? If yes, then how exactly do I find the emission wavelength? Is the analysis similar to the absorption spectra analysis then? I mean using osc. str. etc.? Please guide me.
>>>Also if Kasha's rule holds good, then do I get a high osc. str. at 1st excited state itself in the o/p file?

>>>Regards,

>>>On Thu Sep 25 '14 1:42am, Pavlo Solntsev wrote
>>>----------------------------------------------
>>>>Kasha's rule doesn't always work. For example, if you have redox active group near chromophore.

>>>>Try to figure out absorbtion--emission situation for a porphyrin.

>>>>-Pavlo.
>>>>
>>>>
>>>>
>>>>On Wed Sep 24 '14 9:11pm, Siddheshwar Chopra wrote
>>>>--------------------------------------------------
>>>>>Dear users,
>>>>>Somebody please help me in understanding this concept. I need to know that by setting istate =1 in emission spectra calculation, I am actually setting or calculating the de-excitation from the S1 state. Am I correct? If yes, then the TDDFT o/p gives me a set of transitions possible for every excited state (set by variable nstate).What are these nstates now? Do they hold any significance in S1-->S0 transition? Do we mean that these are the vibrational levels of S0 level? Please explain this in the light of kasha's rule.

>>>>>Kind Regards,
>>>>>
>>>>>
>>>>>n Tue Sep 23 '14 12:21pm, Siddheshwar Chopra wrote
>>>>>---------------------------------------------------
>>>>>>Dear Users,
>>>>>>Please have a look at the attached file which contains a table describing the contributions/transitions from first four states only for both the absorption and emission calculations. For absorption, I understand that the final state can be any of the excited states which has highest osc. strength. Am I right? But my doubt is regarding kasha's rule... If you see the emission data, the first excited state has a very low oscillator strength of 0.004 (HOMO-->LUMO(97‰ at 2678.9 nm. However the higher states have a better contribution than this state. So please tell me how to analyse this properly. To my knowledge emission wavelength is around 600-730nm.
>>>>>>I would be grateful for any help.

>>>>>>Kind Regards,

Fri Sep 26 '14 2:38pm