λ excitation of a fluorescent compound different from its λ absorption, it is obvious as when the compound leave/emit energy it would be less than the absorbed energy, corresponding λ would be more, usually.

Have you taken the excitation wavelength of one of the Q- bands.... instead of the Soret band....?

Thanks Mr. Ghosh for your answer.

λ Q1=565 nm and λQ2=608 mn. This porphyrin was zinc metalated so it has two Q bands instead of four.

As you see λexcitation (=458 nm) does not correspond to Q bands absorption nor to the soret band.

So is it normal that excitation wavelength does not match absorption wavelength ??

As a general rule excitation wavelength and absorption peak need not be same. This is due to the fact the emission is independent of excitation wavelength (Kasha–Vavilov rule).

It's common practice in metal complexes and other systems with large stokes shifts to excite on the low energy side of an absorption band. This minimizes the amount of excess excitation energy and possible photochemical reactions.

As to why the excitation was into the soret band rather than Q bands, maybe the Q bands didn't have a high enough absorbance at the concentration the experiment was performed at.

I think your question needed much more info for a better help, so I'm not sure if that is your answer: Are you considering the dependence of the excitation to the lamp profile output?

For instance, the Xenon lamp output is typically much weaker in the UV. Therefore, for a simple example: consider a sample having two absorption bands with the same absorption intensities, one band is in the UV (~300 nm) and the second one in the near-visible (~450 nm); in a fluorimeter with a Xenon lamp, the visible band will give you the highest excitation, even having similar absorption intensities to the one in the UV.

Some equipments give you the corrected excitation spectrum by dividing your result to the lamp signal (in these equip, the final excitation spectrum is practically identical to the absorption profile), but many others just give you the non-divided value (which differs to the absorption spectrum profile: the difference is in intensity (y axis), BUT the energies of the bands (x axis) should be similar).

Thanks Mr. Zanoni for your answer.

I am using confocal microscope to detect this compound inside cells..

Well, the excitation of an electrons leads to a charge redistribution within the molecule. Therefore the molecular geometry reorganizes (relaxes) to be optimal for the new charge distribution. This relaxation stabilizes the excited state (typically state S1 in a case of a fluorescent compouns). Due to the stabilization of the emitiing state the emission maxima appears at a lower energy (longer wavelength).

Muhammad asked us the classic wrong question.

Of course, if you really want, then you can search the answer for question - why the Earth is flat.

The correct question -

Under what conditions do the absorption and excitation spectra of the luminescence coincide?

In the ResearchGate there is an excellent lecture by Alfons Penzkofer "Fluorescence Excitation Spectroscopy" (https://www.researchgate.net/publication/266181957_Fluorescence_Excitation_Spectroscopy). There is everything to understand the "strange" Muhammad result.

Thanks Mr. Tsaplev for providing that insightful lecture.

Under what conditions do the absorption and excitation spectra of the luminescence coincide?

That what I want to know..

By the way. the flat earth theory is going viral !!!

I hope that you have read the lecture and will now be able to answer your own question.

To add to what has been already said

1. If your samples are thin films then the penetration depths of light at different wavelengths into the sample could be different. Consequently the collection efficiencies could also be different.

2. Is your data corrected for the differences in the sensitivities of the detector at different wavelengths?

3. Is you data corrected for the variations in lamp intensity at different wavelengths?

4. If the films are rough then there could be scattering too which can influence your fluorescence signal.