I have been searching but haven't find any robust material on this topic. Can anyone provide me a paper or book that explains this?
Assymetry can be caused by unresolved shoulders of vibrational sidebands
I started making a drawing of this, and then found a much better representation: https://demonstrations.wolfram.com/FranckCondonPrincipleInVibronicTransitions/
Band broadening in general comes from doppler shifts (translation towards/away from observer) and solvent orientation around dyes (some solvent charges better solvate a particular dye and lower its energy, others worse and raise its energy; you can read about sovatochromism to feel into this). This overlays the vibrational and rotational energy levels.
The link above will show the source of asymmetry. Start reading about the Frank-Condon principle, which is generally taught in the context of an anharmonic oscillator. Basically, you imagine a spring-like force binding two atoms. Force as a function of distance should be parabolic, but bonds can dissociate at larger distances and this flattens the larger radius side of the parabola.
In the excited state, you have more anti-bonding character, so the central distance between atoms is farther. That's why the excited state curve is shifted to the right (larger radial distance) relative to the ground state.
Frank-Condon says electrons are ~1000x lighter than protons, so the nuclei will essentially stay still while the electrons rearrange during absorption or fluorescence. If you now look at the probability distribution curves on each vibrational energy step (these look like normal distributions for the ground vibrational level, then two normal distributions next to each other in the first excited vibrational level, then three...), the probability of transition from the ground vibrational state of the excited electronic state (you fluoresce from here), is proportional to the overlap of that distribution with the distributions of the ground state vibrations.
So... the asymmetry is coming from the shape of the anharmonic curve in the ground state. That's the green in my drawing.
My answer was corrected by Jeremy Pronchik below, but I leave it here for other readers to see what was wrong.
My previous (wrong) answer:
"The bands are assymetric and broader for higher wavelengths if you look at them when on X axis you have nm. They are rather not so if you have on X axis energy or nm^-1. Sometimes you can see additionally overlay of two or more bands which may appear as another assymetry/broadening factor."
Tomasz Fraczyk , you got me curious enough to plot some data. I think Mohamed Ali's question stands irrespective of the plotting method. Here are two plots of the same fluorescence data for Acridine Yellow I took from this website: https://omlc.org/spectra/PhotochemCAD/html/035.html
I just clicked on a random compound, but Acridine Yellow gives a typical spectrum to my eyes. Plotting the data against wavelength (low to high) and wavenumber (high to low), and superimposing the symmetric normal distribution, the fluorescence spectrum is pretty asymmetric on both scales.
I agree with you the asymmetry is a little more exaggerated in the wavelength plot, and plotting in wavenumber makes more physical sense, but I don't want people to get the impression the asymmetry is only an artifact of plotting against wavelength. It is a common feature of fluorescence spectra with a physical explanation.
I believe the basic fundamentals here, that area under curve always depends on nature of molecules and peak intensity is the measure of your absorption or emission value. If we want to check whether it's property of a specific molecule we can draw a relative grap through same process
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