Most importantly - if you are referring to a normal emission spectrum, the wavelength scale is not linear with energy. You need to plot intensity versus wavenumber scale in order to have the x-axis linear with energy.

A Gaussian curve is a histogram that arises for a purely random process. If emission was a result of relaxation from a single vibrational level of a singlet excited state (which it should be following Kasha's Rule) to a single vibrational level of the ground singlet electronic state, then you should have a Gaussian curve whose width would depend on the lifetime of the excited state (due to Heisenberg uncertainty). However, relaxation will occur by relaxing back to many vibrational levels of the ground state. The probability that the molecule will relax back to a particular vibrational level will be dictated by its Franck-Condon factor.

Reference: Chapter 4 of "Modern Molecular Photochemistry" by Nicholas Turro.

I hope that is helpful.

Best regards,

John

Dear questioner

Your question is much too general. Are you referring to the 0-0 band, the so-called origin of the emission spectrum; for what entity? a molecule in solution? what is the width of the band is it fully resolved form the next band?

Broadening can have many causes: in molecules the low-level vibrations and rotations are populated according to the Boltzmann distribution and emission is from all of these levels to their counterparts in the ground electronic level. These have different energies in excited and ground-state, meaning thousands of lines with different wavelengths adding up to a broad band.

Another possibility is that in the 0-0 band there is a hidden weak transition to a low-energy vibrational level that is by definition at the red-side of the origin.

Finally, what is the reason for your question?

Most importantly - if you are referring to a normal emission spectrum, the wavelength scale is not linear with energy. You need to plot intensity versus wavenumber scale in order to have the x-axis linear with energy.

David

This is (again) an example of a questioner in Researchgate who is not even polite enough to react to answers and questions about what he exactly means and just wastes the time of repliers

Harry,

I think that the fact that researchers posted their answers here is definitely not a waste of time since the question is open for all to read and for example I find it useful to read all the answers. I think Research Gate is a very useful place to open scientific discussions and it should be kept like that and discussions should completely avoid our personal opinions about the persons who ask and answer, science should be free from this, facebook and other social networks exist for that.

Best Regards

Valeria

Valeria

I think that you did not fully read my entries here. I answered his question and then asked the questioner for more specific information; this was not given. In this way a real answer cannot be given. I like to mention that ResearchGate asks questioners to ask specific questions so that the questions can be answered in a proper way. Thus my complaint is not a personal opinion but a repeat of the remark by ResearchGate itself

I am very much thankful to

John Hansen, M. Farooq Wahab, David Jameson and Harry ten Brink for their careful effort to answer the question and the answers indeed are very helpful. David's answer sounds interesting. May I ask why it is so ?

Dear Harry ten Brink,

Please don't feel bad as I did not respond to you. I was heavily busy with my thesis as I am at the verge of completion. It was not at all any intentional ignorance nor overlooked. Your answer was very much useful but its not exactly the answer I look for. Probably my query was not clear enough. The question is very basic and was asked by one of the physicist present in my pre-synopsis presentation. This was the only unexplained answer by me. Now I am trying to find out the proper expaination for the same. Here I am reframing the question.

Its about whole steady state emission spectrum (not about a specific band like 0-0) of any fluorophore. If you draw a vertical line from the peak max to x axis the left half of the spectrum does not look like a mirror image of the right half. The question is why it is so? When One could expect a mirror image shape of the left and right half of the spectrum.

Dear Valeria,

Thanks very much

and

Dear Harry

Please don't feel bad

Interestingly there have been efforts over the years to convince researchers to use wavenumber instead of wavelength plots to report emission spectra. But this approach never really caught on, at least in most biophysical and chemical applications. In the earlier days before computers one could understand the use of a wavelength scale since grating steps produced a linear wavelength scale. Nowadays, given computer controlled instrumentation it would be much easier to produce wavenumber scales, indeed some instruments allow one to replot the data using wavenumbers

A word on reploting spectra on an energy scale. For true band shapes this should be done for the reasons laid out by David Jameson, however in doing this the intensity scale must also be corrected for the non linear band pass of the grating (which is measuring in reciprocal energy). So in the red the band pass is much smaller than in the blue and the intensity axis needs to be scaled by lambda squared. Dumas wrote an excellent review on this.