3
$\begingroup$

Question is rather self-explanatory. Came up during a lecture without a concrete answer.

I understand that the differences in emission wavelengths is due to relaxation to the lowest energy level of S1, but why do fluorescent molecules necessarily overlap in their excitation and emission spectra?

$\endgroup$
  • $\begingroup$ If you are asking why there is no overlapping of lines in emission or excitation spectra, the answer to this question also explains why Rutherford model of atom was rejected. If lines were overlapping, it mean that atoms emitted energy (in the form of radiation) in continuous manner, if this was the case, electrons would collapse into the nuclei giving Thomson model of atom. But this never happens. So, atoms can emit energy in discrete manner, which was explained by Bohr. Thus, there will be no overlap of lines in emission or excitation spectra. $\endgroup$ – Immortal Player Dec 7 '13 at 12:20
1
$\begingroup$

Energy-Level Diagram
There is overlap because between the excitation and the emission spectra because there exists a Range of orbitals at different energies. You'll always notice that the energy for excitation is always higher, or the lambda value is always lower (these say the same thing since E=planck's constant * frequency).

I've attached a HE-Ne laser schematic from Verdeyen's Laser Electronics book. For a better visual. What you need to know is let's say your gain medium, in this case He-Ne was excited at it's lowest wavelength (we'll call it Lambda1) then it's emission wavelength may be only 2nm higher (or red shifted), Lambda2.

However, as you'll notice in this schematic there are many wavelengths at which this gain medium (He-Ne) can emit light. These other possible excitation orbitals can have lower energy levels that correspond to higher wavelengths and thus overlap on the spectrum.

$\endgroup$
1
$\begingroup$

This is most easily seen in solution. The overlap in molecules occurs only when the electronically excited states of a molecule have rather similar geometries and so transitions between vibrational levels in the two states occur and the same wavelength. After excitation the excited state rapidly looses energy until only the v=0 level is populated. This takes about $10^{-13}$ sec, far faster than fluorescence.

If the ground state and excited state have very different geometries then there is almost no overlap between absorption and emission. The picture tries to explain the mirror image overlap and features ( the picture is not to scale unfortunately as the length of the red and blue dotted lines should be the same for a couple of transitions) mirror image

In an isolated molecule in the gas phase when only one vibrational level is excited then emission comes only from this level to varoius levels in the ground state; only 1 emission transition overlaps with the absorption line.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.