In fluorometry, when do we take the emission spectra and when do we have to take the excitation spectra of a particular sample....i mean theoretical definition i understand, but practically when and how do i differentiate them?
hi,
when you want to find the emission of particular sample.must you know the excitation(Absorbance) peak of those sample.
Fluorescence is the reemission of longer wavelength (lower frequency) photons (energy) by a molecule that has absorbed photons of shorter wavelengths (higher frequency).
Both absorption and radiation (emission) of
energy are unique characteristics of a particular molecule (structure) during the fluorescence process.
Light is absorbed by molecules in about 10-15 seconds which causes electrons to become excited to a higher electronic state. The electrons remain in the excited state for about 10-8 seconds then, assuming all of the excess energy is not
lost by collisions with other molecules, the electron returns to the ground state. Energy is emitted during the electrons' return to their ground state. Emitted light is always a longer wavelength than the absorbed light due to limited energy loss by the molecule prior to emission.
The emission is more straightforward and commonly used. Gelija has given a thorough explanation.
Excitation spectra are particularly useful in the following cases:
1. When there are two emission bands (or an unnaturally broad one). If the excitation spectra of the two bands are identical, there is an excited state process, e.g. proton transfer, electron transfer. Otherwise, two ground state species are present.
2. If the excitation spectrum deviates a lot from the absorption, emission at the observed wavelength may originate from some strongly-emitting impurities.
3. If the excitation spectrum deviates from the absorption only at shorter wavelengths, there are extra photophysical (e.g. internal conversion) or photochemical (e.g. dissociation) processes going on, which do not go through the usual emitting S1 state.
4. With a strongly scattering sample, such as lipid vesicles, for which the absorption cannot be properly measured.
From a practical experimental standpoint, when setting up the excitation and emission wavelengths for a sample for the first time, if you want to maximize the sensitivity of detection, i.e. get the largest possible signal, you should take both the excitation and emission spectra in order to find the peak wavelengths of both spectra. Then you can set the excitation and emission monochromators to the peak wavelengths in order to get the highest sensitivity.
(1) How to differeniate excitation and emission: when you put the sample in the spectrofluorometer, set it firstly to Emission mode, then you will be asked about the the wanted Exitation wavelength (then choose the peak maxima of the absorption band. Secondly, after measuring the emission, set the instrument to Exitation mode and then you will be asked for the emission, choose the maxima of the emission band you measured in the first step, then you will have the excitaion spectra from which you can define where the excitation exactly occured since the absorption band is broad and not a sharp spectral line. Why should we do these two measurements:Assume a dye absorbes from 610-670 nm with peak maxima at 666 nm, it is not essentially that the dyes get excited at 666 nm, maybe 661 nm. However when you excite it at 666, emission will occur, say at 679. if you measured the excitation at emission of 679 you will find out that the excitation occurs excactly at 661 nm not 666 nm.
Wael, The statement you made at the end of your answer is incorrect. In general, the excitation peak wavelength does not change when the emission monochromator wavelength setting is changed, and vice versa, although this could be observed under non-ideal circumstances.
A researcher may decide to use an excitation wavelength that is shorter than the peak wavelength and/or an emission wavelength that is longer than the peak wavelength in order to separate the excitation and emission wavelengths when the peaks are too close together. This reduces the background caused by direct detection of the excitation light. It comes at the cost of reduced sensitivity.
Mrs/Miss Qasim,
On the base on the intensity is the differentiation