Fluorescence Excitation and Emission Spectra

1. Dec 16, 2013

cardinal72491

I need help understanding the relationship between the excitation and emission spectra for a fluorescent material. This is not homework.

Let's say we have the following graph which plots the two spectra for a given material:

Two curves are shown - the excitation spectra and the emission spectra as well as the stokes shift (the difference in nanometers between the peak amplitudes of both curves).

From looking at the graph, the peak amplitude of the excitation spectrum is around 450, while the peak for the emission spectrum is around 550, making the stokes shift roughly 100 nm.

So my question is this: using this information how would I calculate the the emission wavelength given a particular light source. Below is my interpretation of what to do:

Step 1)
Let's say we have a monochromatic light at 400 nm with a spectral intensity value of 100. I would first look at the graph for the excitation spectra and see what the absorption value is for 400 nm. Let's call this value "A" (in the graph above the y axis is not plotted so let's say its 0.1 for the sake of this example). I would then multiply light intensity * A (100 *0.1) and get an intensity of 10.

Step 2)
Next, to find the corresponding emission wavelength, I shift up the incoming wavelength by stokes shift. So stokes shift in this case is around 100 so we get 400 + 100 = 500 nm for the emitted light.

Step 3)
To find the intensity of the emitted light, I look at the emission spectrum graph and find the value for 500 nm (since the Y axis is unplotted let's say its value is 0.25). I would then modulate the absorbed light intensity by this outgoing value (10 * 0.25) and get an intensity of 2.5.

So based on my understanding a 400nm beam of light with a spectral intensity of 100 would cause this particular material to emit a beam of light at 500 nm at an intensity of 2.5.

Is the above a correct interpretation? The biggest thing I am unsure about here is whether stokes shift uniformly shifts over all wavelengths in the excitation spectra or if it is used only as a reference for measuring difference between peak amplitudes.

2. Dec 16, 2013

Ygggdrasil

The emission wavelength is independent of the excitation wavelength. The emission spectrum looks identical to regardless of the wavelength you use to excite the fluorophore. The only difference is how efficiently your excitation light gets absorbed (for example, since it looks like the fluorophore in the diagram absorbs at ~ 450nm about ten times as strongly as at 350nm, the emitted fluororescence will be ten times weaker when exciting your dye with 350nm light versus 450nm light of the same intensity.

Calculating the amount of fluorescence emission from the amount of excitation light also requires knowledge of the quantum yield of the dye (what fraction of excited electons relax to the ground state via fluorescence versus non-fluorescent relaxation pathways).

3. Dec 16, 2013

cardinal72491

Oh okay, I see. Thank you for your reply.

Just a few more questions though.

- So if the emission spectrum is exactly the same regardless of incoming light/excitation (and just the magnitude changes) - then what accounts for fluorophores that appear different colors under UV lighting vs daylight? Based on what you described, it would seem that a fluorophore would always emit the same exact color just at different brightness levels.

-From an observers point of view, is the light we see from a fluorophore just the light from the emission spectrum or is it excitation + emission?

-If the light you're using to excite your dye is of multiple wavelengths, does the intensity of the emitted fluorescence increase linearly?

Sorry if my questions are very simple

Last edited: Dec 16, 2013
4. Dec 16, 2013

Ygggdrasil

Good question. Under white light illumination, most of the light you see coming from a solution of dye is the transmitted light, not the fluorescence from the dyes. For example, a dye I commonly work with is Cy5 which absorbs at 650nm and emits at 670nm. When you look at a vial containing the dye it appears blue because it is absorbing light from the red end of the visible spectrum, and the light from the blue end of the visible spectrum passes through.

Because you cannot see UV light, however, illuminating a dye with UV in a darkened room will allow you to see the fluorescence (assuming the UV light can excite the dye).

In general, when measuring the fluorescence of a compound, you will shine your light through a solution of the dye in one direction, then place the detector perpendicular to your light path. That way, none of the transmitted light reaches the detector and only the light emitted by fluorescence (along with scattered emission light) reaches the detector. If your detector were on the opposite side as your light source, you would basically just see the absorbance spectrum of your dye, not the emission spectrum.

The intensity of the emitted fluorescence should increase linearly with the intensity of the excitation light no matter what range of wavelengths are used to excite your dye. The intensity of the emitted light will not change linearly, however, as you broaden the range of wavelengths used for excitation (for example, by opening your monochrometer window) because the dye absorbs the different wavelengths with different efficiencies (i.e. the extinction coefficient of the dye changes with wavelength).

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