We usually do it this way:

take a reference dye with a known fluorescence quantum yield, that absorbs in roughly the same spectral region as your sample.

Prepare different dilutions of your sample and the reference dye, and measure their absorption and fluorescence emission (for the emission, it is important that you use the SAME instrument settings for all measurements, meaning same excitation wavelengths, same gain value for the detector...). Take care that you stay within the boundaries of Beer's law.

Now, when you plot the maximum absorption vs. the fluorescence intensity, you should end up with a linear function, both for the reference dye and for your sample. For your reference dye, the gradient of the plot is a measure of the fluorescence quantum yield.

By comparing the gradient of the plot for your sample with the gradient for the reference dye, you should be able to calculate the samples' quantum yield.

E.g if your sample has a gradient that is 1/2 times the gradient of your reference dye, its fluorescence quantum yield also should be 1/2 times that of the reference.

We described that approach, see Figure S2.

http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0058049#s4

We usually do it this way:

take a reference dye with a known fluorescence quantum yield, that absorbs in roughly the same spectral region as your sample.

Prepare different dilutions of your sample and the reference dye, and measure their absorption and fluorescence emission (for the emission, it is important that you use the SAME instrument settings for all measurements, meaning same excitation wavelengths, same gain value for the detector...). Take care that you stay within the boundaries of Beer's law.

Now, when you plot the maximum absorption vs. the fluorescence intensity, you should end up with a linear function, both for the reference dye and for your sample. For your reference dye, the gradient of the plot is a measure of the fluorescence quantum yield.

By comparing the gradient of the plot for your sample with the gradient for the reference dye, you should be able to calculate the samples' quantum yield.

E.g if your sample has a gradient that is 1/2 times the gradient of your reference dye, its fluorescence quantum yield also should be 1/2 times that of the reference.

We described that approach, see Figure S2.

http://www.plosone.org/article/info%3Adoi%2F10.1371%2Fjournal.pone.0058049#s4

can you explain it with William method

Фf(U)= Astd/Au ×Su/Sstd ×[nu/nstd ]^2×Фstd

see the link for more information. There, it is described very well.

The equation you postet describes what I tried to explain. I guess the values A and S refer to something like absorbtion and emission intensities. My approach simply was to use several concentrations to get an error estimate.

Your equation also takes into account the influence of solvent refractive index for the determination of the quatum yield. When your reference dye and your sample can be measured in the same solvent, the last term - [nu/nstd ]^2 - equals 1.

http://www.horiba.com/fileadmin/uploads/Scientific/Documents/Fluorescence/quantumyieldstrad.pdf

Dear Dominik, thanks for your useful reply

One important thing is not explained in the above description, namely what is emission intensity. And by this term a number of photons is assumed. Thus, one needs to calculate integrals under the emission spectra of the reference and studied molecules in proper coordinates.

In the article this is not an issue, since the spectrum does not change with quenching in the particular case, and the integral is directly proportional e.g. to maximum intensity.

Vladimir

I do not understand your second paragraph: in the 1st you correctly mention that the the 2 emission have to be represented in the proper coordinates, meaning wave-number/frequency

Harry, I was replying to the first answer by Dominik (and Abbas' reply). In the Dominik's answer, there is a link to an article, which similarly does not discuss what is emission intensity. My second paragraph is about the linked article.

And the next Dominik's message contains another link with a full discussion of the procedures.