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Topic: How to go about correctly obtaining quantum yield?  (Read 320 times)

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Offline GrimTheCat

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How to go about correctly obtaining quantum yield?
« on: February 21, 2019, 11:00:02 PM »
I am using these equations:

Single point method
Q=QR(I/IR)(AR/A)(n2/n2R)
Where Q is quantum yield, I is the integrated fluorescence intensity, A absorbance, n is the refractive index of the solvent, and finally R is for the reference fluorophore.

Comparative method
Q=QR(m/mR)(n2/n2R)
Where Q is quantum yield, m is the the slope of a line from a plot of integrated fluorescence intensity vs. absorbance, n is the refractive index of the solvent, and finally R is for the reference fluorophore.

My questions are:
1. Do the reference fluorophore(RF) and subject fluorophore (SF) need to be of the same concentration(s)?


2. Do the RF and SF need to be excited at the same wavelength?
This is the main question that confuses me because I figured that they should be excited at their respective max absorbance wavelengths(λ). For example, if I use the RF excitation λ then the Q will not be the SF's optimal Q, which is what this whole experiment is for! However, if I use the SF excitation λ, then the QR will not be what is found in the literature because the lit. QR should be optimal.

Thank you.

Offline Corribus

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Re: How to go about correctly obtaining quantum yield?
« Reply #1 on: February 22, 2019, 08:11:33 AM »
To be clear, these are the same methods of determining a QY- the "comparative method", as you call it, is superior because it provides basically an average over several independent replicates.

The reference and unknown do not have to be the same molar concentration. However, they should have approximately the same optical density at the same wavelength and the optical density at the excitation wavelength should be less than 0.1 for both compounds.

The two compounds do need to be excited at the same wavelength, because the light output of your instrument is not constant at different wavelengths. In principle you could correct for this, but the QY measurement is tricky enough so it's best to remove as many sources of error as you can. For most compounds, the QY does not vary as a function of excitation wavelength (the fluorescence intensity is exactly proportional to the absorptivity at the excitation wavelength), so it does not matter if your reference value was acquired at a different excitation wavelength. Care must be taken though in selecting a reference compound - it is preferable to choose a literature value that was determined by an absolute method, not the comparative method you are using here, because then you are basically making a copy of a copy, and propagating all those errors. Also note that the fluorescence spectral range over which you are integrating should also be reasonably similar for the two compounds.

Not sure what you mean by "optical QY". The QY is the QY.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Offline GrimTheCat

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Re: How to go about correctly obtaining quantum yield?
« Reply #2 on: February 22, 2019, 12:52:05 PM »
Thank you! You have cleared up this process for me! One more thing when you say the same optical density, do you mean exactly the same or is there a range in which they can be within?

Offline Corribus

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Re: How to go about correctly obtaining quantum yield?
« Reply #3 on: February 22, 2019, 01:15:03 PM »
Basically the more things you can keep the same or similar the better.  Of course,  of you opt for your second method,  then you're going to be varying the absorptivity anyway. The most important thing is that you measure at values less than 0.1 to minimize inner filter and other effects.
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

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