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I'm studying the phosphorescence decay of diacetyl in acetonitrile (it has a high quantum yield of intersystem crossing). The fluorimeter than I'm employing has microsecond resolution and, since the decay of the phosphorescence isn't too long, I have to measure the IRF (width of 1-2 $\mu$s) in order to perform a reconvolution fit. However, I get a biexponential fitting with a long time (like 20 $\mu$s) and a very short time (around 30 ns). Does that make sense? I thought that the short lifetime could be due to fluorescence, but I get it even in the decay of the righmost peak while using a long pass filter. Moreover, I don't know if it makes sense to have a lifetime with more resolution that the limit of the aparatus.

Does this make sense or has any meaning? Maybe is an artifact or I'm making some mistake with the measurements?

Edit: I'm using the Spectrofluorometer FS5 from Edimburgh Instruments, and employing its software (Fluoracle) to make to fittings. The light source is a flash lamp. By reconvolution/convolution I mean a fitting that takes into account the IRF (instrument response factor), as the lifetime of the decay approaches the duration of the IRF.

The fittings look like this:

Monoexponential Figure 1: monoexponential fitting

Biexponential Figure 2: biexponential fitting

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    $\begingroup$ What is the fluorimeter’s light source? Is it a flash lamp? If so, maybe the short component of the output is an electrical artifact due to the flash lamp. $\endgroup$
    – Ed V
    Commented Jul 5, 2023 at 15:53
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    $\begingroup$ Please add more details about measuring instrument response function, what is meant by reconvolution? Add a figure that explains what you are doing. $\endgroup$
    – ACR
    Commented Jul 5, 2023 at 16:21
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    $\begingroup$ If response function is 1-2 microseconds, you can't measure & quantify 30 ns decay with any reasonable amount of accuracy. So while one might exist, I don't think your data shows that. It must be instrumental as EdV points out. It's a fitting artifact of some kind, but with so little information available (as AChem and porphyrin point out) it's hard to give much of an answer. However, if it was real I would say it's due to nonradiative recombination which is density dependent - excitations kill each other off quickly until the remaining density is so low that they can decay at their own rate. $\endgroup$
    – uhoh
    Commented Jul 6, 2023 at 7:50
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    $\begingroup$ @uhoh, I added information to the post. I also think its an artifact, but I don't know how to get rid of it. I thought that and artifact of the lamp would be take into account with the IRF. $\endgroup$
    – user78790
    Commented Jul 6, 2023 at 8:20
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    $\begingroup$ Phosphorescent materials, such as “glow in the dark” paints and tape, are readily available. Perhaps make up a test specimen using such a material and see how it behaves in your fluorimeter. If the short lived component is always present, then the evidence for an artifact is stronger. $\endgroup$
    – Ed V
    Commented Jul 6, 2023 at 13:18

1 Answer 1

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Phosphorescence should be at a longer wavelength than any fluorescence so carefully using a filter or grating should remove any fluorescence. Also measuring at different wavelengths will change percent of any fluorescence still getting through a filter. Is there a tiny bit of scattered light being mixed in with the phorphorescence? (Some types of glass filters do fluoresce but usually you need to excite then directly).

How are you doing the convolution, Marquardt type non-linear least squares iterative convolution is normal. What sort of chi-squared are you getting for the fit, what type of weighting is used? How does a 1 exponential vs 2 compare? What do the residuals look like and so on. Are you fitting to a background also? However, 30 ns is really too short to be confident that it actually exists if the irf is 1-2 microsecs.

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    $\begingroup$ I added some information to the post. I'm just doing the convolution with the software (Fluoracle) of the aparatus, so I have no clue about the type of convolution or the parameters. Would it be better to use Origin or another tool? Moreover, I'm not fitting to a background, I just followed the manual of the aparatus (it said I just needed the IRF and the decay). $\endgroup$
    – user78790
    Commented Jul 6, 2023 at 8:17
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    $\begingroup$ Well, now we can see the data there is definitely a lot of a short lifetime there. Have you tried a cell filled with only solvent to check scattering or no cell/empty to check any electrical noise? If all is ok then there is a real short fluorescence / phosphorescence component. $\endgroup$
    – porphyrin
    Commented Jul 6, 2023 at 8:40
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    $\begingroup$ Related to the comment immediately above, the OP could spike a little nitromethane into the solution, to quench both the fluorescence and phosphoresce and see what results. (I assume the solution is oxygen free, more or less.) Also, since a flash lamp is used, how good is the instrument at blocking direct excitation light or “after glow” from the Xe (?) flash lamp? $\endgroup$
    – Ed V
    Commented Jul 6, 2023 at 12:22
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    $\begingroup$ @porphyrin, I monitored a decay with only solvent (at the same wavelenghts and bandwidths than the sample) and I get some signal. Perferming a convolution fit with this signal and the IRF gives me a lifetime of around 50 ns. However, this signal is very weak (5 min of measurement to get a very weak signal vs 1 min to get the strong signal of the sample), so I don't know if that could be the lifetime that I'm seeing. $\endgroup$
    – user78790
    Commented Jul 6, 2023 at 14:49
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    $\begingroup$ So it seems that you do have some scatter coming through which it would be best to totally eliminate if you can, but further to Ed V's comment you could also purge with a heavy atom gas such as Xe or add say iodide ions to cause heavy atom quenching. The short decays will be unaffected if is scatter. All I can now suggest is that you look at, say, Ru trisbpy which has a long excited state lifetime, or some other well measured species to check how you instrument is working. $\endgroup$
    – porphyrin
    Commented Jul 6, 2023 at 15:44

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