The fluorescence quantum yield of a sample $\Phi$ is defined as the fraction of photons emitted per number of photons absorbed. The quantum yield is often calculated by comparing a sample with a reference substance of known quantum yield $\Phi_\mathrm{r}$ such that $$ \Phi=\Phi_\mathrm{r}\times\frac{n^2}{n_\mathrm{r}^2}\times\frac{I}{I_\mathrm{r}}\times\frac{A_\mathrm{r}}{A}, $$ where $n$ ($n_\mathrm{r}$) is the refractive index of the sample (reference), $I$ ($I_\mathrm{r}$) is the integrated emission of the sample (reference), and $A$ ($A_\mathrm{r}$) is the absorbance of the sample (reference) at the excitation wavelength (see here or here for details).

It seems odd to use the integrated emission because it is not proportional to the number of photons emitted unless the emission peak is extremely narrow. In particular, the integrated emission is $$ I=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\,I_\lambda, $$ where $I_\lambda$ is the spectral intensity. Because the number of photons for a given energy changes with wavelength ($E=hc/\lambda$) using a quantity that is proportional to the number of photons seems more sensible. In particular, $$ I_\lambda= n_\lambda \frac{hc}{\lambda}, $$ where $n_\lambda$ is the photon density at wavelength $\lambda$. Consider the quantity $$ N=\int_{\lambda_1}^{\lambda_2}\mathrm{d}\lambda\, \frac{I_\lambda \lambda}{hc}. $$ It is directly proportional to the number of photons even if the emission peak is broad. Why is $I$ used instead of $N$?


1 Answer 1


After some more reading, it turns out that the protocols suggesting to use the integrated intensity are wrong and it is necessary to correct for the variable wavelength across the emission peak. Würth et al. (2013) provide a description of the correct method. First, they state that

all fluorescence spectra have to be converted to a (relative) number of photons per unit time

Second, they describe the conversion as discussed in the question:

the radiometric quantity has to be multiplied with $\lambda/(hc_0)$

  • $\begingroup$ Its probably not important now after a year of asking , but you should convert your spectra to be in wavenumbers and try to get a standard that emits in the same region as your compound because detectors etc have a wavelength dependence. $\endgroup$
    – porphyrin
    Jul 9, 2016 at 15:17

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