# Calculation of fluorescence quantum yield

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 $N$?

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