The fluorescence quantum yield (φ_F) of a species is the measure of its fluorescence efficiency. After absorption of photons, a molecule may get de-excited via different competitive processes, including fluorescence. The fluorescence quantum yield (φ_F) of a molecule can therefore be measured by the ratio of the excited molecules that return to the ground state (S₁→S₀ transition) with emission of photon to the total number of the excited molecules per unit of time and volume. If kr is the fluorescence (radiative) rate constant and Σk is the sum of the rate constants that depopulate the S₁ state (= k_r + k_ic + k_isc + k_q + k_ec + k_pd + k_d + ...., for all the processes like radiative process (fluorescence), internal conversion, intersystem crossing, quenching, external conversion, predissociation, and dissociation, etc. respectively), then the quantum yield (φ_F) can be written as follows

φ_F =k_r /Σk.......................... (1)

Thus, the weaker the competitive phenomena, the higher the fluorescence. In the absence of any competing pathways φF = 1. If fluorescence is sufficiently faster than the other deactivation processes, every molecule will take the fluorescence pathway for losing absorbed energy. In that case, φF = 1. If one of the competing processes is faster than the fluorescence, the amount of emitted light will be negligible. In the intermediate cases, where the deactivation processes have a similar rate to the fluorescence, the extent by which the emitted light is diminished depends on their relative rates.

Radiative (fluorescence) rate constant is related to the radiative lifetime, τr>, as

τ_r =1/ k_r................ (2)

The experimentally observed (real) fluorescence lifetime, τF , is the average time the molecule spends in the excited state and is given by

τ_F =1/ Σk........................ (3)

Therefore, the quantum yield (φ_F) can be determined by measuring the fluorescence lifetimes.

Since the fluorescence quantum yield (φ_F) is the number of photons emitted by the radiative way over that absorbed by the molecule, the absolute method of quantum yield (φ_F) determination requires the number of photons absorbed and emitted to be determined. Using a standard fluorimeter with right-angle illumination of a solution sample, it is difficult to precisely determine the number of photons being absorbed and emitted by the sample. Actually small fraction of the light emitted from the sample reaches the detector of the fluorimeter. Similarly, a small fraction of the exciting light is absorbed by the sample. These absorption and emission fractions are unknown factors. The easiest way to estimate the quantum yield of a fluorophore is by comparison with a reference compound (standard) for which φF has been determined with high degree of accuracy. This method relies on the fact that the unknown fractions should be quantitatively the same for different sample solutions compared under identical conditions of excitation. Ideally, the reference compound should have similar excitation-emission characteristics to the compound of interest.

Thus, one can write quantum yield as

φ_F = [α * (no. of photons detected)] / [β * ( no. of photons absorbed)] .......... (4)

where α and β are constants related to the fraction of emitted light reaching the detector of the fluorimeter and the intensity of the excitation source at the excitation wavelength, respectively. The number of photons detected can be estimated from the wavelength integrated area, ΣA, under the corrected fluorescence spectrum, assuming the counts per second unit of the ordinate as quanta. The number of photons absorbed, i.e., the fraction of exciting light absorbed by the sample is readily determined from a measurement of the absorbance at the excitation wavelength. This converts to the actual number of photons absorbed per unit time with knowledge of the precise intensity (photons per unit time per unit area) of the exciting light source at this wavelength, with the chosen aperture (slit width). Fraction of light absorbed is given by

f = 1 - 10-D ............................... (5)

where D is the absorbance (optical density) at the excitation wavelength. The absorbance is kept below 0.05 to avoid inner filter effects. Hence, for unknown sample, X, one can write:

φ_F,X = [α * ΣA_X] / [β * f_X] ............................... (6)

where ΣAX is the integrated area under the corrected fluorescence spectrum for X. Similarly, for the reference material, R, one can write:

φ_F,R = [α_X * ΣA_R] / [β * f_R] ............................. (7)

By comparing the two measurements on solutions of sample X and a suitable reference, R, one gets the fluorescence quantum yield of the sample, X by dividing the above two equations and rearranging:

φ_F,X / φ_F,R = [ΣA_X / ΣA_R] * [f_R / f_X] .................................. (8)

where all terms on the right hand side of equation and φF,R are known. Here we have assumed that the measurements on solutions of X and R are made under identical conditions of excitation wavelength and aperture settings in the same solvent. If different solvents are used for X and R, then the equation should be further modified to allow for the effect of refractive index on the relative amount of fluorescence collected by the detector. This correction is required since the fluorescence is refracted at the surface separating the solution and air. The flux of fluorescence coming within the aperture of the detection system is inversely proportional to the square of the refractive index. Therefore, the above equation modifies to

φ_F,X / φ_F,R = [ΣA_{X / ΣAR] * [fR / fX] * [nX / nR]2 ............................ (9)}

where the n's are the refractive indices of the solvents used for the two solutions.

If the absorbances of the sample and reference solutions are the same at the exciting wavelength, then

φ_F,X = φ_F,R * [ΣA_X / ΣA_R] * [n_X / n_R]2 ........................... (10)

The measured fluorescence quantum yield is the result obtained from a fluorescence spectrometric measurement when no corrections were made for instrumental response and sample effects.

Therefore, in order to determine fluorophore quantum yield, one needs to measure the absorbances of the fluorophore and of the reference at the excitation wavelength, and needs to calculate the sum of their (i.e., integrated) fluorescence intensities along their fluorescence emission spectra for the unknown as well as for the reference. One can easily determine the area under different emission spectra by integrating the emission spectra by using a computer or by using graph papers. Assuming a uniform thickness/density of paper, the ratio of the areas under the graphs can be also obtained by cutting out the paper under the graphs by using scissors and weighing each paper. Then the ratio of the masses of the two pieces of papers, one for each graph, is the same as the ratio of the areas.

Quinine sulfate, rhodamine 6G, rhodamine 101, fluorescein, β-carboline, perylene, cresyl violet, etc. are used as reference or standard compounds. Strictly speaking, one should obtain correction factors for the emission spectrum of a standard compound by recording it and comparing it to the data for a standard compound. In principle, quantum yield standards should have maximum overlap of the absorption and emission between sample and reference. Here we have chosen to determine the fluorescence quantum yield of Rhodamine B by comparing with Rhodamine 6G which has fluorescence quantum yield of 0.95.