There is a clear relation between the fluorescence yield and the lifetime. For an exponential decay, of a pure compound the yield is equal to the prodcut of the fluorescence lifetime by the radiative rate. But we cant consider the same relation if there is some other non radiative loss mechanism (even a heat transfer) or quenching mechanisms (including solvent relaxations). A lot depends on your samples. There are plenty of work done previously in this domain.

This might help: https://pdfs.semanticscholar.org/4d70/44db095b332f75d9a281c78e263568f7dc7c.pdf

A textbook on the subject

Valeur

Molecular Fluorescence

Principles and Applications

Introduction page and page 42 and following

the simplest answer would be: fluorescence quantum efficiency is defined as the ratio between the radiative decay rate and the total decay rate. The total decay rate is in turn the inverse of the experimental lifetime (obtained in the laboratory), while the rate of radiative decay is the inverse of radiative life time (which is usually difficult to obtain)

The QY can be experimentally obtained by comparing the integrated fluorescence with the integrated fluorescence of a standard molecule of which the QY is known, so you do not need the theoretical approach via the Strickler-Berg formulation:

see pg 42 and following and note 7 on page 47

Dear ARYA,

What do you mean with fluorescence lifetime? Do you means the lifetime resulting from radiative recombination such that radiation rate= dn/ Tau rad, with dn the excess mobile charge carriers dn=dp, and Taurad is radiative recombination lifetime. What do you mean with the fluorescence quantum yield?

Do you mean the photons emitted/ electron hole pairs which is equal to the photon flux/ charge carrier flow rate. After agreeing on the definition we can get an expression relating the quantum yield to the radiative lifetime.

Best wishes

Dear Abdelhalim Zekry Sir,

I have recorded average lifetime of a synthesized fluorescent material using TCSPC (Time Correlated Single Photon Count) method. Relative quantum yield is calculated using Rhodamine 6G as the reference.

Fluorescence quantum yield is the ratio of the number of photons emitted by the fluorophore to the number absorbed.

My question is, is there any direct relation exists between lifetime and quantum yield. Can we tell like as the PL lifetime increases, quantum yield also increases?

Thank you

Dear Arya,

Assume that the material absorbed a photon it will generate an exciton. This exciton will live an average lifetime time T before it recombine. The recombination process may lead to an emission of the a photon or may be nonradioactive.

The quantum yield will be the number of emitted photons to the number of absorbed photons. The radiative recombination rate is proportional 1/ Trad,

The nonradiative recombination rate is proportional to 1/ T nonrad.

So, the quantum yield can be expressed by= 1/Trad/ (1/Trad + 1/Tnonrad),

The quantum yield = Tnonrad/ Tnonrad + Trad= 1/( 1+ Trad/ Tnonrad),

So the quantum yield will increase by reducing the radiation lifetime to the non-radiating lifetime,m which is a logical result.

This is just an estimation not a very elaborated result.

Best wishes

Thank you sir.

@ Abdelhalim Zekry . The quantum yield = Tnonrad/ (Tnonrad + Trad). From this equation, if radiative recombination rate is zero then QY is 1, this is not possible. I think equation should be the quantum yield = Trad/ (Tnonrad + Trad). In this case if nonradiative recombination rate is zero then QY is 1 which is acceptable.

@Arya J S .

The fluorescence lifetime and quantum yield are two important features of a fluorophore. Fluorophores with the largest quantum yields, approaching unity, exhibit the brightest emissions. Fluorescence quantum yield (PLQY) = no. of photons emitted/no. of photons absorbed; 0 ≤ PLQY ≤ 1. It depends on material and conditions e.g. temperature, solvent polarity, and viscosity, rigidity of the local environment, rate of solvent relaxation, probe–probe interactions, probe conformational changes, proton transfer and excited state reactions, changes in radiative and non-radiative decay rates, internal charge transfer.

The fluorescence lifetime determines the time available for the fluorophore to interact with or diffuse in its environment, and hence the information available from its emission. The lifetime of the excited state is defined by the average time the molecule spends in the excited state prior to return to the ground state. The correlation between quantum yield and lifetime can be understood by the well-known Jablonski diagram. The radiative rate constant (kr) and the nonradiative rate constant (knr) both are responsible for depopulation of the excited state. The fraction of fluorophores that decay through emission, and hence the quantum yield, is given by PLQY = kr/(kr +knr); PLQY = kr / tav and tav = 1/(kr+knr). In other forms: kr = PLQY/tav, knr = (1-PLQY)/tav. The quantum yield can be close to unity if the radiationless decay rate is much smaller than the rate of radiative decay, that is knr < kr. It is noteworthy that the energy yield of fluorescence is always less than unity because of Stokes losses. Fluorescence emission is a random process, and few molecules emit their photons at precisely t = average life time. The lifetime is an average value of the time spent in the excited state. The lifetime of the fluorophore in the absence of nonradiative processes is known as intrinsic or natural lifetime, and is given by tn = 1/kr. For a better understanding of fluorescence phenomena, you can follow the "Principles of Fluorescence Spectroscopy" authored by Joseph R. Lakowicz.

Now come to your questions directly:

1. How fluorescence quantum yield related to fluorescence lifetime?

Ans. PLQY = kr/(kr +knr) and tav = 1/(kr+knr). So PLQY = kr*tav. In other forms: kr = PLQY/tav, knr = (1-PLQY)/tav.

2. What happens to quantum yield if lifetime increases?

Ans. PLQY = kr* tav. Following this equation, PLQY increases if lifetime increases, provided kr is constant. When you compare with different systems, it is also to be mentioned that kr value should be same because PLQY depends both on kr and tav.

Regards,

Atanu

Atanu Jana If PLQY=kr/(kr+knr), PLQY=kr*tav right (since 1/(kr+knr)= tav)? so, PLQY increases as lifetime increases!!!

@Arya J S . Yes, you are right. I have corrected the previous equation. You can follow our JPCL paper ( J. Phys. Chem. Lett., 2016, 7 (16), pp 3270–3277 ) page 3273.

Hi Arya JS,

Yes, sometime it is true that lifetime of a material increases if PLQY of the material increases. This fact Indicates that lesser the trap states (due to defective material) material has, higher the radiative recombination occurs. So both PLQY and lifetime of the material increases. I observed this relation from tons of lifetime measurement experiments with perovskite nanomaterials. But if the material has intrinsic trap states, in that case PLQY will not increase if lifetime of the material increases.

Thanks,

Aslam

@Arya J S and Md Aslam Uddin. I haved added more insights on my previous answer. PLQY depends on both kr and tav. If we want to compare tav of different compounds, kr should be constant. kr may vary depending on the trap states. If more trap states are present in the system then knr is higher than the kr beacuse the recombination occurs in trap states is nonradiative. Both the radiative rate constant (kr) and the nonradiative rate constant (knr) are responsible for depopulation of the excited state. Previously, I have mentioned about intrinsic lifetime where nonradiative recombination is absent. The comparison of intrinsic lifetime is much more straightforward.

@Atanu Jana @Md Aslam Uddin, I am confused. If there are lesser number of trap states, radiative recombination increases and thereby increase in PLQY. But, how lifetime increases? More trap states means, it should spend more time in excited state right?

More trap states (e.g. mid gap trap states) means there are high probability that electrons and holes will recombine without emitting photons. In this case, recombination time is short because charge carrier diffusion length is shorter. So, lifetime will decrease. But if there is no trap states, electrons and holes will recombine with emission of photons. In this case, recombination time is long because carrier diffusion length is longer. So, lifetime will increase.

* conversion between diffusion length L and lifetime τ involves the diffusion constant D via L = (Dτ)1/2

@Arya J S . I agree with the information given by Md Aslam Uddin.

@Md Aslam Uddin, Atanu Jana Thank you so much.

Atanu Jana

for bi-exponential fitting of PL decay, is it possible to have higher PLQY but faster decay

Raihana Begum

Please reframe your question or give some more details.

Higher PLQY with respect to which reference (compound or single-exponential fitting)? In the case of bi-exponential fitting, there are two decay channels. The faster decay may occur for first or second decay channel or one can associate it with overall decay.

One can measure the PLQY and radiative recombination rate but not the non-radiative one. By using mathematical formula one can calculate the non-radiative rate constant. Faster decay means the radiative recombinations are mostly taken away by trap states or different non-radiative channels. PLQY = no. of photons emitted/no. of photons absorbed. No of photons emitted can be measured by instrument and it is directly associated with radiative decay.

PLQY = kr* tav; kr = radiative rate constant and tav = average life-time. kr and tav are inversely proportional. If kr value is high than tav will be smaller. So for faster decay tav will be smaller and overall value of (kr* tav) i.e. PLQY will be lower.

Some groups reported PLQY, 100% for perovskite materials where time-resolved photoluminescence data are fitted by bi or triexponential function as there are multiple channels. This is not true. If there are multiple decay channels, PLQY should not be 100%. Fluorescein dye has single decay channel, but still it has not 100% PLQY.

Now come to your question.

for bi-exponential fitting of PL decay, is it possible to have higher PLQY but faster decay.

Ans. PLQY will be lower for faster decay. If there is no trap state then single exponential fitting is ok and the corresponding PLQY will be higher than the system with more trap states which require bi- or tri-exponential fitting.

In short, there is no obvious relation between quantum yield and excited state life time. For example, the opened form of Rhodamine B nucleus is an excellent fluorophore, yet has very fast decay. To sum it up, yopu must consider the fact that quantum yield is related to the population promoted to excited state upon excitation, whereas the life time is a kinetic phenomenon, reflecting the rate at which the molecules are deacivated from S1 state. Hope this helps you out.