[chesterfield]

I have exhausted all hair and I'm now just grasping at the skin of my freshly bald head in desperation.

I'm trying to understand how, from a mechanistic perspective, the presence of a collisional quencher can decrease the excited state lifetime of a fluorophore.

I have a vague understanding of how lifetime is the inverse of the combined decay rate constants and how adding non-radiative decay pathways therefore decreases lifetime. I also have a vague appreciation of the derivation of the Stern-Volmer equation that helps to tie this together.

What I can't wrap my mind around is how this occurs mechanistically based on the fact that we are only measuring electrons that decay via radiative pathways.

I can understand the reduction in fluorescent intensity that would occur, as fewer electrons will decay via the radiative pathway in the presence of a quencher.

But how does the addition of a new non-radiative pathway decrease the measured lifetime of the radiative decay pathway. After all we can't measure the lifetime or intensity of the non-radiative processes as there is nothing for the optode to see.

When we measure the fluorescent lifetime via an optode are we not just measuring k_r and if so how does the addition of k_Q/k_nr affect k_r?

Am I missing something obvious and fundamental? (I sure hope so!)

Thank you in advance,

Chesterfield

[Enthalpy]

Hi chesterfield, welcome here!

[...] we can't measure the lifetime or intensity of the non-radiative processes [...]

We can, it's even common. Have a laser pulse to excite the fluorescence, a fast detector to observe the fluorescence, and a quick oscilloscope to see the decay rate of the fluo light.

Laser pulses can be much shorter that 1ps, detectors few ps fast (perhaps better), oscilloscopes too. This covers the fluorescence time of many substances.

[...] lifetime is the inverse of the combined decay rate constants and how adding non-radiative decay pathways therefore decreases lifetime [...]

Hence additional processes reduce the lifetime. What was the claim: that the radiative decay gets faster, really? Or just that the excited state decays more quickly?

[chesterfield]

Hi Enthalpy,

Thank you for your reply and your kind welcome! I'm clearly having some kind of mental block so I hope you don't mind if I ask you some questions.

We can, it's even common. Have a laser pulse to excite the fluorescence, a fast detector to observe the fluorescence, and a quick oscilloscope to see the decay rate of the fluo light.

We only directly measure the radiative decay rate and intensity. We infer the non-radiative processes from this as we know the intensity of light we've put in. Is that correct?

Hence additional processes reduce the lifetime. What was the claim: that the radiative decay gets faster, really? Or just that the excited state decays more quickly?

So we introduce a new non-radiative decay pathway (quenching). Therefore the quantum yield of fluorescence will decrease. I know the equation for observed lifetime can be expressed as τ₀=Φ_f/k_f but I don't understand the relationship.

How does quantum yield effect the observed lifetime in terms of whats happening to the excited electron and how we're measuring the decay rate of the fluorescence? I thought it would reduce the initial intensity but not the intensity decay rate.

Thanks,

Chester

[Corribus]

So we introduce a new non-radiative decay pathway (quenching). Therefore the quantum yield of fluorescence will decrease. I know the equation for observed lifetime can be expressed as τ₀=Φ_f/k_f but I don't understand the relationship.

How does quantum yield effect the observed lifetime in terms of whats happening to the excited electron and how we're measuring the decay rate of the fluorescence? I thought it would reduce the initial intensity but not the intensity decay rate.

Imagine a big auditorium filled with a very large number of people and only one little exit that can accommodate one person leaving per second. Imagine also that everybody who leaves via this door, and only this door, has to pay you $1. The "quantum yield" of people leaving via that exit is 100%. Which is another way of saying that for every X amount of people that enter the room (the manner of entry doesn't matter), X people leave the room by the available exit. The amount of money you make from exit tolls after the room is completely vacated is $X.

Now suppose someone else opens a big door at the back of the room. From this exit 9 people can leave per second. Assuming the exit chosen by the average person in the room is randomly determined, the total rate of people now leaving the room per second is 10: 1 via the small door and 9 via the large door. The "quantum yield" for the small exit is now 10%. Which is another way of saying that for every X amount of people that enter the room, only 10% of them will leave via the small door. By opening a huge exit that can accommodate a larger rate of people leaving, you vastly increase the rate at which people vacate the room. Moreover, since 9/10 people leave out the rear door without paying, you only make 10% as much money.

Of course, it would be easy to count people leaving via each exit to do this experiment, just as it is easy to count the amount of paying people by counting the dollars in your hand. But suppose for a second you could only observe the amount of people in the room at a given time, with no knowledge of how many people are leaving by each particular exit. You would still notice a vast difference in the amount of people in the room at each moment in time in the second scenario versus the first, and if you were also privy to the amount of money you were making, you could therefore deduce the rate at which people were leaving through both the small door and the large door.

Also, it should be apparent that if we wanted to, we could express the average amount of time people spend in the auditorium in terms of the rates at which they leave. If you have 1000 people in the room initially, and 1 person leaves per second, it is a rather trivial matter to calculate the average amount of time a person spends in the room. This might be called the "lifetime" of staying in the auditorium. It should be readily apparent that once the rear door is opened, resulting in a much faster rate of people leaving, that the average amount of time each person spends in the auditorium is considerably less, and indeed we could calculate is based on the total rate of egress. Note that the average time the person spends in the room depends ONLY on the total number of people who can exit via both exits combined as a function of a time.

So, that's a little analogy that will maybe help you understand how these things work. The people are electrons and the auditorium represents an excited state. Leaving through the small door represents an electron that gives up a photon (pays a dollar), and the rear door represents a non-radiative decay route. When there is a nonradiative decay channel, it competes with the process of fluorescence: there is a probability per unit time that an excited state will decay by each available channel, and any excited state that decays nonradiatively also cannot decay by fluorescence, so it reduces the quantum yield accordingly.

The quantum yield* is the ratio of the amount of excited states that decay by emitting a photon to the total number of electrons in excited states "at time zero". (We can also express it equivalently as the number of photons out to the number of photons in - but this is a somewhat limiting definition because we can also talk about quantum yields of nonradiative processes.) Since all excited states much relax eventually, we can also express this as the ratio of the amount of excited states that decay by emitting a photon to the total number of electrons that decay by all channels. A final, and more experimentally useful, way we can express the quantum yield is in terms of rates. Essentially, if the quantum yield is equal to the total # of radiative events divided by the sum of the total # of radiative events and # of nonradiative events, we can divide both the numerator and denominator by a "unit time" to arrive at rates for the various processes. It is sometimes convention to deal with "lifetimes" rather than "rates"; the former are just the inverse of the latter. There's no particular reason to do this other than that it's more convenient to think in units of time rather than inverse time.**

The equivalence of these various expressions is demonstrated in the analogy above, where we can express/calculate the quantum yield in various ways: dollars collected divided by the amount of people in the auditorium, or the rate at which people are paying divided by the total rate at which they are leaving the auditorium. In both cases, the value is 10%. The only real difference between the auditorium scenario and an ensemble of fluorescing molecules is that that the kinetic profiles are different. In the former, we have stipulated a constant rate of egress, mostly for simplicity. The rate of decay of a body of excited electrons is exponential, varying with time. Fundamentally, this doesn't really change anything conceptual about the analogy, though.

One final thing. Heretofore I've been mostly discussing nonradiative decay in the context of intrinsic relaxation channels. That is, even at infinitely low concentrations, quantum yields are rarely 100% because there are nonradiative relaxation pathways resulting mostly from internal conversion mediated by vibrational molecular states. Quenching due to collisions brings nothing really new other than the fact that the kinetic profiles are different.

Hope that's helpful.

Notes:

*I think it's important for people to understand that although we often refer to the quantum yield as a "quantum efficiency" of a process, it is not an energy efficiency. Even molecules with quantum yields approaching 100% do not have 100% energy efficiency, by which I mean that 100% of the initially absorbed photon energy isn't converted into emitted photon energy. This is because there is almost always a Stokes shift - energy is lost due to excited-state structural rearrangement (the emission band is slightly to the red of the absorption band). Invariably this energy loss results in a miniscule heating of the sample. In fact, it is possible to measure the absolute quantum yield using sensitive calorimetric devices that measure this heating effect.

**It's worth noting that neither radiative nor nonradiative rates are really measured directly. The typical process is to measure the quantum yield and the overall fluorescence lifetime (total rate of decay of the excited state), using time-resolved spectroscopy, from which these other parameters can be then quantified indirectly.