In respect to fluorescence, quantum yield and quantum efficiency are the same thing, i.e. the ratio of photons absorbed to those emitted as fluorescence is referred to as 'fluorescence quantum yield' or 'quantum efficiency'.
Note that the term quantum efficiency is more usually applied to photosensitive devices, e.g. solar cells, for which its definition is different and thus not equivalent to the one for fluorescence quantum yield given above.
By definition, quantum yield is a rate constant that is obtained by dividing the rate of the process by the sum of all rates. Which essentially shows a fraction of times that a specific event will occur in relation to theoretical maximum of events that can occur. Or even more simply and less ‘sciency’ – what are the chances that molecules will behave the way you want/expect them to behave.
In the case of fluorescence, the mentioned ‘events’ are photons emitted and/or absorbed, and the ‘process’ is the emission of those absorbed photons in the form of fluorescence. As you may know (for example from looking at Jablonski’s diagram), a few different things can happen to a molecule (and its atoms) when it absorbs light, for example: intersystem crossing (between different electronic states), change of spin direction, loss of vibrational energy, and etc. Out of all these different things, the one we are interested in is, of course, fluorescence a.k.a. emission of absorbed photons at a longer wavelength a.k.a. the crossing of atoms from singlet state back to the ground state (while preserving the spin direction). An ideal fluorophore will have fluorescence as its dominant process a.k.a “main job”. It would spend all its time and energy on emitting absorbed photons in the shape of fluorescence and not waste time and energy on “side hustles” such as vibrational energy losses and intersystem crossing. And quantum yield shows how “committed” the fluorophore is to do its job. For example, take an ideal non-realistic scenario where fluorophore’s quantum yield (Φ) = 1.0 or 100%. This means that out of ALL the photons absorbed, all 100% of them will be emitted as fluorescence. Hence, pretty much, the higher the quantum yield – the brighter the fluorophore will appear – the higher the efficiency of the process.
Not sure if this was clear enough, but essentially quantum yield measures the rate/pace/speed of the process in some sense. Whereas the efficiency measures how good and reliable that process is when taking into account energy input and the ultimate loss in the aforementioned “side hustles”.
Tl;dr : quantum yield = rate constant of the process, quantum efficiency = how good and reliable the process is.
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