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The other day, I noticed a strip of fluorescent tape when the lights in the room were turned off.

The energy was discharged as a continued, faint glow, rather than one burst or flash of light.

How so? What prevents the molecules from de-exciting all at once?

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    $\begingroup$ Appears to be phosphorescence rather than fluorescence. The former can last minutes or longer, e.g., glow in the dark products, while the latter typically lasts less than milliseconds. Look up phosphorescence in wikipedia. $\endgroup$
    – Ed V
    Commented Oct 27, 2021 at 0:35
  • $\begingroup$ related chemistry.stackexchange.com/questions/31286/… $\endgroup$
    – Mithoron
    Commented Oct 28, 2021 at 15:27

4 Answers 4

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The after-glow is called phosphorescence. Fluorescence stops immediately the moment light is turned off because the process is ultrafast!

Phosphorescence on the other hand requires a change in the spin of the electron in a molecule, such transitions are forbidden, this is why it is slow. You can check the Jablonski diagram to understand this better, see here https://micro.magnet.fsu.edu/optics/timeline/people/jablonski.html

Can you see how many rates and competing processes are involved?

Jablonski diagram

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    $\begingroup$ It is not (strictly) true that fluorescence is instantaneous; many types of molecules fluoresce with lifetimes of hundreds of nanoseconds (although most lifetimes are shorter than this) and phosphorescence in some types molecules is shorter than hundreds of nanosec, but in a given type of molecule fluorescence is always faster than phosphorescence, due to the spin forbidden nature of the latter process. $\endgroup$
    – porphyrin
    Commented Oct 28, 2021 at 7:38
  • $\begingroup$ Yes, I agree. One can see in the figure that there is "delayed" fluorescence too, along with their lifetimes but the OP has not responded at all. $\endgroup$
    – ACR
    Commented Oct 28, 2021 at 15:10
  • $\begingroup$ Yes, but delayed fluorescence is rather rare, eosin etc, as triplets are usually too low in energy and internal conversion between them too fast for this to be significant. However, there has been more research on this latterly to improve fluorescence yields when in devices. There is also T-T annihilation producing fluorescence as well as excimers and so forth. $\endgroup$
    – porphyrin
    Commented Oct 29, 2021 at 7:52
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Phosphorescence is driven by the slow loss of energy from excited electron states

Firstly, the phenomenon you describe is phosphorescence not fluorescence: they have different mechanisms.

Fluorescence happens (to simplify a little) when light kicks electrons from their normal orbitals in some molecules to a higher excited state orbital (often via a multi-step pathway involving multiple orbitals). Those electrons drop back to their low state very quickly, emitting specific colours of light.

Phosphorescence is more complex. The basic idea is the same: light excites some electrons to higher energy orbitals. But, in this case, there is an internal mechanism that rapidly allows the electron to drop into an orbital state where there is a major barrier to the transition back to the ground state (specifically one that involves changes in the spin). These transitions are "forbidden" in some cases for symmetry and quantum reasons. But even "forbidden" transitions can happen but only when the exact molecular shape and circumstances are right. In some cases this happens because of a rare combination of molecular vibrations and these happen far more slowly than normal non-forbidden electronic transitions.

The overall effect is that, in some molecules, electrons can get "stuck" in higher energy orbitals until some much slower internal process breaks the rules that disallow the transition back to the ground state. This means that the decay process is billions or trillions of times slower than fluorescence and can take minutes or hours. So you can often see the light emission from the transition decaying slowly over minutes or hours.

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Although the previous answers correctly explain the different mechanisms of fluorescence and phosphorescence, they do not really touch upon why the light is not emitted all at once. Even in the very fast fluorescence, not all light is emitted in one burst (although one needs fast detectors to see this).

The reason for the slow de-excitation of the molecules is that each of the excited states has roughly the same and independent chance to de-excite under emission of a photon at a certain time. Compare this with tossing a hand of coins, removing all the coins that show head and tossing the coins again that show tail. After the first toss, roughly half of the coins is removed, after two tosses, 75% of the coins is removed etc. Replace the coins with excited molecules and the tosses with increasing time, and you see why not all molecules emit light at the same time: the probability for de-excitation is less than unity and you get an exponential decay of the excited states. The difference between fluorescence and phosphorence rates is that states that can decay by fluorescence have a much higher chance to do so than states that have to decay via phosphorence.

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To answer why the energy is not emitted as a burst we have consider the chance of a photon being emitted.

To match experimental data we have to assume that each event (photon emission) is randomly occurring and is independent of all others and that each occurs in a non-overlapping time interval (i.e. its own). This means that in a time interval $\delta t$ (a) the chance of only one event being recorded is $\lambda \delta t$ where $\lambda$ is a positive number, (b) The probability that zero events happen is then $1-\lambda\delta t$ and (c) the probability that two or more events occur is zero, which means that $ \delta t$ has to be very small. The constant $\lambda$ has units of 1/time and is usually called the rate constant.

The chance of no event occurring at time $t\to t+\delta t$ (where $t$ is fixed) is

$$\displaystyle p(t+\delta t)=(1-\lambda)\delta tp(t)$$

which is the product of the chance of no events up to time $t$ and of no events in time $t\to\delta t$. Rearranged this is

$$\displaystyle \frac{p(t+\delta t)-p(t)}{\delta t}=-\lambda p(t)$$

and in the limit of small time increment is the first order differential equation;

$$\displaystyle \frac{dp(t)}{dt}=-\lambda p(t)$$

Integrating and multiplying by $N_0$ the number of molecules initially present that can react, fluoresce or decay radioactively give the familiar first order decay, $N=N_0e^{-\lambda t}$ where $\lambda$ is the rate constant for the process and its reciprocal the lifetime.

In your example the emission is long lived because the transition is 'forbidden'. Usually this is because the spin of the ground state is different to that of the excited state in which the molecule is now in. The spin forbidden process can be overcome by an interaction that changes the angular momentum such as spin-orbit coupling and if this is weak the excited state decays slowly. Paramagnetic species or heavy atoms will induce spin-orbit coupling.

Initial absorption is usually a spin allowed process but in the excited state interactions can cause a fraction of molecules to cross to a triplet state (spin 1) so a transition is forbidden to the ground state (spin 0) in an aromatic molecule for example. In molecules containing transition metals the spin state can be complicated by the spin of the metal and in general then we would refer to emission as luminescence, and keep fluorescence for spin allowed process and phosphorescence for triplet to singlet transitions.

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