Absorption: up the ladder in a femtosecond
When a photon's energy matches the gap between a molecule's ground state (S0) and one of its excited singlet states (S1, S2, …), it can be absorbed almost instantaneously, promoting one electron upward while its spin stays paired with its ground-state partner — that pairing is what makes it a singlet state. This absorption step, the Franck-Condon transition, happens on a timescale far faster than the nuclei can move, so it launches the molecule into an excited electronic state at essentially its ground-state geometry.
The Jablonski diagram
The Jablonski diagram is the standard map of everything that can happen after absorption: a ladder of electronic states (S0, S1, S2, and a triplet manifold T1) each with their own stack of vibrational sub-levels, connected by arrows for absorption, radiative decay (fluorescence, phosphorescence) and non-radiative decay (internal conversion, intersystem crossing, vibrational relaxation).
Typical excited-state lifetimes: fluorescence (S1 → S0, spin-allowed) τ ~ 1 ns – 100 ns phosphorescence (T1 → S0, spin-forbidden) τ ~ 1 ms – seconds (or longer) τ(phosphorescence) >> τ(fluorescence)
Internal conversion and Kasha's rule
If the molecule was excited to S2 or higher, it almost always drops down to S1 first through fast, non-radiative internal conversion, then relaxes to the lowest vibrational level of S1 through vibrational relaxation — both steps typically complete in picoseconds, dumping their energy as heat into the surroundings before any light is emitted. That near-universal funnelling down to the lowest vibrational level of the lowest excited state before emission is Kasha's rule, and it's why a molecule's emission spectrum barely depends on which wavelength originally excited it.
Fluorescence: the fast, spin-allowed exit
From the relaxed S1 state, the molecule can drop straight back to S0 by emitting a photon — fluorescence. Because both S1 and S0 are singlet states, this transition is spin-allowed and happens quickly, typically within a few to a few hundred nanoseconds, which is why fluorescent glow disappears essentially instantly once the excitation light is switched off.
Phosphorescence: crossing to the triplet
Some molecules instead undergo intersystem crossing from S1 into the triplet state T1, where the excited electron's spin has flipped to become unpaired with the rest of the molecule. That crossing is formally spin-forbidden and only occurs at all because of weak spin-orbit coupling, so it competes inefficiently with fluorescence — but once a molecule is trapped in T1, getting back down to S0 by emitting a photon (phosphorescence) is also spin-forbidden by the same rule, and correspondingly slow. That double spin-forbidden bottleneck is exactly why phosphorescent afterglow — glow-in-the-dark materials being the everyday example — can persist for milliseconds, seconds, or far longer after the excitation source is removed.
The Stokes shift
Because vibrational relaxation bleeds off energy as heat before any photon is emitted, the emitted photon always carries less energy — and therefore a longer wavelength — than the absorbed one. That gap between absorption and emission peaks is the Stokes shift, and it's the practical reason fluorescence microscopy and dye-based sensing work at all: a filter can separate the shorter-wavelength excitation light from the longer-wavelength emitted signal cleanly.
Frequently asked questions
Why is phosphorescence so much slower than fluorescence?
Fluorescence (S1 to S0) is spin-allowed — the electron's spin doesn't need to flip — so it happens quickly, in nanoseconds. Phosphorescence (T1 to S0) requires a spin flip, which is formally forbidden by quantum mechanical selection rules and only occurs weakly through spin-orbit coupling. That weak coupling makes the transition far less probable per unit time, stretching its characteristic lifetime out to milliseconds, seconds, or in some phosphors even longer.
Why does emitted light always have a longer wavelength than the light that was absorbed?
Between absorption and emission, a molecule almost always loses some energy as heat through fast, non-radiative vibrational relaxation and internal conversion down to the lowest vibrational level of its emitting state (Kasha's rule). The emitted photon therefore carries less energy, and correspondingly a longer wavelength, than the absorbed one — the gap between the two is the Stokes shift, and it's essentially universal in fluorescence and phosphorescence.
Does the emission wavelength depend on which wavelength excited the molecule?
Usually not. Kasha's rule says that internal conversion and vibrational relaxation to the lowest vibrational level of S1 (or T1 for phosphorescence) happen so fast — typically picoseconds — that essentially all the excess energy from exciting into a higher state is lost as heat before any emission occurs. As a result the emission spectrum's shape and position are largely independent of the excitation wavelength, even though the absorption spectrum spans a wide range.
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