Fluorescence and Phosphorescence

Glow-in-the-dark stickers keep shining for minutes after the lights go off. A fluorescent highlighter, however, goes dark the instant you stop illuminating it. Both effects start with the same event. A molecule absorbs a photon and climbs to an excited electronic state. What happens next diverges sharply. That divergence is written entirely in spin. Fluorescence and phosphorescence are the two radiative routes back to the ground state. They differ in mechanism, timescale, and spin multiplicity by many orders of magnitude. Understanding why requires the Jablonski diagram. It also requires the distinction between singlet and triplet states. Finally, it requires the quantum-mechanical selection rules that make one process fast and allowed and the other slow and forbidden. This article builds the Jablonski diagram from first principles. It derives the rate laws governing excited-state decay. It works through two fully numeric examples. It assumes you already know the Franck–Condon principle and the basics of absorption and emission. It sets up the vocabulary of spin and relaxation that reappears in magnetic resonance spectroscopy.

Key takeaways

Fluorescence is spin-allowed emission from $S_1 \to S_0$ , occurring in nanoseconds.
Phosphorescence is spin-forbidden emission from $T_1 \to S_0$ , occurring in microseconds to hours.
Intersystem crossing (ISC) converts singlet excitation into triplet excitation via spin-orbit coupling.
Kasha's rule says emission almost always occurs from the lowest excited state of a given multiplicity.
The observed lifetime and quantum yield depend on the competition between radiative and non-radiative rate constants.
Key idea: the huge lifetime gap between fluorescence and phosphorescence is a direct measure of how strongly spin selection rules are broken by spin-orbit coupling.

The core insight

Every excited electronic state decays by competing pathways, and whether that decay is fast fluorescence, slow phosphorescence, or silent heat depends on one variable: whether the transition conserves total electron spin, which sets the radiative rate constant across many orders of magnitude.

The physical picture

When a molecule absorbs a photon, an electron is promoted from a bonding or non-bonding orbital into a higher-energy orbital, but its spin does not flip during this process. The molecule starts in a singlet ground state $S_0$ , with all electron spins paired, and absorption produces an excited singlet state $S_n$ , still with paired spins. This is a spin-allowed transition, governed by the same selection rules and Franck–Condon overlaps that describe ordinary UV-visible absorption. Once in $S_n$ , the molecule almost never emits directly. Instead it cascades down through closely spaced vibrational and electronic levels by internal conversion (IC), a fast, radiationless, spin-conserving process, until it reaches the vibrational ground level of $S_1$ . This relaxation typically takes $10^{-12}$ to $10^{-10}\ \text{s}$ , far faster than emission, so essentially all molecules empty into $S_1$ before doing anything else. This is Kasha's rule: fluorescence emission almost always originates from $S_1$ , regardless of which excited state absorbed the photon. From $S_1$ , three things can compete. The molecule can emit a photon and drop to some vibrational level of $S_0$ : this is fluorescence, spin-allowed and fast, with a natural radiative lifetime typically $1$ – $100\ \text{ns}$ . It can undergo more internal conversion directly to $S_0$ , releasing the energy as heat. Or it can undergo intersystem crossing (ISC) to a triplet state $T_1$ , which requires a spin flip and is formally forbidden by the spin selection rule $\Delta S = 0$ . ISC is made possible by spin-orbit coupling, an interaction between an electron's spin magnetic moment and the magnetic field generated by its own orbital motion, which mixes a small amount of triplet character into nominally singlet states (and vice versa). Once in $T_1$ , the molecule is trapped by the same selection rule that let it in: direct radiative decay $T_1 \to S_0$ also requires a spin flip. This transition, phosphorescence, is allowed only because spin-orbit coupling again mixes a little singlet character into $T_1$ . Because the mixing is weak, the transition dipole is tiny, the radiative rate constant is small, and the triplet state can survive for microseconds to seconds, or even hours in rigid, oxygen-free matrices at low temperature.

Figure

Jablonski diagram: absorption, internal conversion, fluorescence, ISC, and phosphorescence

The second useful picture is a rate-constant scoreboard: it is the relative sizes of $k_r$ , $k_{nr}$ , and $k_{ISC}$ , not just their existence, that determine whether a molecule is a good fluorophore or a good phosphor.

Figure

Relative timescales of excited-state processes on a logarithmic axis

The core relations

\Phi_f = \frac{k_r}{k_r + k_{nr} + k_{ISC}}

where $\Phi_f$ is the fluorescence quantum yield, $k_r$ the radiative rate constant of $S_1 \to S_0$ , $k_{nr}$ the non-radiative internal conversion rate constant, and $k_{ISC}$ the intersystem crossing rate constant.

\tau_f = \frac{1}{k_r + k_{nr} + k_{ISC}}

where $\tau_f$ is the observed fluorescence lifetime, the time for $S_1$ population to fall to $1/e$ of its initial value.

\Phi_{ISC} = \frac{k_{ISC}}{k_r + k_{nr} + k_{ISC}}

where $\Phi_{ISC}$ is the fraction of excited molecules that cross into the triplet manifold instead of returning directly to $S_0$ .

\Phi_p = \Phi_{ISC}\left(\frac{k_r^{T}}{k_r^{T} + k_{nr}^{T}}\right)

where $\Phi_p$ is the phosphorescence quantum yield, $k_r^T$ and $k_{nr}^T$ are the radiative and non-radiative rate constants out of $T_1$ .

I(t) = I_0 e^{-t/\tau}

where $I(t)$ is the emission intensity at time $t$ after excitation and $\tau$ is the corresponding excited-state lifetime (use $\tau_f$ for fluorescence, $\tau_p$ for phosphorescence).

Derivation AFirst-order decay of an excited state and the quantum yield

Goal. Show that excited-state population decays exponentially, and derive the quantum yield and lifetime expressions from the branching ratios of competing first-order pathways.

Step 1. Population balance. Let $N_1(t)$ be the number of molecules in $S_1$ at time $t$ after a short excitation pulse populates $N_1(0) = N_0$ . Each decay channel — fluorescence, internal conversion, intersystem crossing — is a unimolecular, first-order process because it depends only on the excited molecule itself, not on collisions or concentration (in the low-concentration, non-quenched limit).

Step 2. Rate law. The total rate of loss of $S_1$ population is the sum of the individual first-order rates:

-\frac{dN_1}{dt} = (k_r + k_{nr} + k_{ISC})\,N_1 = k_{obs}\,N_1

Step 3. Integrate. Separating variables and integrating from $0$ to $t$ :

N_1(t) = N_0\, e^{-k_{obs} t}

Since emitted intensity is proportional to the instantaneous population undergoing radiative decay, $I(t) \propto N_1(t)$ , so the observed decay is a single exponential with lifetime $\tau_f = 1/k_{obs}$ .

Step 4. Quantum yield as a branching ratio. The quantum yield is the fraction of excited molecules that emit a fluorescence photon rather than decaying by any other channel. Because all channels draw from the same population with fixed relative rate constants, the branching fraction is simply the ratio of that one rate constant to the total:

\Phi_f = \frac{k_r}{k_r + k_{nr} + k_{ISC}} = k_r \tau_f

This last identity, $\Phi_f = k_r \tau_f$ , is extremely useful experimentally: measure $\tau_f$ by time-resolved emission and $\Phi_f$ by comparison to a reference standard, and $k_r$ (and hence $k_{nr}+k_{ISC}$ ) falls out immediately.

Step 5. Extend to the triplet manifold. Once in $T_1$ , the same logic applies with a new pair of rate constants $k_r^T$ and $k_{nr}^T$ (which now includes non-radiative return to $S_0$ via a second, reverse spin-orbit-mediated process). The population reaching $T_1$ in the first place is $\Phi_{ISC} N_0$ , so the phosphorescence quantum yield is that population fraction times the triplet's own branching ratio:

\boxed{\Phi_p = \Phi_{ISC}\left(\frac{k_r^{T}}{k_r^{T}+k_{nr}^{T}}\right), \qquad \tau_p = \frac{1}{k_r^{T}+k_{nr}^{T}}}

Worked examples

Example 1 — Foundational: extracting rate constants from lifetime and quantum yield. Anthracene in ethanol has a measured fluorescence quantum yield $\Phi_f = 0.27$ and a fluorescence lifetime $\tau_f = 4.9\ \text{ns}$ (representative literature values). Find $k_r$ , and the total non-radiative rate constant $k_{nr} + k_{ISC}$ . From the identity derived above,

k_r = \frac{\Phi_f}{\tau_f} = \frac{0.27}{4.9\times10^{-9}\ \text{s}} = 5.5\times10^{7}\ \text{s}^{-1}

The total decay rate constant is

k_{obs} = \frac{1}{\tau_f} = \frac{1}{4.9\times10^{-9}\ \text{s}} = 2.04\times10^{8}\ \text{s}^{-1}

So the combined non-radiative channels account for

k_{nr} + k_{ISC} = k_{obs} - k_r = 2.04\times10^{8} - 0.55\times10^{8} = 1.49\times10^{8}\ \text{s}^{-1}

Sanity check: $k_r$ should be comparable to the inverse of a "natural" radiative lifetime estimated from the integrated absorption cross-section (typically $10^7$ – $10^9\ \text{s}^{-1}$ for strongly allowed $\pi\pi^*$ transitions), and $5.5\times10^7\ \text{s}^{-1}$ falls squarely in that range, consistent with anthracene's intense, allowed absorption band. The non-radiative rate is comparable in magnitude, which is why $\Phi_f$ is well below 1 even for this efficient fluorophore.

Example 2 — Applied: comparing fluorescence and phosphorescence lifetimes and estimating ISC efficiency. A rigid aromatic ketone in a frozen glass at 77 K shows fluorescence with $\tau_f = 8.0\ \text{ns}$ and $\Phi_f = 0.045$ , and phosphorescence with $\tau_p = 1.6\times10^{-2}\ \text{s}$ and $\Phi_p = 0.38$ . Estimate $\Phi_{ISC}$ , and the radiative and non-radiative rate constants of the triplet state, then compare the size of $k_r^T$ to $k_r$ for the singlet.

Step 1 — total singlet decay rate.

k_{obs} = \frac{1}{\tau_f} = \frac{1}{8.0\times10^{-9}\ \text{s}} = 1.25\times10^{8}\ \text{s}^{-1}

Step 2 — singlet radiative rate.

k_r = \frac{\Phi_f}{\tau_f} = \frac{0.045}{8.0\times10^{-9}\ \text{s}} = 5.6\times10^{6}\ \text{s}^{-1}

Aromatic ketones have a low $\Phi_f$ and low $k_r$ because the lowest singlet is often $n\pi^*$ in character, a weakly allowed transition — this is exactly why they are efficient phosphors: weak fluorescence competition leaves more population available for ISC.

Step 3 — estimate $\Phi_{ISC}$ . For a molecule with negligible direct $S_1 \to S_0$ internal conversion (a reasonable approximation for many carbonyl compounds, where ISC dominates over IC), essentially all the non-fluorescing population crosses to the triplet:

\Phi_{ISC} \approx 1 - \Phi_f = 1 - 0.045 = 0.955

Step 4 — triplet branching ratio and rate constants. From $\Phi_p = \Phi_{ISC}\left(\dfrac{k_r^T}{k_r^T+k_{nr}^T}\right)$ ,

\frac{k_r^T}{k_r^T + k_{nr}^T} = \frac{\Phi_p}{\Phi_{ISC}} = \frac{0.38}{0.955} = 0.398

Since $\tau_p = 1/(k_r^T + k_{nr}^T)$ ,

k_r^T + k_{nr}^T = \frac{1}{1.6\times10^{-2}\ \text{s}} = 62.5\ \text{s}^{-1}

k_r^T = 0.398 \times 62.5\ \text{s}^{-1} = 24.9\ \text{s}^{-1}, \qquad k_{nr}^T = 62.5 - 24.9 = 37.6\ \text{s}^{-1}

Step 5 — compare radiative rates.

\frac{k_r}{k_r^T} = \frac{5.6\times10^{6}\ \text{s}^{-1}}{24.9\ \text{s}^{-1}} \approx 2.2\times10^{5}

The singlet radiative rate constant is roughly 200,000 times larger than the triplet's, quantitatively confirming that spin-orbit coupling makes the $T_1 \to S_0$ transition weakly, but not fully, forbidden. This factor of $10^5$ – $10^6$ is the typical scale separating spin-allowed from spin-forbidden radiative transitions, and it is the same physical origin — spin-orbit coupling — that governs hyperfine and relaxation phenomena in electron spin resonance.

Limiting cases & common traps

Situation	What actually happens
Heavy atoms present (Br, I, Pb)	Spin-orbit coupling scales roughly as $Z^4$ ; $k_{ISC}$ and $k_r^T$ both rise sharply (the "heavy-atom effect"), often boosting phosphorescence at the expense of fluorescence
Dissolved O $_2$ present	Triplet oxygen efficiently quenches $T_1$ collisionally; phosphorescence is usually only observed after rigorous degassing or in a rigid glass
Rigid glass or crystal at low $T$	Suppresses vibrational and collisional non-radiative decay, so $k_{nr}^T$ drops and phosphorescence becomes observable on human timescales
Excitation into $S_2$ or higher	Kasha's rule predicts emission still occurs from $S_1$ , with the emission spectrum independent of excitation wavelength
Very small $S_1$ – $T_1$ gap	Enhances ISC (better spin-orbit/vibronic coupling via smaller energy denominator), common in heteroaromatic and carbonyl compounds

The chemist behind it

Alexander Jablonski, a Polish physicist, introduced the energy-level diagrams bearing his name in the 1930s to explain why fluorescence lifetimes, polarization, and quantum yields depend so strongly on solvent viscosity and temperature. His work established that excited states relax through a hierarchy of radiative and non-radiative pathways before emission — precisely the framework used throughout this article to organize $S_0$ , $S_1$ , and $T_1$ and the transitions between them. Gilbert N. Lewis and Michael Kasha later provided the chemical and photophysical evidence identifying the long-lived phosphorescent state as a triplet, cementing the spin-based interpretation still taught today.

Origins · 1 minAlexander Jablonski and how this was discovered

Alexander Jablonski

1898-1980 · Polish physicist who devised the energy-level diagram used to organize absorption, internal conversion, fluorescence, intersystem crossing, and phosphorescence

Ask the tutor: How does the spin-orbit coupling responsible for phosphorescence connect to the electron and nuclear spin phenomena measured in ESR and NMR spectroscopy?

Why is phosphorescence lifetime so variable, from microseconds to hours?

The radiative rate constant $k_r^T$ is fixed by how much singlet character spin-orbit coupling mixes into $T_1$ , but the non-radiative rate $k_{nr}^T$ is extremely sensitive to environment. Collisions, molecular vibrations, and quenchers like oxygen open fast non-radiative channels that dominate $\tau_p = 1/(k_r^T+k_{nr}^T)$ . In a rigid, degassed glass at low temperature these channels are suppressed, so $k_{nr}^T$ shrinks and $\tau_p$ can stretch to seconds or longer; in fluid solution at room temperature, quenching usually wins and phosphorescence is barely detectable.

Can a molecule fluoresce and phosphoresce at the same time?

Yes. After excitation, $S_1$ population splits between fluorescence, internal conversion, and intersystem crossing according to their rate constants. The fraction that crosses to $T_1$ then produces phosphorescence, delayed relative to the fluorescence and typically red-shifted since $T_1$ lies below $S_1$ . In many rigid or heavy-atom-containing samples both emissions are recorded from the same excitation pulse, distinguished by their very different decay timescales and by turning off the excitation and watching which glow persists.

How does the Franck-Condon principle apply to fluorescence and phosphorescence?

Both emissive transitions are vertical in the same sense as absorption: the nuclear geometry does not change during the instant of photon emission, so the emitted photon's vibronic intensity pattern reflects Franck-Condon overlaps between the vibrational wavefunctions of the emitting state and the ground state. Because $S_1$ (or $T_1$ ) has already relaxed to its equilibrium geometry via internal conversion before emitting, fluorescence and phosphorescence spectra are typically mirror images of the absorption spectrum, shifted to lower energy (the Stokes shift).

Why do heavy atoms increase phosphorescence yield?

Spin-orbit coupling strength grows steeply with atomic number, roughly as $Z^4$ for the relevant nuclear charge, because it originates from the interaction between an electron's spin and the magnetic field from its orbital motion around a charged nucleus. Substituting heavy atoms such as bromine, iodine, or heavy metals directly onto a chromophore, or dissolving it in a heavy-atom solvent, increases both $k_{ISC}$ and $k_r^T$ , often converting a purely fluorescent molecule into an efficient phosphor. This same coupling mechanism underlies relaxation processes probed in electron spin resonance.