Lesson 3 · 30 min

Fluorescence: how molecules give light back

Tonic water glows blue under an ultraviolet lamp. The Jablonski diagram shows the routes an absorbed photon can take; a virtual time-resolved fluorometer measures quinine’s lifetime, and the Stern–Volmer equation shows how chloride puts it out.

In this lesson

Follow an absorbed photon on a Jablonski diagram: vibrational relaxation, internal conversion, fluorescence, intersystem crossing and phosphorescence
Calculate excited-state lifetimes and fluorescence quantum yields from rate constants, and explain the Stokes shift
Obtain a quenching rate constant from lifetime data with the Stern–Volmer equation, and tell dynamic from static quenching
Explain why fluorimetry is so sensitive, where its linear range ends, and what interferes

Hold a bottle of tonic water under an ultraviolet lamp and it glows an eerie blue. The quinine in it absorbs invisible ultraviolet light (about 350 nm) and a few nanoseconds later gives out longer-wavelength blue light (about 450 nm). The absorbed photon carries more energy than the emitted one; the difference becomes heat, and the red shift is called the Stokes shift. Fluorescence measures a little light against darkness, not a small dip in a bright beam, which makes it far more sensitive than absorbance — the workhorse for labelling molecules in biochemistry.

Predict first

Add a little table salt (NaCl) to an acidified quinine solution. Under the ultraviolet lamp, what happens to the fluorescence?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Brighter: the ionic strength rises Unchanged: neither Na⁺ nor Cl⁻ absorbs ultraviolet light Dimmer: chloride deactivates the excited state early It turns green

The Jablonski diagram

Absorbing a photon takes about 10⁻¹⁵ s and lands the molecule in some vibrational level of S₁ or a higher singlet. Within about 10⁻¹² s it hands the excess vibrational energy to the solvent and drops to the lowest vibrational level of S₁ — so fluorescence always comes from the same level whatever the excitation wavelength (Kasha’s rule), and at longer wavelength than the absorption. Then several processes compete: fluorescence (a photon comes out), internal conversion (the energy becomes heat on the way back to S₀) and intersystem crossing (a spin flip into the triplet T₁). Getting from T₁ back to S₀ needs another spin flip, which is “forbidden”, so phosphorescence can last milliseconds to seconds.

Every channel is first order, and each takes a share in proportion to its rate constant: the fluorescence quantum yield is the radiative rate constant’s share of the total, and the lifetime is the reciprocal of the total.

\tau = \frac{1}{k_\text{r} + k_\text{ic} + k_\text{isc}}, \qquad \Phi_\text{F} = k_\text{r}\,\tau

Competing first-order decays: all channels together set the lifetime, and a yield is one channel’s share.

$\tau$: excited-state lifetime
$k_\text{r}$: radiative (fluorescence) rate constant
$k_\text{ic}, k_\text{isc}$: rate constants of internal conversion and intersystem crossing
$\Phi_\text{F}$: fluorescence quantum yield

Competing first-order decays · lifetime and quantum yields computed from rate constants

Jablonski diagram: where an absorbed photon goes

radiative k_r = 10^7.5 s⁻¹ internal conversion k_ic = 10^7.3 s⁻¹ intersystem crossing k_isc = 10^4.0 s⁻¹ phosphorescence k_p = 10^1.0 s⁻¹ triplet non-radiative k_T = 10^3.0 s⁻¹

S₁ lifetime τ = 20.4 ns; Φ_F = 0.600, Φ_IC = 0.400, Φ_ISC = 0.000; T₁ lifetime 990 μs, phosphorescence yield Φ_P = 0.000.

Of every 1000 photons absorbed: 600 come back as fluorescence, 0 as phosphorescence, and the rest ends up as heat.

Drag the rate constants and watch the arrow widths (quantum yields) and the lifetime. Speed up internal conversion tenfold and the fluorescence yield and the lifetime fall together — fluorescence and lifetime always move in step. Switch to “mostly intersystem crossing”: nearly every excited molecule ends up in the triplet, and the phosphorescence lasts milliseconds.

Worked example

The radiative rate constant of quinine

Quinine has a fluorescence lifetime of about 20.4 ns and a fluorescence quantum yield of about 0.60. Find the radiative rate constant and the total rate constant of the other channels.

Quenching and the Stern–Volmer equation

Every collision with a quencher Q may send the excited state back to the ground state, adding a decay channel of rate $k_\text{q}[\text{Q}]$ . The intensity and the lifetime then fall in the same proportion: $F_0/F$ and $\tau_0/\tau$ are both straight lines against [Q] with slope $K_\text{SV} = k_\text{q}\tau_0$ . Gutow measured quinine’s decay after laser pulses at several chloride concentrations and found $k_\text{q} = 5.2\times10^9\ \text{M}^{-1}\,\text{s}^{-1}$ ; the diffusion limit $8RT/3\eta$ computed from the solution viscosity is $5.8$ – $6.5\times10^9$ , so nearly every collision quenches.

The other possibility is static quenching: the quencher binds a ground-state molecule into a dark complex. Fewer molecules emit, but those that do keep their full lifetime — intensity alone cannot tell the two mechanisms apart, lifetimes can.

\frac{F_0}{F} = \frac{\tau_0}{\tau} = 1 + k_\text{q}\tau_0[\text{Q}] = 1 + K_\text{SV}[\text{Q}]

The Stern–Volmer equation for dynamic quenching. With static quenching only F₀/F rises with [Q]; τ stays the same.

$F_0, F$: fluorescence intensity without and with quencher
$k_\text{q}$: bimolecular quenching rate constant
$K_\text{SV}$: Stern–Volmer constant

337 nm pulsed excitation · detection at 480 nm · computed with the parameters of Gutow (2005)

Time-resolved fluorometer: quinine and chloride

Sample: quinine in 0.5 M H₂SO₄ with added chloride

log scale

Fluorescence decay

signal
exponential fit to the tail
instrument response

This shot: τ = 20.6 ns (fitted over 420 points)

steady-state fluorescence F/F₀ = 1.000

Stern–Volmer plot

1/τ
diffusion limit 8RT/3η

Measure at least two concentrations to draw a line.

Fire the laser at each chloride concentration: the tail of the decay is fitted for τ and 1/τ goes onto the Stern–Volmer plot. With a few points the slope is

k_\text{q}

— compare it with the diffusion limit (violet band). Then switch to the “dark complex”: the fluorescence dims, yet the lifetime does not budge.

Concept check

A quencher halves a dye’s fluorescence intensity, but its lifetime, measured by time-resolved fluorescence, is unchanged. What kind of quenching is this?

dynamic (collisional) quenching static quenching: a dark complex has formed the inner-filter effect faster intersystem crossing

Try the calculation

Chloride quenches quinine with $K_\text{SV} = 106\ \text{M}^{-1}$ . What chloride concentration (mM) halves the fluorescence?

Key takeaways

An absorbed photon relaxes to the bottom of S₁ and then leaves by competing fluorescence, internal conversion or intersystem crossing; fluorescence lies at longer wavelength than absorption (the Stokes shift).
Lifetime τ = 1/Σk and quantum yield Φ_F = k_r τ: any added decay channel shortens the lifetime and lowers the yield together.
For dynamic quenching F₀/F = τ₀/τ = 1 + k_qτ₀[Q]; chloride quenches quinine with k_q = 5.2 × 10⁹ M⁻¹ s⁻¹, close to the diffusion limit. Static quenching lowers the intensity but not the lifetime.
Fluorimetry is sensitive but linear only at low absorbance (the inner-filter effect), and quenchers must be guarded against.

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Data and references

J. H. Gutow, “Halide (Cl⁻) quenching of quinine sulfate fluorescence: a time-resolved fluorescence experiment for physical chemistry”, J. Chem. Educ. 82, 302–305 (2005).
A. M. Brouwer, “Standards for photoluminescence quantum yield measurements in solution (IUPAC Technical Report)”, Pure Appl. Chem. 83, 2213–2228 (2011).
K. Nawara & J. Waluk, “Goodbye to quinine in sulfuric acid solutions as a fluorescence quantum yield standard”, Anal. Chem. 91, 5389–5394 (2019).
J. R. Lakowicz, Principles of Fluorescence Spectroscopy, 3rd ed. (Springer, 2006), Ch. 1, 8 (Jablonski diagram; quenching).