Photochemistry and Quantum Yield

Point a laser at a bottle of chlorine gas and hydrogen gas, and nothing happens in the dark. Shine light on it, and it explodes. This dramatic contrast is the essence of photochemistry: light does not just heat a reaction, it initiates one, atom by atom, photon by photon. Photochemistry matters because sunlight drives the chemistry that keeps us alive. Photosynthesis, vision, the ozone layer, and industrial polymer curing all hinge on how efficiently absorbed photons convert into chemical change. That efficiency has a name: the quantum yield. This article builds the two foundational laws of photochemistry. It then defines primary and overall quantum yields precisely. Finally, it shows how quantum yields link photon absorption rates to measurable reaction rates, including the classic chain-reaction case where yields exceed one.

Key takeaways

The Grotthuss–Draper law: only light actually absorbed by a molecule can produce a photochemical change.
The Stark–Einstein law: each absorbed photon activates (in the primary step) exactly one molecule.
The primary quantum yield $\Phi_i$ counts molecules undergoing a specific primary photophysical or photochemical event per photon absorbed.
The overall quantum yield $\Phi$ counts final product molecules formed (or reactant molecules consumed) per photon absorbed, and can exceed 1 in chain reactions.
Photochemical rate laws couple the absorbed photon flux, via the Beer–Lambert law, to $\Phi$ to predict $d[\text{P}]/dt$ .
Key idea: quantum yield converts photons absorbed into chemical outcomes, turning a light intensity into a reaction rate.

The core insight

A quantum yield is a molecular efficiency ratio, the number of defined events (a primary process, or a final product molecule) divided by the number of photons absorbed, and it is the single number that connects photophysics to photochemical kinetics.

The physical picture

Photochemistry begins with absorption. A molecule $A$ absorbs a photon of energy $h\nu$ and is promoted to an excited electronic state $A^{*}$ . This is a purely quantum event: the photon must match an allowed transition, and it must actually be absorbed by the reacting species, not merely pass through the solution or be absorbed by an inert component (a solvent, a cell wall, or dust). The Grotthuss–Draper law formalizes this: only radiation absorbed by a system can be photochemically effective. Light transmitted through the sample, or absorbed by species that never react, does nothing chemically. Once $A^{*}$ exists, its fate branches. It can fluoresce back to the ground state, undergo intersystem crossing to a triplet, transfer energy to another molecule, dissociate into fragments, isomerize, or react with a neighboring molecule. The Stark–Einstein law governs the first step of this cascade: absorption of one photon excites exactly one molecule to a higher electronic state. This is a statement about the primary photophysical act, not about the ultimate stoichiometry, because a single excited molecule can trigger a chain reaction that consumes thousands of ground-state molecules afterward. The bookkeeping problem is: how many photons went in, and how many product molecules (or reacted molecules) came out? Quantum yield answers this. The primary quantum yield tracks the first branching step immediately after absorption; the overall quantum yield tracks the final, measurable chemical outcome, summed over every mechanistic pathway including any chain propagation.

Figure

Jablonski-style branching after one absorbed photon: primary quantum yields must sum to 1

Because every excited molecule must do something, the primary quantum yields of every competing decay channel sum to exactly one. The overall quantum yield of a chemical product, however, is a separate quantity built on top of this primary branching, and it depends on everything that happens afterward, including chain propagation steps that consume many ground-state molecules per initiating photon.

Figure

Chain amplification: one absorbed photon can yield Φ ≫ 1 product molecules

The core relations

\Phi_i = \frac{\text{number of molecules undergoing primary process } i}{\text{number of photons absorbed}}

where: $\Phi_i$ is a primary quantum yield, and the sum of all $\Phi_i$ over competing primary pathways equals 1.

\Phi = \frac{\text{number of molecules of product formed (or reactant consumed)}}{\text{number of photons absorbed}}

where: $\Phi$ is the overall quantum yield, measured over the whole illumination period, not just the primary step.

I_{\text{abs}} = I_0\left(1 - 10^{-\varepsilon c l}\right)

where: $I_0$ is incident photon flux (mol photons per unit time), $\varepsilon$ is the molar absorption coefficient, $c$ is concentration, $l$ is path length, and $I_{\text{abs}}$ is the absorbed photon flux (einstein per unit time, per unit volume as needed).

-\frac{d[A]}{dt} = \Phi \, I_{\text{abs}}

where: this is the photochemical rate law, valid when photolysis is the only significant pathway for consuming $A$ .

E_{\text{photon}} = h\nu = \frac{hc}{\lambda}, \qquad E_{\text{mol}} = N_A h\nu

where: $h$ is Planck's constant, $\nu$ is frequency, $\lambda$ is wavelength, and $E_{\text{mol}}$ is the energy of one mole of photons (one "einstein").

Derivation AFrom absorbed photon flux to a photochemical rate law

Goal. Show how the Stark–Einstein law, the Beer–Lambert law, and the definition of quantum yield combine into a rate law for the disappearance of a light-absorbing reactant $A$ .

Step 1. Photon absorption is quantized per molecule. The Stark–Einstein law states that each photon absorbed excites exactly one molecule to $A^{*}$ in the primary photophysical act. The rate at which molecules are excited therefore equals the rate at which photons are absorbed, expressed in moles: if $I_{\text{abs}}$ is the absorbed photon flux in mol photons per liter per second, the excitation rate is

\left(\frac{d[A^{*}]}{dt}\right)_{\text{formation}} = I_{\text{abs}}.

Step 2. Absorbed flux follows Beer–Lambert. For incident flux $I_0$ (same units) hitting a cell of path length $l$ containing absorber at concentration $[A] = c$ , the fraction of light absorbed is $1 - 10^{-\varepsilon c l}$ , so

I_{\text{abs}} = I_0\left(1 - 10^{-\varepsilon c l}\right).

In the optically dilute limit ( $\varepsilon c l \ll 1$ ), a Taylor expansion of $10^{-x}$ gives $1 - 10^{-x} \approx x\ln 10$ , so

I_{\text{abs}} \approx 2.303\,\varepsilon c l\, I_0,

which is first order in $[A]$ . In the optically thick limit ( $\varepsilon c l \gg 1$ ), essentially all light is absorbed and $I_{\text{abs}} \to I_0$ , independent of $[A]$ : a zero-order photochemical regime.

Step 3. Insert the quantum yield. By definition, the overall quantum yield is the number of $A$ molecules consumed per photon absorbed:

\Phi = \frac{-d[A]/dt}{I_{\text{abs}}}.

Rearranging,

-\frac{d[A]}{dt} = \Phi\, I_{\text{abs}}.

Step 4. Combine with Beer–Lambert. Substituting the dilute-limit expression for $I_{\text{abs}}$ gives an explicit rate law,

-\frac{d[A]}{dt} = \Phi \cdot 2.303\,\varepsilon l\, I_0 \cdot [A],

which is pseudo-first-order in $[A]$ at fixed $I_0$ , with observed rate constant $k_{\text{obs}} = 2.303\,\Phi\varepsilon l I_0$ . In the optically thick limit instead,

-\frac{d[A]}{dt} = \Phi I_0 = \text{constant},

a zero-order photolysis, common in thick or concentrated samples where nearly every incident photon is captured.

Boxed result.

\boxed{-\frac{d[A]}{dt} = \Phi\, I_0\left(1-10^{-\varepsilon [A] l}\right)}

This single expression interpolates between first-order photolysis (dilute solution) and zero-order photolysis (concentrated, optically thick solution), with $\Phi$ carrying all the chemical mechanism.

Worked examples

Example 1 — Foundational: quantum yield from a simple photolysis experiment. A solution of a photolabile compound $A$ is irradiated with monochromatic light of wavelength 350 nm. A power meter shows the sample absorbs $2.50 \times 10^{-3}$ J of light energy per second. After 300 s of continuous irradiation, chemical analysis shows that $4.10 \times 10^{-6}$ mol of $A$ has been consumed. Find the overall quantum yield. Step 1: energy per photon.

E_{\text{photon}} = \frac{hc}{\lambda} = \frac{(6.626\times10^{-34}\,\text{J s})(2.998\times10^{8}\,\text{m s}^{-1})}{350\times10^{-9}\,\text{m}} = 5.682\times10^{-19}\ \text{J}.

Step 2: photon absorption rate (photons per second), then convert to mol photons per second.

\text{photons/s} = \frac{2.50\times10^{-3}\ \text{J s}^{-1}}{5.682\times10^{-19}\ \text{J}} = 4.400\times10^{15}\ \text{s}^{-1}.

I_{\text{abs}} = \frac{4.400\times10^{15}\ \text{s}^{-1}}{6.022\times10^{23}\ \text{mol}^{-1}} = 7.307\times10^{-9}\ \text{mol photons s}^{-1}.

Step 3: total photons absorbed over 300 s.

n_{h\nu} = (7.307\times10^{-9}\ \text{mol s}^{-1})(300\ \text{s}) = 2.192\times10^{-6}\ \text{mol photons}.

Step 4: quantum yield.

\Phi = \frac{\text{mol } A \text{ consumed}}{\text{mol photons absorbed}} = \frac{4.10\times10^{-6}}{2.192\times10^{-6}} = 1.87.

Sanity check: $\Phi > 1$ signals that this is not a simple one-photon, one-molecule photolysis; a short chain or a secondary thermal step (for example, a radical intermediate attacking a second $A$ molecule) must be consuming extra reactant beyond the primary photoact. A value near 1.87, not astronomically large, is consistent with a short chain of length roughly 2, which is physically reasonable for many photolabile organics.

Example 2 — Applied: predicting a rate constant and comparing to a chain mechanism. Consider gas-phase photolysis of ozone-relevant Cl2 in the classic H2/Cl2 chain: $\text{Cl}_2 + h\nu \rightarrow 2\text{Cl}\cdot$ (primary quantum yield for dissociation $\Phi_{\text{diss}} = 1.0$ , i.e. essentially every absorbed photon cleaves one Cl2). The chain then propagates:

\text{Cl}\cdot + \text{H}_2 \rightarrow \text{HCl} + \text{H}\cdot, \qquad \text{H}\cdot + \text{Cl}_2 \rightarrow \text{HCl} + \text{Cl}\cdot,

which repeats on average $5.0\times10^{4}$ times before a termination event (e.g., wall recombination) ends the chain. Cl2 is irradiated at 400 nm in a cell absorbing $I_{\text{abs}} = 3.0\times10^{-7}$ mol photons per liter per second, with $[\text{Cl}_2] = 0.050$ mol/L. Estimate (a) the overall quantum yield for HCl formation, and (b) the initial rate of HCl production. (a) Each absorbed photon creates 2 Cl atoms (from $\Phi_{\text{diss}} = 1.0$ giving 1 Cl2 dissociated, i.e., 2 Cl· per photon). Each Cl· initiates a chain of average length $n = 5.0\times10^4$ two-step cycles, and each cycle produces 2 HCl molecules (one from each propagation step). So:

\Phi(\text{HCl}) = 2 \text{ (Cl atoms per photon)} \times n \times 2 \text{ (HCl per cycle)}.

\Phi(\text{HCl}) = 2 \times (5.0\times10^4) \times 2 = 2.0\times10^{5}.

This astronomically large quantum yield is the hallmark of a radical chain reaction: light supplies only the initiation step, and chemistry (not photons) drives the bulk of product formation. This matches the historic observation that H2 + Cl2 photochemistry can proceed almost explosively from a faint light source. (b) Using the photochemical rate law $d[\text{HCl}]/dt = \Phi \, I_{\text{abs}}$ :

\frac{d[\text{HCl}]}{dt} = (2.0\times10^{5})(3.0\times10^{-7}\ \text{mol L}^{-1}\text{s}^{-1}) = 6.0\times10^{-2}\ \text{mol L}^{-1}\text{s}^{-1}.

Sanity check: this rate consumes $[\text{Cl}_2] = 0.050$ mol/L in roughly $0.050/0.060 \approx 0.83$ s, consistent with the qualitative picture of an extremely fast, chain-amplified photochemical reaction triggered by weak illumination. The huge quantum yield, not a huge photon flux, is doing the work.

Limiting cases & common traps

Situation	What happens	Typical $\Phi$
Simple photodissociation, no secondary chemistry	One photon breaks one bond, nothing else reacts	$\Phi \approx 1$
Fluorescence/phosphorescence dominates decay	Excited state re-emits instead of reacting	$\Phi \ll 1$ (photochemical), but $\Phi_f$ near 1 for emission
Radical chain reaction (e.g. H2 + Cl2)	One initiation photon triggers many propagation cycles	$\Phi \gg 1$ , even $10^4$ – $10^6$
Optically thick sample	Beer–Lambert term saturates, rate becomes zero order in $[A]$	$\Phi$ unaffected, but rate law changes order
Photostationary state (two interconverting isomers)	Forward and back photoreactions both active	$\Phi$ measured must isolate net conversion

The chemist behind it

Albert Einstein's 1905 photon hypothesis, developed to explain the photoelectric effect, gave physical meaning to the idea that light is absorbed in discrete quanta rather than continuously. Johannes Stark independently proposed, around the same period, that photochemical activation follows this same one-photon-per-molecule rule. Their joint legacy, the Stark–Einstein law of photochemical equivalence, became the quantitative foundation on which quantum yield is defined: without it, there would be no natural denominator ("photons absorbed") against which to count chemical events.

Origins · 1 minAlbert Einstein and how this was discovered

Albert Einstein

1879–1955 · explained light quantization, giving the Stark–Einstein law its physical basis

Ask the tutor: How does the quantum yield of fluorescence relate to the photochemical quantum yield in a molecule that competes between emitting light and reacting?

Why can the overall quantum yield exceed one, if the Stark–Einstein law says one photon activates one molecule?

The Stark–Einstein law governs only the primary absorption act: one photon excites exactly one molecule to $A^{*}$ . What that excited molecule does next is unrestricted by the law. If $A^{*}$ generates a reactive radical that propagates a chain reaction, consuming many ground-state molecules in subsequent thermal steps, the overall quantum yield (product molecules per photon absorbed) can be far greater than one. Classic examples include H2 + Cl2 and H2 + Br2 photochlorination/bromination chains, where yields of $10^4$ – $10^6$ are observed from chain lengths determined by initiation and termination rates.

What is the difference between primary and overall quantum yield in practice?

The primary quantum yield $\Phi_i$ describes a single, well-defined event immediately following excitation, such as dissociation, intersystem crossing, or fluorescence, and all such competing $\Phi_i$ values sum to 1 because the excited molecule must do exactly one of them. The overall quantum yield $\Phi$ describes the final measured chemical outcome, summed over the entire mechanism, including any secondary thermal or chain steps. They coincide only when the primary process is also the final product-forming step, with no subsequent chemistry.

How is quantum yield actually measured in the laboratory?

Experimenters use chemical actinometry: a reference reaction with a precisely known quantum yield (such as potassium ferrioxalate photolysis) is irradiated under identical conditions to determine the absolute photon flux, $I_{\text{abs}}$ , delivered by the light source. The unknown reaction is then irradiated under the same geometry, and the moles of product formed (or reactant consumed) are measured by spectroscopy or chromatography. Dividing moles of product by moles of photons absorbed, both measured over the same time interval, gives $\Phi$ directly from its definition.

Does quantum yield depend on the wavelength of light used?

Often yes. Different wavelengths can populate different excited electronic states with different decay pathways, so $\Phi_i$ for a specific primary process can vary strongly with excitation wavelength. However, Kasha's rule (a photophysical observation, not covered in depth here) states that emission and much photochemistry typically occur from the lowest excited state of a given multiplicity, regardless of which state absorbed the photon. This often makes quantum yield roughly wavelength-independent above the absorption threshold, though many important exceptions exist.

Why does concentrated or optically thick sample photolysis follow zero-order kinetics?

When $\varepsilon c l$ is large, nearly all incident photons are absorbed within the front layer of the sample, so $I_{\text{abs}} \to I_0$ , a constant set only by the light source, not by $[A]$ . Since the photochemical rate law is $-d[A]/dt = \Phi I_{\text{abs}}$ , the rate becomes constant, independent of concentration, which is the definition of a zero-order process. As the reaction proceeds and $[A]$ eventually drops enough to make $\varepsilon c l$ small, the kinetics smoothly cross over to first order.