GATE Photochemistry — Quantum Yield, Jablonski Diagram and Stern–Volmer

By Aniket Bhardwaj · 18 September 2026 · GATE Chemistry

Photochemistry is a compact, well-defined unit and it rewards preparation. Almost every question reduces to one of four things: the energy carried by a mole of photons, a quantum yield, a photophysical pathway on the Jablonski diagram, or a Stern–Volmer quenching plot. Master those four and you can attempt nearly anything the topic offers. This guide gives the laws, the formulas with their units, three worked numericals with the arithmetic shown, and the errors that most often cost marks.

As with every unit, confirm the exact syllabus and paper pattern from the official GATE notification for your examination year — never from a website.

The two founding laws

Grotthuss–Draper law: only the light that is absorbed by a system can produce a photochemical change. Light that passes through or is reflected does nothing. This is why a photochemical experiment always begins with an absorption spectrum.

Stark–Einstein law of photochemical equivalence: in the primary photochemical act, one molecule absorbs one photon. Note the wording carefully — the law constrains the primary act only. What happens afterwards, in the secondary processes, is ordinary chemistry and can involve any number of molecules. This is the single most misquoted sentence in the topic.

Energy of a mole of photons — the einstein

E (per photon) = hν = hc / λ E (per einstein) = N_A hc / λ

h = 6.626 × 10⁻³⁴ J s (Planck constant)
c = 2.998 × 10⁸ m s⁻¹
N_A = 6.022 × 10²³ mol⁻¹
λ — wavelength, in metres in the formula (convert from nm by multiplying by 10⁻⁹)
One einstein = one mole of photons. It is an amount, not an energy — its energy depends on the wavelength.

Worked example 1 — energy of one einstein at 300 nm

Step 1 — the product hc.
hc = 6.626 × 10⁻³⁴ × 2.998 × 10⁸ = 1.9865 × 10⁻²⁵ J m

Step 2 — energy of one photon. λ = 300 nm = 3.00 × 10⁻⁷ m
E = 1.9865 × 10⁻²⁵ ÷ 3.00 × 10⁻⁷ = 6.622 × 10⁻¹⁹ J

Step 3 — multiply by Avogadro's number.
E = 6.622 × 10⁻¹⁹ × 6.022 × 10²³ = 3.988 × 10⁵ J mol⁻¹ = 399 kJ per einstein

Useful shortcut, and where it comes from. Since N_Ahc = 6.022 × 10²³ × 1.9865 × 10⁻²⁵ = 0.11963 J m mol⁻¹, and 1 m = 10⁹ nm,
E (kJ mol⁻¹) ≈ 1.196 × 10⁵ ÷ λ (nm).
Check: 1.196 × 10⁵ ÷ 300 = 399 kJ mol⁻¹ ✓ — the same answer by a second route.

That shortcut makes the whole ultraviolet–visible range easy to picture:

Wavelength	Region	Energy per einstein (kJ mol⁻¹)	Comparison
200 nm	Far UV	598	Enough to break most single bonds
300 nm	UV-B	399	Comparable to a C−C or C−O bond energy
400 nm	Violet	299	Breaks weak bonds; drives many reactions
700 nm	Red	171	Too weak to break strong bonds directly

This is the honest reason photochemistry usually needs UV or near-visible light: the photon must carry an energy comparable to a bond energy before it can do chemistry.

Quantum yield — the number GATE asks for most

Φ = (number of molecules reacted or formed) / (number of photons absorbed)
equivalently Φ = (moles reacted) / (einsteins absorbed)

Three regimes, each with a physical cause:

Φ ≈ 1 — every absorbed photon produces one reacted molecule. The ideal case implied by the Stark–Einstein law.
Φ < 1 — most absorbed energy is lost by physical deactivation: fluorescence, internal conversion, collisional quenching, or recombination of the products inside a solvent cage before they can escape.
Φ >> 1 — a chain reaction. One photon starts a radical chain and thousands of molecules react per photon. The hydrogen–chlorine reaction is the classic example, with a quantum yield of the order of 10⁴ to 10⁶. This does not violate the Stark–Einstein law, because only the initiation step is photochemical; the propagation steps are thermal.

Worked example 2 — finding a quantum yield

A sample absorbs 1.50 J of radiation at 300 nm and 7.52 × 10⁻⁴ mol of reactant is consumed. Find the quantum yield.

Step 1 — einsteins absorbed. From example 1, one einstein at 300 nm carries 3.988 × 10⁵ J.
einsteins = 1.50 ÷ 3.988 × 10⁵ = 3.762 × 10⁻⁶ mol of photons

Step 2 — divide.
Φ = 7.52 × 10⁻⁴ ÷ 3.762 × 10⁻⁶ = 200

Interpretation: a quantum yield of 200 means each absorbed photon leads on average to 200 molecules reacting, so this must be a chain process. If you compute a quantum yield far above 1, do not assume you have made an error — state that the mechanism is a chain reaction and identify the initiation step.

Practical note: the number of photons absorbed is measured, not assumed. A chemical actinometer — a reaction of accurately known quantum yield, such as the potassium ferrioxalate system — is irradiated under identical conditions, and the photon flux is calculated from how much of it reacts.

The Jablonski diagram in words

Since we cannot draw here, read this table as a vertical energy ladder. S₀ is the singlet ground state at the bottom, S₁ the first excited singlet above it, and T₁ the first triplet, which always lies below S₁ because of the exchange energy favouring parallel spins.

Process	Transition	Radiative?	Spin change?	Typical timescale
Absorption	S₀ → S₁, S₂ …	Yes	No	≈10⁻¹⁵ s
Vibrational relaxation	Within a state	No	No	10⁻¹³–10⁻¹² s
Internal conversion (IC)	S₂ → S₁, S₁ → S₀	No	No	10⁻¹²–10⁻⁶ s
Fluorescence	S₁ → S₀	Yes	No	10⁻⁹–10⁻⁷ s
Intersystem crossing (ISC)	S₁ → T₁	No	Yes	10⁻¹⁰–10⁻⁸ s
Phosphorescence	T₁ → S₀	Yes	Yes	10⁻⁶ s to seconds

Three consequences follow, and all three are examinable:

Phosphorescence is long-lived because it is spin-forbidden. The transition T₁ → S₀ requires a spin flip, so its rate constant is small and the excited state survives far longer. Anything that increases spin–orbit coupling — a heavy atom in the molecule or in the solvent — speeds up both ISC and phosphorescence. That is the heavy-atom effect.
Kasha's rule: emission occurs from the lowest excited state of a given multiplicity, essentially always S₁ (or T₁), whatever state you excited to. Internal conversion from higher states is simply too fast to compete with emission. This is why the fluorescence spectrum does not change when you change the excitation wavelength.
The Stokes shift: fluorescence appears at a longer wavelength (lower energy) than absorption, because vibrational relaxation loses energy in both the excited and the ground state before and after emission.

The Franck–Condon principle underlies the shape of these bands: electronic transitions are so fast that the nuclei do not move during them, so the most intense vibronic transition is the one to the vibrational level whose wavefunction best overlaps the starting one.

Quenching and the Stern–Volmer equation

I₀ / I = 1 + K_SV [Q] where K_SV = k_q τ₀

I₀, I — fluorescence intensity without and with the quencher (the lifetimes τ₀ and τ can be used in the same way for dynamic quenching).
[Q] — quencher concentration in mol dm⁻³.
K_SV — Stern–Volmer constant, units dm³ mol⁻¹.
k_q — bimolecular quenching rate constant, dm³ mol⁻¹ s⁻¹.
τ₀ — excited-state lifetime in the absence of quencher, in seconds.

A plot of I₀/I against [Q] is a straight line of intercept 1 and slope K_SV — one of the neatest linear plots in physical chemistry.

Worked example 3 — Stern–Volmer analysis

Adding 0.010 mol dm⁻³ of a quencher halves the fluorescence intensity of a dye whose unquenched lifetime is 5.0 ns. Find K_SV and k_q.

Step 1 — K_SV. Halving the intensity means I₀/I = 2.
2 = 1 + K_SV × 0.010
K_SV × 0.010 = 1
K_SV = 1 ÷ 0.010 = 100 dm³ mol⁻¹

Step 2 — k_q. τ₀ = 5.0 ns = 5.0 × 10⁻⁹ s
k_q = K_SV ÷ τ₀ = 100 ÷ (5.0 × 10⁻⁹) = 2.0 × 10¹⁰ dm³ mol⁻¹ s⁻¹

What that number tells you. Diffusion-controlled bimolecular rate constants in water at room temperature are of the order of 10¹⁰ dm³ mol⁻¹ s⁻¹. Our value sits right at that limit, so the quenching is essentially diffusion-controlled — every encounter between the excited dye and a quencher molecule leads to quenching. A k_q far above the diffusion limit would be physically impossible and signals static quenching (a ground-state complex) rather than dynamic quenching.

Other processes worth knowing by name

Photosensitisation: a sensitiser absorbs the light and transfers its energy to a molecule that does not absorb at that wavelength. Mercury-sensitised reactions and the generation of singlet oxygen by dye sensitisers are standard examples.
Photostationary state: in a reversible photoreaction, such as cis–trans isomerisation, continued irradiation reaches a steady composition determined by the two quantum yields and the two absorbances — not by thermodynamics. It is generally not the equilibrium composition of the dark reaction.
Chemiluminescence: the reverse situation, where a chemical reaction populates an excited state that then emits light.
Photodissociation, photoisomerisation, photoreduction and photosubstitution are the main reaction types; photochemical reactions can reach products that are thermally forbidden, which is exactly what the Woodward–Hoffmann rules formalise for pericyclic reactions.

Common mistakes that cost marks

Claiming Φ > 1 breaks the Stark–Einstein law. The law governs only the primary act. Chain propagation is thermal.
Leaving λ in nanometres. hc/λ needs λ in metres. Forgetting the 10⁻⁹ shifts the answer by nine orders of magnitude.
Saying phosphorescence is lower in energy so it is slower. The cause is the spin-forbidden character of T₁ → S₀, not the energy gap.
Putting T₁ above S₁. For a given electronic configuration the triplet lies below the corresponding singlet — Hund's rule applied to molecules.
Expecting emission from S₂. Kasha's rule: emission comes from the lowest excited state of that multiplicity.
Treating a curved Stern–Volmer plot as an experimental error. Upward curvature usually means both static and dynamic quenching are operating.
Confusing intensity loss from quenching with the inner filter effect, where the added substance simply absorbs the exciting light. Always check whether the quencher absorbs at the excitation wavelength.

Preparation map

Sub-topic	Question style	Priority
Einstein energy	Energy per mole of photons at a given λ	High — quick, certain marks
Quantum yield	Φ from energy absorbed and moles reacted	High
Jablonski processes	Identify a process from a timescale or a spin change	High — conceptual
Stern–Volmer	K_SV and k_q from intensity data	High
Photosensitisation and photostationary state	Explain a mechanism in one or two lines	Medium
Kasha, Franck–Condon, Stokes shift	Statement-based reasoning	Medium

Photon energy is the calculation this whole unit rests on. Whether you are converting a wavelength into kJ per einstein, checking whether a photon can break a given bond, or comparing UV with visible light, it always comes back to E = hc/λ. The free Photon Energy & de Broglie calculator does that conversion directly, so you can verify the first step of a photochemistry numerical instead of hunting for a stray power of ten.

Open the Photon Energy & de Broglie Calculator →

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