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CSIR-NET Photochemistry — The Essential Rules

By Aniket Bhardwaj · 7 September 2026 · CSIR-NET Chemistry

Photochemistry looks like a memory chapter and is not. Nearly every CSIR-NET question comes from one of three places: a named law, a competition between rate constants, or a quantum-yield calculation. This article states the laws precisely, lays out the Jablonski scheme as a table you can reconstruct from memory, and then does the calculations in full arithmetic.

The two laws that frame everything

Grotthuss–Draper law (first law of photochemistry): only light that is absorbed by a molecule can bring about photochemical change.

Stark–Einstein law (law of photochemical equivalence, second law): in the primary photochemical act, one molecule is activated by one absorbed photon.

Read the second law carefully. It restricts the primary step only. Secondary thermal steps are unlimited, which is why an overall quantum yield can be enormous — and why the law is not violated when it is.

Photon energy — and why one photon is a lot

Ephoton = hc/λ   ·   Eeinstein = NAhc/λ

h = 6.626 × 10−34 J s, c = 2.998 × 108 m s−1, NA = 6.022 × 1023 mol−1. One einstein = one mole of photons.

Energy of one einstein at λ = 300 nm.

hc = (6.626 × 10−34)(2.998 × 108) = 1.9865 × 10−25 J m
Ephoton = 1.9865 × 10−25 ÷ (300 × 10−9) = 1.9865 × 10−25 ÷ 3.00 × 10−7 = 6.622 × 10−19 J
Eeinstein = 6.622 × 10−19 × 6.022 × 1023 = 3.988 × 105 J mol−1 = 399 kJ mol−1

That single number explains most of photochemistry. 399 kJ per mole is comparable to a C–C or C–H bond energy, so near-UV light can break bonds that room-temperature collisions never will.

The Jablonski scheme, written out

A diagram is only a bookkeeping device for these processes. Learn the table instead — the timescales are what decide which process wins.

ProcessStatesRadiative?Spin change?Typical timescale
AbsorptionS0 → S1, S2yesno10−15 s
Vibrational relaxationwithin one electronic statenono10−12–10−10 s
Internal conversion (IC)S2 → S1, S1 → S0nono10−12–10−8 s
FluorescenceS1 → S0yesno10−9–10−7 s
Intersystem crossing (ISC)S1 → T1noyes10−10–10−8 s
PhosphorescenceT1 → S0yesyes10−6–102 s

Three consequences follow directly from that table and are examined constantly:

The Franck–Condon principle explains the band shapes: electronic transitions are so fast that the nuclei do not move during them, so a transition is drawn as a vertical line and its intensity depends on the overlap of the two vibrational wavefunctions.

Selection rules

RuleStatementHow it is broken
SpinΔS = 0; singlet → triplet is forbiddenSpin–orbit coupling; the heavy-atom effect (internal or external) makes ISC and phosphorescence competitive
Laporte (parity)In a centrosymmetric molecule only g ↔ u transitions are allowed, so d–d transitions are forbiddenVibronic coupling, or loss of the centre of symmetry (tetrahedral complexes are far more intensely coloured for this reason)
Symmetry / overlapn → π* in a carbonyl is symmetry-forbiddenVibronic mixing. It still appears, but weakly: ε roughly 10–100, against 103–105 for an allowed π → π*

The molar absorption coefficient ε is therefore diagnostic. A weak, long-wavelength band in a ketone is n → π*; the strong short-wavelength band is π → π*. Solvent polarity separates them further — n → π* shifts blue in polar solvents, π → π* shifts red.

Quantum yield — the definition and the arithmetic

Φ = (number of molecules undergoing the process) ÷ (number of photons absorbed)
  = (moles of product formed) ÷ (einsteins absorbed)

Worked example. A solution absorbs 1.00 J of radiation at 300 nm and produces 2.00 × 10−6 mol of product. Find Φ.

From above, one einstein at 300 nm carries 3.988 × 105 J.
Einsteins absorbed = 1.00 ÷ (3.988 × 105) = 2.508 × 10−6 einstein
Φ = (2.00 × 10−6) ÷ (2.508 × 10−6) = 0.798 ≈ 0.80

Cross-check by the photon route. Photons absorbed = 1.00 ÷ (6.622 × 10−19) = 1.510 × 1018. Molecules formed = 2.00 × 10−6 × 6.022 × 1023 = 1.204 × 1018. Φ = 1.204 ÷ 1.510 = 0.798. The two routes agree.

Two standard cases sit at the extremes. Photolysis of HI has Φ(−HI) = 2: one photon splits one HI into H• and I•, and the hydrogen atom immediately consumes a second HI (H• + HI → H2 + I•). The hydrogen–chlorine chain reaction has a quantum yield of the order of 104–106, because a single photon starts a radical chain that runs many thousands of cycles before termination. Neither breaks the Stark–Einstein law, which governs only the primary act.

Lifetimes, efficiency and quenching

Φf = kf / (kf + kIC + kISC + …) = kf τ0
τ0 = 1 / (kf + Σknr)  ·  natural radiative lifetime τr = 1/kf

Stern–Volmer: Φ0/Φ = 1 + KSV[Q], with KSV = kqτ0

Worked example. A dye has kf = 5.0 × 107 s−1 and total non-radiative decay 5.0 × 107 s−1.

τ0 = 1 ÷ (5.0 × 107 + 5.0 × 107) = 1 ÷ (1.0 × 108) = 1.0 × 10−8 s = 10 ns
Φf = (5.0 × 107) ÷ (1.0 × 108) = 0.50

Cross-check: Φf = kfτ0 = 5.0 × 107 × 1.0 × 10−8 = 0.50. And by the lifetime ratio, τr = 1/kf = 2.0 × 10−8 s = 20 ns, so Φf = τ0r = 10/20 = 0.50. All three routes agree.

Now add a quencher. At [Q] = 0.020 M the emission falls to one third, so Φ0/Φ = 3.0.
KSV = (3.0 − 1) ÷ 0.020 = 100 M−1
kq = KSV0 = 100 ÷ (1.0 × 10−8) = 1.0 × 1010 M−1 s−1

That value sits at the diffusion-controlled limit for a small molecule in water, so this quenching is as fast as encounters allow — a conclusion you can only reach by finishing the arithmetic.

Photochemical reactions worth knowing by name

Orbital symmetry control is examined through the Woodward–Hoffmann rules, and the photochemical row is simply the thermal row reversed:

Reaction typeElectron countThermalPhotochemical
Electrocyclic4nconrotatorydisrotatory
Electrocyclic4n + 2disrotatoryconrotatory
Cycloaddition (suprafacial–suprafacial)4n + 2, e.g. 4 + 2allowedforbidden
Cycloaddition (suprafacial–suprafacial)4n, e.g. 2 + 2forbiddenallowed

Mistakes that cost marks

  • Claiming Φ > 1 breaks the Stark–Einstein law. It does not — the law applies to the primary act only.
  • Confusing τ0 with τr. The measured lifetime includes every decay channel; the natural radiative lifetime is 1/kf alone. Stern–Volmer uses the measured one.
  • Using energy per photon where einsteins are needed. Decide at the start whether you are counting molecules or moles and stay consistent.
  • Expecting phosphorescence in fluid solution at room temperature. T1 lives long enough to be quenched by oxygen and by collisions, which is why phosphorescence is usually observed in a rigid glass at low temperature.
  • Forgetting the first law. Light that passes straight through does nothing. If the compound does not absorb at that wavelength, no photochemistry occurs however intense the lamp.
  • Ignoring the fraction absorbed. With A = 0.30 only about half the incident light is absorbed (1 − 10−0.30 = 0.50); quantum yield is defined per photon absorbed, never per photon supplied.

Where this appears in the exam

ExamTypical demand
CSIR-NET Chemical SciencesQuantum yield arithmetic, Stern–Volmer analysis, Jablonski processes and their ordering, Norrish reactions
GATE ChemistryPhoton energy, laws of photochemistry, fluorescence versus phosphorescence, selection rules
IIT-JAM / CUET-PGhc/λ calculations, Beer–Lambert, absorption versus emission
MSc courseworkActinometry (potassium ferrioxalate is the standard chemical actinometer) and lifetime measurement

Get the powers of ten right. Photochemistry arithmetic is small numbers multiplied by large ones — 10−34 against 1023 — and that is exactly where marks are lost. The ABC Chemistry Calculator Suite keeps the scientific constants, unit converter and calculation tools in one page while you work.

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