Photochemistry and Solar Energy Conversion — Photon Energy, Quantum Yield and Band Gaps
Thermal chemistry gives energy to every molecule in the flask according to a Boltzmann distribution. Photochemistry does something different: it hands a fixed, quantised packet of energy to one molecule at a time. That difference is why light can drive reactions that are thermodynamically uphill — photosynthesis being the obvious example — and it is why solar energy conversion is a chemistry problem and not only an engineering one. Everything in this article follows from two equations you already know.
The equations
Φ = (number of molecules reacting) / (number of photons absorbed)
- h — Planck constant, 6.626 × 10⁻³⁴ J s
- c — speed of light, 2.998 × 10⁸ m s⁻¹
- λ — wavelength in metres in the first form; the second form is a shortcut that already contains the unit conversion, so it takes nanometres and returns electronvolts
- Φ — quantum yield, dimensionless
- One mole of photons is called an einstein; multiply the energy of a single photon by Avogadro's number to get energy per einstein
Two laws frame the subject. The Grotthuss–Draper law: only light that is actually absorbed can cause a photochemical change — light passing straight through does nothing. The Stark–Einstein law: each absorbed photon activates one molecule in the primary act. Note the words "primary act" carefully; they are the reason quantum yield is not automatically 1, and we will return to them.
Worked example 1 — the energy of a green photon
Take λ = 500 nm = 5.00 × 10⁻⁷ m.
hc = 6.626 × 10⁻³⁴ × 2.998 × 10⁸ = 1.986 × 10⁻²⁵ J m
E = 1.986 × 10⁻²⁵ ÷ 5.00 × 10⁻⁷ = 3.973 × 10⁻¹⁹ J per photon
Per mole of photons:
3.973 × 10⁻¹⁹ × 6.022 × 10²³
= 2.392 × 10⁵ J mol⁻¹ = 239.2 kJ mol⁻¹
In electronvolts:
3.973 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹
= 2.480 eV
Cross-check with the shortcut: 1239.8 ÷ 500 = 2.480 eV ✓
239 kJ mol⁻¹ is comparable with the strength of many single covalent bonds. That is the whole point: ordinary visible light carries enough energy per photon to do real chemistry, which is why coloured compounds fade in sunlight and why photodegradation is a formulation problem.
Worked example 2 — band gap sets the longest useful wavelength
In a semiconductor, a photon is only absorbed usefully if its energy is at least the band gap Eg. Rearranging the shortcut gives the threshold wavelength:
For a material with Eg = 1.1 eV:
λ = 1239.8 ÷ 1.1 = 1127 nm, in the near infrared.
For a wide-gap oxide with Eg = 3.2 eV:
λ = 1239.8 ÷ 3.2 = 387 nm, in the ultraviolet.
Read those two results together and you have the central design tension of solar energy conversion:
- A wide band gap absorbs only the blue and ultraviolet end. Since sunlight at ground level is mostly visible and near-infrared, the great majority of the incoming photons are simply not absorbed at all.
- A narrow band gap absorbs almost everything — but every photon whose energy exceeds Eg loses the excess as heat within picoseconds, a process called thermalisation. A 2.48 eV green photon absorbed by a 1.1 eV material delivers only 1.1 eV of usable energy; the other 1.38 eV, more than half, warms the material.
Neither extreme wins, so an optimum sits in between. Every strategy in the field — stacking materials of different gaps, sensitising a wide-gap oxide with a dye that absorbs visible light, tuning composition to shift Eg — is an attempt to escape a trade-off that the two calculations above define exactly.
Worked example 3 — quantum yield from a real measurement
Quantum yield answers the question the two laws leave open: of the photons that were absorbed, how many actually produced chemistry?
A sample absorbs 1.0 W of 400 nm light for 60 s, and 8.0 × 10⁻⁵ mol of product is formed.
Step 1 — energy absorbed: 1.0 W × 60 s = 60 J
Step 2 — energy of one 400 nm photon:
1239.8 ÷ 400 = 3.0995 eV
3.0995 × 1.602 × 10⁻¹⁹
= 4.965 × 10⁻¹⁹ J
Step 3 — how many photons:
60 ÷ 4.965 × 10⁻¹⁹
= 1.208 × 10²⁰ photons
in moles: 1.208 × 10²⁰ ÷ 6.022 × 10²³
= 2.007 × 10⁻⁴ mol (einsteins)
Step 4 — quantum yield:
Φ = 8.0 × 10⁻⁵ ÷ 2.007 × 10⁻⁴
= 0.40
A quantum yield of 0.40 means 60% of the absorbed photons produced no product. They were not wasted in a mysterious way — the excited molecule has several competing exits available: fluorescence, non-radiative decay back to the ground state as heat, intersystem crossing to a triplet state, quenching by a collision partner, or simply relaxing back to the reactants it came from. Photochemistry is largely the study of which of these wins.
Worked example 4 — how much light is actually absorbed
Quantum yield is defined per photon absorbed, so you need the absorbed fraction from Beer–Lambert before Φ means anything:
For a solution of absorbance A = 0.30 at the irradiation wavelength:
transmitted = 10−0.30 = 0.501
absorbed = 1 − 0.501 = 0.499, i.e. about 49.9%
Half the incident light passed straight through and, by the Grotthuss–Draper law, did nothing. If you had divided the product formed by the incident photons instead of the absorbed ones, your quantum yield would have come out roughly half its true value.
Sunlight to fuel — the thermodynamic floor
Splitting liquid water into hydrogen and oxygen has ΔG° = +237 kJ mol⁻¹ under standard conditions. Convert that to a voltage using ΔG = −nFE with n = 2:
E° = 237000 ÷ (2 × 96485) = 237000 ÷ 192970 = 1.23 V
So a photon of at least about 1.23 eV — corresponding to λ = 1239.8 ÷ 1.23 = 1008 nm — carries, on paper, enough energy for the job. In practice a working device must supply appreciably more than 1.23 V, because both the hydrogen-evolving and the oxygen-evolving half reactions have kinetic overpotentials, and the oxygen side in particular is slow because it requires four electrons and the making of an oxygen–oxygen bond. This is why catalysis, not photon energy, is usually the binding constraint in solar fuel research.
Mistakes that cost marks
- Confusing intensity with photon energy. A brighter red lamp delivers more photons, not more energetic ones. If a single red photon cannot cross the threshold, no amount of red light will.
- Leaving λ in nanometres inside hc/λ. That form needs metres. The 1239.8 shortcut is the one that takes nanometres — do not mix them.
- Assuming Φ ≤ 1 always. The Stark–Einstein law applies to the primary photochemical act. If that act starts a chain reaction, one absorbed photon can lead to a great many product molecules, and Φ can be far greater than 1. Conversely a Φ well below 1 is entirely normal.
- Dividing by incident instead of absorbed photons. As Example 4 shows, this quietly corrupts the answer.
- Treating eV and kJ mol⁻¹ as interchangeable. One is per particle, the other per mole. The conversion runs through Avogadro's number.
- Forgetting thermalisation. Absorbing a photon far above the band gap does not deliver its full energy usefully; the excess becomes heat.
Where this appears in your exam
| Exam | How it is asked |
|---|---|
| IIT-JAM | Photon energy and einstein calculations, quantum yield numericals, Beer–Lambert absorbance |
| GATE | Photophysical processes, excited-state deactivation pathways, band gap and threshold wavelength |
| CSIR-NET | Jablonski diagram reasoning, fluorescence versus phosphorescence, quenching, sensitisation and photocatalysis |
| CUET-PG | Direct substitution into E = hc/λ and the two photochemical laws |
Do the photon arithmetic without slips. The Modern Physics tool handles photon energy and de Broglie wavelength directly, so you can convert between wavelength, frequency, joules and electronvolts and check every number in Examples 1 to 3.
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