Solar Radiation and Atmospheric Photochemistry

Introduction

Almost all atmospheric chemistry is driven by sunlight. The Sun delivers energy that splits molecules into reactive fragments — atoms and radicals that then set off chains of reactions. Whether a molecule is split depends on whether it absorbs light of the right wavelength, whether that photon carries enough energy to break a bond, and how many such photons are available at a given altitude. Photochemistry links the physics of radiation to the chemistry of ozone, smog and the cleaning of the air.

Core explanation

The solar spectrum. The Sun radiates approximately as a black body at about 5800 K, peaking in the visible near 500 nm. Outside the atmosphere there is substantial ultraviolet. The atmosphere filters this: O₂ and N₂ absorb almost everything below about 200 nm high in the atmosphere, and stratospheric ozone absorbs most light between 200 and 290 nm. The troposphere therefore receives only wavelengths longer than about 290 nm — the so-called actinic cut-off .

Photon energy. Each photon carries energy E = hc/λ. Per mole of photons, E = N A hc/λ, which is conveniently E (kJ mol⁻¹) ≈ 119 600 / λ(nm). A 300 nm photon therefore carries about 400 kJ mol⁻¹, and a 600 nm photon about 200 kJ mol⁻¹. The O=O bond energy is 498 kJ mol⁻¹, so O₂ can only be split by photons shorter than about 240 nm — which explains why oxygen photolysis happens in the stratosphere and above, not near the ground.

Energy is necessary but not sufficient. A molecule must also absorb at that wavelength. The strength of absorption is described by the absorption cross-section , σ(λ), in cm² molecule⁻¹. It appears in the Beer–Lambert law written for gases: I = I₀ exp(−σnl), where n is number density and l the path length. The product σnl is the optical depth .

Quantum yield. Not every absorbed photon breaks the bond; the excited molecule may instead fluoresce or be quenched by collision. The quantum yield , φ(λ), is the fraction of absorptions that produce a given product. For NO₂ → NO + O, φ is close to 1 below about 395 nm and falls to zero by about 420 nm.

Photolysis rate constant. Photolysis of a molecule A is first order in A:

rate = J[A], with J = ∫ σ(λ) φ(λ) F(λ) dλ

where F(λ) is the actinic flux — photons arriving from all directions, including scattered and reflected light. J depends on the Sun's elevation, altitude, cloud, surface albedo and the ozone overhead. Its reciprocal, 1/J, is the photolytic lifetime .

Key photolyses. NO₂ absorbs strongly across the near-UV and blue; at midday J(NO₂) is roughly 8 × 10⁻³ s⁻¹, giving a lifetime of about two minutes. Ozone photolysis below about 330 nm yields electronically excited oxygen atoms, O(¹D), which can react with water to make OH radicals. Formaldehyde, nitrous acid and peroxides are other important tropospheric radical sources.

Formulae

E = hc/λ; molar photon energy E (kJ mol⁻¹) ≈ 119 600 / λ(nm). Beer–Lambert for gases: I = I₀ e^(−σnl). Photolysis rate constant J = ∫ σ(λ) φ(λ) F(λ) dλ; photolytic lifetime τ = 1/J.

Step-by-step reasoning

To decide whether a molecule will photolyse in the troposphere:

1. Find the bond dissociation energy of the weakest relevant bond. 2. Convert it into a threshold wavelength using λ(nm) ≈ 119 600 / E(kJ mol⁻¹). 3. Check whether the threshold is longer than about 290 nm, so that such photons reach the troposphere. 4. Check that the molecule absorbs with a significant cross-section in that range. 5. Estimate J from σ, φ and the actinic flux, and compare 1/J with other loss lifetimes.

Visual explanation

Imagine three overlapping curves against wavelength: the actinic flux rising steeply from zero at 290 nm, the molecule's cross-section falling as wavelength increases, and the quantum yield dropping to zero at the energy threshold. J is proportional to the area under the product of all three — a narrow peak where they overlap.

Real-world analogy

Opening a coconut needs a hammer blow above a certain force (the energy threshold), but you also have to actually hit the coconut (absorption) and the blow has to crack it rather than bounce off (quantum yield). How many coconuts you open per minute depends on how many swings you make — the actinic flux.

Real-world example

Urban NO₂ concentrations drop around midday, and NO rises, because intense sunlight photolyses NO₂ within minutes. At night photolysis stops, and different chemistry — involving the nitrate radical NO₃ — takes over, since NO₃ is itself destroyed in seconds by visible light during the day.

Why?

Why does chlorofluorocarbon photolysis only occur in the stratosphere? CFCs absorb only below about 220 nm. Such photons are blocked by ozone and oxygen above the troposphere, so CFCs survive until they rise above about 25 km, where short-wavelength UV finally reaches them.

Common misconception

"Any photon with enough energy will break the bond." A molecule transparent at that wavelength cannot use the photon at all. N₂ has plenty of energetic photons available in the stratosphere but essentially no absorption there, so it is not photolysed.

Worked example

Question: The C–Cl bond energy in CF₂Cl₂ is about 320 kJ mol⁻¹. What is the threshold wavelength for breaking it, and why does the molecule still survive in the troposphere?

Reasoning: λ ≈ 119 600 / 320 ≈ 374 nm. Energetically, tropospheric light could break the bond. However, CF₂Cl₂ has a negligible absorption cross-section above about 220 nm, so J is effectively zero in the troposphere.

Answer: About 374 nm; it survives because it does not absorb at wavelengths reaching the troposphere.

Quick check

1. What is the photolytic lifetime of NO₂ if J(NO₂) = 8.0 × 10⁻³ s⁻¹? Answer: τ = 1/J = 1 ÷ (8.0 × 10⁻³) = 125 s, a little over two minutes.

Exam focus

Learn the conversion between wavelength and molar energy and the definitions of σ, φ, actinic flux and J. Examiners often ask you to explain why a molecule does or does not photolyse in a given layer — address both energy and absorption.

Advanced insight

Actinic flux counts photons from all directions, not just the direct beam, which is why it can exceed the downward irradiance. Above bright clouds or snow, reflected light can raise J values by 50% or more, while beneath thick cloud they fall sharply — a major source of variability in modelled radical chemistry.

Summary

Sunlight drives atmospheric chemistry through photolysis. Photons must carry enough energy (E = hc/λ) and be absorbed (cross-section σ), and the excited molecule must dissociate (quantum yield φ). The first-order photolysis rate constant J integrates σ, φ and actinic flux over wavelength. Only λ > 290 nm reaches the troposphere, so many molecules photolyse only in the stratosphere.

Practice questions

1. Calculate the energy in kJ mol⁻¹ of photons of wavelength 240 nm. Answer: E ≈ 119 600 / 240 ≈ 498 kJ mol⁻¹, equal to the O=O bond energy. 2. Define the quantum yield of a photochemical reaction. Answer: The fraction of absorbed photons that lead to the specified product, equal to molecules reacting divided by photons absorbed. 3. Give three factors that make J(NO₂) higher at noon in summer than in winter. Answer: A higher Sun gives a shorter path through the atmosphere, more intense actinic flux, and less attenuation by scattering; longer days also add total photolysis. 4. Explain why the troposphere receives almost no radiation below 290 nm. Answer: Stratospheric ozone and, at shorter wavelengths, oxygen absorb it strongly before it can penetrate to lower altitudes.