B.2 Greenhouse effect

1. 340 W m⁻² — A student who forgets that the scattered sunlight is already leaving picks this, making the infrared alone balance all the incoming radiation. Then Earth would lose 340 + 100 = 440 W m⁻² while receiving 340 W m⁻². Only the absorbed 240 W m⁻² is re-emitted as infrared.
2. 100 W m⁻² — A student who thinks the radiation leaving Earth is simply reflected sunlight takes the 100 W m⁻² scattered back as the outgoing radiation. Scattered sunlight and emitted infrared are separate streams; at equilibrium 340 = 100 + emitted infrared, so 240 W m⁻² of infrared leaves.
3. 390 W m⁻² — A student who assumes the radiation leaving to space is the surface's own emission picks this. Earth would then lose 390 + 100 = 490 W m⁻² while gaining 340 W m⁻². The atmosphere absorbs much of the surface emission and radiates part of it back down, so less leaves to space.
4. 240 W m⁻² — Conservation of energy at equilibrium: incoming 340 W m⁻² = scattered 100 W m⁻² + emitted infrared, so 240 W m⁻² of infrared leaves. The surface emits more (390 W m⁻²) because the atmosphere absorbs much of that emission and radiates part of it back down.

Working At equilibrium the power leaving equals the power arriving (conservation of energy). Leaving = scattered solar radiation + emitted infrared, so emitted infrared = 340 W m⁻² − 100 W m⁻² = 240 W m⁻². The surface emission (390 W m⁻²) is not the outgoing value: most of it is absorbed by the atmosphere, which radiates both upwards and downwards.

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1. It compares the power radiated per unit area with that of a black body at the same temperature. — Emissivity = (power radiated per unit area)/σT⁴, where σT⁴ is the power per unit area radiated by an ideal black surface at the same temperature. It is a ratio with no unit, between 0 and 1.
2. It is the power radiated per unit area of the surface, so it is measured in watts per square metre. — A student who takes emissivity to be the emitted intensity itself picks this. Emissivity is that intensity divided by σT⁴, so the units cancel and it has no unit.
3. It is the fraction of the incident radiation that the surface reflects, so it is equal to its albedo. — A student who treats emissivity and albedo as the same property picks this. Albedo describes scattered incident radiation; emissivity describes the radiation the surface emits because of its own temperature, compared with a black body.
4. It depends on the visible colour of a surface, so white snow has an emissivity close to zero. — A student who judges emission by visible colour picks this. Snow reflects most visible light, but in the infrared, where it emits, its emissivity is close to 1.

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1. The total power absorbed by the planet divided by the total power incident on it — A student who confuses albedo with absorption picks this. This ratio is 1 − α, the fraction absorbed; albedo is the fraction scattered back.
2. The total power scattered by the planet divided by the total power incident on it — Albedo = total scattered power / total incident power. It is a ratio with no unit; Earth's mean value is about 0.30.
3. The power reflected by each square metre of the planet's surface, in W m⁻² — A student who treats albedo as an intensity picks this. Albedo is a ratio of two powers and has no unit; the reflected intensity is albedo × incident intensity.
4. The infrared power the planet emits divided by the power incident on it — A student who thinks Earth's outgoing infrared is reflected sunlight picks this. Albedo counts only incident radiation scattered without absorption; emitted infrared comes from absorbed energy and is described by the emissivity.

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1. Snow and ice cover much of the surface, and low-angle sunlight is reflected more strongly. — Earth's albedo depends on latitude. Snow and ice scatter a large fraction of incident sunlight, and at the low angles at which sunlight reaches high latitudes, surfaces (especially water) reflect a larger fraction than under a high Sun.
2. Snow and ice reflect all of the sunlight that falls on them, whereas ocean water reflects none of it. — A student who pictures reflection as all-or-nothing picks this. No natural surface reflects all or none of the sunlight: fresh snow scatters up to about 0.9 of it and open ocean about 0.06. The high polar albedo comes from the large area of snow and ice together with the low angle at which the rays arrive.
3. The polar regions are further from the Sun, so more of the sunlight that reaches them is reflected. — A student who explains latitude by distance from the Sun picks this. The difference in distance is at most a few thousand kilometres out of 1.5 × 10⁸ km, which is negligible, and distance would not change the fraction reflected anyway.
4. The polar surfaces are colder, and colder surfaces reflect a larger fraction of the radiation on them. — A student who takes the coldness of snow and ice to be the cause of their reflectivity picks this. The fraction reflected depends on the material, its texture and the angle of the rays, not on temperature.

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1. The intensity of the Sun's radiation on a horizontal surface at ground level, averaged over a whole year at the equator — A student who places the solar constant at the ground picks this. The atmosphere scatters and absorbs part of the radiation, and a horizontal surface is not perpendicular to the rays; S is defined above the atmosphere, on a surface facing the Sun.
2. The total power emitted by the Sun in all directions, which stays constant from one year to the next — A student who takes the name to mean a fixed property of the Sun picks this. The Sun's total output is its luminosity, about 3.8 × 10²⁶ W; S is an intensity, the luminosity divided by 4πd² at Earth's distance.
3. The intensity of sunlight above the atmosphere, on a surface perpendicular to the rays, at the mean Earth–Sun distance — This is the definition. S ≈ 1.36 × 10³ W m⁻². It is measured by satellites above the atmosphere, on a surface facing the Sun, at the mean distance of Earth from the Sun.
4. The mean intensity of the Sun's radiation averaged over the whole of Earth's spherical surface, day and night — A student who thinks S falls on every square metre of the surface, and so is also the mean over the whole surface, picks this. The planet intercepts SπR² but has a surface area of 4πR², so the mean over the whole spherical surface, day and night, is S/4; S itself is the intensity on a surface perpendicular to the rays.

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1. Half of the surface is in darkness at any instant, and half of the sunlight reaching the day side is reflected. — A student who looks for two reasons to halve picks this. The factor of 4 is geometric; reflection is accounted for separately by the albedo, as (1 − α)S/4, and Earth's albedo is about 0.30, not 0.50.
2. The atmosphere absorbs or scatters three quarters of the Sun's radiation before it can reach the surface. — A student who links the factor to the atmosphere picks this. S/4 is the mean intensity at the top of the atmosphere, before any absorption; the atmosphere scatters and absorbs much less than three quarters.
3. The planet intercepts power over its cross-section πR², but this power is spread over its total surface area 4πR². — The power intercepted is SπR², set by the planet's projected disc. Averaged over the whole surface area 4πR², the mean incoming intensity is SπR²/4πR² = S/4.
4. The Sun's radiation spreads out over a sphere of area 4πd², so only a quarter of it reaches the planet. — A student who merges the Sun's sphere with the planet's sphere picks this. The 4πd² gives S itself; only a tiny fraction of the Sun's output, πR²/4πd², reaches the planet. The factor of 4 in S/4 is 4πR²/πR².

Working Power intercepted by the planet = S × (area of its projected disc) = SπR². This power is shared over the whole spherical surface, of area 4πR², so the mean incoming intensity = SπR²/4πR² = S/4.

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1. Each of them has natural sources as well as sources in human activity. — Methane, water vapour, carbon dioxide and nitrous oxide all have natural origins (wetlands for CH₄, evaporation for H₂O, respiration and volcanoes for CO₂, soil bacteria for N₂O) and origins in human activity (livestock, irrigation, burning fossil fuels, nitrogen fertilisers).
2. Water vapour is not one of them, since it comes only from natural evaporation. — A student who thinks only human emissions count picks this. Water vapour is one of the main greenhouse gases; it absorbs infrared whatever its source, and human activity such as irrigation and combustion also adds some.
3. Nitrogen and oxygen are the main ones, as they make up about 99% of dry air. — A student who equates abundance with effect picks this. N₂ and O₂ absorb almost no infrared radiation; the main greenhouse gases are CH₄, H₂O, CO₂ and N₂O.
4. They are all pollutants produced only by industry, transport and farming. — A student who equates greenhouse gases with pollution picks this. Each main greenhouse gas also has natural sources, such as wetlands (CH₄), evaporation (H₂O), respiration (CO₂) and soil bacteria (N₂O).

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1. It reflects the photon, so that the infrared radiation is sent straight back down towards the Earth's surface. — A student who pictures greenhouse gases as a mirror picks this. The molecule absorbs the photon and is excited; the later emission is in random directions. This is absorption and emission, not reflection.
2. It moves to a higher vibrational energy level, and the energy is later emitted as infrared in all directions. — The photon energy matches the spacing of the molecule's vibrational energy levels, so the molecule is excited. The energy is later emitted as infrared (often after being shared in collisions) with no preferred direction, so part returns towards the surface and part goes upwards.
3. It moves to a higher energy level and then emits the infrared downwards, back to the surface. — A student who thinks the re-emitted radiation is aimed back at the ground picks this. Emission has no preferred direction; roughly as much goes upwards as downwards.
4. It splits into atoms, because the photon's energy is enough to break the molecule's bonds. — A student who carries over the ultraviolet chemistry of the ozone layer picks this. An infrared photon has energy of order 10⁻²⁰ J, far less than the energy needed to break a bond, so the molecule is only vibrationally excited.

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1. Radiation whose frequency equals a natural frequency of vibration of the bonds drives large vibrations. — In the resonance model the bonds act like springs with natural frequencies. Infrared at a natural frequency drives the vibration at resonance, the amplitude grows, and energy is absorbed strongly; at other frequencies little is absorbed.
2. Infrared photons carry more energy than visible ones, so they shake the molecules more violently. — A student who thinks of infrared as powerful 'heat radiation' picks this. Infrared photons carry less energy than visible photons; absorption depends on matching a natural frequency, not on the photons being energetic.
3. The molecules vibrate at whatever frequency the radiation has, so all are absorbed equally. — A student who equates resonance with forced vibration picks this. A driven molecule does vibrate at the driving frequency, but the amplitude, and so the absorption, is large only near its natural frequency.
4. Radiation at a natural frequency of the bonds makes them vibrate until they break, releasing the energy. — A student who links resonance with objects shaking apart, like a glass shattered by sound, picks this. An infrared photon carries energy of order 10⁻²⁰ J, far less than the energy needed to break a bond, so the molecule is excited into larger vibration, not broken; the energy is later emitted or shared in collisions.

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1. The greenhouse effect itself, which would not occur at all without the greenhouse gases released by human activity — A student who treats the greenhouse effect and the enhanced greenhouse effect as the same picks this. The natural greenhouse effect keeps Earth's mean surface about 33 K warmer than it would otherwise be; the enhanced effect is the human addition to it.
2. Extra warming caused by holes in the ozone layer, which let more solar radiation through to the ground — A student who links ozone depletion with global warming picks this. Ozone depletion lets more ultraviolet reach the surface, a separate problem; the enhanced greenhouse effect is increased absorption of infrared by greenhouse gases.
3. Direct warming of the air by the heat released when fuels are burned in engines and power stations — A student who thinks the heat of combustion causes the warming picks this. That heat is tiny compared with the effect of the CO₂ produced, which increases the absorption of the surface's infrared emission.
4. The augmentation of the natural greenhouse effect by human activities that add greenhouse gases to the air — This matches the guide: the augmentation of the greenhouse effect due to human activities, which raise the concentrations of greenhouse gases, chiefly by burning fossil fuels.

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1. 183 K — A student who leaves out the 333 W m⁻² of back radiation, treating the atmosphere as a blanket that only slows losses, gets σT⁴ = 161 − 97 = 64 W m⁻² and T = 183 K. The atmosphere radiates to the surface, and in this model that input is about twice the absorbed sunlight.
2. 306 K — A student who assumes the surface loses energy only by radiation sets σT⁴ = 494 W m⁻² and gets 306 K. Convection and evaporation remove 97 W m⁻² without radiation, so only 397 W m⁻² is radiated.
3. 289 K — The surface absorbs 161 + 333 = 494 W m⁻² and loses 97 W m⁻² by convection and evaporation, so it radiates 397 W m⁻². σT⁴ = 397 W m⁻² gives T = 289 K, close to Earth's observed mean surface temperature of about 288 K.
4. 255 K — A student who takes the radiation leaving to space as the surface's own emission uses 239 W m⁻² and gets 255 K. That is the temperature of a black body emitting what leaves the top of the atmosphere; the surface radiates more, because it also receives back radiation from the atmosphere.

Working Surface energy balance. Absorbed = 161 + 333 = 494 W m⁻². Non-radiative losses = 17 + 80 = 97 W m⁻². Radiated intensity = 494 − 97 = 397 W m⁻² = σT⁴. T = (397 / 5.67 × 10⁻⁸)^(1/4) = (7.00 × 10⁹ K⁴)^(1/4) = 289 K. (The 239 W m⁻² at the top of the atmosphere is not needed: it is the outgoing radiation from the surface–atmosphere system, not the surface's emission.)

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1. 288 K — Absorbed solar intensity = 0.70 × 1360/4 = 238 W m⁻²; back radiation = 0.780σ(242 K)⁴ = 152 W m⁻². The surface radiates the total: σT⁴ = 390 W m⁻², so T = 288 K.
2. 255 K — A student who does not count the atmosphere's downward radiation as an input to the surface balances σT⁴ against the 238 W m⁻² of sunlight alone and gets 255 K. The atmosphere radiates 152 W m⁻² down to the surface, and the surface absorbs it.
3. 313 K — A student who thinks all the atmosphere's emission comes down to the surface adds both the upward and the downward 152 W m⁻², getting σT⁴ = 238 + 2 × 152 = 541 W m⁻² and T = 313 K. The atmosphere radiates in all directions; the upward part goes to space.
4. 374 K — A student who uses S as the mean intensity over the whole surface gets σT⁴ = 0.70 × 1360 + 152 = 1104 W m⁻² and T = 374 K. The planet intercepts SπR² but has a surface area of 4πR², so the mean incoming intensity is S/4.

Working Solar intensity absorbed by the surface = (1 − 0.30) × (1.36 × 10³ W m⁻²)/4 = 0.70 × 340 = 238 W m⁻². Intensity radiated downwards by the atmosphere = 0.780 × 5.67 × 10⁻⁸ × 242⁴ = 152 W m⁻². Surface equilibrium: σT⁴ = 238 + 152 = 390 W m⁻², so T = (390 / 5.67 × 10⁻⁸)^(1/4) = 288 K. (Check: the atmosphere absorbs 0.780 × 390 = 304 W m⁻² of the surface's emission and emits 2 × 152 = 303 W m⁻², so it is also in balance, and 0.220 × 390 + 152 = 238 W m⁻² leaves to space.)

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1. Nothing changes: energy is conserved, so the planet's energy and hence its temperature must stay the same. — A student who reads 'energy is conserved' as 'the planet's energy stays the same' picks this. Conservation means the 4 W m⁻² absorbed but not emitted is stored in the oceans, land and air. The planet is not isolated, so its energy and temperature can change.
2. The surface and atmosphere warm until the outgoing infrared again equals the solar intensity absorbed. — The 4 W m⁻² imbalance is stored as internal energy, so the temperature rises. A warmer surface and atmosphere emit more infrared, so the outgoing intensity rises until it again equals the absorbed solar intensity, and the planet reaches a new, warmer equilibrium.
3. The planet gains 4 W m⁻² for as long as the extra CO₂ is present, because the gas traps the energy. — A student who pictures greenhouse gases as a lid that keeps energy in picks this. As the planet warms it emits more infrared, so the imbalance shrinks to zero; the gain does not continue once the concentration is held constant.
4. The atmosphere warms because the extra CO₂ absorbs more incoming sunlight; the outgoing infrared is unaffected by CO₂. — A student who thinks greenhouse gases warm the planet by absorbing incoming sunlight picks this. CO₂ is largely transparent to sunlight, which is mostly visible; it absorbs the infrared emitted by the surface. The shortfall in outgoing infrared given in the stem is the effect of the extra CO₂, and the surface and atmosphere warm until the outgoing infrared again balances the absorbed sunlight.

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1. 267 K — A student who multiplies by the emissivity instead of dividing, expecting a poorer emitter to be cooler, gets T⁴ = 0.80 × 360/σ and T = 267 K. A planet that radiates less at a given temperature must be hotter to radiate the 360 W m⁻² it absorbs.
2. 282 K — A student who thinks the albedo already accounts for the surface's radiative properties leaves out the emissivity and gets σT⁴ = 360 W m⁻², T = 282 K. Albedo reduces the absorbed intensity and emissivity reduces the emitted intensity; both are needed.
3. 298 K — (1 − α)S/4 = eσT⁴: 0.60 × 600 = 360 W m⁻² = 0.80 × 5.67 × 10⁻⁸ × T⁴, so T = 298 K. The emissivity below 1 means the planet must be warmer than a black body to radiate the same intensity.
4. 422 K — A student who uses the full 2.40 × 10³ W m⁻² as the mean incoming intensity gets 0.60 × 2400 = 1440 W m⁻² and T = 422 K. The planet intercepts power over πR² but radiates from 4πR², so the mean incoming intensity is S/4.

Working Mean absorbed intensity = (1 − 0.40) × (2.40 × 10³ W m⁻²)/4 = 0.60 × 600 = 360 W m⁻². Emitted intensity = eσT⁴. Equilibrium: 0.80 × 5.67 × 10⁻⁸ × T⁴ = 360, so T⁴ = 360/(4.536 × 10⁻⁸) = 7.94 × 10⁹ K⁴ and T = 298 K.

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1. 0.33 — A student who takes the simple mean (0.60 + 0.06)/2 = 0.33 picks this. The ocean receives 70% of the incident power, so its low albedo must carry more weight.
2. 0.78 — A student who treats albedo as the fraction absorbed calculates 1 − 0.222 = 0.78. That is the fraction of the incident power the region absorbs; the albedo is the fraction scattered, 0.22.
3. 0.30 — A student who assumes cloud reflects all the sunlight and ocean none takes the cloud cover, 30%, as the albedo. The cloud scatters only 0.60 of what falls on it, and the ocean scatters 0.06, so the albedo is 0.22.
4. 0.22 — Weight each albedo by the fraction of the incident power it receives: 0.30 × 0.60 + 0.70 × 0.06 = 0.180 + 0.042 = 0.222 ≈ 0.22.

Working Albedo = total scattered power / total incident power. For incident intensity I on a total area A: scattered power = 0.60 × (0.30IA) + 0.06 × (0.70IA) = (0.180 + 0.042)IA = 0.222IA. Albedo = 0.222IA / IA = 0.22.

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1. The cloud absorbed and held the Sun's energy, and this trapped energy is what raised the albedo. — A student who thinks clouds work by trapping heat picks this. Albedo is scattered power over incident power, so absorption cannot raise it; the rise shows that the cloud scattered more sunlight back to space.
2. The cloud scattered a larger fraction of the sunlight back to space, so less of it reached the forest. — The same incident intensity arrives on both days, and the scattered fraction rises from 0.12 to 0.48. The cloud scatters sunlight back to space, so less reaches and is absorbed by the forest: Earth's albedo varies daily with cloud formation.
3. A region's albedo is a fixed property of its surface, so one of the two readings must be a measurement error. — A student who treats albedo as a constant picks this. The forest has not changed, but the system seen from the satellite now includes thick cloud, and Earth's albedo varies from day to day with cloud formation.
4. More sunlight arrived above the region on day 2, so more was reflected, and more reflected means a larger albedo. — A student who thinks of albedo as the amount of sunlight reflected picks this. The Sun was at the same elevation on both days, so the same intensity arrived at the top of the atmosphere. Albedo is a ratio, scattered power over incident power, so it rose because the cloud raised the fraction scattered, not because more light arrived.

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1. 8.95 × 10² W m⁻² — A student who takes intensity to be inversely proportional to distance gets 1.36 × 10³ × (1.50/2.28) = 895 W m⁻². The power spreads over an area 4πd², so the ratio must be squared.
2. 3.14 × 10³ W m⁻² — A student who writes the distance ratio upside down gets 1.36 × 10³ × (2.28/1.50)² = 3.14 × 10³ W m⁻², more than at Earth. Mars is further from the Sun, so it must receive less: use (d_Earth/d_Mars)².
3. 1.47 × 10² W m⁻² — A student who thinks the factor of 4 belongs in the solar constant divides the correct 589 W m⁻² by 4. That gives the mean incoming intensity over Mars's whole surface, S/4, not the solar-constant equivalent.
4. 5.89 × 10² W m⁻² — The Sun's power spreads over a sphere of area 4πd², so S ∝ 1/d². S_Mars = 1.36 × 10³ × (1.50/2.28)² = 589 W m⁻².

Working Intensity follows an inverse-square law, I = L/(4πd²), so S_Mars = S_Earth × (d_Earth/d_Mars)² = 1.36 × 10³ W m⁻² × (1.50/2.28)² = 1.36 × 10³ × 0.4328 = 589 W m⁻² = 5.89 × 10² W m⁻².

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1. 360 K — A student who uses S as the mean intensity on the whole surface gets σT⁴ = 0.70 × 1360 = 952 W m⁻² and T = 360 K. The power intercepted over πR² is spread over 4πR², so the mean incoming intensity is S/4.
2. 255 K — Absorbed intensity = (1 − 0.30) × 1360/4 = 238 W m⁻² = σT⁴, so T = 255 K. This is about 33 K below Earth's observed mean surface temperature; the difference is the greenhouse effect.
3. 303 K — A student who divides by 2 because half the planet is lit gets σT⁴ = 476 W m⁻² and T = 303 K. Dividing by 2 allows for night but not for the oblique angle of the rays on the day side; the correct mean is S/4.
4. 206 K — A student who treats the albedo as the fraction absorbed uses 0.30 × 340 = 102 W m⁻² and gets T = 206 K. Albedo is the fraction scattered, so the absorbed fraction is 1 − 0.30 = 0.70.

Working Energy balance: (1 − α)S/4 = σT⁴. Absorbed intensity = 0.70 × (1.36 × 10³ W m⁻²)/4 = 238 W m⁻². T = (238 / 5.67 × 10⁻⁸)^(1/4) = (4.20 × 10⁹ K⁴)^(1/4) = 255 K.

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1. Only the methane from farming absorbs infrared; methane from wetlands is part of a natural balance. — A student who thinks natural greenhouse gases have no effect picks this. A molecule does not carry a record of its source; wetland methane absorbs infrared exactly as farm methane does, and it is part of the natural greenhouse effect.
2. Methane from farming absorbs more strongly, because it is released with pollutants that trap heat. — A student who links greenhouse gases with pollution picks this. The infrared absorption of methane is a property of the CH₄ molecule, not of any substances released with it.
3. Neither absorbs much, since carbon dioxide is the only gas in air that absorbs infrared. — A student who knows only CO₂ as a greenhouse gas picks this. Methane is one of the main greenhouse gases, and an additional methane molecule has a larger effect than an additional CO₂ molecule.
4. Methane from both sources absorbs infrared identically, as absorption depends on the molecule. — A CH₄ molecule absorbs infrared at wavelengths set by its vibrational energy levels. These are the same whatever the source, so methane from wetlands and from livestock contributes to the greenhouse effect in the same way.

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1. The photons of 10.0 μm and 15.0 μm — A student who thinks any photon with at least enough energy is absorbed includes the 10.0 μm photons (1.99 × 10⁻²⁰ J). A transition between two levels needs a photon whose energy equals the gap, not exceeds it.
2. Photons of all three wavelengths — A student who treats all infrared as 'heat' that greenhouse gases absorb picks this. Only a photon whose energy matches an energy-level difference is absorbed, so absorption happens at particular wavelengths.
3. None of these infrared photons — A student who thinks greenhouse gases act on incoming ultraviolet sunlight, not on infrared, picks this. The 15.0 μm photon energy, 1.33 × 10⁻²⁰ J, equals the vibrational energy-level gap, so infrared of that wavelength is absorbed.
4. Only the photons of 15.0 μm — E = hc/λ gives 1.99 × 10⁻²⁰ J, 1.33 × 10⁻²⁰ J and 9.95 × 10⁻²¹ J. Only the 15.0 μm photons have an energy equal to the gap between the levels, so only they are absorbed.

Working E = hc/λ. For 10.0 μm: (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(10.0 × 10⁻⁶) = 1.99 × 10⁻²⁰ J. For 15.0 μm: 1.33 × 10⁻²⁰ J. For 20.0 μm: 9.95 × 10⁻²¹ J. Only the 15.0 μm photon energy equals the energy-level difference, so only those photons are absorbed.

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1. They contradict each other: one treats infrared as a wave, the other as photons, so only one is valid. — A student who expects a single 'true' model picks this. The two descriptions agree numerically: the frequency that resonates with the vibration is the frequency whose photons carry energy equal to the level spacing.
2. Only the resonance model is valid, since energy levels belong to electrons in atoms, not to bonds. — A student who ties energy levels to electrons alone picks this. Molecular vibrations are also quantised, with level spacings of order 10⁻²⁰ J, which is why they absorb infrared photons.
3. Both are valid: the radiation's frequency is the natural frequency, and hf is the level spacing. — c/λ = 3.00 × 10⁸/15 × 10⁻⁶ = 2.0 × 10¹³ Hz, the natural frequency, so the resonance model predicts strong absorption. The photon energy hf = 1.33 × 10⁻²⁰ J equals the vibrational energy-level spacing, so the energy-level model predicts absorption at the same wavelength.
4. Both are valid, but the level model predicts absorption at all shorter wavelengths, whose photons have more energy. — A student who thinks a molecule absorbs any photon with at least enough energy picks this. A transition between two vibrational levels needs hf equal to the level spacing, not greater, so photons of shorter wavelength are not absorbed by this transition. The energy-level model therefore agrees with the resonance model: strong absorption at 15 μm, not at every shorter wavelength.

Working f = c/λ = (3.00 × 10⁸ m s⁻¹)/(15 × 10⁻⁶ m) = 2.0 × 10¹³ Hz, equal to the natural frequency (resonance model). Photon energy hf = 6.63 × 10⁻³⁴ × 2.0 × 10¹³ = 1.33 × 10⁻²⁰ J, the spacing of the vibrational energy levels (energy-level model). Both models describe the same absorption.

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1. Volcanoes are the main cause of the rise, because they release more CO₂ than human activity. — A student who believes volcanoes out-emit human activity picks this. The data contradict it: volcanoes release less than 1 × 10⁹ tonnes of CO₂ per year, against more than 35 × 10⁹ tonnes from burning fossil fuels. With the natural exchanges in balance, the small volcanic flow cannot account for the rise.
2. The extra CO₂ has thinned the ozone layer, so more ultraviolet now reaches and heats the surface. — A student who confuses the enhanced greenhouse effect with ozone depletion picks this. CO₂ acts by absorbing outgoing infrared; ozone depletion, caused mainly by CFCs, is a separate problem.
3. Burning fossil fuels is the main cause of the rise, and the extra CO₂ absorbs more of the surface's infrared. — Before industrialisation the large natural flows of CO₂ into and out of the air were approximately in balance, so a sustained rise needs a net addition. Burning fossil fuels adds more than 35 × 10⁹ tonnes per year, over 35 times the volcanic output, so it is the main cause. CO₂ absorbs infrared emitted by the surface, so a higher concentration absorbs more of it and enhances the greenhouse effect.
4. At under 0.05% of the air, CO₂ is too dilute for the rise to change the infrared absorbed. — A student who judges a gas's effect by its proportion picks this. CO₂ absorbs strongly near 15 μm, where Earth emits strongly, so a rise from 316 to 421 ppm increases the infrared absorbed even though CO₂ is a small fraction of the air.

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