It is often given in units of cm -1 : watch out for those unit conversions! (1 cm -1 = 100 m -1, not 0.01 m -1 ).

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1

Energy of one quantum of radiation (photon) E = hv h = Planckʼs constant, 6.626 10-34 Js v = frequency of radiation Wave equation: vλ = c c = speed of light, 299 792 485 m/s in vacuum, less in other media (you can usually use the vacuum value for air). λ = wavelength In spectroscopy, the wavenumber is often used instead. It is often given in units of cm -1 : watch out for those unit conversions! (1 cm -1 = 100 m -1, not 0.01 m -1 ). Objects that absorb 100% of incoming radiation are called blackbodies. For blackbodies, the emissivity at wavelength λ, φ λ, depends only on the temperature, and is given by the Planck function: Φ = στ 4 Kirchhoffʼs law Function of T only! Often denoted B(λ,T) σ = 2π 5 k 4 /15c 2 h 3 = 5.67x10-8 W m -2 K -4 Stefan-Boltzmann constant λ max = hc/5kt Wienʼs law λ max D.J. Jacob 2

causes the incoming (solar) and outgoing radiation spectra to deviate significantly from the blackbody curves. Absorption is governed by the resonance condition: if the molecule is at energy level E 1, for absorption at frequency v to occur there has to be an energy level E 2 such that (E 2 -E 1 ) = hv. Whether or not the absorption then actually occurs depends on crosssections, transition moments etc (= complicated quantum mechanics). As a first approximation, molecules have four types of energy levels: Translational Rotational Vibrational Electronic E total = E trans + E rot + E vib + E el Translational energy levels are so closely spaced that they can be considered to be continuous, but rotational, vibrational and electronic energy levels are quantized: absorption can only occur at some specific wavelengths. Translational motion (and e.g. collisions) cause broadening: absorption happens in wavelength bands Absorption of solar radiation by gases Ultra-violet visual infrared D.J. Jacob 3

Scene over Niger valley, N Africa real radiation spectrum (not black body) D.J. Jacob i. Electronic transition: UV radiation (<0.4 µm), sometimes also visible light Jump of electron from ground state to a higher-energy molecular orbital. Concerns mainly incoming solar radiation. ii. Vibrational transition: near-ir (0.7-10 µm) Increase in vibrational energy of a given vibrational mode of a molecule. (Vibrational mode = some combination of bond stretching, angle bending and torsional motion.) For the absorption to be strong, a change in dipole moment of the molecule is required. Concerns mainly outgoing radiation. iii. Rotational transition: far-ir (10-100 µm) Increase in angular momentum around rotation axis. The Greenhouse effect involves absorption of near-ir terrestrial radiation by molecules undergoing vibrational and vibrational-rotational transitions. I.e. by preventing the high energetic light escape, the atmosphere serves as a cover for the Earths surface! D.J. Jacob 4

ozone water vapour CO 2 N 2 O, methane D.J. Jacob Scene over Niger valley, N Africa troposphere surface top of stratosphere D.J. Jacob 5

Example of a GG absorbing at 11 µm 1. 1. Initial state 2. 2. Add to atmosphere a GG absorbing at 11 µm; emission at 11 µm decreases (we donʼt see the surface anymore at that λ, but the atmosphere) D.J. Jacob 3. At new steady state, total emission integrated over all λʼs must be conserved Emission at other λʼs must increase The Earth must heat! 3. E The efficient GGs are the ones that absorb in the atmospheric window (8-13 µm). Gases that absorb in the already-saturated regions of the spectrum are not so efficient GGs. D.J. Jacob 6

= the interaction of molecules with (visible and UV) radiation that leads to chemical reactions. Activation energies for most atmospheric chemical reactions are too high for them to occur at ambient temperatures without an extra input of energy from somewhere. Solar radiation provides this source of energy, e.g. by splitting chemical bonds to produce radicals. Absorption of radiation also produces (either directly or via bond splitting) electronically excited atoms and molecules (which may or may not also be radicals); they also possess excess energy which speed up reactions. Most important example: excited oxygen atoms O( 1 D). Photochemical reactions occur via two steps. First, a molecule AB absorbs a photon, and is excited to an excited electronic state (marked here with a *): AB + hv AB* This can be thought of as a bimolecular reaction involving a molecule and a photon. (But usually the photon flux is included in an effective unimolecular rate constant J or j.) Next, the excited molecule undergoes one of many possible processes Quenching (via collisions) back to the ground state Dissociation to A + B Chemical reaction Ionization (loss of an electron) Luminescense (re-emission of radiation) back to the ground state or some lower-energy excited state Energy transfer within the molecule to some other state, or energy transfer to another molecule Often, e.g. ionization or dissociation occurs almost immediately, but sometimes the excited molecule AB* lives long enough to react with other species. 7

AB + M + M Chemically different species quenching chemical reaction AB + CD* Intermolecular energy transfer + CD AB* dissociation A + B ionization AB Intramolecular energy transfer luminescence AB + + e - AB + hv Energy requirement for AB + hv AB* A + B: hv must be more than the energy ΔE required to break the A-B bond. The wavelength limit is not always sharp as internal energy from vibrational and rotational modes can assist dissociation a little. Excess energy is not usually a problem as it can be distributed into translational, rotational and vibrational modes ( hot products ). However, the precise mechanism of dissociation ( optical dissociation, pre-dissociation ) can lead to complicated structures (peaks, bands, etc.) in the dissociation spectrum (dissociation efficiency vs v or λ). Sometimes also the products can be electronically excited, e.g. O( 1 D). Most covalent bonds have energies of several hundred kj/mol λ < 200-300 nm or so. O 2 absorbs very strongly for λ < 200 nm, with O 2 + hv O 2 * O + O being the major channel. O 2 also absorbs in the 200-230 nm range. All radiation with λ < 200 nm is absorbed high in the atmosphere, and is called vacuum UV because it can only be found outside the atmosphere. Only a few molecules in the atmosphere have bonds that can easily be split by the longer-wave (near UV + visible) radiation reaching the troposphere & lower stratosphere, e.g.: O 3 + hv O 2 + O (bond energy 105 k/mol). NO 2 + hv NO + O (bond energy 300 kj/mol) 8

O 3 + hν O 2 + hν Only this part reaches the Earth s surface (λ > 290 nm). For a polyatomic molecule, there may be many different bonds that can be split, leading to different photolysis channels. ABC + hv ABC* AB + C ABC + hv ABC* A + BC Even for a relative simple molecule like O 3, the products can be formed in different electronic states, also leading to different channels. Different channels for the O 3 + hv => O 2 + O photolysis reaction. (Wayne pg 105.) 9

The quantum yield of a photolysis channel is the fraction of photon absorption events that actually lead to that particular reaction. Quantum yield for channel ABC + hv ABC* AB + C: The quantum yield generally depends on a lot of things: temperature, concentration of other reactants (especially of quenchers [M] p tot ) but most importantly the wavelength. Quantum mechanical selection rules may lead to some channels having zero (or very close to zero) yields, e.g. spin-forbidden channels. Confusingly, even spin-forbidden channels may have non-zero quantum yields, these can sometimes be quite important! The quantum yield for the formation of O( 1 D) from ozone photolysis is of particular interest for atmospheric chemistry. X + hν X* Y + Z The rate coefficient for photochemical reactions is usually denoted as J: d[x]/dt = -J [X] k or J = q x σ x I q x is the quantum yield [molecules photon -1 ], σ X is the absorption cross section [cm 2 molecule -1 ] and I the actinic flux [photons cm -2 s -1 ]. Since all parameters are dependent on the wavelength λ of the radiation we should write: J = λ q x (λ)σ X (λ)i λ dλ I λ σx You can usually look up values for J, qx or σ X at the given conditions. Effectively, photolysis reactions are thus written as if they were unimolecular (with the photon concentration Included in the rate coefficient J). 10

Can be considered a special case of photodissociation, where the other product is a free electron. Energies required for ionization are typically higher than those required for (normal) photolysis. mainly important in the uppermost regions of the atmosphere The electrons released by photoionization can cause electronic excitation via collisions: N 2 + e - N 2 * + e - Galactic cosmic rays (GCR; gamma radiation) and radioactive decay of radon produce some ions also in the troposphere; typical ion production rates are on the order of 1-10 ions/cm 3 s. In the troposphere, ions rapidly recombine: typical steady-state ion concentrations are on the order of 10 3 ions /cm 3. Ions may participate in nucleation (aerosol particle formation), though recent experiments indicate that their role is minor whenever basic or condensing organic molecules are around. Note: highly controversial topic due to possible CGR climate connections! The electronic state of an atom or molecule is usually denoted by a term symbol. 2S+1 L 2S+1 + / Λg / u atoms diatomic molecules E.g. O( 3 P), O( 1 D), O( 3 Σ) O( 1 Δ) L or Λ denote the total orbital angular momentum of the atom's or molecule's electrons, respectively. These are given using latin (atoms) or greek (molecules) letters, analogously to electron shells: S/Σ corresponds to an angular momentum of 0 P/π corresponds to an angular momentum of 1 D/Δ corresponds to an angular momentum of 2, etc. S is the total spin angular momentum of the electrons. 2S+1 is called the spin multiplicity because it gives the different number of ways that the spins can be arranged for a certain value of S. 11

2S+1 L 2S+1 + / Λg / u atoms diatomic molecules There are also other elements in the term symbols (e.g. Ω, +/-, g/u for molecules, possible subscripts like J or M J for atoms) which are related to various wavefunction symmetries or couplings between orbital and spin momenta; these will not be treated here. For polyatomic molecules, the orbital angular momentum becomes difficult to define, so either the states are labeled by various symmetry designations, or an empirical notation is used instead: 2S+1 X is the ground (lowest-energy) state Excited states with the same spin multiplicity as the ground state are labelled (in order of increasing energy) 2S+1 A, 2S+1 B, 2S+1 C... Excited states with a different spin multiplicity are labelled (in order of increasing energy) 2S+1 a, 2S+1 b, 2S+1 c... Term symbols are useful because they can be used to quickly check if reactions follow various selection rules (most importantly, those related to spin) If the total spin angular momentum is S, the component of this spin along an arbitrary axis (usually designated the z-axis) can take the values S z = {S, S-1... -S+1, -S} This gives 2S + 1 different possible values for S z. Heisenberg uncertainty principle: if S z is known exactly, S x and S y are totally unknown: see "cone" visualization on the next page. The possible values of the total spin depend on the number of electrons. The spin of a single unpaired electron is ½ A system with 0 electrons naturally has a total spin of S=0, giving a multiplicity of 1 ("singlet"). A system with 1 electron has a total spin of S=½, and a multiplicity of 2 ("doublet"). (Simple visualization: the z-component of the electron spin can be either "up" or down", or ). A system with 2 electrons can have a total spin of either 0 or 1, giving multiplicities of either 1 ("singlet") or 3 ("triplet"). Why? 12

13

One might naively expect that the four different ways that two electrons can combine their spins are:, and and, giving two states with total spin 0 and two states with total spin 1. However, this is not the case. Why? At least three different ways of explaining - pick the one you understand best and remember that one: 1. the " " and " " states are not allowed due to the Pauli Principle (wavefunction must be antisymmetric with respect to the exchange of two fermions), and the actual allowed "pairings" are: where the notation is S,S z > 2. Another way to think about this is the rule for allowed values of S z : For S=0 there is only one possible value for S z, so there can only be one state, while for S=1 there are three values: -1, 0 and +1, so there have to be three states. 3. Or then you can just look at the "cone & vector" pictures spins add like vectors, not like (simple) numbers. 14

Similar reasoning can be applied to larger numbers of electrons, e.g. 3 electrons gives rise S=1.5 (quartet) and S=0.5 (doublet) terms, etc. However, these rules are needed relatively seldom in atmospheric chemistry, since electrons in closed shells don't count (i.e. they always combine to give S=0): it's only the unpaired electrons that need to be considered. Most compounds with an even number of electrons have all their electrons in closed shells in their ground state, and are therefore singlets. Exceptions: ground states of O and O 2 are triplets. Most compounds with an odd number of electrons have all but one electron paired in closed shells. Exception: free nitrogen atoms have a quartet ground state N( 4 S) (important very high up in the atmosphere). Most stable molecules do not have unpaired electrons in their ground state. This implies that the total spin angular momentum S=0, and the multiplicity 2S + 1 = 1. The ground states are therefore called singlets (or singlet states) The main exceptions in the atmosphere are the oxygen molecule and free oxygen atom, which have two unpaired electrons in their ground state, implying S = 2 ½ =1, and 2S+1=3. Their ground states are labeled O 2 ( 3 Σ ḡ ), O(3 P) These states are called triplets (or triplet states). The first excited states of O 2 and O are singlets: O 2 ( 1 Δ), O( 1 D). 30 15

The spin selection rule is simple: ΔS=0 For unimolecular reactions, this implies that transitions between singlet and triplet states are spin-forbidden, and hence slow. This includes radiative transitions between singlet and triplet states. For reactions with more than one reactant or products, checking spin-allowedness is a bit more complicated but usually not more difficult than 1+2 (or at worst, 2-2) Definition: a reaction A + B C + D is spin-allowed, if the lists S tot, reactants = S A + S B, S A + S B 1,... S A S B S tot, products = S C + S D, S C + S D 1,... S C S D have at least one element in common. If the lists do not have common elements, the reaction is spin-forbidden, and hence slow. 31 Unlike O and O 2, the ground state of O 3 is a singlet. Let's take a look at the photolysis reaction O 3 + hν O 2 + O Using the rule of spin conservation. For the reactant (O 3 ), we know that S = 0 S tot, reactants = {0}. If both products are singlets (S=0), we get: S tot, products = S O2 + S O, S O2 + S O 1,... S O2 S O = 0 + 0,..., 0 0 = {0} allowed If one product is a singlet (S=0) and the other a triplet (S=1): S tot, products = 1 + 0,..., 1 0 = {1} forbidden If both products are triplets (S=1), we get S tot, products = 1 + 1,..., 1 1 = {2,1,0} allowed Thus, the photolysis products of O 3 must either both be in an excited state (singlets), or both be in their ground state (triplets). 16

The only spin-allowed mechanism for O( 1 D) production is: O 3 + hν O 2 ( 1 Δ) + O( 1 D) (λ < 310 nm) This limits the formation of O( 1 D) in the troposphere severely! In practice we see some O( 1 D) production also at longer wavelengths Thermal excitation of vibrational and rotational states may allow photolysis even if the photon energy is slightly lower. The spin-forbidden channel O 3 + hν O 2 ( 3 Σ ḡ ) + O(1 D) also has a small contribution (via spin-orbit coupling). 33 D.J. Jacob 17

Dashed line: older recommendation by data panels Solid line: recent experiments Theoretical wavelength limit for spin-allowed dissociation channel is 310 nm internal energy and spinforbidden channels both contribute. O( 1 D) is about 190 kj/mol higher in energy than O( 3 P) O( 3 P) + H 2 O OH + OH: requires 70 kj/mol energy (endothermic) O( 1 D) + H 2 O OH + OH: releases 120 kj/mol (exothermic) The latter reaction is the main primary source of OH (the most important oxidant) in the troposphere and stratosphere. Note: electronically excited species may be more reactive not only due to energy related reasons (lower activation energies) but also due to structural reasons (higher pre-exponential factors in the Arrhenius equation). The pre-exponential factor for the O( 1 D) + H 2 O reaction is about 20 times higher than that of the O( 3 P) + H 2 O reaction. But how can a single atom (typically visualized as a hard sphere ) have structure? Answer: the wavefunction (both in real-space and spin co-ordinates ) 18

The ground state of O 2 is a triplet. The ground state of almost all other stable molecules are singlets. Let's consider e.g. a simple oxidation reaction where oxygen reacts with methane to give formaldehyde and water: O 2 + CH 4 CH 2 O + H 2 O This reaction is thermodynamically favorable. Why does it not happen there is plenty of O 2 around? Answer: CH 4, CH 2 O and H 2 O are singlets. Thus, the possible total spins of the reactants and products are: S tot, reactants = 1 + 0, 1 + 0 1,... 1 0 = {1} S tot, products = 0 + 0, 0 + 0 1,... 0 0 = {0} Since the two lists do not have common elements, the reaction is spin-forbidden, and hence extremely slow. This is the reason why the reaction of O 2 with non-radical molecules is so slow, despite being thermodynamically very favorable. Radicals have one unpaired electron (s=½), so their total spin is S=0.5, and their multiplicity is 2 they are doublets. Let's look at the reaction of O 2 with a typical organic radical, CH 3 : O 2 + CH 3 CH 3 OO O 2 is still a triplet (S=1), CH 3 and CH 3 OO are doublets (S=0.5). Thus, the possible total spins of the reactants and products are: S tot, reactants = 1 + 0.5, 1 + 0.5 1,... 1 0.5 = {1.5, 0.5} S tot, products = {0.5} Since the two lists have a common elements (S=0.5), the reaction is spin-allowed, and can thus occur reasonably fast. This is why an initial radical step is needed before O 2 can start participating in oxidation chains! 19