Lecture 6: Spectroscopy and Photochemistry II

Required Reading: FP Chapter 3 Suggested Reading: SP Chapter 3

Atmospheric Chemistry CHEM-5151 / ATOC-5151 Spring 2005 Prof. Jose-Luis Jimenez

Outline of Lecture • The Sun as a radiation source • Attenuation from the atmosphere – Scattering by gases & aerosols – Absorption by gases • Beer-Lamber law • Atmospheric photochemistry – Calculation of photolysis rates – Radiation fluxes – Radiation models

1 Reminder of EM Spectrum

Blackbody Radiation Linear Scale Log Scale

From R.P. Turco, Earth Under Siege: From Air Pollution to Global Change, Oxford UP, 2002.

2 Solar & Earth Radiation Spectra • Sun is a radiation source with an effective blackbody temperature of about 5800 K • Earth receives circa 1368 W/m2 of energy from solar radiation From Turco From S. Nidkorodov

• Question: are relative vertical scales ok in right plot?

Solar Radiation Spectrum II

From F-P&P

•Solar spectrum is strongly modulated by atmospheric scattering and absorption From Turco

3 Solar Radiation Spectrum III

UV Photon Energy ↑ C B A

From Turco

Solar Radiation Spectrum IV • Solar spectrum is strongly

O3 modulated by atmospheric absorptions O 2 • Remember that UV photons have most energy

–O2 absorbs extreme UV in mesosphere; O3 absorbs most UV in stratosphere – Chemistry of those regions partially driven by those absorptions – Only light with λ>290 nm penetrates into the lower troposphere – Biomolecules have same bonds (e.g. C-H), bonds can break with UV absorption => damage to life • Importance of protection From F-P&P provided by O3 layer

4 Solar Radiation Spectrum vs. altitude

From F-P&P

• Very high energy photons are depleted high up in the atmosphere • Some photochemistry is possible in stratosphere but not in troposphere • Only λ > 290 nm in trop.

Solar Zenith Angle

• Aside form the altitude, the path length through the atmosphere critically depends on the time of day and geographical location. • Path length can be calculated using the flat atmosphere approximation for zenith angles under 80º. Beyond that, Earth curvature and atmospheric refraction start to matter. Solar Flux 1015

Actual pathlength L "Air Mass" = m = ≈ = secθ ]

-1 Vertical pathlength h nm

-1 14 s 10 -2 SZA = 0º SZA = 86º

1013 At large SZA very little UV-B radiation

Solar Flux [Photons cm reaches the troposphere

1012 300 400 500 700 1000 W avelength [nm ]

5 Direct Attenuation of Radiation

-t×m From F-P&P & S. Nidkorodov I = I0 × e

t = tsg + tag + tsp + tap I ≡ radiation intensity (e.g., F)

I0 ≡ radiation intensity above atmosphere m ≡ air mass t ≡ attenuation coefficient due to – absorption by gases (ag) – scattering by gases (sg) – scattering by particles (sp) – absorption by particles (ap)

Rayleigh scattering t ∝λ-n -4 sp tsg ∝λ much

more Deep UV – O, N2, O2 complex Mid UV & visible – O3 Near IR – H2O Infrared – CO2, H2O, others tag ∝ σ

Scattering by Gases • Purely physical process, not absorption • Approximation: 5 2 4 tsg =1.044⋅10 ⋅(n0λ −1) / λ • Strongly increases as λ decreases • Reason why “sky is blue” during the day From Turco

6 Scattering & Absorption by Particles • Particles can scatter From Jacob and absorb radiation • Scattering efficiency is very strong function of particle size – For a given wavelength • Visible: λ ~ 0.5 µm – Particles 0.5-2 µm are most efficient scatterers! • Will discuss in more detail in aerosol lectures

Gas Absorption: Beer-Lambert Law I

I = I0 exp(−σ ⋅ L⋅ N) • Allows the calculation of the decay in intensity of a light beam due to absorption by the molecules in a medium

Definitions:

• A = ln(I0/I) = Absorbance = σ⋅L⋅N (also “optical depth”) • σ ≡ absorption cross section Solve in class: Show that in the small [cm2/molec] absorption limit the relative change in light intensity is approximately equal to • L ≡ absorption path length [cm] absorbance. • n ≡ density of the absorber [molec/cm3]

From F-P&P & S. Nidkorodov

7 Beer-Lambert Law II Pitfalls: • Other units are frequently used to express absorbance, for example:

A =ln(I0/I) = ε×L×CA=ln(I0/I) = α×L×P ε ≡ extinction coefficient [L mol-1 cm-1] α ≡ absorption coefficient [atm-1 cm-1] C ≡ density of the absorber [mol L-1] P ≡ partial pressure [atm] • Base 10 is used in most commercial spectrometers instead of the natural base:

Abase 10 = log(I0/I) = Abase e/ln(10)

Physical interpretation of σ • σ , absorption cross section (cm2 / molecule) – Effective area of the molecule that photon needs to traverse in order to be absorbed. – The larger the absorption cross section, the easier it is to photoexcite the molecule.

– E.g., pernitric acid HNO4

Collisions Light absorption σ≈10-15 cm2/molec σ≈10-18 cm2/molec From S. Nidkorodov

8 Measurement of Absorption Cross Sections

Measurement of absorption cross sections is, in principle, trivial. We need a light source, such as a lamp (UV), a cell to contain the molecule of interest, a spectral filter (such as a monochromator) and a detector that is sensitive and responds linearly to the frequency of radiation of interest:

Gas cell Filter I0 I Detector n [#/cm3] L Measurements are repeated for a number of concentrations at each wavelength of interest. Although seemingly trivial, in practice such measurements are difficult because of impurities, especially when it comes to very small cross sections (< 10-20 cm2/molec)

Solve in class: Sample contains 1 Torr of molecules of interest with σ=3x10-21 cm2/molec and 1 mTorr of impurity with σ=2x10-18 cm2/molec. What is the total absorbance in a 50 cm cell? From S. Nidkorodov

Example: UV Attenuation by O3 and O2

Attenuation coefficient is dominated by O3 absorption in the 200-300 nm window. Therefore, direct attenuation can be easily calculated from known absorption cross

sections of O3. Similar formulas apply to attenuation by O2 in 120-180 nm window.

-σ (λ )×A×m I(λ) = I0 (λ)× e where A ≡ column density ∞ A = [O (z)]dz ∫ 3 z Alternatively written : From S. Nidkorodov -τ (λ,z) Solve in class: Using barometric law estimate I(λ) = I0 (λ)× e column density of O2 in the atmosphere. By how where τ ≡ optical depth much does atmospheric O2 attenuate solar radiation at around 170 nm (σ ≈ 10-17 cm2/molec) ∞ at noon (m = 1)? Assume that O fraction (21%) is τ (λ, z) = σ (λ)× m×[O (z)]dz 2 ∫ 3 independent of altitude and T = 270 K. z Ans: 4x1024; by exp(-107)

9 Solar Radiation Intensity

To calculate solar From F-P&P spectral distribution in any given volume of air at any given time and location one must know the following: – Solar spectral distribution outside the atmosphere – Path length through the atmosphere – Wavelength dependent attenuation by atmospheric molecules – Amount of radiation indirectly scattered by the earth surface, clouds, aerosols, and other volumes of air

Surface Albedo Reflected Radiation(λ) Albedo(λ) = Incident Radiation(λ) • Wavelength dependent! • Question: for the same incident UV solar flux, will you tan faster over snow or over a desert?

From F-P&P

10 Total vs. Downwelling Radiation

From Warneck • If atmosphere was completely transparent and surface completely absorbing (albedo = 0)

–FU = 0

–FD =FT= 1 • Due to gas + aerosol scattering and surface reflection

–FU can be large

–FT > solar flux!

Calculation of Photolysis Rates I Generic reaction: A + hν → B + C d[A] = −J [A] dt A • A “first-order process”

• What does JA depend on?

• JA depends on – Light intensity from all directions • “Actinic flux” – Absorption cross section (σ) – Quantum yield for photodissociation (φ) – All are functions of wavelength

11 Calculation of Photolysis Rates II Generic reaction: A + hν → B + C d[A] = −J [A] = − σ (λ)φ (λ)F(λ)dλ ×[A] dt A ∫ A A -1 JA – first order photolysis rate of A (s ) 2 σA(λ) – wavelength dependent cross section of A (cm /#)

φA(λ) – wavelength dependent quantum yield for photolysis F(λ) – spectral actinic flux density (#/cm2/s)

So, what are the smallest cross sections that matter? The solar actinic flux (photons cm-2 s-1 nm-1) is of order 1014. In many cases, we need to know whether the photolytic lifetime of a molecule is 10 days (J=10-6 s-1), or 100 days (J=10-7 s-1). This means that cross sections as small as 10-20 cm2 or even smaller are potentially interesting. Such small cross sections are very challenging to measure with sufficient accuracy.

Radiation Fluxes Definitions Quantity Description Units Actinic flux density: energy crossing a unit area per unit time without F J m-2 s-1 consideration of direction (we do not care where photons come from) F(λ) Spectral actinic flux density: same as flux but per unit wavelength J m-2 s-1 nm-1 E Irradiance: same as flux but for a unit area with a fixed orientation J m-2 s-1 E(λ) Spectral irradiance: same as radiance but per unit wavelength range J m-2 s-1 nm-1

L(θ, ϕ) Radiance: radiant flux density per unit solid angle J m-2 s-1 sr-1 J m-2 s-1 nm-1 sr-1 L(θ, ϕ, λ) Spectral radiance: same as radiance but per unit wavelength range

2ππ E = ∫ L(θ,ϕ)cosθdω = ∫∫L(θ,ϕ)cosθ sinθdθdϕ ω 00 2ππ F = ∫ L(θ,ϕ)dω = ∫∫L(θ,ϕ)sinθdθdϕ ω 00 Radiance as a function of direction gives a complete description of the radiation field. When L is independent of direction, the field is called isotropic, in which case E = π L and F = 2π L.

From F-P&P Solve in class: There are 109 photons flying into a 0.01 cm diameter opening every & S. Nidkorodov second. What is F with respect to this opening in units of #/cm2/s?

12 Radiation Mesurements Flat Plate → Irradiance 2π→½ of Actinic Flux

• Radiation does not just come directly from the sun – scattered radiation is just as important – Measure total or spectrally-resolved flux

– Use models From F-P&P

Radiation Models • Predict radiation intensity – As f(time, altitude, latitude, λ) – Results of Madronich (1998) described in text • Will use extensively in homeworks & exams – Typical model results: From F-P&P

13 Example model results

From F-P&P Q: summer/winter solstices intensity at 445 nm: @ 8 am? @ noon?

Examples of Photolysis Rates

From F-P&P

14 Example: Photolysis of CH3CHO

CH CHO + hν → CH + HCO (a) λi 3 3 J = ∫σ (λ)φ(λ)F(λ)dλ ≈ ∑σ (λ)φ (λ)F (λ)∆λ → CH4 + CO (b) λ=290nm

From F-P&P

15