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

Atmospheric ChemistryCHEM-5151 / ATOC-5151

Spring 2005Prof. Jose-Luis Jimenez

Lecture 6: Spectroscopy and Photochemistry II

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

2

Reminder of EM Spectrum

Blackbody RadiationLinear Scale Log Scale

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

3

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

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

From TurcoFrom S. Nidkorodov

Solar Radiation Spectrum II

From Turco

From F-P&P

•Solar spectrum is strongly modulated by atmospheric scattering and absorption

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Solar Radiation Spectrum III

From Turco

UVC B A Photon Energy ↑

O3

O2

Solar Radiation Spectrum IV• Solar spectrum is strongly

modulated by atmospheric absorptions

• 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 provided by O3 layerFrom F-P&P

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Solar Radiation Spectrum vs. altitude

• 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.

From F-P&P

Solar Zenith Angle

sechL

pathlength Verticalpathlength Actualm Mass"Air " θ=≈==

S o la r F lu x

W a v ele n g th [n m ]1 0 0 0

Sola

r Fl

ux [P

hoto

ns c

m-2

s-1 n

m-1

]

1 0 1 2

1 0 1 3

1 0 1 4

1 0 1 5

S Z A = 0 ºS Z A = 8 6 º

3 0 0 4 0 0 5 0 0 7 0 0

At large SZA very little UV-B radiation

reaches the troposphere

• 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.

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Direct Attenuation of Radiation

I ≡ radiation intensity (e.g., F)I0 ≡ radiation intensity above atmospherem ≡ air masst ≡ attenuation coefficient due to

– absorption by gases (ag)– scattering by gases (sg)– scattering by particles (sp)– absorption by particles (ap)

apspagsg

m-t

ttttt eI I

+++=×= ×

0

Rayleigh scatteringtsg ∝ λ-4

Deep UV – O, N2, O2Mid UV & visible – O3

Near IR – H2OInfrared – CO2, H2O, others

tag ∝ σ

tsp ∝ λ-n

much more

complex

From F-P&P & S. Nidkorodov

Scattering by Gases• Purely physical

process, not absorption

• Approximation:

• Strongly increases as λdecreases

• Reason why “sky is blue” during the day

From Turco

420

5 /)1(10044.1 λλ −⋅⋅= ntsg

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Scattering & Absorption by Particles• Particles can scatter

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

From Jacob

Gas Absorption: Beer-Lambert Law I

• 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

[cm2/molec]• L ≡ absorption path length [cm]• n ≡ density of the absorber

[molec/cm3]

)exp(0 NLII ⋅⋅−= σ

Solve in class: Show that in the small absorption limit the relative change in light

intensity is approximately equal to absorbance.

From F-P&P & S. Nidkorodov

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Beer-Lambert Law IIPitfalls:• Other units are frequently used to express absorbance,

for example:A = ln(I0/I) = ε×L×C A = ln(I0/I) = α×L×Pε ≡ 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

Collisionsσ ≈ 10-15 cm2/molec

Light absorptionσ ≈ 10-18 cm2/molec From S. Nidkorodov

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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:

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)

Gas cell

n [#/cm3]DetectorFilter

L

I0

L

I

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?

Measurement of Absorption Cross Sections

From S. Nidkorodov

dzzmz

eI I

dzzA

A eI I

z

z-

z

mA-

∫

∫

∞

∞

××

××=

≡×=

=

≡×=

)](O[)(),(

depth optical where)()(

:nely writteAlternativ

)](O[

densitycolumn where)()(

3

),(0

3

)(0

λσλτ

τλλ

λλ

λτ

λσ

Attenuation coefficient is dominated by O3absorption 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.

Solve in class: Using barometric law estimate column density of O2 in the atmosphere. By how

much does atmospheric O2 attenuate solar radiation at around 170 nm (σ ≈ 10-17 cm2/molec) at noon (m = 1)? Assume that O2 fraction (21%) is

independent of altitude and T = 270 K.Ans: 4x1024; by exp(-107)

Example: UV Attenuation by O3 and O2

From S. Nidkorodov

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Solar Radiation IntensityTo calculate solar

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

From F-P&P

Surface Albedo

• Wavelength dependent!

• Question: for the same incident UV solar flux, will you tan faster over snow or over a desert?

)Radiation(Incident )Radiation( Reflected)(

λλλ =Albedo

From F-P&P

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Total vs. Downwelling Radiation• 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!

From Warneck

Calculation of Photolysis Rates I

• 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

][][ AJdtAd

A−=

Generic reaction: A + hν → B + C

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Calculation of Photolysis Rates II

][)()()(][][ AdFAJdtAd

AAA ×−=−= ∫ λλλφλσ

Generic reaction: A + hν → B + C

JA – first order photolysis rate of A (s-1)

σA(λ) – wavelength dependent cross section of A (cm2/#)

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.

J m-2 s-1Irradiance: same as flux but for a unit area with a fixed orientationE

J m-2 s-1 nm-1Spectral irradiance: same as radiance but per unit wavelength rangeE(λ)

J m-2 s-1 sr-1Radiance: radiant flux density per unit solid angleL(θ, ϕ)

J m-2 s-1 nm-1 sr-1Spectral radiance: same as radiance but per unit wavelength rangeL(θ, ϕ, λ)

J m-2 s-1 nm-1Spectral actinic flux density: same as flux but per unit wavelengthF(λ)

J m-2 s-1Actinic flux density: energy crossing a unit area per unit time without consideration of direction (we do not care where photons come from)F

UnitsDescriptionQuantity

∫ ∫∫

∫ ∫∫

==

==

π π

ω

π π

ω

ϕθθϕθωϕθ

ϕθθθϕθωθϕθ

2

0 0

2

0 0

sin),(),(

sincos),(cos),(

ddLdLF

ddLdLE

Radiance as a function of direction gives a complete descriptionof the radiation field. When L is independent of direction, the field is called isotropic, in which case E = π L and F = 2π L.

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

Radiation Fluxes Definitions

From F-P&P& S. Nidkorodov

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Radiation Mesurements

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

Flat Plate → Irradiance 2π → ½ of Actinic Flux

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

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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

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Example: Photolysis of CH3CHO∑∫

=

∆≈=i

nmFdFJ

λ

λ

λλλφλσλλλφλσ290

)()()()()()(CH3CHO + hν → CH3 + HCO (a)

→ CH4 + CO (b)

From F-P&P