Lecture 6: Spectroscopy and Photochemistry II

Transcription

1 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

2 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,

3 Solar & Earth Radiation Spectra From S. Nidkorodov Sun is a radiation source with an effective blackbody temperature of about 5800 K Earth receives circa 1368 W/m 2 of energy from solar radiation From Turco Question: are relative vertical scales ok in right plot? Solar Radiation Spectrum II From Turco Solar spectrum is strongly modulated by atmospheric scattering and absorption 3

4 Solar Radiation Spectrum III UV C B A Photon Energy From Turco Solar Radiation Spectrum IV O 2 O 3 Solar spectrum is strongly modulated by atmospheric absorptions Remember that UV photons have most energy O 2 absorbs extreme UV in mesosphere; O 3 absorbs most UV in stratosphere Chemistry of those regions partially driven by those absorptions Only light with >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 O 3 layer 4

5 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 > 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 Solar Flux [Photons cm -2 s -1 nm -1 ] SZA = 0º SZA = 86º At large SZA very little UV-B radiation reaches the troposphere Actual pathlength L " Air Mass" = m = = secθ Vertical pathlength h W avelength [nm ]

6 Direct Attenuation of Radiation t = t I = I sg + t I radiation intensity (e.g., F) I 0 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) 0 ag e + t -t m sp + t ap & S. Nidkorodov t sp -n much more complex Rayleigh scattering t sg -4 Deep UV O, N 2, O 2 Mid UV & visible O 3 Near IR H 2 O Infrared CO 2, H 2 O, others t ag σ From Turco Scattering by Gases t sg Purely physical process, not absorption Approximation: = n ( 0 1) / Strongly increases as decreases Reason why sky is blue during the day 6

7 Scattering & Absorption by Particles Particles can scatter and absorb radiation Scattering efficiency is very strong function of particle size For a given wavelength Visible: ~ 0.5 µm Particles µm are most efficient scatterers! Will discuss in more detail in aerosol lectures From Jacob 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 Solve in class: Show that in the small absorption limit the relative change in light intensity is approximately equal to absorbance. Definitions: A = ln(i 0 /I) = Absorbance = σ L N (also optical depth ) σ absorption cross section [cm 2 /molec] L absorption path length [cm] n density of the absorber [molec/cm 3 ] & S. Nidkorodov 7

8 Beer-Lambert Law II Pitfalls: Other units are frequently used to express absorbance, for example: A =ln(i 0 /I) = ε L C A=ln(I 0 /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: A base 10 = log(i 0 /I) = A base e /ln(10) Physical interpretation of σ σ, absorption cross section (cm 2 / 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 HNO 4 Collisions σ cm 2 /molec Light absorption σ cm 2 /molec From S. Nidkorodov 8

9 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: From S. Nidkorodov Filter I 0 Gas cell n [#/cm 3 ] I Detector 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 (< cm 2 /molec) Solve in class: Sample contains 1 Torr of molecules of interest with σ=3x10-21 cm 2 /molec and 1 mtorr of impurity with σ=2x10-18 cm 2 /molec. What is the total absorbance in a 50 cm cell? Example: UV Attenuation by O 3 and O 2 Attenuation coefficient is dominated by O 3 absorption in the nm window. Therefore, direct attenuation can be easily calculated from known absorption cross sections of O 3. Similar formulas apply to attenuation by O 2 in nm window. I( ) where A column density I( ) τ (, z) = - = I ( ) e A = z 0 z [O ( z)] dz - = I ( ) e σ ( ) A m Alternatively written : 0 τ (, z) whereτ optical depth 3 σ ( ) m [O ( z)] dz 3 From S. Nidkorodov Solve in class: Using barometric law estimate column density of O 2 in the atmosphere. By how much does atmospheric O 2 attenuate solar radiation at around 170 nm (σ cm 2 /molec) at noon (m = 1)? Assume that O 2 fraction (21%) is independent of altitude and T = 270 K. Ans: 4x10 24 ; by exp(-10 7 ) 9

10 Solar Radiation Intensity To 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 Surface Albedo Albedo( ) = Reflected Radiation( ) Incident Radiation( ) Wavelength dependent! Question: for the same incident UV solar flux, will you tan faster over snow or over a desert? 10

11 Total vs. Downwelling Radiation From Warneck If atmosphere was completely transparent and surface completely absorbing (albedo = 0) F U = 0 F D =F T = 1 Due to gas + aerosol scattering and surface reflection F U can be large F T > solar flux! Calculation of Photolysis Rates I Generic reaction: A + hν B + C A first-order process What does J A depend on? J A depends on Light intensity from all directions Actinic flux d[ A] = J dt [ A] Absorption cross section (σ) Quantum yield for photodissociation (φ) All are functions of wavelength A 11

12 Calculation of Photolysis Rates II Generic reaction: A + hν B + C d[ A] = J A [ A] = A( ) A( ) F( ) d [ A] dt σ φ J A first order photolysis rate of A (s -1 ) σ A () wavelength dependent cross section of A (cm 2 /#) φ A () wavelength dependent quantum yield for photolysis F() spectral actinic flux density (#/cm 2 /s) So, what are the smallest cross sections that matter? The solar actinic flux (photons cm -2 s -1 nm -1 ) is of order 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 cm 2 or even smaller are potentially interesting. Such small cross sections are very challenging to measure with sufficient accuracy. Quantity F F() E E() Radiation Fluxes Definitions Description Actinic flux density: energy crossing a unit area per unit time without consideration of direction (we do not care where photons come from) Spectral actinic flux density: same as flux but per unit wavelength Irradiance: same as flux but for a unit area with a fixed orientation Spectral irradiance: same as radiance but per unit wavelength range Units J m -2 s -1 J m -2 s -1 nm -1 J m -2 s -1 J m -2 s -1 nm -1 L(θ, ϕ) L(θ, ϕ, ) Radiance: radiant flux density per unit solid angle Spectral radiance: same as radiance but per unit wavelength range J m -2 s -1 sr -1 J m -2 s -1 nm -1 sr -1 E = ω L( θ, ϕ)cosθdω = F = ω L( θ, ϕ) dω = 2ππ 0 0 2ππ 0 0 L( θ, ϕ)cosθ sinθdθdϕ L( θ, ϕ)sinθdθdϕ 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. & S. Nidkorodov Solve in class: There are 10 9 photons flying into a 0.01 cm diameter opening every second. What is F with respect to this opening in units of #/cm 2 /s? 12

13 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 Radiation Models Predict radiation intensity As f(time, altitude, latitude, ) Results of Madronich (1998) described in text Will use extensively in homeworks & exams Typical model results: 13

14 Example model results Q: summer/winter solstices intensity at noon? Examples of Photolysis Rates 14

15 15 Example: Photolysis of CH 3 CHO = = i nm F d F J φ σ φ σ 290 ) ( ) ( ) ( ) ( ) ( ) ( CH 3 CHO + hν CH 3 + HCO (a) CH 4 + CO (b)