Lecture 6: Spectroscopy and Photochemistry II

Lecture 6: Spectroscopy and Photochemistry II

<p>Lecture 6: Spectroscopy and <br>Photochemistry II </p><p><em>Required Reading: FP Chapter 3 Suggested Reading: SP Chapter 3 </em></p><p>Atmospheric Chemistry <br>CHEM-5151 / ATOC-5151 <br>Spring 2005 <br>Prof. Jose-Luis Jimenez </p><p>Outline of Lecture </p><p>• The Sun as a radiation source • Attenuation from the atmosphere </p><p>– Scattering by gases &amp; aerosols – Absorption by gases </p><p>• Beer-Lamber law </p><p>• Atmospheric photochemistry </p><p>– Calculation of photolysis rates – Radiation fluxes – Radiation models </p><p>1</p><p>Reminder of EM Spectrum <br>Blackbody Radiation </p><p>Log Scale <br>Linear Scale </p><p>From R.P. Turco, Earth Under Siege: From Air Pollution to Global Change, Oxford UP, 2002. </p><p>2</p><p>Solar &amp; Earth Radiation Spectra </p><p>• Sun is a radiation source with an effective blackbody temperature of about 5800 K <br>• Earth receives circa 1368 W/m<sup style="top: -0.29em;">2 </sup>of energy from solar radiation </p><p>From Turco <br>From S. Nidkorodov </p><p>• Question:&nbsp;are relative vertical scales ok in right plot? </p><p>Solar Radiation Spectrum II </p><p>From F-P&amp;P </p><p>•Solar spectrum is strongly modulated by atmospheric scattering and absorption </p><p>From Turco </p><p>3</p><p>Solar Radiation Spectrum III </p><p>UV </p><p><strong>Photon Energy ↑ </strong></p><p><strong>C </strong>B A </p><p>From Turco </p><p>Solar Radiation Spectrum IV </p><p>• Solar&nbsp;spectrum is strongly modulated by atmospheric absorptions <br>• Remember&nbsp;that UV photons have most energy </p><p>O<sub style="top: 0.21em;">3 </sub><br>O<sub style="top: 0.21em;">2 </sub></p><p>– O<sub style="top: 0.21em;">2 </sub>absorbs extreme UV in mesosphere; O<sub style="top: 0.21em;">3 </sub>absorbs most UV in stratosphere <br>– Chemistry&nbsp;of those regions partially driven by those absorptions <br>– Only&nbsp;light with λ&gt;290 nm penetrates into the lower troposphere <br>– Biomolecules&nbsp;have same bonds <br>(e.g. C-H), bonds can break with UV absorption =&gt; damage to life </p><p>• Importance&nbsp;of protection provided by O<sub style="top: 0.21em;">3 </sub>layer </p><p>From F-P&amp;P </p><p>4</p><p>Solar Radiation Spectrum vs. altitude </p><p>From F-P&amp;P </p><p>• Very high energy photons are depleted high up in the atmosphere • Some photochemistry is possible in stratosphere but not in troposphere </p><p>• Only λ &gt; 290 nm in trop. </p><p>Solar Zenith Angle </p><p>••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. </p><p><strong>Solar F&nbsp;lu x </strong></p><p><strong>10</strong><sup style="top: -0.17em;"><strong>15 </strong></sup><strong>10</strong><sup style="top: -0.165em;"><strong>14 </strong></sup><strong>10</strong><sup style="top: -0.165em;"><strong>13 </strong></sup><strong>10</strong><sup style="top: -0.17em;"><strong>12 </strong></sup></p><p>Actual pathlength Vertical pathlength <br>Lh<br>"Air Mass"= m = </p><p></p><ul style="display: flex;"><li style="flex:1">≈</li><li style="flex:1">= secθ </li></ul><p></p><p><strong>S Z A&nbsp;= 0º </strong></p><p><strong>S Z A&nbsp;= 86º </strong></p><p>At large SZA very little UV-B radiation reaches the troposphere </p><p></p><ul style="display: flex;"><li style="flex:1"><strong>300 </strong></li><li style="flex:1"><strong>400 500 </strong></li></ul><p></p><p><strong>W avelength&nbsp;[nm ] </strong></p><p><strong>700 </strong></p><p><strong>1000 </strong></p><p>5</p><p>Direct Attenuation of Radiation </p><p><em>I </em>= <em>I</em><sub style="top: 0.25em;">0 </sub>× <em>e</em><sup style="top: -0.445em;"><em>-t</em>×<em>m </em></sup></p><p>From F-P&amp;P &amp; S. Nidkorodov </p><p><em>t </em>= <em>t</em><sub style="top: 0.245em;"><em>sg </em></sub>+ <em>t</em><sub style="top: 0.245em;"><em>ag </em></sub>+ <em>t</em><sub style="top: 0.245em;"><em>sp </em></sub>+ <em>t</em><sub style="top: 0.245em;"><em>ap </em></sub></p><p><em>I </em>≡ radiation intensity (e.g., <em>F</em>) <em>I</em><sub style="top: 0.21em;"><em>0 </em></sub>≡ radiation intensity above atmosphere </p><p><em>m </em>≡ air mass </p><p><em>t </em>≡ attenuation coefficient due to <br>– absorption&nbsp;by gases (ag) – scattering&nbsp;by gases (sg) – scattering&nbsp;by particles (sp) – absorption&nbsp;by particles (ap) </p><p>Rayleigh scattering </p><p><strong>t</strong><sub style="top: 0.21em;"><strong>sp </strong></sub><strong>∝ λ</strong><sup style="top: -0.21em;"><strong>-n </strong></sup><strong>t</strong><sub style="top: 0.21em;"><strong>sg </strong></sub><strong>∝ λ</strong><sup style="top: -0.21em;"><strong>-4 </strong></sup></p><p><em>much more complex </em></p><p>Deep UV – O, N<sub style="top: 0.21em;">2</sub>, O<sub style="top: 0.21em;">2 </sub>Mid UV &amp; visible – O<sub style="top: 0.21em;">3 </sub><br>Near IR – H<sub style="top: 0.21em;">2</sub>O <br>Infrared – CO<sub style="top: 0.21em;">2</sub>, H<sub style="top: 0.21em;">2</sub>O, others </p><p><strong>t</strong><sub style="top: 0.21em;"><strong>ag </strong></sub><strong>∝ σ </strong></p><p>Scattering by Gases </p><p>• Purely&nbsp;physical process, not absorption </p><p>• Approximation: </p><p><em>t</em><sub style="top: 0.265em;"><em>sg </em></sub>=1.044⋅10<sup style="top: -0.48em;">5 </sup>⋅(<em>n</em><sub style="top: 0.265em;">0λ </sub>−1)<sup style="top: -0.48em;">2 </sup>/ λ<sup style="top: -0.48em;">4 </sup></p><p>• Strongly increases as λ </p><p>decreases <br>• Reason&nbsp;why “sky is blue” during the day </p><p>From Turco </p><p>6</p><p>Scattering &amp; Absorption by Particles </p><p><em>From Jacob </em></p><p>• Particles&nbsp;can scatter and absorb radiation </p><p>• Scattering&nbsp;efficiency is very strong function of particle size </p><p>– For&nbsp;a given wavelength </p><p>• Visible:&nbsp;λ ~ 0.5 µm </p><p>– Particles&nbsp;0.5-2 µm are most efficient scatterers! </p><p>• Will&nbsp;discuss in more detail in aerosol lectures </p><p>Gas Absorption: Beer-Lambert Law I </p><p><em>I </em>= <em>I</em><sub style="top: 0.38em;">0 </sub>exp(−σ ⋅ <em>L</em>⋅ <em>N</em>) </p><p>• Allows&nbsp;the calculation of the decay in intensity of a light beam due to absorption by the molecules in a medium </p><p><strong>Definitions: </strong></p><p>• <em>A </em>= ln(<em>I</em><sub style="top: 0.21em;"><em>0</em></sub><em>/I</em>) = <em>Absorbance </em>= σ⋅<em>L</em>⋅<em>N (also “optical depth”) </em></p><p>• σ ≡ absorption cross section </p><p>Solve in class: Show that in the small </p><p>[cm<sup style="top: -0.25em;">2</sup>/molec] <br>• <em>L </em>≡ absorption path length [cm] </p><p>absorption limit the relative change in light intensity is approximately equal to </p><p>• <em>n </em>≡ density of the absorber </p><p>absorbance. </p><p>[molec/cm<sup style="top: -0.2501em;">3</sup>] </p><p>From F-P&amp;P &amp; S. Nidkorodov </p><p>7</p><p>Beer-Lambert Law II </p><p><strong>Pitfalls: </strong></p><p>• Other&nbsp;units are frequently used to express absorbance, for example: </p><p></p><ul style="display: flex;"><li style="flex:1"><em>A </em>= ln(<em>I</em><sub style="top: 0.25em;"><em>0</em></sub><em>/I</em>) = ε×<em>L</em>×<em>C </em></li><li style="flex:1"><em>A </em>= ln(<em>I</em><sub style="top: 0.25em;"><em>0</em></sub><em>/I</em>) = α×<em>L</em>×<em>P </em></li></ul><p></p><p>ε ≡ extinction coefficient [L mol<sup style="top: -0.29em;">-1 </sup>cm<sup style="top: -0.29em;">-1</sup>] α ≡ absorption coefficient [atm<sup style="top: -0.29em;">-1 </sup>cm<sup style="top: -0.29em;">-1</sup>] <em>C </em>≡ density of the absorber [mol L<sup style="top: -0.29em;">-1</sup>] <em>P </em>≡ partial pressure [atm] </p><p>• Base&nbsp;10 is used in most commercial spectrometers instead of the natural base: </p><p>A<sub style="top: 0.25em;">base 10 </sub>= log(<em>I</em><sub style="top: 0.25em;"><em>0</em></sub><em>/I</em>) = A<sub style="top: 0.25em;">base e</sub>/ln(10) </p><p>Physical interpretation of σ </p><p>• σ , absorption cross section (cm<sup style="top: -0.415em;">2 </sup>/ molecule) </p><p>– Effective area of the molecule that photon needs to traverse in order to be absorbed. </p><p>– The larger the absorption cross section, the easier it is to photoexcite the molecule. </p><p>– E.g., pernitric acid HNO<sub style="top: 0.2899em;">4 </sub></p><p>Collisions σ ≈ 10<sup style="top: -0.21em;">-15 </sup>cm<sup style="top: -0.21em;">2</sup>/molec <br>Light absorption σ ≈ 10<sup style="top: -0.21em;">-18 </sup>cm<sup style="top: -0.21em;">2</sup>/molec </p><p>From S. Nidkorodov </p><p>8</p><p>Measurement of Absorption Cross Sections </p><p>Measurement of absorption cross sections is, in principle, trivial.&nbsp;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: </p><p><strong>Gas cell </strong></p><p><strong>I</strong><sub style="top: 0.21em;"><strong>0 </strong></sub></p><p><strong>I</strong></p><p><strong>Detector </strong><br><strong>Filter </strong></p><p><strong>n [#/cm</strong><sup style="top: -0.21em;"><strong>3</strong></sup><strong>] </strong></p><p><strong>L</strong></p><p>Measurements are repeated for a number of concentrations at each wavelength of interest. </p><p><em>Although seemingly trivial, in practice such measurements are difficult because of impurities, especially when it comes to very small cross sections (&lt; 10</em><sup style="top: -0.2101em;"><em>-20 </em></sup><em>cm</em><sup style="top: -0.21em;"><em>2</em></sup><em>/molec) </em></p><p>Solve in class: Sample contains 1 Torr of molecules of interest with σ=3x10<sup style="top: -0.21em;">-21 </sup>cm<sup style="top: -0.21em;">2</sup>/molec and 1 mTorr of impurity with σ=2x10<sup style="top: -0.21em;">-18 </sup>cm<sup style="top: -0.21em;">2</sup>/molec. <br>What is the total absorbance in a 50 cm cell? </p><p>From S. Nidkorodov </p><p>Example: UV Attenuation by O<sub style="top: 0.415em;">3 </sub>and O<sub style="top: 0.415em;">2 </sub></p><p>Attenuation coefficient is dominated by O<sub style="top: 0.21em;">3 </sub>absorption in the 200-300 nm window. Therefore, direct attenuation can be easily calculated from known absorption cross sections of O<sub style="top: 0.21em;">3</sub>. Similar formulas apply to attenuation by O<sub style="top: 0.21em;">2 </sub>in 120-180 nm window. </p><p><em>I</em>(λ) = <em>I</em><sub style="top: 0.2199em;">0 </sub>(λ)× <em>e</em><sup style="top: -0.395em;"><em>-</em>σ (λ)×<em>A</em>×<em>m </em></sup></p><p>where <em>A </em>≡ column density </p><p>∞</p><p><em>A </em>= [O<sub style="top: 0.2199em;">3 </sub>(<em>z</em>)]<em>dz </em></p><p>∫</p><p><em>z</em></p><p>From S. Nidkorodov </p><p>Alternatively written : </p><p>Solve in class: Using barometric law estimate column density of O<sub style="top: 0.165em;">2 </sub>in the atmosphere. By how much does atmospheric O<sub style="top: 0.165em;">2 </sub>attenuate solar radiation at&nbsp;around 170 nm (σ ≈ 10<sup style="top: -0.21em;">-17 </sup>cm<sup style="top: -0.21em;">2</sup>/molec) at noon (<em>m </em>= 1)? Assume that O<sub style="top: 0.165em;">2 </sub>fraction (21%) is independent of altitude and T = 270 K. </p><p><em>I</em>(λ) = <em>I</em><sub style="top: 0.2199em;">0 </sub>(λ)× <em>e</em><sup style="top: -0.39em;"><em>-</em>τ (λ,<em>z</em>) </sup></p><p>whereτ ≡ optical depth </p><p>∞</p><p>τ (λ, <em>z</em>) = σ (λ)×<em>m</em>×[O<sub style="top: 0.22em;">3 </sub>(<em>z</em>)]<em>dz </em></p><p>∫</p><p><em>Ans: 4x10</em><sup style="top: -0.21em;"><em>24</em></sup><em>; by &nbsp; exp(-10</em><sup style="top: -0.21em;"><em>7</em></sup><em>) </em></p><p><em>z</em></p><p>9</p><p>Solar Radiation Intensity </p><p>To calculate solar </p><p>From F-P&amp;P </p><p>spectral distribution in any given volume of air at any given time and location one must know the following: </p><p>– Solar&nbsp;spectral distribution outside the atmosphere <br>– Path&nbsp;length through the atmosphere <br>– Wavelength&nbsp;dependent attenuation by atmospheric molecules <br>– Amount&nbsp;of radiation indirectly scattered by the earth surface, clouds, aerosols, and other volumes of air </p><p>Surface Albedo </p><p>Reflected Radiation(λ) Incident Radiation(λ) </p><p><em>Albedo</em>(λ) = </p><p>• Wavelength dependent! </p><p>• Question:&nbsp;for the same incident UV solar flux, will you tan faster over snow or over a desert? </p><p>From F-P&amp;P </p><p>10 </p><p>Total vs. Downwelling Radiation </p><p>From Warneck </p><p>• If&nbsp;atmosphere was completely transparent and surface completely absorbing (albedo = 0) </p><p>– F<sub style="top: 0.25em;">U </sub>= 0 – F<sub style="top: 0.25em;">D </sub>=F<sub style="top: 0.25em;">T</sub>= 1 </p><p>• Due&nbsp;to gas + aerosol scattering and surface reflection </p><p>– F<sub style="top: 0.25em;">U </sub>can be large – F<sub style="top: 0.25em;">T </sub>&gt; solar flux! </p><p>Calculation of Photolysis Rates I </p><p>Generic reaction:&nbsp;A + <em>h ν </em>→ B + C </p><p><em>d</em>[<em>A</em>] <br>= −<em>J</em><sub style="top: 0.29em;"><em>A</em></sub>[<em>A</em>] </p><p><em>dt </em></p><p>• A&nbsp;“first-order process” </p><p>• <em>What does J</em><sub style="top: 0.3349em;"><em>A </em></sub><em>depend on? </em></p><p>• <em>J</em><sub style="top: 0.335em;"><em>A </em></sub>depends on </p><p>– Light intensity from all directions </p><p>• <em>“Actinic flux” </em></p><p>– Absorption cross section (σ) – Quantum yield for photodissociation (φ) – All are functions of wavelength </p><p>11 </p><p>Calculation of Photolysis Rates II </p><p>Generic reaction:&nbsp;A + <em>h ν </em>→ B + C </p><p><em>d</em>[<em>A</em>] <br>= −<em>J</em><sub style="top: 0.29em;"><em>A</em></sub>[<em>A</em>] = −&nbsp;σ <sub style="top: 0.29em;"><em>A </em></sub>(λ)φ<sub style="top: 0.29em;"><em>A </em></sub>(λ)<em>F</em>(λ)<em>d</em>λ ×[<em>A</em>] </p><p>∫</p><p><em>dt </em></p><p><em>J</em><sub style="top: 0.25em;"><em>A </em></sub>– first order photolysis rate of A (s<sup style="top: -0.29em;">-1</sup>) <em>σ</em><sub style="top: 0.25em;"><em>A</em></sub><em>(</em>λ) – wavelength dependent cross section of A (cm<sup style="top: -0.29em;">2</sup>/#) φ<sub style="top: 0.25em;"><em>A</em></sub><em>(</em>λ) – wavelength dependent quantum yield for photolysis </p><p><em>F(</em>λ) – spectral <em>actinic flux </em>density (#/cm<sup style="top: -0.29em;">2</sup>/s) </p><p>So, what are the smallest cross sections that matter? The solar actinic flux (photons cm<sup style="top: -0.25em;">-2 </sup>s<sup style="top: -0.25em;">-1 </sup>nm<sup style="top: -0.25em;">-1</sup>) is of order 10<sup style="top: -0.25em;">14</sup>. In&nbsp;many cases, we need to know whether the photolytic lifetime of a molecule is 10 days (<em>J</em>=10<sup style="top: -0.25em;">-6 </sup>s<sup style="top: -0.25em;">-1</sup>), or&nbsp;100 days (<em>J</em>=10<sup style="top: -0.25em;">-7 </sup>s<sup style="top: -0.25em;">-1</sup>). This means that cross sections as small as 10<sup style="top: -0.25em;">-20 </sup>cm<sup style="top: -0.25em;">2 </sup>or even smaller are potentially interesting. Such small cross sections are very challenging to measure with sufficient accuracy. </p><p>Radiation Fluxes Definitions </p><p></p><ul style="display: flex;"><li style="flex:1"><strong>Quantity </strong></li><li style="flex:1"><strong>Description </strong></li><li style="flex:1"><strong>Units </strong></li></ul><p></p><p><strong>Actinic flux density: </strong>energy crossing a unit area per unit time without consideration of direction (we do not care where photons come from) </p><p><strong>F</strong></p><p>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup></p><p><strong>Spectral actinic flux density: </strong>same as flux but per unit wavelength </p><p>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup>nm<sup style="top: -0.21em;">-1 </sup>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup></p><p><strong>F(λ) E</strong></p><p><strong>Irradiance: </strong>same as flux but for a unit area with a fixed orientation <strong>Spectral irradiance: </strong>same as radiance but per unit wavelength range </p><p>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup>nm<sup style="top: -0.21em;">-1 </sup></p><p><strong>E(λ) </strong></p><p><strong>Radiance: </strong>radiant flux density per unit solid angle </p><p>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup>sr<sup style="top: -0.21em;">-1 </sup></p><p><strong>L(θ, ϕ) </strong></p><p>J m<sup style="top: -0.21em;">-2 </sup>s<sup style="top: -0.21em;">-1 </sup>nm<sup style="top: -0.21em;">-1 </sup>sr<sup style="top: -0.21em;">-1 </sup></p><p><strong>Spectral radiance: </strong>same as radiance but per unit wavelength range </p><p><strong>L(θ, ϕ, λ) </strong></p><p>2π π </p><p></p><ul style="display: flex;"><li style="flex:1"><em>E </em>= </li><li style="flex:1"><em>L</em>(θ,ϕ)cosθ sinθ<em>d</em>θ<em>d</em>ϕ </li></ul><p></p><p>∫</p><p><sup style="top: -0.2549em;"><em>L</em>(θ,ϕ)cosθ<em>d</em>ω = </sup>∫∫ </p><p>ω</p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>2π π </p><p></p><ul style="display: flex;"><li style="flex:1"><em>F </em>= </li><li style="flex:1"><em>L</em>(θ,ϕ)sinθ<em>d</em>θ<em>d</em>ϕ </li></ul><p></p><p>∫</p><p><sup style="top: -0.255em;"><em>L</em>(θ,ϕ)<em>d</em>ω = </sup>∫∫ </p><p>ω</p><p></p><ul style="display: flex;"><li style="flex:1">0</li><li style="flex:1">0</li></ul><p></p><p>Radiance as a function of direction gives a complete description of the radiation field. When <em>L </em>is independent of direction, the </p><p>field is called <em>isotropic</em>, in which case <em>E = </em>π <em>L </em>and <em>F </em>= 2π <em>L. </em></p><p>Solve in class: There are 10<sup style="top: -0.165em;">9 </sup>photons flying into a 0.01 cm diameter opening every </p><p>From F-P&amp;P &amp; S. Nidkorodov </p><p>second. What is <em>F </em>with respect to this opening in units of #/cm<sup style="top: -0.165em;">2</sup>/s? </p><p>12 </p><p>Radiation Mesurements </p><p><strong>Flat Plate → Irradiance </strong><br><strong>2π → ½ of Actinic Flux </strong></p><p>• Radiation&nbsp;does not just come directly from the sun </p><p>– scattered&nbsp;radiation is just as important – Measure&nbsp;total or spectrally-resolved flux – Use&nbsp;models </p><p>From F-P&amp;P </p><p>Radiation Models </p><p>• Predict&nbsp;radiation intensity </p><p>– As f(time, altitude, latitude, λ) – Results of Madronich (1998) described in text </p><p>• Will&nbsp;use extensively in homeworks &amp; exams </p><p>– Typical model results: </p><p>From F-P&amp;P </p><p>13 </p><p>Example model results </p><p>From F-P&amp;P </p><p>Q: summer/winter solstices intensity at 445 nm: @ 8 am? @ noon? </p><p>Examples of Photolysis Rates </p><p>From F-P&amp;P </p><p>14 </p><p>Example: Photolysis of CH<sub style="top: 0.46em;">3</sub>CHO </p><p>λ</p><p><em>i</em></p><p>CH<sub style="top: 0.21em;">3</sub>CHO + <em>h</em>ν </p><p></p><ul style="display: flex;"><li style="flex:1">→ CH<sub style="top: 0.21em;">3 </sub>+ HCO </li><li style="flex:1">(a) </li></ul><p></p><p></p><ul style="display: flex;"><li style="flex:1"><em>J </em>= σ (λ)φ(λ)<em>F</em>(λ)<em>d</em>λ ≈ </li><li style="flex:1">σ (λ)φ (λ)<em>F </em>(λ)∆λ </li></ul><p></p><p>∑<br>∫</p><p>→ CH<sub style="top: 0.21em;">4 </sub>+ CO <br>(b) </p><p>λ=290<em>nm </em></p><p>From F-P&amp;P </p><p>15 </p>

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