1. Spectrometry and Photochemistry Theodore S. Dibble Chemistry Department SUNY-Environmental Science and Forestry Syracuse, NY

2. Role of Spectrometry and Photochemistry Light flux, F( ), vs. wavelength, altitude, etc. Photochemistry as fate of a molecule Photolysis as radical source Greenhouse gas absorbances Concentration Measurement

3. Beer-Lambert Law Absorbance A= ln (I o /I) base e not base 10 A =  lc  = absorption cross-section (per molecule) cm 2 /molecule (  =   3.8  10 -21 ) c = concentration (molecules cm -3 ) l in cm IoIo I

4. Example Between 40 and 50 km, [O 3 ] ~ 3 x 10 11 molecules cm -3  254 nm = 1.1 x 10 -17 cm 2 molecule -1 Calculate Absorbance over the 10 km (10 6 cm) A = 3.3

5. Light Intensity Solar Zenith Angle – angle from perpendicular (season, time of day, latitude: see Spreadsheet)Spreadsheet Other Factors Clouds Albedo (reflectivity) Eccentricity

6. Why SZA Matters- Pathlength lolo SZA=0  SZA=40  l = l o /cos(SZA) Absorbance & scatter

7. Ozone UV Spectrum Wavelength in nm

9. Photolysis Rate Constant, J Solar flux: F( ) Absorption cross section:  ( ) Quantum yield for photolysis:  ( ) (fraction of photons absorbed that cause decomposition) Ozone Photolysis Rate = J O3 [O 3 ]

10. Numerical Integration Spreadsheet: www.esf.edu/chemistry/dibble/fch511/calculateJ.xlswww.esf.edu/chemistry/dibble/fch511/calculateJ.xls Exercise: Calculate J for O 3 or HOOH at ground level Use absorption cross-sections from JPL Data Evaluation #14 Assume  =1

11. Photolytic Production of Radicals O 3 + h → O 2 ( 3  )+ O( 3 P) ground state products O 3 + h → O 2 ( 1  )+ O( 1 D) excited state products O( 1 D) much more reactive than O( 3 P) Rate of production of O( 1 D) = Rate of production of O( 3 P) =

12. Quantum Yield for O ( 1 D) from O 3 Explain the altitude dependence of J(O( 1 D)) vs. J(O( 3 P))

13. Key Points Ozone UV absorption dominates F( ) F( ) depends on SZA Photolysis rate constants readily calculable