Spectrometry and Photochemistry
Theodore S. Dibble
Chemistry Department
SUNY-Environmental Science and Forestry Syracuse, NY
Role of Spectrometry and Photochemistry
• Light flux, F(l), vs. wavelength, altitude, etc.
• Photochemistry as fate of a molecule
• Photolysis as radical source
• Greenhouse gas absorbances
• Concentration Measurement
Beer-Lambert Law
Absorbance A= ln (Io/I)
base e not base 10
Io
I
A = slc
s= absorption cross-section (per molecule)
cm2/molecule (s= e  3.8  10-21)
c = concentration (molecules cm-3)
l in cm
Example
Between 40 and 50 km, [O3] ~ 3 x 1011 molecules cm-3
s254 nm = 1.1 x 10-17 cm2 molecule-1
Calculate Absorbance over the 10 km (106 cm)
17
1.110 cm
3 10 molecules
6
A
10 cm
molecule
cm 2
A = 3.3
2
11
Light Intensity
Solar Zenith Angle – angle from perpendicular
(season, time of day, latitude: see Spreadsheet)
Other Factors
Clouds
Albedo (reflectivity)
Eccentricity
Why SZA Matters- Pathlength
Absorbance & scatter
lo
l = lo /cos(SZA)
SZA=0
SZA=40
Ozone UV Spectrum
Wavelength in nm
Ground Level Solar Flux, F(l)
1E+15
1E+14
Photons /
(cm^2 sec^1)
SZA=60
SZA=0
1E+13
1E+12
1E+11
280
300
320
Wavelength (nm)
340
360
Photolysis Rate Constant, J
Solar flux: F(l)
Absorption cross section: s(l)
Quantum yield for photolysis: f(l) (fraction of
photons absorbed that cause decomposition)
               J   F (l )s (l )f (l )dl
1
tim e

photons
area  tim e  wavelength
cross  section m oleculesdissociated
wavelength
molecule
photon
Ozone Photolysis Rate = JO3[O3]
concentration
1

[concentration ]
time
time
Numerical Integration
J   F (l )s (l )f (l )l
J  F (l )l s (l )f (l )
Spreadsheet: www.esf.edu/chemistry/dibble/fch511/calculateJ.xls
Exercise: Calculate J for O3 or HOOH at ground level
Use absorption cross-sections from JPL Data Evaluation #14
Assume f =1
Photolytic Production of Radicals
O3 + hn → O2 (3S)+ O(3P) ground state products
O3 + hn → O2 (1)+ O(1D) excited state products
O(1D) much more reactive than O(3P)
Rate of production of O(1D) =
[O3 ] F (l )s (l )fO(1 D) (l )dl
Rate of production of O(3P) =
[O3 ] F (l )s (l )fO(3P ) (l )dl
Quantum Yield for O (1D) from O3
fO ( D )  fO ( P )  1
1
3
Explain the altitude dependence of J(O(1D)) vs. J(O(3P))
Key Points
• Ozone UV absorption dominates F(l)
• F(l) depends on SZA
• Photolysis rate constants readily calculable