1. lecture 4 photochemistry What is Photochemistry? Photochemistry: a chemical interaction involving radiation Why mention photochemistry? Photochemistry plays an important role in atmospheric processes. The wavelengths in important atmospheric photochemical reactions are shortwave – radiation from the sun

2. lecture 4 photochemistry Nuclear Energy = neutrons + protons minus binding energy Atomic Weight/(Neutrons+Protons) does not equal 1 because of the binding energy. More binding energy, lower ratio AW/(N+P), lower total energy. If two light nuclei join to form a heavier nucleus with higher binding energy (lower total energy), the extra energy is released as radiation – fusion. What is the Sun’s Source of Energy?

3. lecture 4 photochemistry Fusion is a thermonuclear reaction. It releases energy but needs a high enough temperature to bring the two nuclei together. 2 H+ 2 H  4 He+E 2x2.014  4.0026+0.0254(as energy) Requirement, high temperature for activation ~25,000,000 o C This is the temperature in the sun’s interior, but not the temperature at the sun’s surface (the temperature at which the sun emits). Can calculate T surface based on a steady state assumption (heat flow from interior balances heat loss from surface): dT/dt=C 1 M s T interior -C 2 (S.A.) s T surface =0 4/3  R 3 C’ 1 T interior =4  R 2 C’ 2 T surface T surface =C”RT interior =~6000 o K Solar surface temperature is determined from the ratio of surface to volume. Fusion

4. lecture 4 photochemistry Energy per photon: E=h =hc/ h = 6.626 x 10 -34 J sc = 3 x 10 8 m s -1 Energy of Radiation

5. lecture 4 photochemistry Energy Per One Mole of Photons (Einstein) Energy per one mole of photons at: 100nm = 290 kcal/mole 400nm = 72.5 kcal/mole 700nm = 41.5 kcal/mole 1000nm = 29 kcal/mole Comparison to chemical bond strength: N-N = 225 kcal/mole Very strong O-O = 120 kcal/mole Strong C-Cl =75 kcal/mole Intermediate O-O2 = 35 kcal/mole Weak HO…..H=5 kcal/mole Very weak

6. lecture 4 photochemistry What Happens When Radiation Hits a Molecule? We learned in radiative transfer that two possible outcomes are: 1.scattering (no chemical interaction) 2.absorption: Following absorption, there are a number of possibilities.

7. lecture 4 photochemistry Pathways Following Absorption

8. lecture 4 photochemistry i.Dissociation/photolysis: breaking a chemical bond in the molecule Energy of radiation must be greater than bond energy. =100-1000nm is sufficient to break any chemical bond. ii.Ionization: removing an electron from the molecule In general ionization energy is greater than chemical bond strength: He = 552 kcal/mole =  52.6 nm} N 2 = 398 kcal/mole =  79.6 nm} Na = 120 kcal/mole =  250 nm} Pathways Following Absorption cont.

9. lecture 4 photochemistry Table of Ionization Energies

10. lecture 4 photochemistry Pathways Following Absorption cont. iii.Luminescence: re-emission of photon In atoms, re-emited photon is of same energy as excitation: em = ex. In molecules, it can be less: em > ex. Flourescence: visible wavelengths Phosphorescence: non- visible wavelengths

11. lecture 4 photochemistry Pathways Following Absorption cont. iv.Intramolecular energy transfer: conversion of the absorbed energy to several forms of lower energy (vibration, rotation and eventually to heat – typical for large molecules). § v.Intermolecular energy transfer vi.Quenching vii.Reaction: conversion to more active state and undergo selective chemical reactions

12. lecture 4 photochemistry What Determines the Pathway? 1.wavelength – whether or not its possible 2.population of excited states – whether or not its probable 3.conservation of orbital angular momentum and spin – whether or not its probable

13. lecture 4 photochemistry rate of formation of AB* = J for a photochemical reaction is the equivalent of a rate constant for a chemical reaction J can be treated as a first order rate constant (units of time -1 ) but it depends on light intensity and spectral distribution. Rate of a Photochemical Reaction

14. lecture 4 photochemistry How is J Calculated? For a given wavelength: J{ }=PF{ }  { } Y{ } PF{ }: PhotoFlux  { } – Absorption cross section (population of excited states) Y{ } – Quantum yield (conservation of orbital angular momentum and spin) Y{ } = The quantum yield is sometimes also symbolized  For a range of wavelengths (solar range): J =  PF{ }  { } Y{ } d

15. lecture 4 photochemistry NO 2 +h  NO + O( 3 P) J{NO 2 } The rate of O( 3 P) formation d[O( 3 P)]/dt = J{NO 2 } [NO 2 ] The rate of O formation will change diurnally even at constant NO 2. Photolysis Rate Example

16. lecture 4 photochemistry J=J max cos(  ) cos(  J max  SRI/R 2 SRI: solar radiation intensity   t  x  x   – Geographical Latitude  – Seasonal motion of the earth  cos(2  {JD} /365) JD: Julian Day How Does J Vary With Latitude and Season?

17. lecture 4 photochemistry The Solar Spectrum