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Spectroscopy & Photochemistry I Spectroscopy & Photochemistry I Required Reading: FP Chapter 3B, 3C, 4 Required Reading: Jacob Chapter 7 Atmospheric Chemistry CHEM-5151 / ATOC-5151 Spring 2013 Jose-Luis Jimenez Importance of Spectroscopy and Photochemistry I • Most chemical processes in the atmosphere are initiated by photons – Photolysis of O3 generates OH – the most important atmospheric oxidizer: 1 O3 + hv O2 + O( D) 1 O( D) + H2O 2 OH – Solar photodissociation of many atmospheric molecules is often much faster than any other chemical reactions involving them: NO2 + hv O + NO (source of O3 in the troposphere) CF2Cl2 + hv CF2Cl + Cl (photolysis of CFCs in the stratosphere) HONO + hv OH + NO (source of OH in the troposphere) NO3 + hv O2 + NO or O + NO2 (removal of NO3 generated at night) Cl2 + hv Cl + Cl (source of Cl radicals) H2CO + hv H2 + CO or H + HCO (important in hydrocarbon oxidation) etc. 1 Importance of Spectroscopy and Photochemistry II • Absorption of solar and earth radiation by atmospheric molecules directly influences the energy balance of the planet – Greenhouse effect (CO2, H2O, N2O, CFCs) – Stratospheric temperature inversion (O3 photochemistry) • Spectroscopy of atmospheric molecules is used to detect them in situ – OH is detected via its electronic transition at 310 nm –NH3 is detected via its fundamental vibrational transition at 1065 cm-1, etc. Blackbody Radiation Linear Scale P/A=T4 Log Scale T= b Interactive demo: http://phet.colorado.edu/en/simulation/blackbody-spectrum From R.P. Turco, Earth Under Siege: From Air Pollution to Global Change, Oxford UP, 2002. 2 Solar & Earth Radiation Spectra • Sun is a radiation source with an effective blackbody temperature of about 5800 K • Earth receives circa 1368 W/m2 of energy from solar radiation From Turco • Clicker Question: are relative vertical scales correct in right plot? A. Yes B. Somewhat off C. Completely wrong D. I don’t know Adapted from S. Nidkorodov Solar Radiation Spectrum UV Photon Energy C B A IR From Turco 3 Solar Radiation: Initiator of Atmos. Reactions Average thermal energy of collisions: Each degree of freedom ~ ½ kT (per molecule) Collision energy ~ RT = 8.3 J mol-1 K-1 x T (per mole) RT = 2.5 kJ mol-1 @ 300 K Energy of photons (E = hv): 300 nm photon = 380 kJ mol-1 600 nm photon = 190 kJ mol-1 Typical bond strengths: -1 D0(O2) = 495 kJ mol -1 D0(Cl2) = 243 kJ mol C-H, O-H, C-O ~ 400 kJ mol-1 Atmospheric chemistry on Earth is driven by photolysis, not by thermal excitation!!! Adapted from S. Nidkorodov What is light? • Dual nature – Photon: as particle • Energy but no mass – As wave: electric and magnetic fields oscillating in space and time • Wavelength, From F-P&P frequency • c ~ 3 x 109 m/s Discuss in class: at a fundamental physical level, why are molecules capable of absorbing light? 4 The Electromagnetic Spectrum • Units used for photon energies and wavelengths: -1 c – 1 eV = 8065.54 cm = 96.4853 kJ/mol = 23.0605 kcal/mol – 1 Å = 0.1 nm = 10-10 m; – 1 micron (m) = 10-6 m = 1000 nm E h -34 2 – Planck's constant = 6.626068 × 10 m kg / s 1 • Clicker Q: the energy of a green photon ( = 530 nm) is: v A. 1 kJ/mol B. 230 kJ/mol C. 230 kcal/mol (wavenumber) D. 6 x 10-7 kJ/mol E. I don’t know Reminder of EM Spectrum 5 Types of radiation important in lower atmosphere • Ultraviolet and visible radiation ( = 100-800 nm) – Excites bonding electrons in molecules – Capable of breaking bonds in molecules photodissociation) – Ultraviolet photons ( = 100-300 nm) have most energy, can break more and stronger bonds. We will pay special attention to them. • Infrared radiation ( = 0.8 - 300 m) – Excites vibrational motions in molecules – With very few exceptions, infrared radiation is not energetic enough to break molecules or initiate photochemical processes • Microwave radiation ( = 0.5 - 300 mm) – Excites rotational motions in molecules http://phet.colorado.edu/en/simulation/molecules-and-light Spectroscopy & Photochemistry II Required Reading: FP Chapter 3B, 3C, 4 Required Reading: Jacob Chapter 7 Atmospheric Chemistry CHEM-5151 / ATOC-5151 Spring 2013 Jose-Luis Jimenez 6 Fundamentals of Spectroscopy • Molecules have energy in translation, vibration, rotation, and electronic state – Translation (= T) cannot be changed directly with light – We will focus on the other 3 energy types v', J', … • Molecule can absorb radiation efficiently if: photon – The photon energy matches the energy E spacing between molecule’s quantum levels v", J", … – Optical transition between these quantum levels is allowed by “selection rules” – “Forbidden” transitions can occur but are weaker Vibrational Energy & Transitions • Bonds can be viewed as “springs” • Energy levels are quantized, – Ev = hvvib(v+1/2) – vvib is constant dependent on molecule – v = 0, 1, 2… is vibrational quantum number From F-P&P 7 Vibrational Energy Levels • Ideally: Harmonic Oscillator – Restoration force of “spring” follows Hooke’s law: F= k x – Ev = hvvib(v+1/2), v = 0, 1, 2… – Energy levels are equally spaced • Really: Anharmonic oscillator – Restauration force rises sharply at small r, bond breaks at large r 2 3 E hν v 1 hx v 1 hy v 1 ... vib 2 e 2 e 2 – Vibrational quantum levels are more closely spaced as v increases From F-P&P http://phet.colorado.edu/en/simulation /energy-skate-park Example: Ground Electronic State of HF Etotal Erot Evib Erot BJ(J 1) Evib hνv Rotational level manifolds for Possible different vibrational quanta rovibrational overlap with each other transition: v=0 v=1 J=14 J=15 HF molecular constants -1 Bv=0 = 20.557 cm (rotational constant) = 4138.32 cm-1 (harmonic frequency) xe = 89.88 cm-1 (anharmonicity) From S. Nidkorodov 8 Vibration-rotation of HCl • Molecules vibrate and rotate simultaneously From F-P&P Electronic Transitions (ETs) • Molecules can undergo an ET upon absorption of an appropriate photon – Simultaneous vibrational and rotational transitions – No restriction on v, many vib. trans. can occur – J = -1, 0, +1 • P, Q, and R branches • Frank-Condon principle – Time for ET so short (10-15 s) that internuclear distance cannot change – “vertical” transitions From F-P&P 9 Pathways for Loss of e- Excitation From Wayne • Photophysical processes – Lead to emission of radiation – Energy converted to heat • Photochemical processes – Dissociation, ionization, reaction, isomerization Business Items • Piazza details – Student’s answer, instructor answer, comments – Email alerts digest 10 Repulsive States • No minima in PE vs r curves • Dissociation occurs immediately after absorption of light From F-P&P Molecules & Light Simulation http://phet.colorado.edu/en/simulation/molecules-and-light 11 Quantum Yields () • Relative efficiency of various photophysical and photochemical processes: Number of excited molecules proceeding by process i i Total number of photons absorbed * • E.g.: NO3 + hv NO3 (3) * NO3 NO2 + O (4a) NO + O2 (4b) NO3 + hv (4c) Number of NO molecules formed 2 4a Total number of photons absorbed and so on • i Are wavelength dependent, above all important at different Quantum Yields II From F-P&P 12 Spectroscopy and Photochemistry III Required Reading: FP Chapter 3B, 3C, 4 Required Reading: Jacob Chapter 7 Atmospheric Chemistry ATOC-5151 / CHEM-5151 Spring 2013 Prof. Jose-Luis Jimenez On Beer’s Law “The taller the glass, the darker the brew, Clicker prediction: The less the amount of light that comes through” If I set up (1) 200 M concentration & 1 cm path (2) 100 M concentration and 2 cm path The transmittance will be: A. Same B. More C. Less D. A lot less E. I don’t know http://phet.colorado.edu/en/simulation/beers-law-lab 13 Gas Absorption: Beer-Lambert Law I I I0 exp( L N) • Allows the calculation of the decay in intensity of a light beam due to absorption by the molecules in a medium Definitions: • A = ln(I0/I) = Absorbance = LN (also “optical depth”) • absorption cross section Solve in class: Show that in the small [cm2/molec] absorption limit the relative change in light intensity is approximately equal to • L absorption path length [cm] absorbance. • n density of the absorber [molec/cm3] From F-P&P & S. Nidkorodov Strength of the bands Quantify w/ absorption cross-section for one molecule: cm2 17 2 What is a large a. 10 cm 2 (but still realistic) b. 10 cm -12 2 value of for c. 10 cm -17 2 molecules? d. 10 cm e. 10-20 cm2 From Tolbert 14 Physical interpretation of • , absorption cross section (cm2 / 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 HNO4 Collisions Light absorption 10-15 cm2/molec 10-18 cm2/molec From S. Nidkorodov 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: Gas cell Filter I0 I Detector n [#/cm3] L Measurements are repeated for a number of concentrations at each wavelength of interest. From S. Nidkorodov 15 Clicker Q A sample contains 1 mbar of molecules of interest (A) with =2x10-21 cm2/molec and 1 bar of impurity (I) with =2x10-18 cm2/molec, both at =530 nm. Which contributes the most to the total absorbance in a 50 cm cell with 1012 photons cm-2 s-1 at the wavelength of interest? A) Molecules A contribute most B) Impurity I contributes most C) They contribute equally D) They are both negligible E) I don’t know Oxygen Spectrum •O2 photolysis in the 200- 240 nm range is the major source of O3 in the stratosphere •O2 can absorb nearly all radiation with = 10- 200 nm high up in the atmosphere Hole in the Clicker Q: Estimate the spectrum coincident with length of air column at P Lyman line of = 0.01 mbar and T= 200 H-atom K (characteristic of 80 km altitude) required to reduce the radiation flux at 150 nm by 10 orders of magnitude.
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