Lecture 6: Spectroscopy and
Photochemistry II

Required Reading: FP Chapter 3 Suggested Reading: SP Chapter 3

Atmospheric Chemistry
CHEM-5151 / ATOC-5151
Spring 2005
Prof. Jose-Luis Jimenez

Outline of Lecture

• The Sun as a radiation source • Attenuation from the atmosphere

– Scattering by gases & aerosols – Absorption by gases

• Beer-Lamber law

• Atmospheric photochemistry

– Calculation of photolysis rates – Radiation fluxes – Radiation models

1

Reminder of EM Spectrum
Blackbody Radiation

Log Scale
Linear Scale

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
From S. Nidkorodov

• Question: are relative vertical scales ok in right plot?

Solar Radiation Spectrum II

From F-P&P

•Solar spectrum is strongly modulated by atmospheric scattering and absorption

From Turco

3

Solar Radiation Spectrum III

UV

Photon Energy ↑

C B A

From Turco

Solar Radiation Spectrum IV

• Solar spectrum is strongly modulated by atmospheric absorptions
• Remember that UV photons have most energy

O₃
O₂

– O₂ absorbs extreme UV in mesosphere; O₃ absorbs most UV in stratosphere
– Chemistry of those regions partially driven by those absorptions
– Only light with λ>290 nm penetrates into the lower troposphere
– Biomolecules have same bonds
(e.g. C-H), bonds can break with UV absorption => damage to life

• Importance of protection provided by O₃ layer

From F-P&P

4

Solar Radiation Spectrum vs. altitude

From F-P&P

• Very high energy photons are depleted high up in the atmosphere • Some photochemistry is possible in stratosphere but not in troposphere

• Only λ > 290 nm in trop.

Solar Zenith Angle

••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.

Solar F lu x

10¹⁵ 10¹⁴ 10¹³ 10¹²

Actual pathlength Vertical pathlength
Lh
"Air Mass"= m =

• ≈
• = secθ

S Z A = 0º

S Z A = 86º

At large SZA very little UV-B radiation reaches the troposphere

• 300
• 400 500

W avelength [nm ]

700

1000

5

Direct Attenuation of Radiation

I = I₀ × e^-t×m

From F-P&P & S. Nidkorodov

t = t_sg + t_ag + t_sp + t_ap

I ≡ radiation intensity (e.g., F) I₀ ≡ radiation intensity above atmosphere

m ≡ air mass

t ≡ attenuation coefficient due to
– absorption by gases (ag) – scattering by gases (sg) – scattering by particles (sp) – absorption by particles (ap)

Rayleigh scattering

t_sp ∝ λ^-n t_sg ∝ λ^-4

much more complex

Deep UV – O, N₂, O₂ Mid UV & visible – O₃
Near IR – H₂O
Infrared – CO₂, H₂O, others

t_ag ∝ σ

Scattering by Gases

• Purely physical process, not absorption

• Approximation:

t_sg =1.044⋅10⁵ ⋅(n_0λ −1)² / λ⁴

• Strongly increases as λ

decreases
• Reason why “sky is blue” during the day

From Turco

6

Scattering & Absorption by Particles

From Jacob

• Particles can scatter and absorb radiation

• Scattering efficiency is very strong function of particle size

– For a given wavelength

• Visible: λ ~ 0.5 µm

– Particles 0.5-2 µm are most efficient scatterers!

• Will discuss in more detail in aerosol lectures

Gas Absorption: Beer-Lambert Law I

I = I₀ exp(−σ ⋅ 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 = σ⋅L⋅N (also “optical depth”)

• σ ≡ absorption cross section

Solve in class: Show that in the small

[cm²/molec]
• L ≡ absorption path length [cm]

absorption limit the relative change in light intensity is approximately equal to

• n ≡ density of the absorber

absorbance.

[molec/cm³]

From F-P&P & S. Nidkorodov

7

Beer-Lambert Law II

Pitfalls:

• Other units are frequently used to express absorbance, for example:

• A = ln(I₀/I) = ε×L×C
• A = ln(I₀/I) = α×L×P

ε ≡ extinction coefficient [L mol^-1 cm^-1] α ≡ absorption coefficient [atm^-1 cm^-1] C ≡ density of the absorber [mol L^-1] P ≡ partial pressure [atm]

• Base 10 is used in most commercial spectrometers instead of the natural base:

A_{base 10} = log(I₀/I) = A_{base e}/ln(10)

Physical interpretation of σ

• σ , 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 σ ≈ 10^-15 cm²/molec
Light absorption σ ≈ 10^-18 cm²/molec

From S. Nidkorodov

8

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

I₀

I

Detector
Filter

n [#/cm³]

L

Measurements are repeated for a number of concentrations at each wavelength of interest.

Although seemingly trivial, in practice such measurements are difficult because of impurities, especially when it comes to very small cross sections (< 10^-20 cm²/molec)

Solve in class: Sample contains 1 Torr of molecules of interest with σ=3x10^-21 cm²/molec and 1 mTorr of impurity with σ=2x10^-18 cm²/molec.
What is the total absorbance in a 50 cm cell?

From S. Nidkorodov

Example: UV Attenuation by O₃ and O₂

Attenuation coefficient is dominated by O₃ absorption in the 200-300 nm window. Therefore, direct attenuation can be easily calculated from known absorption cross sections of O₃. Similar formulas apply to attenuation by O₂ in 120-180 nm window.

I(λ) = I₀ (λ)× e^{-σ (λ)×A×m}

where A ≡ column density

∞

A = [O₃ (z)]dz

∫

z

From S. Nidkorodov

Alternatively written :

Solve in class: Using barometric law estimate column density of O₂ in the atmosphere. By how much does atmospheric O₂ attenuate solar radiation at around 170 nm (σ ≈ 10^-17 cm²/molec) at noon (m = 1)? Assume that O₂ fraction (21%) is independent of altitude and T = 270 K.

I(λ) = I₀ (λ)× e^{-τ (λ,z)}

whereτ ≡ optical depth

∞

τ (λ, z) = σ (λ)×m×[O₃ (z)]dz

∫

Ans: 4x10²⁴; by   exp(-10⁷)

z

9

Solar Radiation Intensity

To calculate solar

From F-P&P

spectral distribution in any given volume of air at any given time and location one must know the following:

– Solar spectral distribution outside the atmosphere
– Path length through the atmosphere
– Wavelength dependent attenuation by atmospheric molecules
– Amount of radiation indirectly scattered by the earth surface, clouds, aerosols, and other volumes of air

Surface Albedo

Reflected Radiation(λ) Incident Radiation(λ)

Albedo(λ) =

• Wavelength dependent!

• Question: for the same incident UV solar flux, will you tan faster over snow or over a desert?

From F-P&P

10

Total vs. Downwelling Radiation

From Warneck

• If atmosphere was completely transparent and surface completely absorbing (albedo = 0)

– F_U = 0 – F_D =F_T= 1

• Due to gas + aerosol scattering and surface reflection

– F_U can be large – F_T > solar flux!

Calculation of Photolysis Rates I

Generic reaction: A + h ν → B + C

d[A]
= −J_A[A]

dt

• A “first-order process”

• What does J_A depend on?

• J_A depends on

– Light intensity from all directions

• “Actinic flux”

– Absorption cross section (σ) – Quantum yield for photodissociation (φ) – All are functions of wavelength

11

Calculation of Photolysis Rates II

Generic reaction: A + h ν → B + C

d[A]
= −J_A[A] = − σ _A (λ)φ_A (λ)F(λ)dλ ×[A]

∫

dt

J_A – first order photolysis rate of A (s^-1) σ_A(λ) – wavelength dependent cross section of A (cm²/#) φ_A(λ) – wavelength dependent quantum yield for photolysis

F(λ) – spectral actinic flux density (#/cm²/s)

So, what are the smallest cross sections that matter? The solar actinic flux (photons cm^-2 s^-1 nm^-1) is of order 10¹⁴. In many cases, we need to know whether the photolytic lifetime of a molecule is 10 days (J=10^-6 s^-1), or 100 days (J=10^-7 s^-1). This means that cross sections as small as 10^-20 cm² or even smaller are potentially interesting. Such small cross sections are very challenging to measure with sufficient accuracy.

Radiation Fluxes Definitions

• Quantity
• Description
• Units

Actinic flux density: energy crossing a unit area per unit time without consideration of direction (we do not care where photons come from)

F

J m^-2 s^-1

Spectral actinic flux density: same as flux but per unit wavelength

J m^-2 s^-1 nm^-1 J m^-2 s^-1

F(λ) E

Irradiance: same as flux but for a unit area with a fixed orientation Spectral irradiance: same as radiance but per unit wavelength range

J m^-2 s^-1 nm^-1

E(λ)

Radiance: radiant flux density per unit solid angle

J m^-2 s^-1 sr^-1

L(θ, ϕ)

J m^-2 s^-1 nm^-1 sr^-1

Spectral radiance: same as radiance but per unit wavelength range

L(θ, ϕ, λ)

2π π

• E =
• L(θ,ϕ)cosθ sinθdθdϕ

∫

^{L(θ,ϕ)cosθdω =} ∫∫

ω

• 0
• 0

2π π

• F =
• L(θ,ϕ)sinθdθdϕ

∫

^{L(θ,ϕ)dω =} ∫∫

ω

• 0
• 0

Radiance as a function of direction gives a complete description of the radiation field. When L is independent of direction, the

field is called isotropic, in which case E = π L and F = 2π L.

Solve in class: There are 10⁹ photons flying into a 0.01 cm diameter opening every

From F-P&P & S. Nidkorodov

second. What is F with respect to this opening in units of #/cm²/s?

12

Radiation Mesurements

Flat Plate → Irradiance
2π → ½ of Actinic Flux

• Radiation does not just come directly from the sun

– scattered radiation is just as important – Measure total or spectrally-resolved flux – Use models

From F-P&P

Radiation Models

• Predict radiation intensity

– As f(time, altitude, latitude, λ) – Results of Madronich (1998) described in text

• Will use extensively in homeworks & exams

– Typical model results:

From F-P&P

13

Example model results

From F-P&P

Q: summer/winter solstices intensity at 445 nm: @ 8 am? @ noon?

Examples of Photolysis Rates

From F-P&P

14

Example: Photolysis of CH₃CHO

λ

i

CH₃CHO + hν

• → CH₃ + HCO
• (a)

• J = σ (λ)φ(λ)F(λ)dλ ≈
• σ (λ)φ (λ)F (λ)∆λ

∑
∫

→ CH₄ + CO
(b)

λ=290nm

From F-P&P

15