MArk Claire ATM chem

Question 1

Layers of the Atmosphere:

Answer 1

– Atmospheric Boundary
Layer
– Troposphere
– Stratosphere
– Mesosphere
– Thermosphere

Question 2

Layer Separations:

Answer 2

• Layers separated by:
– Tropopause
– Stratopause
– Mesopause

Question 3

‘Mixing ratio’

Answer 3

gas density divided by the total density, reported as %, ppm, ppb etc.

Question 4

Equation for energy

Answer 4

E=hv
E = Energy ; h is a constant (Planck’s constant)
Energy is proportional to frequency

Question 5

Equation for frequency

Answer 5

v = c/λ

v = frequency ; c is a constant (speed of light)
λ is wavelength.

Energy
inversely proportional to wavelength

Question 6

breaking of O3

Answer 6

O3 + hν →O+O2

O3 has a bond energy of ~450 kJ
mol-1
so can break if λ < 290 nm

(Breaking chemical bonds
requires energy:
• Sunlight has energy
• If sufficient energy is deposited
in the bond, then it will break)

Question 7

Photolysis

Answer 7

depends on:

1. Wavelength of Absorbed light
2. Temperature
3. Pressure

(often called the initiators of the atmospheric photochemistry reactions)

Question 8

how would you work out the change in concentration of O in these two reactions?

O3 + hν→O + O2

O+ HO2→OH + O2

Answer 8

reaction 1 has reaction rate = j1

reaction2 reaction rate= k1
Total change in O concentration is:
Δ[O] = PRODUCTION - LOSS
Δ[O] = j1[O3 ] - k1[O][HO2 ]

Question 9

What make sup the photochemical reaction rate constant? (j1)

Answer 9

1. what photons are available
   (actinic flux)
2. How do the photons interact
   (absorption cross section)

(number of photons absorbed by the molecule at a particular wavelength TIMES the density of photons in the atmosphere at a particular wavelength)

Absorption cross section, σ1 *Spectral actinic flux, I:
J1(λ,z)= σ 1(λ) × I(λ,z)
Over all wavelengths:
j1(z) =Σσ A(λ) × I(λ,z) Δλ

Question 10

Actinic flux=

Answer 10

flux at a given height

Question 11

Absorption cross section

Answer 11

How strongly a molecule absorbs as a function of wavelength #(cm2/per molecule

Question 12

Absorption of light depends on:

Answer 12

the concentration of the gas, N ,
its absorption cross section,
σ, &
the path length,ℓ, through the gas

Beer Lambert Law:
I(ℓ) = I0 e(−σNℓ)

Question 13

Optical Depth

Answer 13

σNℓ = “optical depth” = τ

unit less measure of absorbance and is wavelength dependent

τ(λ)&raquo_space; 1 implies complete absorption of that
wavelength in a column of gas

Question 14

Column density

Answer 14

term used for the total amount of gas in a vertical column collapsed into an effective area

ΣN × Δℓ ≅ Nℓ , units molecules/cm2

Question 15

At 250 nm, the O3 cross-section is 10-17 cm2.
The modern O3 column density is measured as
8.1x1018 cm-2, what is the optical depth of O3 at 250
nm?

Answer 15

τ = 8.1x1018 x 10-17 = 81

Question 16

Radiocarbon

Answer 16

stable carbon isotopes are 12C and 13C

When cosmic rays slam into largest component of atm (14N) then we get :
14N + n -> 14C + p
14C is radioactive

one in every 10 ^12 carbon atoms in the atm are C14

Question 17

use of radioactive material

Answer 17

measuring the radionucleotides allowed us to see the transportation and the atmospheric dynamics in place

Question 18

Planetary energy balance

Answer 18

Energy absorbed (Fin)= Energy Emitted (Fout)

S (1-A) πR2 = σSBTeff
4 4πR2
S(1-A)/4 = σSBTeff
4

S is incoming sunlight

A is planetary albedo, the fraction of this light that is
reflected

σ SB = 5.67x10-8 W m-2 K-4
The Stefan-Boltzmann
constant

R is the radius of the
Earth