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A little spectroscopy Homework…

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Page 1: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

A little spectroscopy

Homework…

Page 2: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 3: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Dipole Moments An electric dipole is a separation of positive & negative charges.

A dipole moment is a vector pointing from the negative to the positive charge with a magnitude equal to the strength of each charge times the separation between the charges. The overall dipole moment of a molecule may be approximated as a vector sum of bond dipole moments. Pm=picometre=1012 m

Page 4: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Which molecule has the highest & lowest dipole moment?

Potassium bromide carbon dioxide: carbon monoxide: ozone: water vapor: hydrogen cyanide:

Page 5: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Which molecule has the highest & lowest dipole moment?

Potassium bromide 10.41 D carbon dioxide: 0 carbon monoxide: 0.112 D ozone: 0.53 D water vapor: 1.85 D hydrogen cyanide: 2.98 D

1 Debye = 1×10−18 statcoulomb-centimetre 1 statC =dyn1/2 cm1 & 1newton = 105 dyne

Page 6: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Permanent dipoles: One atom attracts electrons more than another, becoming more negative, while the other atom becomes more positive. These molecules are called polar molecules, e.g. water. Instantaneous dipoles: These occur when electrons become concentrated in one place than another in a molecule, creating a temporary dipole. Induced dipoles: These can occur when one molecule with a permanent dipole repels another molecule's electrons, inducing a dipole moment in that molecule. A molecule is polarized when it carries an induced dipole. This process adds to Titan’s greenhouse effect.

Page 7: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotation Spectra: (Rigid Rotor)

Classical Mechanics:

Where I is the moment of inertia and ω the angular momentum

Quantum Mechanics:

Quantize L with a value of ħJ where ħ = h/2π and J is an integer specifying the state

The energy difference between 2 successive orders: = ħ/I ~ 10-3 to 10-4 eV Compare to translational energy at room temp 3/2 k T = ~ 10-2 eV

I for 2 masses a distance R : I = m1 m2/(m1+m2) R2

Page 8: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotational transitions: wavelength spacing

What is the wavenumber (or frequency) spacing between rotational lines?

Note:

Page 9: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotational transitions: wavelength spacing

Selection Rule: ΔJ = ± 1

Degeneracy of rotational states: 2 J + 1

Page 10: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Centrifugal Distortion

When a molecule rotates, the centrifugal force pulls the atoms apart, which causes the moment of inertia to increase, thus decreasing the rotation constant, B, used for the rigid rotor.

For a diatomic molecule for example one adds more terms to account for this effect.

The spacing of the lines is no longer constant with frequency, and decreases with increasing quantum number.

Page 11: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotational transitions: wavelength spacing

What is the wavelength spacing between rotational lines?

=

Taking the derivative:

Page 12: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Example

Titan

Page 13: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Vibrational states

For the lowest states – can approximate as a simple harmonic oscillator. Quantum Mechanics:

Where νk is the mode frequency and v (an integer) is the vibration quantum number

What is the separation between vibrational states?

Page 14: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Consider CO

Lowest Energy State: E = hν = hv1/2 Wavenumber (ν/c) = 2127 cm-1 or 4.7 um

The 1-> 0 and 2 -> 1 transition lines emit radiation at this wavelength

The 2->0 lines emit radiation at what frequency?

CO has only 1 mode

Page 15: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Consider CO

Lowest Energy State: E = hν = hv1/2 Wavenumber (ν/c) = 2127 cm-1 or 4.7 um

The 1-> 0 and 2 -> 1 transition lines emit radiation at this wavelength

CO has only 1 mode

The 2->0 lines emit at 2 x 2127 cm-1 = 4254 cm-1 or 2.35 um

Page 16: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CO spectrum

Page 17: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotation transitions: Selection rules ---

Also: ΔJ = 0 For spherical top molecules Not diatomic molecules

Rovibrational Transitions

R P

Page 18: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Overtones

For Δv=1 à4.7 um or 2127 cm-1 For Δv=2 à2.3 um or 4254 cm-1 For Δv=3 à1.6 um or 6381 cm-1

Calculations of cross sections for hot gasses, from UCL ~ 10 million transitions

Page 19: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

How many vibrational modes of an N>2 molecule?

Page 20: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

How many vibrational modes of an N>2 molecule?

A molecule with N atoms has a total of 3N degrees of freedom (each nucleus can be independently displaced in three perpendicular directions). Three of these degrees of freedom correspond to translational motion of the center of mass. For a nonlinear molecule, three more degrees of freedom determine the orientation of the molecule in space, and thus its rotational motion. This leaves 3N - 6 vibrational modes. For a linear molecule, there are just two rotational degrees of freedom, which leaves 3N - 5 vibrational modes.

3N

3N-3

3N-3-3

3N-3-2

Page 21: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Summary vibrational Transitions

ν1 mode of CO

0->1, 1à2, 2à3...

Page 22: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 23: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CO2

How many vibrational bands does CO2 have? Are we missing any?

Do they all interact with radiation?

Page 24: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CO2

Symmetric Stretch ν1 No electric dipole

No radiative interaction

Asymmetric Stretch ν2 4.26 μm 2349 cm-1

Bend ν2 & ν4 15.0 μm 667 cm-1

Greenhouse warming

Page 25: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 26: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Earth’s IR emission spectrum

Page 27: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CH4

ν1 3.2 μm

3025.5 cm-1

ν3 3.3 μm

3156.8 cm-1

ν2 6.3 μm

1582.7 cm-1

ν4 7.3 μm

1365.4 cm-1

Whooping feature

Page 28: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CH4

ν1 3.2 μm

3025.5 cm-1 (inactive)

ν3 3.3 μm

3156.8 cm-1

ν2 6.3 μm

1582.7 cm-1

ν4 7.3 μm

1365.4 cm-1

Page 29: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 30: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Methane ν3 spectrum

Page 31: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 32: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Optical reflection spectra

Page 33: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 34: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 35: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

H2O

Page 36: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

H2O

ν3 2.7 μm

3755.9 cm-1

ν2 6.3 μm

1594.8 cm-1

ν1 2.7 μm

3657.1 cm-1

For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um)

Page 37: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Burrows 2013

ν1 ν3

ν3,1

ν3,1

Page 38: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Fp/F

s

Page 39: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

CO model of an exoplanet

Fp/F

s

Page 40: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 41: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Solar Abundance of Elements

Page 42: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Main Molecules in H2 atmopheres

Page 43: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Thermochemical Equilibrium!Chemistry driven by the thermal energy of the system.

In equilibrium, the Gibbs free energy, G, is a minimum.

Definitions:

H= enthalpy, S=Entropy, T=temperature

Then:

Substituting for dH:

Isothermal process in which the pressure is changed alters G by:

i.e.

Page 44: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Composition!

VENUS gas! Abundance! EARTH gas! Abundance!

CO2! 96.4 %! N2! 78.1 %!N2! 3.5 %! O2! 21.0 %!SO2! 0.015 %! H2O! 0.01-4 %!Ar! 0.007 %! Ar! 0.93 %!H2O! 0.002 %! CO2! 0.40 %!

Page 45: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Volcanic Gases from Kilauea!

Ninety-nine percent of the gas molecules emitted during a volcanic eruption at Kilauea volcano are water vapor (H2O), carbon dioxide (CO2), and sulfur dioxide (SO2). The remaining one percent is made up of hydrogen sulfide (H2S), carbon monoxide (CO), hydrogen chloride (HCl), hydrogen fluoride (HF). There is also N2 and Ar.!

VENUS gas!

Abundance!

CO2! 96.4 %!N2! 3.5 %!SO2! 0.015 %!Ar! 0.007 %!H2O! 0.002 %!

Page 46: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Carbon: Earth vs Venus!

Venus CO2: equivalent to 0.88 km of calcium carbonate (CaCO3) !This is 2 times the calcium carbonate in Earth’s crust.!Origin of CO2 is volcanism. !!Earth CO2: 25% is anthropogenic and 75% biological. Volcanism: minor source!Most of the carbon is in calcium carbonate, mainly skeletal fragments of marine organisms such as coral, forams and molluscs!

Page 47: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Which if any molecule is stable in Earth’s

atmosphere?

Page 48: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

N2 cycle "(even triply bonded N2!)!

N2 fixation:! N2 + 8H+ + 8e- è 2NH3 + H2 !

Through enzymes (nitrogenase) !produced by prokaryotes !(both bacteria and archaea) !

Nitrification! Bacteria convert NH4

+ to NO2−!

then (with other bacteria) to NO3−!

Ammonification ! Bacteria & Fungi produce NH4

+!

from plant & animal remains!

Denatrification! Bacteria convert NO3

− to N2! !

Page 49: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Earth

Page 50: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Processes that establish composition

Page 51: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 52: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

n  Earth, Venus & Mars require surface interaction to explain the atmospheres !!n  Cache elements (rust and oxygen on Mars) !n  Supply elements such as sulfur (volcanism of S02)!n  Participate in the molecular changes (N cycle) !

n  Small & hot planets are susceptible to volatile loss. !n  Mars’ lack of volatiles!n  Small hot Jupiters!

n  Large and cool Jovian Planets are simpler*.!n  Thermochemical equilibrium !n  Photochemistry!n  Disequilibrium from transport in conditions out of equilibrium!

Summary: "Lower Atmosphere !

Earth’s biosphere is totally dependent on life.!

Page 53: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Rotational and Vibrational

Probe the molecular abundances of lower atmospheres Span near-IR to microwave wavelengths. With pure rotational bands at long wavelengths. And Rovibrational bands at IR wavelengths

Page 54: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 55: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Strength of spectroscopic lines

�  Probability of a transition

�  Population of states

Page 56: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Absorption Coefficients

The intensity of a spectral line depends on: 1) the probability that the molecule or atom makes the transition 2) the occupation of the initial state

The latter depends on the temperature

Line strengths are generally determined at room temperature, and listed in large data bases. The strengths at different temperatures are determined by considering the occupation of states.

Page 57: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

DO 600 KL=1,NLAY ! Print *, 'gettin all silly with layer ', kl, ' ,yo!' Do 29 i=1,ngas do 30 ialfa=1,100 alfa=0.001*ialfa

Call Profiles(nw,np,w0,mw(i),t(kl),p(kl),alfa,ialfa,w1,f1) ! nw=25 do k=1,nw width(k,ialfa,i) = w1(k,ialfa) ! Voigt profile for Lorentz with of ialfa/1000 ff(k,ialfa,i) = f1(k,ialfa) ! Voigt profile at width for a=iafa/1000 end do

do 20 k=1,nw-1 k1=k+1 ffww3(k,ialfa,i)=(ff(k,ialfa,i)-ff(k1,ialfa,i))/ * (width(k1,ialfa,i)-width(k,ialfa,i)) 20 continue 30 continue Do j=1,ng(i) If (e(j,i).eq.-1.) then

e(j,i)=550. End if ! Print *, 'First intensity ', j, T(kl), S(j,i) ! Print *, 'Energy ', H2*E(j,i), E(j,i) ! Print *, 'Temp and term ', T(kl),(1./296.-1./T(kl)) Sl(j,i)=S(j,i)*((T0/T(kl))**xgeo(i))* ! xgeo=molecule geometry * (EXP(H2*E(j,i)*(1./T0-1./T(kl)))) ! linear =1/nonlin=1.5 ! Print *, 'After partition function ', Sl(j,i)

Sl(j,i)=Sl(j,i)*( 1-exp(-H2*f(j,i)/t(kl)) )/ . ( 1-exp(-H2*f(j,i)/T0) ) ! stimulated emission ! Print *, 'After stimulated emission ', Sl(j,i) Sl(j,i)=Sl(j,i)*2.68E+19*1.E+05 ! intensity (km*amag)-1 ! Print *, 'Units to km amagat-1 ', Sl(j,i) ! Read(5,*) istop Sl(j,i)=Sl(j,i) * mix(i,kl) ! times mixing ratio ! If (kl.eq.1) print *, j, Sl(j,i), a(j,i) End do 29 End do ! Print *, 'Finished reading intensities ' ! Read(5,*) istop

Page 58: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 59: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Summary

Page 60: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

A few questions

�  Why are absorption bands of solid material broad?

�  For what kind of observations/studies do we need high spectral resolution data?

�  What information can we gather about planetary formation, surface composition, surface processes, atmospheric composition, atmospheric dynamics, chemistry and evolution through spectroscopy?

Page 61: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin
Page 62: Homework…griffith/PTYS517/Chem1.pdf · 2.7 μm 3657.1 cm For Δv=2 à3190 (3.1 um), For Δv=3 à4784 (2.1 um) For Δv=2 à7512 (1.3 um) Burrows 2013 ... Chem1.pptx Author: Caitlin

Absorption Coefficients