banistdo Figure 10.2 Tr 1.0 0.8 0.6 0.4 Tr 1.0 0.8 0.6 - 0.4 0.2 - Tr 1.0 0.8 0.6 0.4 0.0 + 2 للللللل Tr 1.0 0.2 - 0.0 0.8 0.6 2 4 0.4 2 Tr 1.0 0.8 0.6 2 Tr 1.0 0.8 0.6 - 0.4 - 0.2 - 0.0 2 0.4 0.2 - 0.0 4 p 7 4 4 4 6 6 6 6 6 H command option 8 8 V 8 10 12 8 8 10 12 14 16 10 12 14 10 10 14 16 12 12 14 16 14 16 16 16 www H₂O (g) 18 20 22 λ/μm CO₂ (g) 18 20 22 λ/μm O, (g) 18 20 22 λ/μm CH, (g) 18 20 N₂O (g) 18 22 2/μm 20 22 2/μm CF₂Cl₂ (g) 20 22 λ/μm 18 20 2 468 10 12 14 IR spectra of some greenhouse gases243 Abscissa: wavelength, Ordinate: relative transmittance. The partial pressures are 0.03-0.26 bar diluted to 1 bar with N₂(g). The peak at 42 um in the O, spectrum is due to a trace impurity of CO₂.

Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter20: Molecular Spectroscopy And Photochemistry
Section: Chapter Questions
Problem 4P
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Compare the emission spectrum of Earth (figure 10.1) with the absorption spectra of the greenhouse gases shown in figure 2. At what wavelength(a) can earthshine escape to space without being absorbed by atmospheric gases?
Figure 10.2
Tr 1.0
0.8
0.6
0.4
Tr 1.0
0.8
0.6 -
0.4
0.2 -
0.0 +
2
Tr 1.0
T 1.0
0.8
0.6
0.4 -
0.2 -
0.0
0.8
للللللل
0.6
0.4
Tr 1.0
0.8
0.6
2 4
Tr 1.0
0.8
0.6
0.4
0.2
0.0
2
0.4 -
0.2 -
0.0
2
2
7
4
4
4
6
4
6
prom
6
6
H
6
command option
8
8
8
10 12
my
V₁
10 12 14 16
10
12
14
12 14 16
10
8
10
8
16
14
12 14
16
16
H₂O (g)
ww
18 20 22 λ/μm
CO₂ (g)
18 20 22 λ/μm
O, (g)
-
18 20 22 λ/um
CH, (g)
18
18
20 22 λ/um
N₂O (g)
20 22 2/μm
CF₂Cl₂ (g)
20 22 2/μm
18 20
2 468
10 12 14 16
IR spectra of some greenhouse gases243
Abscissa: wavelength, Ordinate: relative transmittance.
The partial pressures are 0.03-0.26 bar diluted to 1 bar with N₂(g).
The peak at 4.2 μm in the O3 spectrum is due to a trace impurity of CO₂.
Transcribed Image Text:Figure 10.2 Tr 1.0 0.8 0.6 0.4 Tr 1.0 0.8 0.6 - 0.4 0.2 - 0.0 + 2 Tr 1.0 T 1.0 0.8 0.6 0.4 - 0.2 - 0.0 0.8 للللللل 0.6 0.4 Tr 1.0 0.8 0.6 2 4 Tr 1.0 0.8 0.6 0.4 0.2 0.0 2 0.4 - 0.2 - 0.0 2 2 7 4 4 4 6 4 6 prom 6 6 H 6 command option 8 8 8 10 12 my V₁ 10 12 14 16 10 12 14 12 14 16 10 8 10 8 16 14 12 14 16 16 H₂O (g) ww 18 20 22 λ/μm CO₂ (g) 18 20 22 λ/μm O, (g) - 18 20 22 λ/um CH, (g) 18 18 20 22 λ/um N₂O (g) 20 22 2/μm CF₂Cl₂ (g) 20 22 2/μm 18 20 2 468 10 12 14 16 IR spectra of some greenhouse gases243 Abscissa: wavelength, Ordinate: relative transmittance. The partial pressures are 0.03-0.26 bar diluted to 1 bar with N₂(g). The peak at 4.2 μm in the O3 spectrum is due to a trace impurity of CO₂.
M₂/(Wm')
8x10¹3
4x10¹3
500
Figure 10.1
(10.8)
where a is the fraction of the irradiance that is converted into heat and the ensuing
thermal radiation. Kirchhoff also noted that a system with a = 1 was "a completely
blackbody,"33
T=5780 K
dλ
M = a I
M₂/(W m³)
1.4x10'
1.0x10²
0.6x10²
0.2x10²
1000 1500 2000 λ/nm
Blackbody emission spectra of the Sun and the Earth
Left panel: Spectrum of the Sun, To = 5780 K: Amax
Mo = 63.29 MW m2; P = 3.853x1026 W.
10
20
Right panel: Spectrum of the Earth, T = 255 K: Amax
Me = 239 W m-2; P = 1.22×10¹7 W.
T=255 K
30 40 50
λ/μm
501 nm = 0.501 μm;
= 11.4 µm;
Planck's radiation law (1900)
The determination of the formula for spectral excitance was called "Kirchhoff's
challenge." It was a significant problem for theoretical physicists for 40 years, until
Planck solved it.
According to Planck, the spectral excitance of a hlaoid
AL01
Inbrig
Transcribed Image Text:M₂/(Wm') 8x10¹3 4x10¹3 500 Figure 10.1 (10.8) where a is the fraction of the irradiance that is converted into heat and the ensuing thermal radiation. Kirchhoff also noted that a system with a = 1 was "a completely blackbody,"33 T=5780 K dλ M = a I M₂/(W m³) 1.4x10' 1.0x10² 0.6x10² 0.2x10² 1000 1500 2000 λ/nm Blackbody emission spectra of the Sun and the Earth Left panel: Spectrum of the Sun, To = 5780 K: Amax Mo = 63.29 MW m2; P = 3.853x1026 W. 10 20 Right panel: Spectrum of the Earth, T = 255 K: Amax Me = 239 W m-2; P = 1.22×10¹7 W. T=255 K 30 40 50 λ/μm 501 nm = 0.501 μm; = 11.4 µm; Planck's radiation law (1900) The determination of the formula for spectral excitance was called "Kirchhoff's challenge." It was a significant problem for theoretical physicists for 40 years, until Planck solved it. According to Planck, the spectral excitance of a hlaoid AL01 Inbrig
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