Published on *METEO 300: Fundamentals of Atmospheric Science* (https://www.e-education.psu.edu/meteo300)

Atmospheric radiation plays a critical role in life on Earth and in weather. Without solar heating, Earth would be a dead frozen ball hurtling through space. Luckily, the energy that Earth receives from solar radiation is sufficient to produce liquid water on its surface, thus enabling life to thrive. In this lesson we will look at solar radiation and its changes over time. Radiation is just another form of energy and can be readily converted into other forms, especially thermal energy, which is sometimes called "heat." In this lesson, we will use the word "radiation" to mean all electromagnetic waves, including ultraviolet, visible, and infrared. We will introduce some unfamiliar terms like "radiance" and "irradiance" and will be careful with our language to prevent confusion.

An important concept in studying atmospheric radiation is that all objects emit and absorb radiation. For a perfect emitter, the radiation emitted by an object, called the irradiance, is determined by the Planck function, which depends only on temperature and wavelength. The higher the temperature, the greater the radiation emitted at all wavelengths and the shorter the wavelength of the peak energy. The Sun emits in the visible while Earth and its atmosphere emit in the infrared. No object is really a perfect emitter at every wavelength; the unitless number emissivity measures how good or poor an emitter is. At each wavelength, a good emitter is a good absorber.

How well an object absorbs at different wavelengths of radiation, called its absorptivity, depends on its chemical composition and the rules of quantum mechanics. As a result, some absorption is strong and some is weak; some is in sharp lines while some is in broad features in the wavelength spectrum; some is in the UV, particularly due to O_{2} and O_{3}, little is in the visible, and much absorption, including broad bands and sharp lines, occurs in the infrared, particularly by H_{2}O, CO_{2}, and O_{3}.

The radiation that is not absorbed by a gas, liquid or solid is either transmitted or scattered. The amount of transmitted radiation depends on the absorption and scattering cross sections of the gas, liquid, or solid components of matter, so that the larger the cross section and the distance through the matter, the less radiation passes through the matter. The decay of the transmitted light with distance through the matter is exponential, as is described by Beer’s Law.

Earth’s surface and atmospheric gases can emit and absorb radiation at the same wavelengths. Most of Earth’s emissions are at infrared wavelengths, whether the emission be from the surface, clouds, or atmospheric gases.

Scattering of atmospheric radiation complements absorption and is even more difficult to track through the atmosphere than absorption is. The wavelength of the radiation and the size, shape, and composition of the scattering particle together determine the scattering efficiency and scattering pattern. Many of the skies that we remember best are due to the scattering and absorption of sunlight.

By the end of this lesson, you should be able to:

- identify the causes of changing solar radiation on Earth
- calculate properties of the spectrum of solar and earth radiation in terms of the Planck function
- calculate the absorption between you and a light source
- explain why the sky looks blue and hazy in the summer

Please see Canvas for a list of required assignments, due dates, and submission instructions.

If you have any questions, please post them to the Course Questions discussion forum in Canvas. I will check that discussion forum daily to respond. While you are there, feel free to post your own responses if you, too, are able to help out a classmate.

Everything radiates—the Sun, the Earth, the atmosphere, and you. The energy provided by the Sun is reused in the Earth system to provide the energy that drives weather and climate. But ultimately, the infrared radiation radiated by Earth into space must balance the solar visible radiation coming into the Earth system. From the point-of-view of the Earth system, we are most concerned about how atmospheric radiation interacts with matter. Matter is simply molecules and atoms and the structures that they build, such as the air, the clouds, the Earth, and the Sun.

When radiation encounters matter, three things can happen. The radiation can be *transmitted through* matter; it can be *absorbed by* the matter; it can be *scattered by* the matter. One of these three things must happen, so we can sum them up to one:

$$a+t+s=1$$

where *t* is the transmissivity, the fraction transmitted; *a* is the absorptivity, the fraction absorbed; and *s* is the reflectivity, the fraction that is scattered or reflected.

Scattering and reflecting are related but are different because reflection is scattering *in a particular direction* whereas scattering tends to go *in a range of directions*.

Credit: W. Brune

Here is another chance to earn **one point of extra credit**: **Picture of the Week**!

- You take a picture of some atmospheric phenomenon—a cloud, wind-blown dust, precipitation, haze, winds blowing different directions—anything that strikes you as interesting.
- Add a short description of the processes that you think are causing your observation. A Word file is a good format for submission.
- Use your name as the name of the file. Upload it to the
**Picture of the Week Dropbox**in this week's lesson module. To be eligible for the week, your picture must be submitted by 23:59 UT on Sunday of each week. - I will be the sole judge of the weekly winners. A student can win up to three times.
- There will be a Picture of the Week dropbox each week through Lesson 11. Keep submitting entries!

Solar radiation drives the Earth system and makes life possible. Solar radiation is absorbed and then put to use to increase the surface temperature, to change the phase of water, and to fuel atmospheric chemistry. The uneven distribution of solar radiation on Earth’s surface drives atmospheric dynamics.

The total amount of solar energy per unit time and unit area, also called the solar irradiance, is 1361 W m^{–2} at the top of the atmosphere (Stephens et al., 2012, Nature Geoscience 5, p. 691). It is distributed unevenly over Earth’s surface. That distribution changes over the course of the seasons (see next two figures). The seasons result primarily from the Earth's rotation axis not being perpendicular to the plane of the Earth's orbit around the Sun.

Credit: National Weather Service [1]

Credit: National Park Service [2]

Earth’s spin and its orbit around the sun are not constant, but instead, change with time, like a spinning top. The orbit’s eccentricity (i.e., how different it is from circular) varies with a 100,000-year period. The tilt of Earth’s rotation axis with respect to the perpendicular from its orbit, which is called its obliquity, varies from 22.1^{o} to 24.5^{o} during a 41,000-year cycle. It is currently 23.4^{o} and decreasing. Finally, the precession of Earth’s orbit, which is the orientation of the rotation axis with respect to Earth’s orbital position, also varies with a period of about 26,000 years, although the eccentricity of the orbit is also rotating around the sun, so that the effective period of precession is about 21,000 years. These motions, when taken together, slowly and periodically change the distribution of solar irradiance on Earth’s surface and are described by the **Milankovitch Theory** (see next two figures).

Credit: CO2CRC [3]

Credit: The Azimuth Project [4]

Changes in Earth’s orbit and spin are not the only ways that solar irradiance changes—the Sun’s energy output also changes. It has been increasing slightly (0.05–0.10%) over the past 300 years and varies by another ~0.1% over the course of the 11-year solar cycle. The ultraviolet (200–300 nm) irradiance has increased about 3% in the past 300 years and varies by ~1.5% between solar maximum and solar minimum. This increased UV leads to greater stratospheric ozone production, which increases stratospheric heating, leading to poleward displacement of the stratospheric meridional wind.

We can think of radiation either as waves or as individual particles called photons. The energy associated with a single photon is given by $$** **** ****E**** =**** h****ν*** * , where

The energy of a single photon that has the wavelength *λ* is given by:

$$E\text{}=\text{}\frac{hc}{\lambda}=\text{}\frac{1.986\times \text{}{10}^{-16}{\text{Jnmphoton}}^{-1}}{\lambda}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamyraiaabckacqGH9aqpcaqGGcWaaSaaaeaacaWGObGaam4yaaqaaiabeU7aSbaacqGH9aqpcaqGGcWaaSaaa8aabaWdbiaaigdacaGGUaGaaGyoaiaaiIdacaaI2aGaey41aqRaaeiOaiaaigdacaaIWaWdamaaCaaaleqabaWdbiabgkHiTiaaigdacaaI2aaaaOGaaeiOaiaabQeacaqGGaGaaeOBaiaab2gacaqGGcGaaeiCaiaabIgacaqGVbGaaeiDaiaab+gacaqGUbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiabeU7aSbaaaaa@5AA2@

Note that as the wavelength of light gets shorter, the energy of the photon gets greater. The energy of a mole of photons that have the wavelength *λ* is found by multiplying the above equation by Avogadro's number:

$${E}_{m}\text{}=\text{}\frac{hc{N}_{A}}{\lambda}=\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{\lambda}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamyra8aadaWgaaWcbaWdbiaad2gaa8aabeaak8qacaqGGcGaeyypa0JaaeiOamaalaaabaGaamiAaiaadogacaWGobWdamaaBaaaleaapeGaamyqaaWdaeqaaaGcpeqaaiabeU7aSbaacqGH9aqpcaqGGcWaaSaaa8aabaWdbiaaigdacaGGUaGaaGymaiaaiMdacaaI2aGaaeiOaiabgEna0kaabckacaaIXaGaaGima8aadaahaaWcbeqaa8qacaaI4aaaaOGaaeiiaiaabQeacaqGGaGaaeOBaiaab2gacaqGGcGaaeyBaiaab+gacaqGSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiabeU7aSbaaaaa@5A32@

Credit: Wikimedia Commons

In the lesson on atmospheric composition, you saw how solar UV radiation was able to break apart molecules to initiate atmospheric chemistry. These molecules are absorbing the energy of a photon of radiation, and if that photon energy is greater than the strength of the chemical bond, the molecule may break apart.

Consider the reaction O_{3} + UV → O_{2} + O*. If the bond strength between O_{2} and O* (i.e., excited state oxygen atom) is 386 kJ mol^{–1}, what is the longest wavelength that a photon can have and still break this bond?

ANSWER: Solve for wavelength in equation [6.2b]

$$\lambda =\text{}\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{{E}_{m}}\text{}=\frac{1.196\text{}\times \text{}{10}^{8}{\text{Jnmmol}}^{-1}}{386\text{}\times \text{}{10}^{3}{\text{Jmol}}^{-1}}=309\text{nm}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdWMaeyypa0JaaeiOamaalaaapaqaa8qacaaIXaGaaiOlaiaaigdacaaI5aGaaGOnaiaabckacqGHxdaTcaqGGcGaaGymaiaaicdapaWaaWbaaSqabeaapeGaaGioaaaakiaabckacaqGkbGaaeiOaiaab6gacaqGTbGaaeiOaiaab2gacaqGVbGaaeiBa8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaGcpaqaa8qacaWGfbWdamaaBaaaleaapeGaamyBaaWdaeqaaaaak8qacaqGGcGaeyypa0ZaaSaaa8aabaWdbiaaigdacaGGUaGaaGymaiaaiMdacaaI2aGaaeiOaiabgEna0kaabckacaaIXaGaaGima8aadaahaaWcbeqaa8qacaaI4aaaaOGaaeiOaiaabQeacaqGGcGaaeOBaiaab2gacaqGGcGaaeyBaiaab+gacaqGSbWdamaaCaaaleqabaWdbiabgkHiTiaaigdaaaaak8aabaWdbiaaiodacaaI4aGaaGOnaiaabckacqGHxdaTcaqGGcGaaGymaiaaicdapaWaaWbaaSqabeaapeGaaG4maaaakiaabckacaqGkbGaaeiOaiaab2gacaqGVbGaaeiBa8aadaahaaWcbeqaa8qacqGHsislcaaIXaaaaaaakiabg2da9iaaiodacaaIWaGaaGyoaiaabckacaqGUbGaaeyBaaaa@817C@

Credit: NASA Goddard Space Flight Center [6] via flickr

The Sun emits radiation from X-rays to radio waves, but the irradiance of solar radiation peaks in the visible wavelengths (see figure below). Common units of irradiance are Joules per second per m^{2} of surface that is illuminated per nm of wavelength (e.g., between 300 nm and 301 nm), or W m^{–2} nm^{–1} for the plot below. These units are the units of spectral irradiance, which is also simply called irradiance, but as a function of wavelength. To get the total irradiance in units of W m^{–2}, the spectral irradiance should be integrated over all the wavelengths.

Note the following for the solar spectrum:

- About half of the energy is in the visible wavelengths below 0.7 μm. We can tell this by doing a quick integration.
- O
_{3}and O_{2}absorb much of the UV irradiance below 300 nm high in the atmosphere. - About 70% of the visible irradiance makes it all the way to sea level.
- O
_{3}absorbs a little of the visible irradiance. - A significant fraction of the visible irradiance is scattered by clouds and aerosol. Some is reflected back out into space so that this portion never deposits energy in the Earth system.
- There are large wavelength bands in which water vapor, CO
_{2}, and O_{3}absorb infrared irradiance.

For solar wavelengths at which the absorptivity is high, the solar irradiance at sea level is small. Note that the big absorbers of infrared irradiance are water vapor, carbon dioxide, and ozone.

Credit: Nick84 [CC BY-SA 3.0 [7]], via Wikimedia Commons

- Find
**Practice Quiz 6-1**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 6-1**. You will be allowed to take this quiz only**once**. Good luck!

Recall that molecules have a wide range of speeds and thus a wide range of energies. The **Maxwell–Boltzmann Distribution**, which gives the distribution of molecules as a function of energy, is given approximately by the equation:

$$f(E)=\frac{2\sqrt{E}}{\sqrt{\pi}{\left(kT\right)}^{3/2}}\mathrm{exp}\left(-E/kT\right)$$

where *f* is the probability that a molecule has an energy within a small window around *E*, *T* is the absolute temperature, and *k* is the Boltzmann constant. The above equation, when integrated over all energies, gives the value of 1.

The functional form of this distribution is shown below:

All objects—gas, liquid, or solid—emit radiation. If we think of radiation as photons, we would say that these photons have a distribution of energies, just like molecules do. However, photons cannot have continuous values of photon energy; instead, the photon energy is quantized, which means that it can have only discrete energy values that are different by a very very small amount of energy. When this quantized distribution is assumed, then the distribution of spectral irradiance leaving a unit area of the object’s surface per unit time per unit wavelength interval into a hemisphere is called the **Planck Distribution Function of Spectral Irradiance**:

$${P}_{e}(\lambda )=\frac{2\pi h{c}^{2}}{{\lambda}^{5}}\frac{1}{\mathrm{exp}(hc/\lambda \text{\hspace{0.05em}}k\text{\hspace{0.05em}}T)-1}$$

where *h* is Planck’s constant, *c* is the speed of light, *k* is the Boltzmann constant (1.381 x 10^{–23} J K^{–1}), *T* is the absolute temperature, and *λ* is the wavelength. The integral of this function over all wavelengths leads to the **Stefan–Boltzmann Law Irradiance**, which gives the total radiant energy per unit time per unit area of the object’s surface emitted into a hemisphere.

*P _{e}*(

The spectral irradiance is the amount of energy that is emitted from or falling on a unit area in space per unit time per unit wavelength (W m^{–2} µm^{–1}). So the m^{2} in this case indicates a surface area on which energy is leaving or falling. To get the Sun’s spectral irradiance at the top of Earth’s atmosphere, we must multiply the spectral irradiance emitted by the Sun's surface by the Sun's surface area to get the total energy emitted per unit time and per unit wavelength by the Sun and then divide by the surface area of the sphere that has a radius equal to the Earth–Sun distance to get the energy per time per unit wavelength per unit area of the surface located at the top of Earth's atmosphere.

All the radiation that is emitted from the Sun’s surface continues to move outward at the speed of light until it hits objects, but there are very few objects between the Sun and Earth, except occasionally the Moon. Thus, in the absence of any objects, the total amount of solar irradiance striking the planets decreases as the square of the distance between the center of the Sun and the surface of the planet.

The Planck distribution function spectral irradiance for an object at a temperature of 5777 K (the Sun's surface temperature) is shown in the figure below. Note the rapid rise at the shorter wavelengths, the peak value, and then the slower decline at longer wavelengths. Look at the peak value of the irradiance, which is about 25 million W m^{–2} nm^{–1}! That's a lot of energy being radiated from the Sun's surface, but of course the Earth is 150 million km away from the Sun and thus intercepts less than half a billionth of the solar irradiance.

Credit: W. Brune

The peak of this distribution as a function of wavelength can be found by taking the derivative of *P _{e}*(

$${\lambda}_{max}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{T}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdW2damaaBaaaleaacaWGTbGaamyyaiaadIhaaeqaaOWdbiabg2da9maalaaapaqaa8qacaaIYaGaaGioaiaaiMdacaaIWaGaaGPaVlabeY7aTjaab2gacaaMc8Uaae4saaWdaeaapeGaamivaaaaaaa@46FA@

For the sun with a photospheric temperature of about 5780 K, *λ _{max}* ~ 0.500 μm or 500 nm, which is the color green. However, for Earth with a mid-tropospheric temperature of about 260 K, the peak wavelength is closer to 11 μm, well into the infrared (see below).

Credit: W. Brune

If the Planck distribution function spectral irradiance is integrated over all wavelengths, then the total irradiance emitted into a hemisphere is given by the **Stefan–Boltzmann Law**:

$${F}_{s}=\sigma \text{\hspace{0.05em}}{T}^{4}$$

where *σ* is called the Stefan-Boltzmann constant (5.67 x 10^{–8} W m^{–2} K^{–4}). *F _{s}* has SI units of W m

The Stefan–Boltzmann law (total) irradiance applies to an object that radiates according to the Planck distribution function spectral irradiance. If we look at the figure below, we see that the solar spectrum at the top of the atmosphere is similar to the Planck distribution function but does not follow it perfectly. However, the Planck distribution function with the same total irradiance as the sun has a temperature of 5777 K, as in the second figure.

Credit: Nick84 [CC BY-SA 3.0 [7]], via Wikimedia Commons

Credit: W. Brune

Clouds radiate. Assume two spherical clouds, one with a radius of 100 m and a temperature of 275 K and a second with a radius of 100 m and a temperature of 230 K. Assuming that they both radiate according to the Planck distribution function, calculate the emission for each cloud in W m^{–2} and in W. Which cloud is radiating more total energy and by how much?

ANSWER:

Cloud T (K) | Cloud radius (m) | F_{s} (W m^{–2}) |
F_{s} x 4πR_{c}^{2} (W) |
---|---|---|---|

275 | 100 | 324 | 4.1 x 10^{7} |

230 | 100 | 159 | 2.0 x 10^{7} |

The warmer cloud is radiating about twice as much energy as the cooler cloud. These little clouds are radiating quite a lot of energy in all directions, but some of it is down toward Earth’s surface. If we make the simple assumption that half the radiation goes up and the other half goes down, the amount of energy radiated toward Earth’s surface per second is approximately 10 million W. If the clouds are not too far from the surface, this downward radiation could contribute a few hundred W m^{–2} of heating at Earth’s surface. Thus clouds can act like additional heat sources for Earth’s surface, keeping its temperature higher than it would be on a clear night. The image below is an infrared photograph of the sky above Ogden, Utah. Infrared radiation detected by the camera has been converted to temperature, with higher temperatures indicating more infrared emission.

Credit: activerain [8]

Remember that when radiation encounters matter it may be absorbed or transmitted or scattered (including reflected). For an object acting as a perfect Planck distribution function, it must absorb all radiation completely with no scattering and no transmission. Some objects absorb very well at some wavelengths but not at others. For instance, water vapor absorbs little visible radiation but absorbs infrared radiation at some wavelengths very well.

At the same time, the sun, like other objects, does not radiate perfectly according to the Planck distribution function spectral irradiance, but instead radiates at a fraction of it at some wavelengths. This fraction, which goes from 0 to 1, is called the emissivity and is denoted by ε. How is an object’s emissivity related to its absorptivity?

**Kirchhoff’s Law** states that at any given wavelength, an object’s emissivity *ε* is equal to its absorptivity, that is:

$$\epsilon \left(\lambda \right)=\alpha \left(\lambda \right)$$

Thus, if an object has some wavelengths at which radiation is scattered or reflected, then the object will have an emissivity less than 1 at the wavelength, and the fraction that is absorbed will be equal to the emissivity at each wavelength.

Thus, when we integrate the Planck distribution function spectral irradiance over wavelength to obtain the irradiance emitted by the object, it first has to be multiplied by the wavelength-dependent emissivity, thus leading to the modified form of the Stefan-Boltzmann law:

$$F=\epsilon \text{\hspace{0.17em}}\sigma \text{\hspace{0.05em}}{T}^{4}$$

where we understand that *ε* is some form of averaged emissivity.

Watch the following video (1:07), where the Stefan–Boltzmann Law is described in greater detail:

Some typical average emissivities are listed in the table below. These are emissivities averaged over all wavelengths. At any particular wavelength, the emissivity may be greater or less than the average.

Material | Emissivity, ε |
---|---|

ice | 0.97 |

pure water | 0.96 |

snow | 0.8-0.9 |

trees (oak, beech, maple, pine) | 0.97-0.98 |

grass | 0.98 |

soil | 0.93 |

aluminum foil | 0.03 |

asphalt | 0.88-0.94 |

What about gases? Gases absorb and thus emit like all other matter. To know more about the emissivity of all objects, we need to know more about the absorption of objects.

- Find
**Practice Quiz 6-2**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 6-2**. You will be allowed to take this quiz only**once**. Good luck!

We see that there is a significant amount of absorption of radiation in the infrared but rather little in the visible. Also, we see that gases absorb strongly at some wavelengths and not at others. Why is this?

To answer this question, we need to look at the configurations of the electrons that are zooming around atoms and molecules. More than 100 year ago, scientists began using prisms to disperse the light from the sun and from flames containing different elements. While the sun gave the colors of the rainbow, the flames had light in very distinct lines or bands. This puzzle was finally resolved a little more than 100 years ago with the invention of quantum mechanics, which basically says that the electrons zooming around atoms and molecules and the vibrations and rotations of molecules can have only discrete energies that are governed by rules of conservation of angular momentum.

Credit: Dominic Alves [9] via flickr

The following bulleted list is a crash course in absorption by the electrons in atoms and molecules. Refer to the figure below the box.

- Chemical bonds and quantum mechanics together determine the energy levels that any electron, atom, or molecule can be in.
- The molecular energy is a sum of the energy related to the position of the electron relative to the stable ground electronic state, the molecular vibration, and the molecular rotation.
- Absorption occurs when the energy of the photon matches the difference between two energy levels in a molecule, Δ
*E*=*E*–_{final}*E*=_{initial}*hc*/*λ*. - Rules set by conservation of angular momentum and electron spin determine which transitions are allowed.
- The amount of absorption, called simply the absorption cross section,
*σ*, comes from many factors, but varies significantly from molecule to molecule and from transition to transition. The cross section has dimensions of area and commonly has units of cm^{2}. - Electronic transitions occur when the electrons actually jump into other orbits around the nuclei. They have energies that are equivalent to radiation (i.e., photons) in the ultraviolet to visible wavelengths.
- Vibrational transitions occur when the molecule vibrates at a different frequency or in a different way. Diatomic molecules (e.g., O
_{2}and N_{2}) have only one way to vibrate—back and forth along the chemical bond that binds them. But more complicated molecules (e.g., H_{2}O and CO_{2}) can vibrate not only with the nuclei going toward and away from each other, but also by bending in three directions. These vibrational transitions, accompanied with motions that combine vibration and rotation, have energies equivalent to the near and middle infrared radiation (i.e., photons). - Rotational transitions occur when a molecule changes its rate of rotation. These transitions have energies equivalent to radiation (i.e., photons) in the far infrared to radio wave wavelengths.
- Translational energies of molecules in Earth's atmosphere, ~
*kT*, are generally a little larger than the energy required to move from one rotational level to another, 10–100 times less than the energy required to go from one vibrational level to another, and hundreds to thousands of times less than what is required to go from one electronic level to another. - We have already seen that some molecules and atoms have more kinetic energy than others. All molecules with kinetic energy larger than the energy difference between rotational levels can collide with the molecule and give it enough rotational energy to change to a higher rotational level. Thus, we see that rotational energy is distributed over many rotational levels, but that vibrational and electronic levels are usually the ground (lowest energy) levels.
- Atoms do not have molecule-like vibrations and rotations because they have only one nucleus, so their spectra consist only of electronic transitions.
- A sharp transition from one discrete level to another, which appears as a line in a spectrum, occurs in a narrow range of energies about the transition energy difference. The resulting radiation occurs in a narrow band of wavelengths about the line's central wavelength. The width of this line (measured at half the line's maximum height) is called the linewidth.
- This natural linewidth can be broadened by the molecule’s motion, called
**Doppler broadening**, or by collisions, called**pressure broadening**. - High in the atmosphere, Doppler broadening is dominant because the pressure is low, but lower in the atmosphere, pressure broadening becomes dominant even though Doppler broadening also increases. So the absorption lines are broader near Earth’s surface than they are higher up in the atmosphere.

The absorption cross section, *σ*, varies significantly over the width of the absorption line. So it is possible for all the radiation to be absorbed in the middle of the line but very little absorbed in the “wings.”

Credit: UC Davis chemwiki [10]

Atoms and molecules can absorb radiation (a photon) only if their structure has an energy difference between levels that matches the photon’s energy (*hc*/*λ*). Otherwise, the atom or molecule will not absorb the light. Once the molecule has absorbed the photon, it can either lose a photon and go back to its original lower energy level; or it can break apart if the photon energy is greater than the chemical bond holding the molecule together; or it can collide with other molecules, such as N_{2} or O_{2}, and transfer energy to them while it goes back to its lower energy level. Collisions happen often, so the energy of the absorbed photon is often transferred to thermal energy.

Credit: Robert A Rohde, Global Warming Art, via Wikimedia Commons

Note that Earth's upgoing infrared irradiance is limited to a few atmospheric "windows" and the irradiance at all other wavelengths is strongly absorbed, mostly by water vapor, but also by carbon dioxide, ozone, nitrous oxide, methane, and other more trace gases that aren't shown in the figure above.

Gases, liquids, and solids can all absorb radiation. Since we are most interested in gases, we will use a gas to develop the equation that we need for absorption.

Suppose we have a volume of uniformly distributed gas that absorbs. Let's suppose that there is a beam of radiation that passes into this medium. Let's narrow this beam down so that all the radiation is traveling in essentially the same direction, so now we are interested in the irradiance that is traveling in a narrow range of angles, which is called the radiance, which is denoted by *I* and has units of W m^{–2} steradian^{–1} nm^{–1}. Refer back to Lesson 6.5 [11] to refresh your memory about what steradians are.

In an infinitesimal slab of that medium, a certain fraction *dI* of the radiation with radiance *I* (W m^{–2}) is absorbed. *dI* is equal to the absorption cross section, *σ* (m^{2}), of a single molecule or particle multiplied by the number density, *n* (# m^{–3}) of absorbers, the length of the light-path through the slab, *ds* (m), and the amount of radiance *I* itself :

$$dI\left(\lambda \right)\text{}=\text{}-I\left(\lambda \right)\sigma \left(\lambda \right)n\text{}ds$$

Note that both *I* and *σ* are functions of wavelength. If there is a beam of radiation with a certain wavelength, then the chance of absorbing the radiation depends on the cross-sectional area of each molecule or particle, which quantifies the efficiency with which the molecule absorbs the radiation with that particular wavelength. The absorption cross sections are largest if the energy of the radiation's wavelength matches the transition energy of an electron, atom, or molecule.

Integrating this expression over the total thickness of the volume, we get the **Beer–Lambert Law**:

$$I\left(\lambda \right)={I}_{o}\left(\lambda \right)\mathrm{exp}\left[-\sigma \left(\lambda \right)ns\right]$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacqGH9aqpcaWGjbWdamaaBaaaleaapeGaam4BaaWdaeqaaOWdbmaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaciGGLbGaaiiEaiaacchadaWadaqaaiabgkHiTiabeo8aZnaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaacaWGUbGaam4CaaGaay5waiaaw2faaaaa@4DF0@

where *I _{o}* is the radiance at the front edge of the path and

When *σns = 1, I(λ)/I _{o}(λ) = exp(-1) = 0.368*.

For a uniform gas, the optical depth (a.k.a., optical thickness) equals *σns*. We often use *τ* (tau) as the symbol for the optical depth. If the optical depth gets large, then very little radiation gets through.

Credit: Wikipedia [12]

Ozone has an absorption cross section of 1.0 x 10^{–19} cm^{2} at a wavelength of 310 nm. If the average ozone concentration in the ozone layer is 2 x 10^{12} molecules cm^{–3} and the thickness of the ozone layer is 25 km, what is the fraction of sunlight that gets through when the sun is directly overhead?

ANSWER:

$$\frac{I\left(\lambda \right)}{{I}_{o}\left(\lambda \right)}=\mathrm{exp}\left(\mathrm{\u20131}\times {10}^{-19}{\text{cm}}^{2}\cdot 2\times {10}^{12}{\text{cm}}^{-3}\cdot 25\times {10}^{5}\text{cm}\right)=0.6$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6A4A@

At 310 nm, 60% of the solar UV gets through to Earth’s surface. Yet at 290 nm, where *σ* = 1.09 x 10^{–18} cm^{2}, only 0.4% of the solar UV gets through.

Once we know the absorption coefficients of all of the absorbing molecules between the sun and the air volume of interest and the number densities of the absorbers, we can determine how much radiation will be absorbed at every wavelength.

The angle of the Sun with respect to the zenith (directly overhead) is called the solar zenith angle (SZA) and is 0^{o} overhead and 90^{o} on the horizon. The SZA affects the total path through absorbers, and thus is important. We often make an assumption that the atmosphere is a parallel plane (see figure below).

The result is that we can write $ds=\frac{1}{\mathrm{cos}\left(SZA\right)}dz=\mathrm{sec}\left(SZA\right)dz$ in the differential form of Beer's Law:

$$\frac{dI\left(\lambda \right)}{I\left(\lambda \right)}=-\sigma \left(\lambda \right)n\text{sec}\left(\text{SZA}\right)dz$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeWaaSaaa8aabaWdbiaadsgacaWGjbWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaaWdaeaapeGaamysamaabmaapaqaa8qacqaH7oaBaiaawIcacaGLPaaaaaGaeyypa0JaeyOeI0Iaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gacaqGGaGaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaaiaadsgacaWG6baaaa@5168@

which gives:

$$I\left(\lambda ,z\right)={I}_{0}\left(\lambda \right)\mathrm{exp}\left[-\text{sec}\left(\text{SZA}\right){{\displaystyle \int}}^{\text{}}\sigma \left(\lambda \right)n\left(z\right)dz\right]$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaamysamaabmaapaqaa8qacqaH7oaBcaGGSaGaamOEaaGaayjkaiaawMcaaiabg2da9iaadMeapaWaaSbaaSqaa8qacaaIWaaapaqabaGcpeWaaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiGacwgacaGG4bGaaiiCamaadmaabaGaeyOeI0Iaae4CaiaabwgacaqGJbWaaeWaa8aabaWdbiaabofacaqGAbGaaeyqaaGaayjkaiaawMcaa8aadaqfGaqabSqabeaacaaMb8oaneaapeGaey4kIipaaOGaeq4Wdm3aaeWaa8aabaWdbiabeU7aSbGaayjkaiaawMcaaiaad6gadaqadaWdaeaapeGaamOEaaGaayjkaiaawMcaaiaadsgacaWG6baacaGLBbGaayzxaaaaaa@5DB3@

Often, the integral is called the optical thickness, or optical path, where *s _{1}* is one point along the path and

$$\tau ({s}_{1},{s}_{2})\equiv {\displaystyle \underset{{s}_{1}}{\overset{{s}_{2}}{\int}}\sigma (s)\cdot n(s)\cdot ds}$$

For matter that absorbs and does not scatter, the absorptivity, *a = 1 – t*. Note that in this case the emissivity

Usually, but not always, *σ* is a function of altitude because it often is a function of pressure and temperature, which vary for different altitudes.

Here is a video (2:03) further explaining Beer's Law:

Absorption and scattering apply to infrared radiation, just as they do to UV, visible, and microwave radiation. However, all matter that is at Earth-like temperatures is radiating in the infrared but isn’t hot enough to radiate significantly in the visible.

Remember that the peak in the Planck distribution function for a body at *T* = 300 K is given by Wien's Law:

$${\lambda}_{\mathrm{max}}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{T}=\frac{2898\text{\hspace{0.17em}}\mu \text{m}\text{\hspace{0.17em}}\text{K}}{300\text{K}}=10\text{\hspace{0.17em}}\mu \text{m}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpi0dc9GqpWqaaeaabiGaciaacaqabeaadaabauaaaOqaaabbaaaaaaaaIXwyJTgapeGaeq4UdW2damaaBaaaleaapeGaciyBaiaacggacaGG4baapaqabaGcpeGaeyypa0ZaaSaaa8aabaWdbiaaikdacaaI4aGaaGyoaiaaiIdacaaMc8UaeqiVd0MaaeyBaiaaykW7caqGGaGaae4saaWdaeaapeGaamivaaaacqGH9aqpdaWcaaWdaeaapeGaaGOmaiaaiIdacaaI5aGaaGioaiaaykW7cqaH8oqBcaqGTbGaaGPaVlaabccacaqGlbaapaqaa8qacaaIZaGaaGimaiaaicdacaqGGaGaae4saaaacqGH9aqpcaaIXaGaaGimaiaaykW7cqaH8oqBcaqGTbaaaa@5D95@

To examine how extinction and emission work together, consider the case of an absorbing atmosphere with no scattering at a fixed wavelength. Assume that there is an absorbing gas between us and some radiation source which has a radiance *I*. Then *I* will be reduced as it travels through the absorbing gas:

$$\begin{array}{l}d{I}_{absorption}\text{=}-\sigma \text{\hspace{0.17em}}n\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds=-{\kappa}_{a}\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds\hfill \\ {\kappa}_{a}\equiv \text{absorption}\text{\hspace{0.17em}}\text{coefficient,}\text{\hspace{1em}}{\text{SIunits:m}}^{-\text{1}}\text{\hspace{0.17em}}\hfill \end{array}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@84E7@

But, at the same time, the layer with width *ds* is emitting:

$$d{I}_{emission}\text{=}{\kappa}_{a}\text{\hspace{0.17em}}{P}_{e}\text{\hspace{0.17em}}ds$$

Normally, we would multiply *P _{e}* by the emissivity for that wavelength. But because the absorptivity = emissivity for any given wavelength and type of matter by Kirchhoff’s Law, we can replace the emissivity, ε, with the absorptivity,

$$\begin{array}{l}dI=d{I}_{absorption}+d{I}_{emission}=-{\kappa}_{a}\text{\hspace{0.17em}}I\text{\hspace{0.17em}}ds+{\kappa}_{a}\text{\hspace{0.17em}}{P}_{e}\text{\hspace{0.17em}}ds={\kappa}_{a}\text{\hspace{0.17em}}\left({P}_{e}-I\right)\text{\hspace{0.17em}}ds\hfill \\ \begin{array}{l}\text{or}\\ \frac{dI}{ds}={\kappa}_{a}\text{\hspace{0.17em}}\left({P}_{e}-I\right)\text{\hspace{0.17em}}\end{array}\hfill \end{array}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@8546@

This equation is called **Schwarzschild’s Equation**. Please watch the video below (2:29).

Credit: NOAA [13]

At every infrared wavelength, the change in the irradiance is due to the absorption of irradiance coming from some source, *F*, and the emission of irradiance by the absorbing gas itself. If the source is relatively strong (i.e., it produces a spectral irradiance greater than the Planck distribution function spectral irradiance from the gas), then the change in *F* will decrease mainly according to the Beer–Lambert Law. However, if the source is relatively weak, then the irradiance that gets to us will be due mostly to the gas itself.

Let’s look at this problem another way. Suppose we have an infrared sensor and are pointing it at the Sun. The Sun radiates some energy in the infrared, but at some wavelengths the water vapor and carbon dioxide are also radiating and the sensor may be picking up their irradiance. So what is the infrared sensor seeing?

Break the atmosphere into very thin layers, so thin that the temperature and amount of water vapor and carbon dioxide is almost constant over the thickness of each layer. Start with the layer closest to the sensor. It radiates both down to the sensor but also up to the layers above it. Since this first layer is just above us, the sensor measures the infrared radiation from it.

The next layer may be at a different temperature, but it too is radiating both down and up. Some of the irradiance in this second layer might get absorbed in the first layer depending upon what the first layer’s absorptivity is. If this absorptivity is a lot, then very little of the infrared radiation from the second layer will get through the first layer, according to the Beer–Lambert Law, but at the same time, the emissivity of the first layer will be great, so the irradiance of the first layer will be close to that of the Planck distribution function spectral irradiance at the layer’s temperature. As a result, the sensor will see mostly the first layer and not the second. If, on the other hand, the first layer has low absorptivity and thus low emissivity, then irradiance from the first layer to the sensor will be weak and irradiance from the second layer will get to the sensor along with irradiance from the first layer.

If we add a third layer, it too will be radiating the same amount of energy both down and up. Some or all of its irradiance might be absorbed in the first and second layers, so depending on the absorptivity of the first two layers, the sensor may or may not see any irradiance from the third layer.

You can see how we can build a model of atmospheric radiation by stacking individual layers with distinct amounts of water vapor and carbon dioxide on top of each other, one at a time, and as we add each one, use the Beer–Lambert Law to determine how much radiation from one layer is absorbed by the layers below or above it.

- Find
**Practice Quiz 6-3**in Canvas. You may complete this practice quiz as many times as you want. It is not graded, but it allows you to check your level of preparedness before taking the graded quiz. - When you feel you are ready, take
**Quiz 6-3**. You will be allowed to take this quiz only**once**. Good luck!

Besides being absorbed or transmitted, radiation can be scattered. The scattering can be by particles of all sizes and by molecules. We can talk about the extinction of radiation by a particle that both absorbs and scatters.

$$\text{extinction=absorption+scattering,or}{\sigma}_{e,\lambda}={\sigma}_{a,\lambda}+{\sigma}_{s,\lambda}$$[5]@5@5@+=faaagCart1ev2aaaKnaaaaWenf2ys9wBH5garuavP1wzZbItLDhis9wBH5garmWu51MyVXgaruWqVvNCPvMCaerbdfwBIjxAHbqee0evGueE0jxyaibaieYlf9irVeeu0dXdh9vqqj=hHeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@6EA4@

When we have absorption and scattering by particles or molecules, we simply replace the absorption cross section with the extinction cross section in the equations above.

Scattering is messy!!!! While absorption simply removes photons, or energy, from a beam of radiation, scattering can redirect the beam of radiation into other directions, where it can be absorbed or scattered some more. Furthermore, the intensity of the scattered radiation in different directions depends strongly on the radiation's wavelength and on the size, shape, and composition of the “particle,” where we can consider a molecule to be a particle.

A very important parameter is called the **size parameter**:

$$x\equiv \frac{2\pi r}{\lambda}$$

where *r* is the radius of a spherical particle and *λ* is the radiation’s wavelength.

Credit: W. Brune (after Grant Petty)

The effectiveness of scattering depends on the size parameter, which separates different types of scattering. For *x* > 2000, we have** geometric optics**. Geometric optics is when the particles get so big relative to the wavelength that we can calculate the interaction of radiation with them using optical principles (think of a lens or prism). For *x* = 0.2 to 2000, we have **Mie scattering**. Mie scattering is when the particle size is comparable to the size of the radiation wavelength, such as many atmospheric particles and light, from UV to microwave. For *x* between 0.002 and 0.2, we have **Rayleigh scattering**.

Rayleigh scattering occurs for particles that are smaller than the radiation’s wavelength. This is the regime for molecules for the UV into the visible and for drizzle and raindrops for radar. Let’s look at its properties.

Credit: W. Brune (after Born and Wolf)

Without going through the math or the reasoning, I will state only that the scattering cross section for a molecule, particle, or raindrop in the Rayleigh regime (x << 1) is proportional to the sixth power of the particle radius and fourth power of the radiation wavelength.

$${\sigma}_{s}\propto \frac{{r}^{6}}{{\lambda}^{4}}$$

In the Rayleigh regime, this scattering is equally strong in the backward and forward directions, with slightly less scattering in other directions (see figure above).

This strong wavelength dependence of scattering by small particles includes molecules at the wavelengths of visible radiation. This wavelength dependence of scattering leads to the preferential scattering of blue solar radiation compared to red solar radiation, which contributes to the blue of the sky in directions away from the sun and the deep orange-to-red colored sun as it sets. The image below shows both effects in one picture—a yellow-orange sunset below the cloud and a bright blue sky above the cloud.

For radar wavelengths, the r^{6} dependence of scattering by cloud and rain drops leads to much stronger back scattering of the radar beam with increasing drop size, thus allowing for estimates of raindrop sizes and thus rainfall rates.

Credit: W. Brune

See below for a video (2:09) explaining scattering in greater depth:

When you feel you are ready, take **Quiz 6-4**. You will be allowed to take this quiz only **once**. Good luck!

This lesson provided you with a basic understanding of atmospheric radiation. You should now be familiar with the following basic concepts:

- when radiation encounters matter, it can only be transmitted, absorbed, or scattered
- solar variability
- all objects are emitting radiation all the time
- emissivity = absorptivity at the same wavelength
- the molecular cause of absorption of different wavelengths of radiation by objects of different chemical composition
- Beer’s Law, which quantifies how much radiation will be absorbed as atmospheric radiation travels through an absorbing medium
- Schwarzschild’s equation, which quantifies the change in radiant energy per distance traveled through an emitting medium due to emission by layers and absorption of those emissions by layers above and below it, which are also emitting
- the dependence of the behavior of scattering on the wavelength of the radiation and the size of the scattering object

We will use many of these concepts in the next lesson on Applications of Atmospheric Radiation, in which we will consider two cases important to the atmosphere and weather: Earth’s radiant energy balance and Earth observed in the infrared by satellites in space.

You have reached the end of Lesson 6! Double-check that you have completed all of the activities before you begin Lesson 7.

**Links**

[1] http://www.weather.gov/cle/Seasons

[2] http://www.nature.nps.gov/views/KCs/Milankovitch/Milankovitch.htm#

[3] http://www.co2crc.com.au

[4] http://www.azimuthproject.org/azimuth/show/HomePage

[5] mailto:MathType@MTEF

[6] https://www.flickr.com/photos/gsfc/6244140977/in/photolist-avLRz4-8KEdvU-ojzHqP-4Z8oZE-q6v5Rf-crKhqo-aEXWPZ-5s4nN4-qknztW-4Z8oZC-jyRs-nKjSHJ-tFhDP-d5UHAf-owwsV3-jfYzAS-6eL1Hz-c2CzVb-5PE822-nVqp6G-cm3MMb-biTqPV-qMhtqo-d5UJiq-8tgtsy-8YV188-dPqRnn-8n3GQi-d6S7eU-8n6Pub-6fuYxZ-7v8ckb-6TpvWX-p2tH3K-qH4Btm-4REMVG-edpJDh-9gMRuB-d85XtJ-c2vJ5f-9QeRrF-4owLKU-qie4jv-dcr6LN-kkLDUD-9q5WqG-7LtFqE-qEyb2w-fq2Gw4-bJQKHX

[7] https://creativecommons.org/licenses/by-sa/3.0/deed.en

[8] http://activerain.trulia.com/blogsview/3694424/detecting-moisture-within-the-clouds-of-ogden-utah-using-infrared

[9] https://www.flickr.com/photos/wonderferret/290014648/in/photolist-rCpfy-6sMdtf-9NfGPi-oT1WAq-oT1Rm7-r2M9Dt-2KNgcD-dd7CNi-ekUVgh-bpVG8i-dzhZTs-pXPG3t-8JXTvh-ap1y1-oE2Vqt-8YmzB3-4YtvoL-4JDFzx-c3kak7-rVamEr-oE2Vrv-4YtpRm-4YtwiE-dPjhU6-5ZUyE-4DCEt6-4DCEta-6MF6VH-6MKiaA-6MKgXh-6MF7pe-6MF6vv-6MKgQG-6MKmFG-6MKmAq-6MF65z-6MF6r6-4Ypdex-4Ypg5v-9h8EbF-4YtsB5-c3kaAW-c3kaHo-c3kas3-4YoLSk-4YtqWs-4Ytyxj-4YpeaH-4YtrYW-4YpfrX

[10] http://chemwiki.ucdavis.edu/Theoretical_Chemistry/Symmetry/Group_Theory%3A_Application

[11] https://www.e-education.psu.edu/meteo300/node/684

[12] http://en.wikipedia.org/wiki/Beer%E2%80%93Lambert_law

[13] http://www.weather.gov/satellite#wv