Blackbody emission and absorption

In summary, the radiation from a material is broadened by the vibration of the atoms in the wall of the container.
  • #1
Hypatio
151
1
I am trying to understand radiative transport of thermal energy in materials from first (or close to first) principles.

I do not understand the systematics/statistics of how photons are emitted and absorbed in a medium. How is it that photons can be emitted at so many wavelengths and atoms can absorb photons at these different wavelengths (http://panda.unm.edu/Courses/Finley/P262/ThermalRad/PlanckRadHotT.gif ) when we know from spectroscopy that photons are emitted and absorbed only at very specific wavelengths (http://www.astro.cornell.edu/share/sharvari/websiteV7/images/absorption.gif ).

Is it the case that this emission and absorption changes with kinetic energy (thermal energy) of each individual atom? Does an absorbing atom have to have exactly the same thermal energy as the emitter? If there are so many quantized energy levels, how is it not made statistically impossible for a randomly emitted photon to encounter an atom at a temperature necessary to absorb it?

As you can see my understanding is muddled. Let me know what you think.
 
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  • #2
Spectroscopy is usually done with gases, where you see energy levels of electrons in individual atoms or molecules. In other phases, you can have more energy levels and more possible transitions and methods to absorb/emit photons. In solid objects, the energy levels can be so dense that a broad range of photons can get absorbed (or reflected).

Does an absorbing atom have to have exactly the same thermal energy as the emitter?
No.
 
  • #3
Hypatio said:
I am trying to understand radiative transport of thermal energy in materials from first (or close to first) principles.

I do not understand the systematics/statistics of how photons are emitted and absorbed in a medium. How is it that photons can be emitted at so many wavelengths and atoms can absorb photons at these different wavelengths (http://panda.unm.edu/Courses/Finley/P262/ThermalRad/PlanckRadHotT.gif ) when we know from spectroscopy that photons are emitted and absorbed only at very specific wavelengths (http://www.astro.cornell.edu/share/sharvari/websiteV7/images/absorption.gif ).

Is it the case that this emission and absorption changes with kinetic energy (thermal energy) of each individual atom? Does an absorbing atom have to have exactly the same thermal energy as the emitter? If there are so many quantized energy levels, how is it not made statistically impossible for a randomly emitted photon to encounter an atom at a temperature necessary to absorb it?

As you can see my understanding is muddled. Let me know what you think.
Absorption and remission of radiation does not have to be at specific frequencies. Only the bound states of the electrons have quantized energies relative to the center of mass of the atom. However, the atoms are allowed to vibrate.

You are dismissing the contribution of the atoms in the wall of the container. The basic assumption is that the atoms are vibrating with a quantized amplitude. This was first calculated by Planck. According to Planck, the main mechanism of broadening is the vibration of the atoms in the walls of the container.


In order for the black body cavity to be at equilibrium, the walls of the cavity have to be at the same temperature as the photons inside the cavity. This means that the atoms in the walls of the container have to be vibrating with the probability distribution of the amplitudes are determined by the Planck distribution of energy.

Suppose that the electrons in the atoms did not absorb any energy from the electromagnetic field. Suppose the light always bounced elastically off the atoms. The reflected photons would have a shift in energy due to the Doppler effect. The size of the Doppler effect would be influenced by the Planck distribution of vibration energies in the atoms that comprise the walls of the container. So even if there was no absorption per se, the Doppler effect would broaden the energy distribution of the photons.

Suppose an atom in the ground state does absorb a photon. The atom is connected to other atoms in the walls of the container, so that it can't go very far. It can only vibrate.

The energy would be distributed among different vibration states of the electron. When photons are remitted, the energy range of the photons will be broadened by the vibration states of the atom.

If there is more than one bound state of electrons in the atom, the absorbed energy would be distributed among these bound states. Multiple photons could be emitted, broadening the photons.

Note it also works the other way around. Suppose the walls of the container were not vibrating, or vibrating coherently with one amplitude. The reflection and absorption of photons which have many frequencies would broaden the amplitude distribution of the atoms in the wall.

So photons and phonons broaden each other. The amplitude distribution of the vibrating atoms is broadened by the collision of photons. The energy distribution of photons is broadened by collisions with vibrating atoms. This is how the container and the photon gas come into thermal equilibrium with each other.

If all else fails, Doppler shift !-)
 
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FAQ: Blackbody emission and absorption

1. What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by an ideal object that absorbs all radiation incident upon it. It is considered to be a perfect emitter and absorber of radiation.

2. What is the relationship between temperature and blackbody radiation?

The intensity and wavelength distribution of blackbody radiation emitted by an object is determined by its temperature. As the temperature increases, the intensity of radiation also increases and shifts towards shorter wavelengths.

3. How does the Stefan-Boltzmann law relate to blackbody radiation?

The Stefan-Boltzmann law states that the total energy radiated per unit surface area of a blackbody is proportional to the fourth power of its absolute temperature. This means that as the temperature of a blackbody increases, its total energy emission also increases significantly.

4. What is the greenhouse effect and how does it relate to blackbody radiation?

The greenhouse effect is the trapping of heat in the Earth's atmosphere by certain gases, such as carbon dioxide and water vapor. These gases absorb and re-emit infrared radiation, which is a form of blackbody radiation. This process helps to keep the Earth at a habitable temperature for living organisms.

5. How is blackbody radiation used in everyday life?

Blackbody radiation is used in many everyday technologies, such as incandescent light bulbs, infrared heaters, and thermal imaging cameras. It is also used in scientific research and industries that involve high temperatures, such as metallurgy and material processing.

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