Thermal radiation and emission spectra

In summary: The red coated piece would emit most of its radiation in the infrared spectrum because the red pigment would be excited and give off radiation. The black coated piece would only emit radiation in the mid-infrared spectrum because the black pigment would be the only molecule that would be excited and give off radiation at that frequency. Individual molecules have very low levels of emissivity or absorptivity.
  • #1
klawlor419
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I have some questions regarding thermal radiation spectra of different objects. Is thermal radiation of objects due to electromagnetic radiation alone? Does my desk which is at room temperature emit as a blackbody with the proper factor for emissivity? Where is the "cavity" that the radiation is trapped inside for my desk? Is it the intermolecular spacing of the atoms in the desk? I have been thinking about this and getting confused or missing concepts. I think the theory has kind of clouded my understanding of the way that objects radiate, or just shown how little intuition I have of the way in which things thermally emit.

Also, this is probably a really dumb question but why is it that we consider a blackbody to a perfect absorber of electromagnetic radiation? Why should black be any different than for example red? It is technically a color in the "visible" spectrum

Thanks ahead of time for any help with my confusion
 
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  • #2
The exact thermal spectrum of an object is determined by molecular properties, like which functional groups it has. In the infrared spectrum specifically, many molecules have a distinct emission spectrum. Absorption and emission are essentially the same. "Absorptivity" means the fraction of the radiation falling on an object is absorbed vs. reflected. "Emissivity" means the fraction of radiation that is emitted vs. that which goes to heating up the object. A black body has an absorptivity and emissivity of 1. Radiation will fall on to some theoretical black body at all wavelengths, but will be emitted as a smooth spectrum (poisson distribution probably) centered on some frequency much lower than the light falling on it, which is probably white light with equal intensities at all frequencies. The "cavity" is just a thought experiment to help you understand what a black body is- it's just so that the radiation and the temperature of the black body can reach equilibrium before the radiation is re-emitted. Your desk might appear red because it absorbs all frequencies of light but is not capable of emitting very much blue or green light at its normal temperature. If you heated your desk to a few thousand degrees, it would probably emit light in blue and green because more of the light coming out of it is coming out as a black body spectrum (not including fire of course, it would be on fire too by the way). Black bodies appear black at normal ambient temperatures but are really putting out radiation at lower frequencies (infrared) and would become bright if you heated them up.
 
  • #3
colliflour said:
The exact thermal spectrum of an object is determined by molecular properties, like which functional groups it has. In the infrared spectrum specifically, many molecules have a distinct emission spectrum. Absorption and emission are essentially the same. "Absorptivity" means the fraction of the radiation falling on an object is absorbed vs. reflected. "Emissivity" means the fraction of radiation that is emitted vs. that which goes to heating up the object. A black body has an absorptivity and emissivity of 1. Radiation will fall on to some theoretical black body at all wavelengths, but will be emitted as a smooth spectrum (poisson distribution probably) centered on some frequency much lower than the light falling on it, which is probably white light with equal intensities at all frequencies. The "cavity" is just a thought experiment to help you understand what a black body is- it's just so that the radiation and the temperature of the black body can reach equilibrium before the radiation is re-emitted. Your desk might appear red because it absorbs all frequencies of light but is not capable of emitting very much blue or green light at its normal temperature. If you heated your desk to a few thousand degrees, it would probably emit light in blue and green because more of the light coming out of it is coming out as a black body spectrum (not including fire of course, it would be on fire too by the way). Black bodies appear black at normal ambient temperatures but are really putting out radiation at lower frequencies (infrared) and would become bright if you heated them up.

So are you saying that an objects emissions spectra are the collective effect of the way in which its microscopic constituents emit as well as the macroscopic emissivity or absorptivity of the material. Is that correct? Do individual molecules have emissivity or absorptivity? Or is the emissivity and absorptivity a property that is seen only in macroscopic matter, i.e. chunks of metal or my desk?

Lets say that we have two spherical pieces of metal that are exactly same size and composition. One is coated in red pigment and the other is coated in black pigment. Let's say that they are heated by a torch and brought up to a very high temperature well below their melting point. What are the differences in the way these would emit? Is the emissivity a function of frequency for pigments like this? Would there be a certain range of frequencies missing in the $$u_w=\text{spectral radiancy function}$$?

I can see that effecting the way in which the two spheres emit because it would decrease the total energy emitted by the red pigmented object. Still though it seems really non-intuitive to me that those spheres would emit differently. Maybe they don't? Thanks for the help
 
  • #4
The emission spectra is measured as a collective effect, but each molecule has a distinct spectra and has its own microscopic sort of absorptivity and emissivity. If you take a sample of this molecule and turn it into a gas and measure the spectrum, you might actually get different values than if it was looked at when it was a liquid or gas. Odd scattering effects can happen with thin layers of a substance, like oil on water, or with a microscopic texture on a substance that makes it reflective vs a flat looking color. These are all optics phenomena and happen because the temperature is low compared to the black body temperature required to produce visible light, the object's absorption and emission not being in equilibrium. I hadn't really thought about this before, but the microscopic geometry can affect emissivity and absorptivity, apparently pyramidal divots make an object have increased emissivity...

As to your thought experiment comparing spheres heated by a torch, i see what you're getting at, but a better example would be two different filament wires in an incandescent light bulb. Once the metals are heated to glow in the visible spectrum, yes, each type of metal will have a distinct and different emission spectrum and have different overall emissivity. Now suppose you took two of the same incandescent bulb with an ideal black body for a filament. The light coming from the filaments and going to the surface of the bulbs would be a smooth spectrum. If the coatings were thick enough, the red coating would be reflecting light from the room, not emitting visible light from the filament inside. The black coated bulb would be reflecting no visible light, but both would be radiating infrared because of the radiation absorbed from the filaments inside. Assuming you could put enough wattage into both bulbs, the red and black bulb would both glow red hot (or white hot) depending on how hot they got exactly. The peaks and troughs in the emission spectra, the function you talked about, would be missing some frequencies, but to the unaided human eye, they would look like all the frequencies were coming through. You'd need to look through a spectroscope to see dark bands in the frequency spectrum. Once you looked through the spectroscope, you'd see that the red and black pigments do in fact have different spectra. Which means, of course, emissivity as a function of frequency. To my knowledge, emissivity is always a function of frequency, but when someone says they have determined an emissivity for an object in general, it's an average over the part of the electromagnetic spectrum that's of interest. You would certainly have a different emissivity and absorptivity for radio waves vs visible light.
 
  • #5


Thermal radiation is the process by which objects emit electromagnetic radiation due to their temperature. This radiation is a result of the movement of charged particles within the object, such as atoms and molecules. This movement creates electromagnetic waves that are emitted from the object.

To answer your first question, thermal radiation is not solely due to electromagnetic radiation. It also includes other forms of radiation, such as infrared, visible light, and ultraviolet. These forms of radiation have different wavelengths and energies, and are emitted by objects at different temperatures.

Your desk, being at room temperature, will emit thermal radiation in the form of infrared and possibly some visible light. The amount and type of radiation emitted will depend on the composition and surface properties of the desk. The emissivity factor takes into account the efficiency of an object in emitting thermal radiation, and can vary depending on the material.

The "cavity" that the radiation is trapped inside for your desk is the space between the atoms and molecules of the desk itself. This space allows for the movement of charged particles and the creation of electromagnetic waves, which are then emitted as thermal radiation.

The concept of a blackbody as a perfect absorber of electromagnetic radiation is a theoretical construct used in thermodynamics and radiation physics. It is considered to be a perfect absorber because it absorbs all incoming radiation without reflecting or transmitting any of it. This allows for the blackbody to reach thermal equilibrium with its surroundings, making it a useful model for understanding thermal radiation.

The color of an object does not affect its ability to emit or absorb thermal radiation. The color of an object is determined by the wavelengths of light that it reflects, but it does not affect the emission or absorption of thermal radiation. Therefore, black is not inherently different from red in terms of thermal radiation.

I hope this helps to clarify some of your confusion about thermal radiation and emission spectra. It is a complex topic, but with some further study and understanding of the underlying principles, you will be able to grasp it better. Keep asking questions and seeking answers, as that is the essence of scientific inquiry.
 

1. What is thermal radiation and how does it differ from other types of radiation?

Thermal radiation is the type of electromagnetic radiation that is emitted by matter due to its temperature. It differs from other types of radiation, such as light and radio waves, because it is solely based on the temperature of the emitting object and does not require a medium to propagate.

2. How does the emission spectrum of an object relate to its temperature?

The emission spectrum of an object is directly related to its temperature. As the temperature of an object increases, the intensity of its emitted thermal radiation increases, and the peak wavelength of the emission spectrum shifts to shorter wavelengths.

3. What is the difference between a continuous and a discrete emission spectrum?

A continuous emission spectrum is produced by an object that emits radiation at all wavelengths, while a discrete emission spectrum is produced by an object that only emits radiation at specific wavelengths. A discrete emission spectrum is often referred to as a line spectrum.

4. How does the color of an object relate to its thermal radiation?

The color of an object is determined by the wavelengths of light that it reflects. As an object's temperature increases, the wavelengths of its emitted thermal radiation shift to shorter wavelengths, which can change the apparent color of the object. For example, a heated piece of metal may appear red hot due to the high intensity of shorter wavelength thermal radiation it is emitting.

5. Can thermal radiation be harmful to humans?

Thermal radiation can be harmful to humans if the intensity is high enough. For example, prolonged exposure to intense thermal radiation from the sun can cause sunburn and other skin damage. In addition, exposure to high levels of thermal radiation, such as in industrial settings, can cause burns and other serious injuries.

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