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Black body radiation problem


kos

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Does the absorption and emission spectrum for some element are the same .Do the molecules and atoms emit in the same spectrum they absorb em radiation?

If yes how we can explain black color since black materials obviously absorb all the incoming light and re emit it in some other wavelenght ?

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Does the absorption and emission spectrum for some element are the same .Do the molecules and atoms emit in the same spectrum they absorb em radiation?

If yes how we can explain black color since black materials obviously absorb all the incoming light and re emit it in some other wavelenght ?

 

 

Molecules and atoms are not blackbodies.

 

If you aren't in thermal equilibrium, then the absorption and emission spectra will not be the same for a blackbody — the peak emission wavelength depends on the temperature. IOW, if the incoming radiation is not from a source at the same temperature or not from a thermal source, then the emission spectrum won't match.

 

Atomic and molecular states and transitions are identical for absorption and emission, but there are ways for light to be absorbed but then have a nonradiative relaxation, or have multiple emission pathways, so the emission will occur at different wavelengths than the absorption. e.g. if you excite an atom to a high level, it can de-excite in multiple steps of lower-energy photons.

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Does the absorption and emission spectrum for some element are the same .

I think I read this question differently from swansont, since he seemed not to answer it the way I read it. I read the question as, "Do different elements sometimes have the same emission spectrum?' IDK.

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I think I read this question differently from swansont, since he seemed not to answer it the way I read it. I read the question as, "Do different elements sometimes have the same emission spectrum?' IDK.

 

 

You're right — my interpretation was comparing absorption and emission for the same element. I don't see a comparison of elements in the wording. But yes, there are undoubtedly some accidental matches - light from one element can be absorbed by another. Helped by the fact that transitions are not infinitely narrow on their own, and you have broadening mechanisms.

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Do the molecules and atoms emit in the same spectrum they absorb em radiation?

The spectrum emitted by any material due to its temperature only is exactly the same as its absorption spectrum.

 

This is consistent with the second law of thermodynamics.

 

If a body emits E% as much as the blackbody does (the blackbody is the most efficient emitter possible) then it also absorbs E% of the incoming radiation (the blackbody absorbs everything). Since light converts from and to light in this process, and since no net heat passes from one body to an other at identical temperatures, the ability to emit and absorb light is equal.

 

Because filters exist that let some wavelengths pass to the destination and reflect the other wavelengths to the source, the absorptivity and emissivity must match at all wavelengths.

 

Because filters can let only one polarization pass through, the absorptivity and emissivity must match at all polarizations.

 

This fertile idea can be extended. For instance, a surface coated with pyramids absorbs light better thanks to its multiple bounces. Such a surface emits better too. An antireflective coating too emits as efficiently as it absorbs at any wavelength.

 

One should use this idea with caution (...as always with the second law).

 

It holds for each wavelength. Absorbing much visible light does not imply emitting infrared easily. The values differ, and for instance satellites use coatings whose absorptivity (understand: for solar light, mainly visible and near infrared) differs from the emissivity (understand: for far infrared, around 10µm at 300K) to adjust the satellite's temperature.

 

It holds for one temperature. Both the absorbtivity and the emissivity vary with the temperature, slowly but a lot. The emissivity of tungsten at room temperature is no indication to its behaviour in a light bulb.

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