According to Planck's law, all matter emits radiation at all wavelengths but is this statement true for gases and pure elements? Gases like hydrogen and helium have specific emission spectrums and I am having difficulties understanding how they can emit radiation at all wavelengths. Why does Planck's law include all matter and not just blackbodies?
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$\begingroup$ see the plot here physics.stackexchange.com/questions/744489/… $\endgroup$– anna vCommented Jan 9, 2023 at 5:50
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$\begingroup$ You do not state in you profile your level of physics, and this is a duplicate question to the one you asked before. look at this plot. researchgate.net/figure/… . can you understand what it depicts? that wavelength is limited within the curve and is improbable outside it? $\endgroup$– anna vCommented Jan 9, 2023 at 5:59
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$\begingroup$ We aren't being deliberately unhelpful by closing your questions, it's just that we can't see what you're asking that's different from the previous questions. The previous questions explain that gases do not emit black body radiation because they aren't dense enough. Can you edit your question to make it clearer what exactly you're asking that isn't covered by a previous question? $\endgroup$– John RennieCommented Jan 9, 2023 at 6:56
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1 Answer
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Planck's law does not say "all matter emits radiation at all wavelengths". While generally true, that's not what Planck's law is about.
Planck's law is specifically about spectrum of EM radiation in thermodynamic equilibrium, or spectrum of beam of radiation coming from inside a region where radiation is in (approximately) thermodynamic equilibrium.
Real piece of radiatively uninsulated matter in vacuum never produces exactly radiation with such spectrum (blackbody radiation). This includes solid matter, liquids and gases. They all have frequency bands where they radiate less than blackbody at the same temperature, and in some cases, they may have bands where they radiate more than the blackbody.
It takes a good radiation insulation around a real emitting body to establish equilibrium radiation with Planck's spectrum in the enclosed region.
Atoms and molecules have sharp emissions lines or bands. Interaction between many such particles can broaden and shift the lines, and produce "blurred" spectrum where no lines or holes are easily detectable.
This happens mostly with liquids and solids, because particles are close to each other and interaction between them is strong.
Gases are usually different; interaction between the particles is weak, emission spectrum is usually made of sharp lines (but still with finite width) similar to those of single atom/molecules and between those lines there is little (but not zero) emission of radiation.
answered Jan 8, 2023 at 19:35
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1$\begingroup$ Does this mean that gases like single atoms/molecules can never emit radiations of all wavelengths considering the fact that they have sharp lines in their emission spectrum? $\endgroup$ Commented Jan 8, 2023 at 21:34
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$\begingroup$ No, because those sharp lines have finite thickness and thus intensity is non-zero typically everywhere. "Amount of radiation of definite wavelength" is not an objective property of radiation, it is a result of some variant of Fourier analysis which assumes some (arbitrary) base frequency, or Fourier integration. In general, we can expect non-zero intensity at every multiple of the base frequency, even if there are some sharp lines where the intensity is much higher. Zero intensity everywhere except at infinitely sharp lines would happen only for unrealistic radiation, like infinite sinusoid. $\endgroup$ Commented Jan 9, 2023 at 20:46