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This might be a stupid question but I could not find the answer in my textbook or on the internet with a few searches.

So I believe when an atomic electron moves down to a lower energy level it emits radiation in the process. However since the energy levels are discrete, the photons released have specific energies and hence wavelength which results in the line spectra.

However apparently this is only true for hot gases and not liquids or solids, which have continous emission spectrum. Why is this?

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In liquids and solids the difference in energy between energy levels becomes very small, due to the electron clouds of several atoms bein in very close proximity of one another. These similar energy levels will form 'bands' of indistinguishable spectral lines.

In gases however, atoms will be spaced loosely enough such that the interaction between atoms will be minimal. This allows the energy levels to have sufficient difference in energy for distinct lines to be formed.

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  • $\begingroup$ Ok that makes sense. So in liquids and solids, do the energy levels of several atoms of different elements merge? Does this mean electrons are free to move from the energy levels of one atom to the energy levels of another atom? $\endgroup$ – IK-_-IK Jan 28 '18 at 2:21
  • $\begingroup$ Does that mean that you can't really see spectral lines in rainbow created by prism? New Cosmos series with Tyson suggested they should be visible $\endgroup$ – Lope Jan 28 '18 at 5:57
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    $\begingroup$ Rainbows have little to do with spectral lines. They're caused by diffraction. $\endgroup$ – Simon Jan 28 '18 at 8:52
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    $\begingroup$ @Lope. NDGT showed the analysis of the light via the prism. The prism is not the source. If the light you dispersed had lines inside the spectrum the lines will show up and made visible by magnifying the attained rainbow. Is not related to the Q. At least not at prisma level. $\endgroup$ – Alchimista Jan 28 '18 at 12:13
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    $\begingroup$ @Gregory25 Rainbows have a lot to do with spectral lines. A traditional rainbow is a "spread" spectrum of white light from the sun (and the sun has a continuous spectrum in that range, basically, so you don't see lines). If the source of light being put through a prism has spectral lines in the visible spectrum, it will appear when you make a "rainbow" of it with a prism. Or is that not what you mean? $\endgroup$ – Yakk Jan 29 '18 at 16:52
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You see line spectra usually only in gases because there the interaction between the atoms can be neglected. In gases with high pressures you get the so-called collision broadening of the lines which eventually become bands. Similarly, in liquids and solid the atoms are so close that the interaction between them leads to the discrete spectral lines becoming bands.

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It's a good question, it's not stupid. Actually, this phenomena can be observed with liquids and solids as well. Each element has its own distinct spectral line and this fact can and has been used to identify an element. However, it is much more difficult to observe the spectral lines of liquids and solids due to how close together the atoms are. Also, tables of the spectral lines of elements only seem to go up to the 99th element, Einsteinium (not including Astatine (At, 85) and Francium (Fr, 87).

I could not find any data as to why this may be, however, I believe it is simply because we could not test for the spectral lines of the heavier elements due to their instability and scarcity. It's incredible because some of the heavier and more unstable elements have insanely short half lives ranging from 100.5 days (the most stable isotope of fermium (Fm, 100)), to 0.69 microseconds (0.00069 milliseconds) (Oganesson (Og, 118)). This would make measuring their spectral lines nearly impossible. This isn't even considering how much this would cost. These heavier elements likely have their own spectral lines, however, due to all I stated above, it isn't exactly possible to measure.

I hope this helped,

You can see a list of of all the known spectral lines of elements on Wikipedia since it seems to have the most updated table. mostly all textbooks on the spectral lines of elements tend to only go up to uranium, however, textbooks go into much more detail for each element.

https://en.wikipedia.org/wiki/Spectral_line

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    $\begingroup$ Just because an element is liquid or solid at STP and you find a atomic spectrum for it does not mean that the spectrum was taken while the substance was in solid or liquid state. I don't know for certain that it doesn't happen, but I do know that often a portion of a sample is converted into a rarified phase phase for spectrography. $\endgroup$ – dmckee --- ex-moderator kitten Jan 28 '18 at 1:22
  • $\begingroup$ If that were true? how in world did they find this data for carbon? with a boiling point at 5100 K (that's possible, but still extremely hot) and carbon would have burned and formed a bond with another atom (like oxygen or hydrogen) way before boiling. Plus, Do you really think they boiled Rhenium which has a boiling point of 5,870 K (pretty darn hard to boil) and one of the a rarest elements on earth. $\endgroup$ – Clandestinity Jan 28 '18 at 3:39
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    $\begingroup$ In modern machines they typically vaporize a bit of the sample with a pulsed laser. That's commercial-off-the-shelf technology. $\endgroup$ – dmckee --- ex-moderator kitten Jan 28 '18 at 3:43
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    $\begingroup$ I understand, I guess that way, they're able to vaporize carbon in a vacuum using the laser, furthermore preventing it from forming a bond. My apologies, I should have done more research. thank you dmckee for clarifying this. @IK-_-IK, do not rely on my statement, for it is false. $\endgroup$ – Clandestinity Jan 28 '18 at 3:52
  • $\begingroup$ @Anthony You imply here that the spectral lines for each element are only known through measurement. That's a little surprising to me, does our model for atomic structure not allow us to reliably predict the emission spectra? $\endgroup$ – Samuel Jan 29 '18 at 18:05
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However apparently this is only true for hot gases and not liquids or solids

Rather than the phase of the material, you should be evaluating the optical thickness.

If the material is roughly transparent (like a thin gas), then the discrete transition radiation can be directly received and you see the line spectrum.

If the material is roughly opaque, then the discrete transition radiation is more likely to interact with the material. This interaction thermalizes the radiation and generates the continuous spectrum.

http://www.physics.usyd.edu.au/~helenj/SeniorAstro/lecture04.pdf

Note that the important point here is that the radiation is interacting with the material, not whether the material has any interaction with itself.

The base of photosphere (where the continuous spectrum of the sun comes from), is not particularly dense at around $3\times 10^{-4}\text{kg/m}^3$. But the total amount of material is sufficient to block radiation produced deeper.

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  • $\begingroup$ What about glass or monomolecular transparent crystals that can be seen through? Like Mica or a soap bubble. I think they will be thin enough. A nitrogen laser is said to be self limiting in length (about 1m) because it will absorb the lasing light more than generating it if the laser pulse intensity is amplified beyond a certain threshold. I suppose it will loose its lines if path is longer than 1m as well. $\endgroup$ – KalleMP Jan 28 '18 at 20:40
  • $\begingroup$ Just google line spectra in solids and you'll see lots of examples. Here is a nice one: researchgate.net/publication/… $\endgroup$ – JohnS Apr 14 '18 at 18:30
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    $\begingroup$ @JohnScales, Your example shows IR/Visible transmission spectra and x-ray diffraction graphs. Neither are emission spectra. $\endgroup$ – BowlOfRed Apr 14 '18 at 18:39
  • $\begingroup$ That's not the question is it? $\endgroup$ – JohnS Apr 14 '18 at 18:46
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    $\begingroup$ @JohnScales, since the OP is asking about radiation when electrons move to lower energy levels, I would say yes it is. $\endgroup$ – BowlOfRed Apr 14 '18 at 18:49

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