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I was reading about the development of the quantum theory when I got to the explanation for spectral lines. It's a topic that I've revisited many times but I came up with a question. I know that in some interactions with matter the amount of energy absorbed or transmitted in the form of photons is quantized but then what processes lead to the wide spectrum of frequencies that we observe in an object's electromagnetic spectrum? For example in a star what is going on in the nucleus that we get radiation in most frequencies?

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marked as duplicate by Thomas Fritsch, stafusa, Jon Custer, John Rennie, Kyle Kanos Aug 9 at 11:23

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  • $\begingroup$ What you are talking about is atomic emission, which is a purely quantum mechanical process. But there is also a "classical" way through which charged particles (e.g., electrons) emit light, this happens when the particle is accelerated. As you probably know, the matter in the sun is a plasma which is comprised of charged particles (ions and electrons, not bound to each other) and when these accelerate they can emit a photon, whose frequency depends on the acceleration, not on the spectra of atomic states. $\endgroup$ – S V Aug 8 at 18:56
  • $\begingroup$ @SV Sun's spectrum is close to a Planck spectrum, which doesn't reflect whether or not the particles are ionized, but simply reflect its temperature. In fact, although the interior of the Sun is ionized, its surface — which is what we see — is mostly neutral. $\endgroup$ – pela Aug 8 at 19:33
  • $\begingroup$ Is there reason to think that when free electrons are accelerated and emit light, that this light is quantized into discrete photons? $\endgroup$ – J Thomas Aug 8 at 20:54
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    $\begingroup$ @SV Don't thank me, it seems I'm wrong :) $\endgroup$ – pela Aug 9 at 8:26
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    $\begingroup$ @Pela Mostly neutral is correct. But if it were entirely neutral the photosphere would be much hotter, because we would see to greater depths. Hydride ions start to form below about 10,000K. The hydride opacity also has a very steep temperature dependence, which is why the photosphere is thin. $\endgroup$ – Rob Jeffries Aug 9 at 11:03
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The main source of opacity that defines the visible photospheric continuum of the Sun is that due to H$^{-}$ ions. This has been understood for 80 years (Wildt 1939). These ions are in an equilibrium with neutral hydrogen atoms, where the additional electron can be ejected due to the absorption of photons (bound-free absorption) with a continuum of energies and vice-versa can be formed via the emission of a continuum (free-bound emission) of photon energies.

Talk of atomic transitions is a red herring. Of course these occur, but are responsible for the emission of photons at relatively discrete energies and from higher up in the atmosphere (that's why they appear as "dark" absorption lines).

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All the material in the outer region of the Sun has thermal motion, which includes vibration and continuous excitation/de-excitation of atoms and ions. This goes on at all frequencies from zero to high, with a peak around the visible spectrum and tailing off above that. This motion produces light through various mechanisms, including electrons changing energy levels in atoms and ions, and also electrons simply accelerating and bumping into things during their random motion. Any acceleration of a charged particle will result in electromagnetic radiation.

This process is similar in many respects to what goes on in the filaments of light bulbs working by the glowing filament method. The main difference is that the white-hot filament of a light bulb is solid metal whereas the Sun is plasma and gas. But the thermal motion producing acceleration and oscillation at all frequencies is similar.

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