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Black body is an object, hence it's made of atoms. Depending on what atom it consists of, that's how the emitted spectrum should be in my opinion. If it contains hydrogen, absorbed light on black body would excite electrons and electrons of hydrogen can only produce certain wavelengths. Hence, there must be dark lines in the final, emitted radiation(which is called - black body radiation).

how does perfect black body have continues spectrum (with no dark lines ?) That means when we talk about perfect black body, we shouldn't talk about atoms inside it as there's no atom existing that absorbs or emits light at any wavelength.

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Black body is an object, hence it's made of atoms.

No, a black body is not made of atoms. It is an imaginary object which has the special (but physically unrealizable) property that it is a perfect absorber - any incident electromagnetic radiation is absorbed, with zero reflection and zero transmission. Under certain conditions, many objects in nature (stars, rocks, people) can be modeled as black bodies (or some modification thereof) to varying degrees of accuracy.

Depending on what atom it consists of, that's how the emitted spectrum should be in my opinion. If it contains hydrogen, absorbed light on black body would excite electrons and electrons of hydrogen can only produce certain wavelengths.

The electromagnetic radiation emitted by an object is not determined exclusively by the electronic transitions which are possible in its constituent atoms. Solid objects have electronic energy levels which form broad, continuous bands, not discrete spectral lines. Stars are comprised of plasma, not neutral atoms; photons are produced via the scattering of charged particles (bremsstrahlung) and their energies are further randomized via Compton scattering.

That means when we talk about perfect black body, we shouldn't talk about atoms inside it as there's no atom existing that absorbs or emits light at any wavelength.

Yes, that's right.

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  • $\begingroup$ Even if it is a mathematical idealisation, the pre-idealisation prototype necessarily must be made of atoms, and then we take the limit towards idealisation. And no, it is not needed to be instantaneously absorbed. In fact, the original conception is specifically allowing infinite time to slowly absorb. $\endgroup$ Apr 24, 2023 at 7:58
  • $\begingroup$ @naturallyInconsistent I disagree. The blackbody radiation spectrum can be obtained by maximizing the entropy of a photon gas. If you want to ask how such a gas could possibly come to equilibrium in a real physical system, then sure you can make reference to atomic oscillators in the cavity walls, but even then taking the requisite idealized limit strips away any traces of atomic structure. I will remove "instantly" from my answer because I did not mean to make reference to any particular time scale, given the assumption of thermal equilibrium. $\endgroup$
    – J. Murray
    Apr 24, 2023 at 14:44
  • $\begingroup$ Is it not the case that if you assumed that the cavity walls obeyed classical mechanics, you could derive Rayleigh-Jeans law in the idealised limit? This part is one of the atomic structure traces that gets left intact in the limit and has to be taken correctly at the start. $\endgroup$ Apr 24, 2023 at 14:49
  • $\begingroup$ @naturallyInconsistent The assumption is required for the derivation of the Planck distribution is that the energy content of a given electromagnetic mode is some integer multiple of $\hbar \omega$, an assumption already built into the photon gas. It's true that classical oscillators do not provide a suitable model for bringing the (classical) radiation in the cavity to equilibrium, which provided historical evidence for the existence of photons in the first place. $\endgroup$
    – J. Murray
    Apr 24, 2023 at 15:03
  • $\begingroup$ Thanks for the answer. If we say that star is very close to black body, then it means it's able to absorb radiation incident on it very close number to 100%. Question 1: what radiation does the sun absorb that's incident on it and coming from outside the sun ? Question 2: How would the sun for example absorb all wavelenghts(not all, but since it's close, let's go with all) ? when light gets incident on the sun's surface, sun's atoms can only absorb the specific wavelengths. What's the logic here ? $\endgroup$
    – Matt
    Apr 25, 2023 at 0:10
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A blackbody must be capable of absorbing (and emitting) light at all wavelengths. Anything in nature that approximates to an ideal blackbody (a tiny hole into a large cavity) has this property.

You can find out what processes lead to this absorption in the linked question.

What are the various physical mechanisms for energy transfer to the photon during blackbody emission?

In your example of a hydrogen gas (and I guess to are talking about stars), the "continuum absorption" in the visible part of the spectrum is provided by photoionisation of the the H$^{-}$ ion, which has an ionisation energy of just 0.7 eV and absorbs photons across the visible part of the spectrum. This means that stars like the Sun have spectra approximating to blackbodies in the visible part of the spectrum.

The Sun does of course have "dark lines" (Fraunhofer lines) in its spectrum. This is because both: (a) absorption is enhanced at the wavelengths of atomic transitions as you suggest; and (b) there is a temperature gradient in the Sun which then means we see to depths occupied by cooler plasma at those wavelengths. Thus the lines are not totally dark, they are just less bright than the surrounding continuum. The non-isothermality of the Sun's atmosphere is why it doesn't emit an ideal blackbody spectrum.

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  • $\begingroup$ Re (a) both absorption and emission are enhanced, re (b) the gas temperature gradient is not the reason for the Fraunhofer absorption lines. They are present because the gas causing them via absorption/scattering is optically too thin to produce enough radiation to fill in the absorbed/scattered intensity. This would be the case even if it had the same temperature everywhere, but its density was too small. To produce radiation with intensity close to that of a black body radiation, the emitting matter has to be dense enough. $\endgroup$ Apr 24, 2023 at 17:20
  • $\begingroup$ The solution of the radiative transfer equation for a very thick slab (with any density profile) is $I_\lambda = \int^{\infty}_{0} B_\lambda \exp(-\tau')\ d\tau' $ @JánLalinský If the slab is isothermal then this just equals $B_\lambda$ at that temperature. $\endgroup$
    – ProfRob
    Apr 24, 2023 at 18:23
  • $\begingroup$ In that result, in addition to uniform temperature, you're assuming Kirchhoff's law is valid for arbitrarily thin gas and also you're assuming that observation of radiation happens at a distance big enough so that the atmosphere column becomes opaque. This seems to be a specific case; for thin enough gas, scattering is relevant and Kirchhoff's law is not necessarily accurate. Length of the gas column is finite and it is not clear it can be approximated by infinity. $\endgroup$ Apr 24, 2023 at 23:19
  • $\begingroup$ Are we sure Fraunhofer lines are entirely formed in the deep layers of the atmosphere where the assumption of LTE and Kirchhoff's law are valid? I would expect some lost intensity in those lines is due to diluted gas, possibly out of LTE. $\endgroup$ Apr 24, 2023 at 23:22
  • $\begingroup$ Thanks for the answer. so, by photoionisation of the H− ion, you mean that previously it was H-(with 1 proton, 2 electrons), but photon came, hit it, hydrogen atom absorbed and absorbtion caused 1 electron to be removed. I understand that, but if we know that atoms can only absorb specific wavelengths(photon energies), how in your example would it be able to absorb continous spectrum - that means all wavelengths. Not all wavelengths would have enough energy to cause electron removal and not discrete energy to cause electron shell change, hence not every photon would be absorbed. Thoughts ? $\endgroup$
    – Matt
    Apr 25, 2023 at 0:21

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