15

As you point out, this is a difficult task. Nonetheless, it can be done. Key to the process is having on hand the absorption line spectra (Fraunhofer lines) for as many of the (pure, unmixed) elements as possible in advance; this then lets the investigator determine if (in your example) the combined spectra for A+B is 1) actually another element C, or 2) ...


8

Here is the solar spectrum. Despite what you have been taught, it shows a great deal of structure, dominated by lots of dark (but not black) absorption lines. These "Fraunhofer lines" are indeed the signature of what chemical elements (and a few molecules) are present in the solar photosphere - that region in the outer part of the Sun from whence light ...


3

However, shouldn't also the sun have the footprints of its components? Shouldn't it's spectrum be similar to the hydrogen and helium ones? (it isn't, but why?) The light emitted by the sun is heat energy that's being re-radiated from hot gasses. The actual electromagnetic energy generated in the sun's core is in the form of x-rays; it gets absorbed fairly ...


3

Yes and no. It is indeed true that for every emission line, there is a corresponding absorption transition that, under suitable conditions, can be observed in an absorption spectrum. However, it is not true that if you take an emission spectrum and an absorption spectrum from the same atom, the lines in the two must always match. The reason is simply that ...


2

It means that each atom sees the incident radiation in its own frame. So suppose the atom has an absorption line at 500 THz. And you shine a laser at it at 500.1 THz (as measured in your frame). Then the atoms that are moving toward you are going to see a slightly higher frequency than you measured (say 500.2 THz). So they won't absorb this light. But the ...


1

The electron eventually falls back to its original energy level, reemitting the absorbed light. That is not true for an object that absorbs light. Typically, the electrons will relax to the ground state by making atoms vibrate more. That is why a black object becomes warmer. It does not emit light.


1

That is a very good question! To answer it, let's focus on a mirror. What surface of a mirror have unique compared to other objects? It is very smooth. If otherwise, it would scatter the incoming light in random direction. And that is what other objects do! However that is just a macroscopic approximation. To dig in deeper, we need to think what happens ...


1

Not in general (though it is very common). Under some conditions you can have states that are reached by a single absorption event (creating a high energy absorption line) that decay (with a non-trivial branching ratio) through several steps (creating several lower energy emission lines). Indeed this kind of process is the usual scheme for setting a ...


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