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As far as I've understood, the color of an object depends on which wavelenghts from the more or less "white" sun light it reflects and absorbs.

I think that an atom can only absorb the light that has exactly the right wavelenght to exite an electron to a higher energy level. All light that doesn't is reflected and determines the color of the atom (or, more precisely, of a very big group of atoms as individual ones are way to small to be seen by the naked eye).

Is this right? And how does it compare to emission lines which are basically the wavelenghts being absorbed and then re-emitted? Wouldn't it be kind of the same thing?

However, that doesn't make a lot of sense since oxygen, for example is invisible but has emission lines in the visible spectrum.

Thanks in advance

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  • $\begingroup$ This is a very common question. See, for instance, physics.stackexchange.com/questions/211388/… $\endgroup$ Commented Jan 8, 2020 at 20:58
  • $\begingroup$ @QuantumBrick it is similar, but doesn't include the difference to spectroscopy $\endgroup$
    – Fraser
    Commented Jan 8, 2020 at 21:14

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Light is scattered by material surfaces. That means, absorption, and emission of some wavelengths in several directions.

Green leaves for example absorb the incoming light, and scatter wavelengths in the range that our eyes perceive as green.

The spectral lines are signatures of isolated atoms. Small amounts of iron for example can be heated until vaporize, and its typical spectral lines be identified from the emitted radiation.

But the colour of a cold iron bar is another subject. Its behaviour regarding to light scattering depends on the crystal ordering of the atoms, and band structure of its electrons. If the surface is oxidized (and it is always partly oxidized), its colour depends also on the type and thickness of the oxide layer.

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