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When you shine light on an atom, it absorbs some of the wavelenghts; the rest determine its color. However, as far as I've understood, the absorbed wavelenghts exite the electrones moving them to a higher energy level.

The electron eventually falls back to its original energy level, reemitting the absorbed light.

So why doesn't the object appear white? I understand that it can emit the absorbed wavelenghts in any direction, but shouldn't even in this case some of this light go in the same direction? So shouln't absorption lines not be completely black but instead like 99% black?

I hope you understand my question :)

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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 when a photon 'collides' with an atom.

As you know, shining light on an atom can excite it's electron and make them emit photons while they transition to the lower energy levels. This is correct, but not the only scenario that can happen. The photon can also get scattered if it's energy is higher than required, or insufficient. It can also transfer it's energy into vibration of an atom and produce effectively heat, or both can happen simultaneously.

The object will therefore absorb the wavelength that closely matches it's electron energy levels, some will get scattered and some transformed into heat. Some light interactions with materials may be so negligible that photons act as if the object was not there. The light we see from the object is usually the light reflected from it. But it can also emit it's own light, 'contributing' to the color we perceive. (as hydrogen isotope; tritium).

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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.

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What determines the colour of a sample is the light reflected and/or emitted by it. As you say, certain light is absorbed and could then be re-emitted isotropically, this means that we will detect the colour of this emitted wavelength. If we sent something through the sample, the emitted light from the sample would be what we're detecing right? The originally incident light would have passed through and interacted/reflected somewher else. This emitted light has discrete energy levels as you say and these are what spectrometers detect.

Further, if we radiate a sample with white light which reflects only red light, it will appear red to us. If this reflection is scattering (classical reflection) or an absortion followed by re-emission is not important for this example here.

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