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If atom is hit by a photon that is the precise wavelength/energy like one of its absorption lines, will it put the electron inside the atom to a higher excited orbital level, in that short timeframe before the electron falls down to lower energy level,is that atom less absorptive at that photon wavelength?

To further expand my question, if truly does the absorption line disappear, do lower absorption lines below the one that was hit disappear, too? My thinking is that each absorption line represent one energy level of atom. It goes from lowest in infrared to highest levels in ultraviolet. If the atom is hit by an UV photon, then all the lower energy levels are for a moment not avaliable for "activation", so the corresponding lower energy/longer wavelenght absorption lines disappear too. Or is this wrong?

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  • $\begingroup$ Yes, the possible absorptions (better absorption probabilities) of an atom depend on its electronic state in the way you suggest: To calculate them you compute transition probabilities between the initial and the final states you're interested in. Note that absorption lines correspond to differences between electronic energy levels, not the actual electronic states. $\endgroup$
    – Adomas Baliuka
    Commented Feb 11, 2017 at 1:52
  • $\begingroup$ Adomas Baliuka what is different between electronic energy level and electronic state? I thought its the same thing! $\endgroup$
    – wav scientist
    Commented Feb 11, 2017 at 1:54
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    $\begingroup$ For starters, two different electronic states may have the same energy. Imagine a state for a Hydrogen atom where the electron is in a region centered around the z axis (say a p-orbital, you can find pictures of the probability distribution). This state has a certain energy. The state where the same shape occurs rotated in any direction has the same energy.(If you apply an external electric field, this will change) If this were not so, the atom would have a preferred direction and very interesting electric and magnetic properties. Experiments suggest that atoms don't have preferred directions. $\endgroup$
    – Adomas Baliuka
    Commented Feb 11, 2017 at 2:03

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The absorption lines of an atom doesn't show its energy levels. They show all possible differences between its energy levels. Thus an atom having $n$ excited states will have $\frac{n(n-1)}{2}$ lines in its spectrum.

The spectrum isn't measured on a single atom, it is measured on a lot of atoms. To see many spectrum lines, we need a soup of atoms having all of the excited states, and continously jumping between them. In most cases, this can be reached in a gas heated to some thousands K.

If a single atom gets a photon, it goes into an excited state. From now, it can decay, or get another photon, going into another state. In all of the states it has a different absorption cross-section distribution for the different photons. The experimental result of the spectrum of a heated gas is the mix of these distributions.

The interaction between the excited atoms and the photons can be more complex, for example it is possible that an atom gets a photon, emit a new one and they moves into another (ground or excited) state. This is how lasers work, check for induced emission.

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