If we have 2 electrons in the 1st energy level named X & Y, and one electron in the 2nd energy level named Z.

  • If the electron X was excited and transferred to the higher energy level, why didn't Z replace it in the 1st energy level?

  • Why don't we say that there is a type of nuclear gravity, and it increases when the atom becomes unstable and that gravitates Z electron towards the nucleus instead of the excited one?

  • $\begingroup$ Z might well replace it. But, you put energy in to the system to get X in to another level, so the system will have to release energy to drop Z. I wouldn't quite call this an Auger process, but... $\endgroup$
    – Jon Custer
    May 30, 2017 at 13:13

1 Answer 1


On the first part: it will, or won't, in a subsequent, separate process.

If you excite the system and promote an electron from the lowest orbital to the next orbital, the immediate result is that you have a lowest orbital populated by one electron and a higher orbital populated by two electrons. It is an excited state, and will tend to relax back to the fundamental state if there is an available mechanism to do so - but that will be a second distinct event after the excitation. In fact, the excited state can, depending on the circumstances, last long.

The second part makes no sense. The nature of the interaction between nucleus and electron is not gravitational (it's electromagnetic), and the lowest energy level isn't "below" in any spatial sense. The gravitational analogy is misleading.

  • 1
    $\begingroup$ Regarding the gravity part, the electron doesn't fall into the nucleus because of its kinetic energy. $\endgroup$ May 30, 2017 at 13:24
  • $\begingroup$ If we look at this figure: en.wikipedia.org/wiki/Atomic_orbital#/media/… we can see that with same probability the electron in 1S orbital occupies smaller volume than the electron in 2S orbital. We can can say that 2S is above 1S. And electrostatic and gravitational forces shares the very same Faraday's and Newton's equation with appropriate constants and measures. The gravitational analogy is quite good in educational terms. $\endgroup$
    – Crowley
    May 30, 2017 at 14:59
  • $\begingroup$ its most probably a typo but the upper orbital will have one electron not two as you write in line 3. $\endgroup$
    – porphyrin
    May 31, 2017 at 9:02
  • $\begingroup$ @porphyrin If I'm not mistaken, OP asks about a system where the second orbital is already half-occupied (he calls that electron Z). $\endgroup$
    – user41033
    May 31, 2017 at 15:33
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    $\begingroup$ @Vic Lineal , yes you are correct, I misread . And as you state the excited state will last for some time until the energy is released and the ground state is recovered. As electrons are indistinguishable we cannot know which one returns. $\endgroup$
    – porphyrin
    May 31, 2017 at 20:19

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