Quantum jumps inside atoms always have the same energy, at least in a hydrogen atom when jumping from $n=1$ to $n=2$, like from a 1s1 to a 2s1 state. My question is, if an electron can be anywhere in the orbital, and jumping from $n=2$ to $n=1$ always has the same quantum of energy, does this mean the electron was somewhere on the radius particular to that energy when it left the orbital or can it jump from way out in left field and land in the $n=1$ area without ever hitting the $n=2$ radius immediately prior to releasing its energy? I assume quantum theory rules this question with uncertainty, but I'm not sure.

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    $\begingroup$ There is no fixed radius to an orbital. $\endgroup$ – ACuriousMind Nov 5 '14 at 13:20
  • $\begingroup$ If you like this question you may also enjoy reading this Phys.SE post. $\endgroup$ – Qmechanic Nov 5 '14 at 14:20
  • $\begingroup$ I beg to differ, since the energy released is always the same exact wavelength, then the electron itself must be measuring itself against the distance from the nucleus for it to be so precise in the released or absorbed wavelength, right? $\endgroup$ – William David Pope Nov 9 '14 at 21:20

The idea of circular orbits with fixed energies for the hydrogen atom's electron is outdated.

As was mentioned in the commentary, once you deal with the Atom at the quantum level, there is no fixed radius to a given orbital. Orbitals become probability distributions - for the shapes just put "Orbitals" in a google image search.

As you will see, there is quite some overlap of orbitals on different shells. This is not a contradiction to them having different energies though.


The electron checks NOTHING. But your question is more profound. The electron jumps from a higher energy level to a lower one if a photon can be emitted. It is the tendency of the atom to get rid of the excess energy, that drives the decay. There are related topics to your question, as "tunneling", "decay from resonant levels", "the nuclear reaction theory". In what regards your question, the atom de-excitation and the nuclear decay, are similar processes.

Please know that an excited level (usually called resonant level) has NO sharp energy, and NO sharp radius. A resonant atom-level can emit photons of different energies. The range of energies may be small, and in this case it takes a very long time for the electron to jump, (i.e. for the photon to leave the atom), or, the range may be big, and in this case the electron falls quickly from the higher level to the lower one.

You will learn these things if and when you study the nuclear reaction theory.

About the uncertainty principle, indeed, an atomic level doesn't have an absolutely sharp radius, it has a radial width, and correspondingly, the uncertainty principle allows an imprecision in the radial linear momentum. If the uncertainty in the linear momentum is small, the radial width of the orbital is very big. In some cases it can cover even the nucleus, see in Wikipedia "conversion electrons".

With pleasure, Sofia

  • $\begingroup$ The question refers to an electron in the hydrogen atom. Afaik even when considering the atom coupled to the EM field the excitation energies are sufficiently well-defined as to speak of sharp energy (the natural line width is very very small compared to the separation of adjacent lines). $\endgroup$ – Neuneck Nov 5 '14 at 14:56
  • $\begingroup$ My answer is rigorous. Everybody knows the hydrogen series, Lyman, Balmer, etc. If the Hydrogen excited-levels were sharp, there were no de-excitation and the series didn't exist. What you try to ADD is that in practice there are narrow resonances which take a long time to decay. $\endgroup$ – Sofia Nov 8 '14 at 13:26
  • $\begingroup$ Even for the Lyman etc. series the natural line width is small compared to the separation of the resonances. For me, this then qualifies as a sharp energy from which the state decayed. Of course, if you only accept truly stable states, then there is only a handful in all of physics. You might even consider not believing in quarks then - any of those more heavy than the u or d anyways. $\endgroup$ – Neuneck Nov 10 '14 at 6:21

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