Take the 2-minute tour ×

Is there an upper limit to the number of orbits an electron can have around say a proton? Arent there states that are unstable(for n!=1) with corresponding mean/half lives and therefore uncertainty in energy.

So how do we differentiate between 2 n values (say v1,v2) esp if the uncertainty associated with their energy levels gets larger to the order of the energy difference associated with the transition?

Also is it known that we can predict the following data with only computation resources being a bottleneck?

The half lives of each state(decay rate). The transition rate(from v2 to v1 given v2>v1)

share|improve this question
It's really not clear what you're asking here. What is n? What do you mean by orbits - did you mean to say orbitals? What kind of atom are you talking about? –  David Z Jun 22 '11 at 4:15
David,as described above..this is a simple hydrogen nuclues with an electron whose states we want to differentiate..yes maybe the correct term would be orbital.. –  Ajay Jun 22 '11 at 4:18
Okay, then what sorts of states are you talking about? Why do you bring up uncertainty in energy when talking about the half-life? Are you asking how you could predict the transition rate of an arbitrary quantum state of the electron in a hydrogen atom? There are formulas for that, so I'm not sure why you're talking about computation resources being a bottleneck. –  David Z Jun 22 '11 at 4:22
Look up "Rydberg atoms" and "Rydberg states". I know there's research on these going on at Colorado State, so you might also visit their physics dept's site. –  DarenW Jun 22 '11 at 5:48
One source of possible confusion here is that the multiple "orbits" of a hydrogen-like atom (i.e. only one electron) do not remain solutions when you add additional electrons, because there are interactions between the electrons as well as between the nucleus and the individual electrons. –  dmckee Jun 22 '11 at 15:00

2 Answers 2

Interesting question! This article says that "The natural lifetime of an undisturbed Rydberg atom increases as $n^3$ for a given electronic angular momentum." That means that each state has a natural width $\Delta E_{w}\propto n^{-3}$. The energy levels go like $n^{-2}$, and differentiation tells us that the spacing between successive levels goes as $\Delta E_s\propto n^{-3}$. This means that $E_w/E_s$ is approximately independent of $n$ for large $n$. Since we know that $\Delta E_w/\Delta E_s << 1$ for small $n$, it sounds to me like this continues to be true for large $n$. In other words, the states are still well defined and non-overlapping for arbitrarily large $n$. (Of course there will still be degeneracy within each $n$ value.) This is all for a hydrogen atom in free space with zero external electric or magnetic field and no background of blackbody photons.

share|improve this answer

The decay rate from a certain state depends on the environment. If the atom is in an optical cavity which is strongly detuned from all of the decay modes of the atom, it will stay in the excited state much longer. Because how much longer it remains excited depends on the properties of the cavity, I don't think there's a hard "states with $n>something$ cannot be distinguished from other states" limit, at least in non-relativistic QM.

Density of states factors pop up in statistical physics, so you can still tell there's a bunch of states there even if you can't distinguish them.

share|improve this answer

Your Answer


