It seems like you're confused about zero-temperature quantum mechanics. Even at absolute zero, the electrons do not 'fall into the core'---they sit in the orbitals---except at extraordinarily high pressures/densities where (as in neutron stars) they merge with the protons, turn into neutrons, emitting a neutrino.
Note that neutrons are heavier than protons, and when alone undergo beta decay with a lifetime of about 15 minutes, turning into a proton, emitting an electron and an antineutrino. So a hydrogen atom, left alone, cannot undergo the process you suggest---there simply isn't enough energy for it. It's only at very high density that the coulomb interaction provides enough energy to perform this conversion. There may be new physics, such as proton decay, that changes this story, but as far as anybody knows, there is not.
Perhaps more precisely, at absolute zero temperature atoms with orbitals are mathematically what quantum mechanics predicts. I of course don't know how to produce an atom that cold in the lab :) Except... of course... I do know how to produce an atom where the electron is in its ground state. If I've just got one atom, that's easy-peasy. Then, what is its temperature? Because it's just one atom, it's not well defined.
There's a more interesting question you ask, though: can the electron have a tempature? The different energy eigenstates are populated with different amplitudes. The way those states are populated can have a well-defined distribution in terms of a superposition. But if we restrict ourselves to a classical case, as OP was thinking about, superpositions are not allowed. Then, again, you at least have a definite energy.
Moreover, just a single electron in a superposition of energy eigenstates... it's not clear to me that that thing has a well-defined temperature either, because 'average' energy means something other than an ensemble average.
> at absolute zero, the electrons do not 'fall into the core'--
well, actually I think nobody has shown this empirically. To add to my point, I don't think there can't be any closed system but one and it seems reasonable to assume that any substem could be at zero only if all are. Maybe energy just cant retrivably convert to other forms and loose all heat in the process when it's environment is already at rest. At best you could maybe get singularities before the environment exited by the process feeds heat back into the subsystem.
Anything else would be as good as a papetuum mobile which you'd need to cool down this far.
Note that neutrons are heavier than protons, and when alone undergo beta decay with a lifetime of about 15 minutes, turning into a proton, emitting an electron and an antineutrino. So a hydrogen atom, left alone, cannot undergo the process you suggest---there simply isn't enough energy for it. It's only at very high density that the coulomb interaction provides enough energy to perform this conversion. There may be new physics, such as proton decay, that changes this story, but as far as anybody knows, there is not.
Perhaps more precisely, at absolute zero temperature atoms with orbitals are mathematically what quantum mechanics predicts. I of course don't know how to produce an atom that cold in the lab :) Except... of course... I do know how to produce an atom where the electron is in its ground state. If I've just got one atom, that's easy-peasy. Then, what is its temperature? Because it's just one atom, it's not well defined.
There's a more interesting question you ask, though: can the electron have a tempature? The different energy eigenstates are populated with different amplitudes. The way those states are populated can have a well-defined distribution in terms of a superposition. But if we restrict ourselves to a classical case, as OP was thinking about, superpositions are not allowed. Then, again, you at least have a definite energy.
Moreover, just a single electron in a superposition of energy eigenstates... it's not clear to me that that thing has a well-defined temperature either, because 'average' energy means something other than an ensemble average.