If a nucleus undergoes a change in Z or Mass due decay or absorption, could this disrupt the electrons from their orbital/shell energy levels?

If so, could the electrons that were previously in the original orbitals have insufficient energy for the new orbitals?

What would happen to these electrons if this occurs?

For example:

  • Could the electrons formerly in the k shell or s1 orbitals have insufficient energy to remain in those orbitals after a change in the nucleus?
  • What happens to them if this occurs since there is no lower energy orbital for them to enter but they remain trapped in the potential well?
  • Would they remain in an unstable state (i.e. not in a stable orbital) or reacquire the energy level due to rearrangement of other atomic states?
  • I suppose higher energy level electrons would also not move to the new energy level due to shielding of the lower ones, even if they were not in a stable orbital?
  • If these electrons are in an unstable state, would they remain paired as in the original orbitals?
  • 2
    $\begingroup$ The reaction can't proceed if there is zero phase space for the final state... $\endgroup$ Jul 15, 2015 at 23:15

1 Answer 1


The typical way to handle such things is the "sudden approximation". The time scale of the decay/capture process is assumed to be much smaller than the time scale of the evolution of the electron shell. The probabilities of the new states will then just be the projections of the old state to the new stationary states. (The typical analytically solvable example is a charged harmonic oscillator suddenly subjected to an electric field, giving a Poisson distribution of the probability of finding the oscillator in the excited states if it was in the ground state before).

No, it is impossible they have "insufficient energy" to be in the new orbitals. When the charge of the nucleus changes the added/missing charge has to go somewhere, thus changing the potential energy of the electrons.

You can also easily show this by considering the energy functional under the sudden approximation. We have a Hamiltonian $H_0$ and a Hamiltonian $H_1$ with changed parameters, correspondingly we have two functionals $E_0$ and $E_1$. The electrons be in a state $\left|\psi\right>$ prior to the change of parameters. The sudden approximation says, that the electronic configuration remains $\left|\psi\right>$ during the decay/capture, so the energy afterwards will be $E_1[\left|\psi\right>] \ge E_1^\text{ground}$ which is greater than the ground state energy after the parameter change, so the energy of the electron configuration will be greater than the lower bound in all cases.

Note that the energy of the electrons will typically not be preserved in such a process! If charge enters/leaves the atom suddenly, the electrostatic energy of the electrons will change without a change of their spatial configuration!

  • $\begingroup$ Thanks a lot Sebastian I have been wondering about this for a while and I think you have given me a good answer. It's an intelligent answer that has get me thinking about the physics so I'm reading up about it and sudden approximation is something new for me to look into. I suppose that reduced electron mass effects and other fine structure effects would also be similarly affected by this approach. $\endgroup$
    – Stephen
    Jul 16, 2015 at 20:17
  • $\begingroup$ The variational argument is general, it even works if we do not assume the sudden approximation, just by the fact, that the electrons will be in some state after the nuclear reaction (even if this is not the initial state). (Side note: Of course total energy will be conserved, but not the energy of the electrons by themselves). $\endgroup$ Jul 17, 2015 at 14:41

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