How do accelerated ions with remaining electrons in the shell produce synchrotron radiation?

Question

Since ions are charged, they should produce synchrotron radiation in an accelerator. I am interested the characteristics of the emitted radiation when there are still electrons in the ion's shell (not just fully stripped nuclei). If the ion has $n$ remaining electrons in its shell, is it the sum of $n$ times the synchrotron radiation of a single electron (plus the synchrotron radiation of the nucleus, which is probably negligible) or is the ion to be considered as a whole? If the latter is the case I'd expect the synchrotron radiation to be strongly suppressed because of the large total mass of the ion compared to the masses of the electrons.

EDIT: The Relativistic Heavy Ion Collider is a good example that I just found. The linked Wikipedia article describes the acceleration process as follows:

(...) Gold nuclei leaving the EBIS have a kinetic energy of 2 MeV per nucleon and have an electric charge Q = +32 (32 of 79 electrons stripped from the gold atom). The particles are then accelerated by the Booster Synchrotron to 100 MeV per nucleon, which injects the projectile now with Q = +77 into the Alternating Gradient Synchrotron (AGS), before they finally reach 8.86 GeV per nucleon and are injected in a Q = +79 state (no electrons left) into the RHIC storage ring (...)

The Gold ions still have electrons left in their shell while they are accelerated to relativistic speeds. Each time when they are transfered to another stage they emit synchrotron radiation when the beam is bent (even though this is probably an undesired side effect in this case). Before they are completely stripped away, do the remaining electrons influence the emitted radiation?

Answer 1

I have the feeling that the (partially stripped) ion should be considered as a single particle. The residual electrons should influence the emitted radiation just by altering the charge state of the ion, therefore the $C_\gamma$ parameter which enters in the calculation of the total emitted power and power spectrum.

The external bending field should be considered just as a perturbation for the electrons which are immersed in the field from the nucleus. Plus how could an electron radiate a (several keV) photon being bounded to a nucleus? Where would that energy come from? Where would the electron end up being?

A similar argument could be raised for any composite particle: would the quarks in a proton radiate independently? By direct measures at the LHC, we know that they do not.

Answer 2

I totally agree with @DarioP. People many times accelerate ions with singly/doubly charged states to several 100s of MeV in cyclotrons and no appreciable radiation is detected. If the electrons radiate as if they are free then the energy loss will be significant and that will alter their orbits, but it is not observed.

For example uranium atom is accelerated to ~80 GeV (345 MeV/nucleon$\times$238) in cyclotron if electrons radiate this would be impossible. Due to radiation de-acceleration there will be phase slip in the cyclotron and one will not get good acceleration gradient. This is one of the reasons why we can not accelerate electrons in the cyclotrons.

It may be noted here that the charge state is +72 to +88, however there are still some electrons attached with the uranium ion (Z=92).