I have read that the mechanism behind internal conversion, in which a nuclear transition leads to the ejection of an electron in one of the lower atomic orbitals, is related to the fact that the wavefunctions of some of these electrons penetrate the nucleus. Because of this, there is a finite probability that the electron will be in the nucleus, and, when this occurs, the electron may couple to the excited state of the nucleus nuclear transition when it happens.

Is it conjecture that the electron must penetrate the nucleus in order for internal conversion to occur, or is this a straightforward result? I would have figured that since internal conversion is mediated by the electromagnetic force (which has long reach), for any electron within range of a virtual photon there will be a nonzero cross section for this type of decay. What details am I missing in thinking this?

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    $\begingroup$ the electron may couple to the excited state of the nucleus. This part is incorrect. There is no excited state involved. $\endgroup$
    – user4552
    Commented Sep 1, 2014 at 16:32

2 Answers 2


Is it conjecture that the electron must penetrate the nucleus in order for internal conversion to occur, or is this a straightforward result?

1) it is an experimental result that K etc electrons are ejected from atoms that have radioactive nuclei

2)The probabilities of the electrons to overlap with the nucleus are given by the atomic solutions of the quantum mechanical equations.

3)To be ejected a photon must "hit" the electron so that it gets the necessary energy to go free

So at this level one has to go to the nuclear level and the probability of a virtual photon to hit an electron in conjuction with the probability of the electron to be available as a "target".

Virtual photons can be available from the charged field of protons and the quarks that contain them. When the orbital of the electron is outside the nuclear volume, the probabilities of interaction of charges falls with the distance . The radius of the nucleus is of order Fermi, 10^-15 meters, the atom is in angstroms , 10^-10 meters. On this roughly the 1/r^2 fall of intensity ( classically but it is a rough measure) reduces probabilities too. Thus the probability of a random electron around the atom to interact with a virtual photon from a quark ( which are the final carriers of the charges in the nucleus) is very much smaller due to geometry. In S states the probability is quantum mechanically enhanced by the overlap of the loci of probabilities.

  • $\begingroup$ Thank you. I note that the Wikipedia article mentions that the electron interacts with the field of the nucleus as a whole rather than individual protons or quarks. (This might not qualify anything you've said.) $\endgroup$ Commented Jun 25, 2014 at 8:49
  • $\begingroup$ Well, it is possible to be the dominant approximation, a collective phenomenon, but that field is built up by the individual charged quarks . In any case I think the energy must come from the available energy for a nuclear gamma decay. The opposite, electron capture , also happens because of overlap probabilities , and there the Z changes, a proton turning into a neutron. $\endgroup$
    – anna v
    Commented Jun 25, 2014 at 9:20
  • $\begingroup$ Is the fall in field strength also a function of the energy of the transition, in addition to 1/r^2? $\endgroup$ Commented Jun 25, 2014 at 16:01
  • $\begingroup$ Solving the quantum mechanical problem energies come up as eigen values, the fields are input to the differential equation. The classical argument is for order of magnitude. $\endgroup$
    – anna v
    Commented Jun 25, 2014 at 16:17
  • $\begingroup$ This answer doesn't address the question, which is about internal conversion. $\endgroup$
    – user4552
    Commented Sep 1, 2014 at 16:33

"The process of imparting energy from the nucleus to an orbital electron is a quantum process and may be seen as taking place by means of a virtual photon. In that sense the photon involved can be considered as a "virtual gamma ray"..." (Wikipedia)

The electromagnetic force is indeed a long range force - virtual photons bind the electrons to the nucleus. The virtual gamma involved here has a much higher "energy", and therefore much shorter range. This is why the process only occurs when the electron is in the nucleus. Probably the range of this interaction is smaller than the diameter of the nucleus.

Remember that all the quarks in the nucleus are "point particles", so the electron can never be truly "inside" them. "Point particles" can never "collide" - but they nevertheless interact with finite cross-sections. High energy interactions with large momentum transfer like this one have very small cross-sections.

This discussion will also make you realise that the overlap of the electron wave function with the nucleus is not optional - the nucleus is also "mostly empty space" in the overused phrase.

  • $\begingroup$ Thank you for the interesting reply. I will wait a while before accepting any replies, since this topic is far from my what I know. A related question -- is it a requirement that the energy be transmitted to a single electron by way of a single virtual photon with large energy? Is the probability of an interaction involving a large number of smaller virtual photons negligible? $\endgroup$ Commented Sep 2, 2014 at 20:58
  • $\begingroup$ Yes, I believe the chance of an event always declines the more particles are involved. In reality all possible ways for the event to take place contribute to the amplitude for it, and therefore to its probability, but the contribution from a more complex process corresponds to a term which is higher order in the fine structure constant (~0.007). $\endgroup$
    – akrasia
    Commented Sep 2, 2014 at 22:04
  • $\begingroup$ Is there a textbook that covers EC in some detail at an intermediate level that you can recommend? $\endgroup$ Commented Sep 2, 2014 at 22:14
  • $\begingroup$ Sorry -- not EC, internal conversion! $\endgroup$ Commented Sep 2, 2014 at 22:33

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