I'm in the process of trying to better understand internal conversion and
it's cross section vis-a-vis inner shell electrons and sources of charge in
the far field.  I'm hoping someone (Robin?) can help me to get the
terminology right and point me to further reading.

Here is my understanding so far.  Internal conversion is a process in which
an inner shell electron is expelled from an atom as the result of a nuclear
transition.  It is mediated by the electromagnetic force (in contrast to
the weak or strong interactions) and results from electromagnetic coupling
between the electron and an excited nucleus.  The kind of nuclear
transition that leads to internal conversion is generally an isomeric
transition; i.e., the deexcitation of a metastable isomer to a lower energy
level.  Internal conversion competes with gamma emission, and there is an
internal conversion coefficient, which for a competing pair of branches,
one IC and the other gamma photon emitting, gives you the ratio of internal
conversion electrons to photons.  In some cases the internal conversion
coefficient can be quite high, meaning that IC is greatly favored over
gamma photon emission.  There are a number of factors that are thought to
go into a high IC coefficient -- when the energy of the transition is
small, when the nucleus is large, and when the daughter nucleus has zero
spin, for example.  Unlike in the case of beta emission, the energies for
internal conversion electrons are not broadband and show up in line spectra
as sharp peaks.  This is because unlike in the case of beta decay there is
no neutrino to take part of the energy of the decay away from the emitted
electron.

Here's where my understanding starts to get fuzzy.  The above description
talked about isomeric transitions, which involve the decay of a metastable
isomer to the ground state of the isotope.  Metastable isomers are
long-lived excited states of nuclei, ones that have significant half-lives.
 Similar, shorter-lived nuclei are not considered metastable and are
instead referred to as compound nuclei.  For example, in dd fusion, the
short-lived compound nucleus [dd]* is not an isomer of 4He because it
decays quite rapidly in one of three ways -- to p+t, n+3He and 4He+ɣ.  (The
4He+ɣ branch is orders of magnitude less likely than the p+t and n+3He
branches, whose likelihoods are roughly split 50-50.)  I understand from
reading around that the emission of a gamma photon during the deexcitation
of a metastable isomer can be on the order of 10E-9 seconds, and that the
time required for the emission is something that depends upon the spin of
the excited nucleus.  Excited nuclei with certain spins will take
significantly longer to emit a gamma photon than nuclei with other spins.
 Am I correct in thinking that the same principles apply to a compound
nucleus such as [dd]*?  I.e., in the gamma photon branch, the [dd]*
de-excitation is on the order of 10E-9 seconds, or perhaps longer?  Also, I
haven't found a reference that gives the approximate times needed for the
other branches (p+t and n+3He), in which the compound nucleus splits up
into fragments.

Returning to internal conversion, one explanation for it focuses on the
fact that inner shell electrons have a high probability of passing through
the nucleus.  The idea is that during the time that the electron is within
the nucleus there is a nontrivial probability that it will interact with
the excited state, which, if this happens, will result in the energy of the
excited state being passed on to the electron.  The implication is that the
less likely an electron is to be found within the nucleus, the less likely
that the electron will be ejected as a result of internal conversion.  So
the probability of IC is highest with K-shell electrons and decreases the
further you go out.  One question I have about this explanation is that IC
is mediated via the electromagnetic interaction; my understanding of this
is that there is a virtual photon that passes from the nucleus to the
electron.  I do not see why the electron would particularly need to be
passing through the nucleus for such a virtual photon to reach it, for the
electromagnetic interaction is long-range.  Anywhere a virtual photon can
reach, it seems, there would be a nontrivial probability for an internal
conversion decay to occur.  Another challenge I have with this explanation
is that I think the de Broglie wavelength of an orbital electron is going
to be far larger than the nucleus or any than particular nucleon; my
understanding is that this is a problem because the de Broglie wavelengths
have to be roughly comparable for an interaction of some kind to be
probable.

One question I have has to do with the energy of the virtual photon.
 Internal conversion is less likely, other things being equal, if the
energy of the transition is large (e.g., on the order of MeVs).  Not having
read this detail, I would have thought that the energy of the decay would
factor into the distance at which the electron would need to be in order
for an interaction to be likely.  At low energies (in the keV), the
electron would need to be nearby, and at higher energies (MeV), the
electron would need to be further away.  What details of the underlying
mechanics am I missing in thinking this?

Eric

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