In reply to Eric Walker's message of Sat, 5 Jul 2014 23:28:13 -0700: Hi Eric, [snip] >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+?. (The >4He+? 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.
By first order approximation, metastable nuclei can't decay. In order to do so, I think they need some form of external interaction. That's why they have comparatively long lifetimes. > 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. You can get an idea of this from the HUP where delta E x delta t >= h_bar/2. If delta E is the energy of the reaction (about 4 MeV), then you get a time of at least 8E-23 sec. (I think this is the way it is normally calculated.) In any event it is obvious that a process that only takes order 1E-22 seconds is far more likely to occur than one which takes 1E-9 seconds (I think this is actually more like 1E-17 BTW, but I'm not sure whether these times can be measured or this is just calculated.) > >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. Yes but they are only present for a very short time. >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. A photon is only virtual if the separation distance is less than the wavelength of the photon (actually I suspect that this should be wavelength/2*Pi). >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. ...but it isn't very probable, that's why it can barely compete with "slow" gamma emission. However I don't know what part of the probability is due to duration of the interaction, and what part due to Be Broglie wavelength mismatch. Perhaps you also need to take into consideration the be Broglie wavelength of the nucleons. Also consider that by the time an orbital electron passes through the nucleus it has gained considerable kinetic energy from the electric field, so it's De Broglie wavelength is much shorter. (Still too long, but not by many orders of magnitude. :) > >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). Perhaps because high energy reactions often decay via other faster paths? The HUP logic applied here above would indicate that gamma decay times decrease with increasing energy, while IC times would be unaffected, because the chance of an electron being in the nucleus is not affected by the excited state thereof, so gamma decay has a better chance of winning the race at higher energies. >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 Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/project.html
