Are there any textbooks you can recommend that touch on some of these areas in some detail?
Eric On Sun, Jul 6, 2014 at 3:41 PM, <mix...@bigpond.com> wrote: 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 > >
