Can someone explain nuclear isomers to me, and in particular what the energy involved is? I understand generally that we're talking about moving from a less to more stable configuration of nuclear particles, but that's about as far as my understanding goes.
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asked Sep 19, 2011 at 20:26
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$\begingroup$ en.wikipedia.org/wiki/Nuclear_isomer $\endgroup$– Ron MaimonCommented Sep 20, 2011 at 23:10
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$\begingroup$ Hey, look! I win! $\endgroup$– dmckee --- ex-moderator kittenCommented Sep 21, 2011 at 1:01
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$\begingroup$ Yeah, I've read the wikipedia article, but I think it could use a lot more depth. It skips straight from a very superficial treatment (like my understanding above) to details, without ever stopping to explain principles, like I feel like dmckee did below. $\endgroup$– Chris B. BehrensCommented Sep 21, 2011 at 13:50
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2 Answers
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Given that we don't have an answer yet, I'll chime in to say that I've not heard this term before despite being a nuclear physicist in grad school.
Isotopes are atoms with the same number of protons, but different numbers of neutrons. Accordingly they have the same chemical properties, but differing atomic masses. (Actually there is a small difference in some chemical properties such as vapor pressures that can be used to extract enriched isotopic samples, but these differences are small enough that the processes tend to be expensive.)
Isomers in chemistry are molecules with the same atomic content, but differing structure. The simplest example that comes to mind are the left- and right-handed versions of some large organic molecules.
From that we deduce that if we found some isotope that had two stable configurations we would call them "nuclear isomers" by the obvious extension. Which leads to the speculation that meta-stable nuclear states might be given that name.
Take that with a grain of salt. I've no references and no experience with the term in the mouths of practicing scientists.
answered Sep 20, 2011 at 19:15
dmckee --- ex-moderator kittendmckee --- ex-moderator kitten
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$\begingroup$ So basically, it's neutrons and protons in a less stable configuration. And when the nucleus collapses to a more stable configuration, it's releasing the previously potential weak nuclear force? $\endgroup$ Commented Sep 20, 2011 at 20:39
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$\begingroup$ If this conjecture is correct, yes a rearranging of the nucleons (how familiar are you with the nuclear shell model?). However, the potential released will be primarily related to the strong nuclear force (as manifested at nuclear lengths scales i.e. meson exchange potentials). $\endgroup$ Commented Sep 20, 2011 at 20:43
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$\begingroup$ Ah...that's useful. Sounds like I have some more reading to do. $\endgroup$ Commented Sep 20, 2011 at 20:45
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$\begingroup$ dmckee - your deduction is spot on. I find it interesting that you've not encountered this term - Technetium-99m (used as a medical tracer) is a fairly well known 'nuclear isomer'. Perhaps there are other names used for these states? $\endgroup$ Commented Sep 21, 2011 at 4:20
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1$\begingroup$ @Richard, well I was a JLAB medium-energy, hard-probe kind of nuclear physicist rather than a xray-spectrum, energy-level-forest, multipole-transition listing kind of nuclear physicist. So I have broad regions of ignorance in what would have been a traditional nuclear physicist core competency. $\endgroup$ Commented Sep 21, 2011 at 14:14
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same atomic number and same mass number but differ in their structure and energy.
