Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I was reading a paper today claiming observation of a new $2^+$ state in $^{12}\mbox{C}$, which would correspond to a sort of rotationally excited Hoyle state. Looking at NuDat reveals that this region of excitation energy seems well probed, with many states known above and below the new state. This got me to wondering:

What fraction of excited states are known in the light nuclei? (Presumably, heavier nuclei have very many states and must be less completely known. Also, let's assume we stay below the single-nucleon ejection energy so that we have a definite upper bound.)

Are there some regions where we know that all excited states are known? (Probably "yes" for systems such as the deuteron: there aren't any.)

Are there enough too-broad states such that we can never really know if we've got them all? (This new state is 0.8 MeV wide at 10.0 MeV above ground.)

share|improve this question
    
Just taking the title seriously, of course you have to answer "We don't know because we don't know how many unknown ones there are", but I appreciate that this is not very useful. If you had a good estimate of the number of scientist days going into this problem from year to year you could try to project to the asymptote. Beyond that I am at a loss. –  dmckee Mar 19 '13 at 22:52
    
I do note that study of nuclear structure continues where there is a reason for it, but isn't really considered hot for itself anymore; notwithstanding that there is a lot that could be done if people and money were directed at it. The structure of carbon is of great interest to people doing late life and high mass stars. I was involved in a study of $^{13}\mathrm{C}(\alpha,n)$ recently and the need for it had triggered a higher precision cross-section study. –  dmckee Mar 19 '13 at 22:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.