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The decay mode of Carbon-8 is listed as 'XP' in this table. None of the references I looked at listed XP as a decay mode. What is it?

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    $\begingroup$ half-life in keV is also suspicious $\endgroup$
    – Pygmalion
    Commented May 23, 2012 at 15:35
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    $\begingroup$ @Pygmalion: That's a width. It means a exceedingly short halflife, determined by $\Delta t \Delta E > \hbar/2$. Nothing surprising there as the isotope is very far from stability. $\endgroup$ Commented May 23, 2012 at 15:37
  • $\begingroup$ @dmckee Yes, this was my guess too, but all other halflifes are specified in second (including on Wikipedia). $\endgroup$
    – Pygmalion
    Commented May 23, 2012 at 15:52
  • $\begingroup$ @dmckee Is that notation used because the half life is so short (2*10^-21s) that measuring it is impractical at rest energy? $\endgroup$ Commented May 23, 2012 at 15:55
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    $\begingroup$ Dan, Two things. Firstly it is the width of the reconstructed energy that we use to measure these decays, and secondly the uncertainty in the width is roughly symmetric (i.e. you can safely write $\pm$), but when you invert it you get an asymmetric range unless the width uncertainty is very small. $\endgroup$ Commented May 23, 2012 at 16:11

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According to http://en.wikipedia.org/wiki/Isotopes_of_carbon#Table it emits two protons

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    $\begingroup$ Ah. Nice. The answer I was writing was "I note that the detailed listing spells this as 'Xp' which suggest multiple proton emission to me", which may have some value to the OP. $\endgroup$ Commented May 23, 2012 at 15:39
  • $\begingroup$ Thanks. My residual question stemming from this is if the wiki page only listed the most common number of protons released; or if the table from Berkley is using an notation that's more generic than necessary. $\endgroup$ Commented May 23, 2012 at 15:47
  • $\begingroup$ @DanNeely: The low precision of the width measurement suggest that this isotope has not been investigated very closely; as does the fact that there is only one reference and it dates to 1988. A really detailed understanding may not be available. $\endgroup$ Commented May 23, 2012 at 15:50
  • $\begingroup$ @dmckee LOL, this is the first time today that I succeeded to answer something before John Rennie did :) $\endgroup$
    – Pygmalion
    Commented May 23, 2012 at 15:59
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    $\begingroup$ Oops, sorry! I'm doing a really boring project at the moment and I'm desparately trying to avoid doing any real work :-) $\endgroup$ Commented May 23, 2012 at 17:48

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