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I'm using this link to search for particular energies in which gammas may be emitted (for nuclide identification on a gamma spectrum).

If on the above link you go down to the "γ condition #1" line, and put the energy between 2241 an 2243, and click search, you get the list of Nuclides which emit gammas between these energies. The second in the list (24Mg), in the half life column, has a halflife given in keV.

screenshot of results

This isn't the only time this occurs, this is just one example. To my mind there's no way that you can have a half life given in keV.

Can you, in fact, give a half life in keV (or any unit of energy)? How would that make sense?

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    $\begingroup$ Commenting on my own question to ask David Z how he did that - Putting in the picture of exactly what I was talking about - and to politely thank him for tidying it up :) $\endgroup$
    – Matt
    Commented Jan 8, 2016 at 11:53
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    $\begingroup$ (re: question) I'm pretty sure it's not an error, because the search form allows you to put conditions on the half life using energy units. I know one can express a time as the reciprocal of an energy in natural units, but I'm not sure if that's what's going on here. FWIW, $\hbar/(2.5\ \mathrm{keV}) = 2.6\times 10^{-4}\ \mathrm{fs}$. (re: comment) I just did the search and took a screenshot, saved it to my desktop, and uploaded it like any other image. $\endgroup$
    – David Z
    Commented Jan 8, 2016 at 11:58
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    $\begingroup$ Related: physics.stackexchange.com/q/162467 physics.stackexchange.com/q/21282 $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jan 8, 2016 at 15:43

2 Answers 2

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I think I found your answer in the glossary section (slightly edited):

$T_{1/2}$ [the half-life] is related to (...) the width $\Gamma$ by $$ T_{1/2} = \frac{\hbar \ln 2}{\Gamma}$$ where $\hbar$ is the reduced Planck's constant. Please note that when the half-life of a given nuclear level is smaller than $10^{-15}$ seconds, it is customary to list the width of the level instead.

$\Gamma$ has units of energy, so plugging in $2.5\, \mathrm{keV}$ I get a half life of $1.82\times 10^{-19}\,\mathrm{s}$ or $1.82\times 10^{-4}\,\mathrm{fs}$.

What "width" refers to is the width of the energy curve: when measuring decays you can plot the probability of getting a particle with a specific energy as a function of said energy. This plot is bell shaped; it is centered around the average energy you get, and its width at half-maximum is called $\Gamma$. Measuring $\Gamma$ is usually how these very short half lives are determined.

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    $\begingroup$ Excellent find. We see the uncertainty principle at work here. The shorter the duration of an event, the less well defined the energy is. So very short duration (half life) events have a spread on the energy. Not sure if it would be Gaussian, but "bell shaped" is good. $\endgroup$
    – Floris
    Commented Jan 8, 2016 at 13:23
  • $\begingroup$ I think the relevant distribution is the Breit-Wigner distribution. At least, that's the one we use in elementary particle physics, but I believe it also applies to radioactive decay - I'm not sure though. (edit) This suggests it's actually the Lorentzian a.k.a. Cauchy-Lorentz distribution, which is related but simpler. $\endgroup$
    – David Z
    Commented Jan 8, 2016 at 14:08
  • $\begingroup$ The other key insight here is that as the time gets shorter, and harder to measure, the line width gets larger and easier... $\endgroup$
    – Floris
    Commented Jan 8, 2016 at 14:29
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It is in natural units that time and space may be described in energy units.

![units

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  • $\begingroup$ True - but the reasons for this (why use time sometimes, and energy at other times) is the interesting Cruz of this question. See the answer by @Javier and my associated comments $\endgroup$
    – Floris
    Commented Jan 8, 2016 at 14:31
  • $\begingroup$ @Floris It may be the reason for using energy units, but not calling it width and instead calling it half life is only justified mathematically in the natural units framework $\endgroup$
    – anna v
    Commented Jan 8, 2016 at 15:29
  • $\begingroup$ Fair enough. Your answer adds something valuable to Javier's answer, but I think the other answer is the better one. I'll upvote this all the same because I really agree with your comment. $\endgroup$
    – Floris
    Commented Jan 8, 2016 at 15:33

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