How does the Gamma decay of Cobalt-60 occur?

Motivation: A research team led by D. Habs made contributions to our understanding of the gamma decay of Ca-40 and Zr-90: http://prl.aps.org/abstract/PRL/v53/i20/p1897_1. You don't need to own this article to answer, but does anyone have any idea how to describe a similiar process for Cobalt-60?

For those of you who don't have the article, I am asking about the perturbation background during the gamma decay of Cobalt-60.

  • $\begingroup$ The paper you link seems to be an experimental paper with some interpretation. As such, using the same technique for another isotope means running the experiment again...check out the authors on Inspire to see if they've done it already. (The "citing articles" tab doesn't show any one having followed up with a similar paper for Co-60.) Hmmm...from 1984, too. There presumably has been some theoretical follow up, but the article is lightly cited. $\endgroup$ – dmckee --- ex-moderator kitten May 17 '12 at 21:56
  • $\begingroup$ One of the citations is a review article from 2004: doi:10.1238/Physica.Regular.069a000C1, which makes no mention of Co-60. Not promising. $\endgroup$ – dmckee --- ex-moderator kitten May 17 '12 at 22:05
  • $\begingroup$ So there is no way to theoretically predict or conceptually describe the double gamma decay of Cobalt 60? Can someone elaborate on what double gamma decay even means in this context because it seems to be unavoidable in confronting the problem of how Cobalt-60 decays. $\endgroup$ – James Yu-tai May 17 '12 at 22:22
  • $\begingroup$ I don't know, but the paper you link does not seem to be a good starting place for hunting down the answer you want. Nor do the papers that cite it in turn. $\endgroup$ – dmckee --- ex-moderator kitten May 17 '12 at 22:41
  • $\begingroup$ I note that the Jp of Co-60 is 5+ and it's predominate decay is $\beta^-$ to Ni-60 with a Jp of 0+. That's a big change of angular momentum. Too much to be spanned with the just the spin on a electron and one photon (or even $\beta \gamma \gamma$), so there must be some nuclear rearrangement going on in there. $\endgroup$ – dmckee --- ex-moderator kitten May 17 '12 at 22:47

The paper that you cite describes decays in calcium-40 and zirconium-90 by emission of two photons at once. Both of these nuclides have a first excited state with spin-parity $0^+$, the same as their ground state. Since a single photon must carry away at least one unit of spin, these excitations cannot decay by one-photon emission. Mostly they decay by emitting a "virtual" photon, which produces a real positron-electron pair in the field of the nucleus. The paper you have linked measures a rare mode where two real photons are produced, and a surprising observation that in the double decay $E$-type photons are produced at the same rate as $M$-type photons. Usually, in nuclear decays, magnetic-dipole transitions are suppressed compared to electric-dipole transitions.

Since cobalt-60 has ground state spin-parity $5^+$, first excited state $2^+$, a single photon can mediate the transition. That photon must carry lots of orbital angular momentum, in addition to its spin, so the first excited state of cobalt is a relatively long-lived isomer (about ten minutes).

Cobalt-60 is used as a gamma source because it decays to an excited state of the nickel-60 nucleus, which then cools by emitting a sequence, or a "cascade," of photons. The most common path seems to be

\begin{align*} ^{60}\mathrm{Co}(5^+) &\to {}^{60}\mathrm{Ni}(4^+) + \beta + \bar\nu & \Delta E &= 0.316\,\mathrm{MeV} \\ ^{60}\mathrm{Ni}(4^+) &\to {}^{60}\mathrm{Ni}(2^+) + \gamma & \Delta E &= 1.173\,\mathrm{MeV} \\ ^{60}\mathrm{Ni}(2^+) &\to {}^{60}\mathrm{Ni}(0^+) + \gamma & \Delta E &= 1.332\,\mathrm{MeV} \end{align*}

These sorts of cascades are the bread and butter of nuclear physics; the double-photon decay paper you found is much rarer.


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