Decay of Cobalt-60 isotope 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.
 A: 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.
