# Why does carbon-12 decay take so much longer to occur than alpha decay?

In this specific question, I'm talking about why there is such a large difference in the time taken to emit a carbon nucleus, than an alpha nucleus.

In a recent lecture, my professor discussed the fact that heavy elements can emit a carbon nucleus as well as an alpha nucleus, but I missed out on the part when he discussed the reasons why we don't see carbon decay very often/it takes such a long time.

Any clarification on this would be appreciated.

• The Wikipedia article on 'Cluster Decay' would be a good place to start... May 5, 2017 at 13:46
• @JonCuster The wiki page explains that it has a very small branching ratio, but it doesn't explain the reason. May 5, 2017 at 14:27
• @peterh - sure, but the 27 references provided in the Wiki article cover it pretty well... May 5, 2017 at 15:39

The decay rate for alpha emission can be approximated by modelling the process as due to the particle quantum-tunneling out of a potential. The logarithm of the half-life is an additive constant (dependent on how we nondimensionalise time) plus $$2c\pi\alpha Z_1 (Z-Z_1)\sqrt{2m/E}$$, where the emitted particle has mass $$m$$, kinetic energy $$E$$ and nucleon number $$Z_1$$, and $$Z$$ is the nucleon number of the parent nucleus. While $$Z-Z_1$$ switches from $$Z-4$$ for alpha decay to $$Z-12$$ for carbon-12 decay, which is a little smaller than $$Z-4$$ for the kind of large nuclei susceptible to carbon-12 decay, the $$Z_1\sqrt{m}$$ factor is proportional to $$Z_1^{3/2}$$ so is $$5.2$$ times larger in the carbon-12 decay. The log half-life is thus somewhat larger in this case, making the half-life itself much longer than for alpha emission.

In https://en.wikipedia.org/wiki/Cluster_decay - pointed out by @JonCuster one finds

$\lambda = \ln 2 /T_c = \nu S P$

Comparing with $\alpha$ decay, $S$ - formation probability for heavier fragments can be much lower than $\alpha$, because they have more complicated structure, while $\alpha$ is a very good cluster in nucleus in many cases (in Li, C, ...).

Another term is $P$ - penetrability, which translates to possibility to overcome a Coulomb barrier. What is the "penetrability factor"? - but from elsewhere you find this equation:

$P \sim \exp(-2\pi z_1 z_2 e^2/\hbar v)$

From this one can see that (if you consider Z=92 nucleus) $z_1z_2$ is 180 for $\alpha$ (2*90) and 516 for carbon (6*86). Then you must have higher velocity $v$ for a similar $P$, which means a lower fraction of the the wavefunction can participate. And exponential makes it even more important.

One is used to a fact, that the Coulomb barrier plays a role when particle is approaching the nucleus from outside, but it is actually similar when the particle resides inside, because it is low in a potential well created by some field of all particles.

A simple (if hand-wavy) answer is that one part of the wavefunction for a carbon nucleus is a cluster of three alpha particles. Therefore, while you're waiting around for your carbon nucleus to form so that it can possibly tunnel out to cause a cluster decay, you're necessarily forming alpha particles that are more likely to tunnel out themselves than to cluster together inside the nucleus.

• It's not really true that a carbon nucleus is even partly a cluster of three alphas (such literal clustering violates the exclusion principle), and in any case this answer misses the huge factor of the lower penetrability of the Coulomb barrier.
– user4552
May 13, 2019 at 18:10