nuclear physics - Why are alpha particles such a prominent form of radiation and not other types of nucleon arrangement?

It is said in many textbooks that alpha decay involves emitting alpha particles, which are very stable. Indeed, the binding energy (~28.3 MeV) is higher than for $Z$-neighboring stable isotopes. But the binding energy is lower than, for example, ${}^9\mathrm{Be}$ (~58.2 MeV). My question is why aren't other nuclear compounds ejected from heavy nuclei, e.g. ${}^9\mathrm{Be}$?

The Gamow factor $$e^{-\frac{4\pi}{\hbar}\frac{Ze^2}{4\pi\epsilon_0}\frac{1}{v_\alpha}}$$ decreases exponentially in $Z$, so it explains intuitively why lower-$Z$ particles would tunnel more often. Specifically, it would explain why we would see ${}^9\mathrm{Be}$ emission ${e^{-2}}\simeq 0.14$ times as often compared to ${}^4\mathrm{He}$ emission. Also, the particles need to form in the nucleus prior emissions; but with a similar binding energy per nucleon (~7.08 MeV for ${}^4\mathrm{He}$ vs. 6.47 MeV for ${}^9\mathrm{Be}$) and higher total binding energy for the ${}^9\mathrm{Be}$ nucleus, I would expect that its formation in the same order of prevalence as the alpha particle (according to Ohanian, between 0.1 and 1 alpha particles are in existence at any moment in time).

Can anyone explain this? A reference to an article/textbook would be preferable.

EDIT Same goes for ${}^{16}\mathrm{O}$ which is also a double magic isotope, as 'anna v' pointed out. For it, the Gamow factor is smaller by $e^{-4}\simeq 0.02$, and emission should still be viable.

Answer

When people say that the decay rate depends critically on the $Q$ value, they're talking about alpha decays compared to other alpha decays. When you compare alpha decay to emission of other small clusters, the dependence on the atomic number $Z_c$ of the emitted cluster is much more prominent. The reason is as follows.

In the Gamow model of beta decay, we assume that the decay rate is the product of three factors: (1) a hand-wavy probability of preformation of the cluster; (2) the frequency with which a cluster assaults the Coulomb barrier; and (3) the probability of transmission through the barrier. (Re #1, don't take Ohanian too seriously when he says this factor is 0.1 to 1. Actually the literal existence of any cluster bouncing around inside an atomic nucleus would violate the exclusion principle. The whole thing is just a model.)

The critical factor is the tunneling probability $P$, which can be estimated using the WKB approximation, which looks like $\exp(-\int\ldots)$, where the integral is over the classically forbidden region. The integral depends on the Q value, because a higher Q value both shrinks the classically forbidden region and reduces the value of the integrand within that region. However, the height of the Coulomb barrier is proportional to the product $Z_cZ_d$ of the atomic numbers of the cluster and the daughter nucleus. If you want all the gory details, you can google the Geiger-Nuttall equation. But the result turns out to be of the form $\ln P=a-b$, where the $Z$-dependence is dominated by the term $a=(1/\hbar)\sqrt{32Z_cZ_d m_c R ke^2}$. For the alpha decay of uranium, we have $a\approx 74$. In Yval's example, decay by emission of 9Be basically doubles the value of $a$, which reduces the decay rate by a factor of $e^{-74}\approx10^{-32}$.

In the question, Yuval estimated that emission of Be should only be down by a factor of $e^{-2}$ relative to emission of alphas. This was an algebra mistake. We have an expression of the form $e^{-Zu}$, where $u$ is a constant. Changing this expression from $e^{-2u}$ to $e^{-4u}$ doesn't just reduce its value by a factor of $e^{-2}$, it reduces it by $e^{-2u}$, which is a huge factor.

Actually, as pointed out by JoeHobbit, the real mystery isn't why we don't emit larger clusters, it's why we don't emit lighter objects like protons or deuterons. A proton doesn't have to worry about preformation, and its tunneling probability would be much higher. Possibly this is due to the lower $Q$ value for proton emission. This comes into the Geiger-Nuttall equation because $b\propto Z_cZ_d/\sqrt{Q}$. In fact proton emission does occur, but it's only competitive for extremely proton-rich nuclei. There is also neutron emission, which doesn't involve any Coulomb barrier at all; as you'd expect, its half-life is very short (on the order of the assault frequency) when it's energetically allowed.