# 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 Gamov factor is smaller by $e^{-4}\simeq 0.02$, and emission should still be viable.

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I realized this is much like physics.stackexchange.com/q/23615, but the answer there is my assumption here, so I think it's still a different question. –  Yuval Apr 20 '13 at 13:46
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## 4 Answers

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.

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+1 for catching the algebraic error –  anna v Apr 21 '13 at 16:32
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To use a rather brain dead picture of the nucleus imagine that the nucleons can be modeled as a bunch of billiard balls zipping around (they can't but I'll discuss a way in which it the model is useful in a moment). In order for a large nucleus to emitted you have get all the nucleons that will make up that fragment moving in roughly the same direction and with roughly the same speed (in order that the fragment will be stable itself).

Not surprisingly it turns out that this is an exceedingly rare if the nucleon momenta are uncorrelated for the case of, say, eight protons and eight neutrons so that a $^{16}\mathrm{O}$ could be emitted.

Now from theory and models it is known that there are strong correlations in momentum space (which is not dissimilar from that "billiard ball" view as long as we only concentrate on momenta) for small numbers of nucleons but no evidence for correlations of large numbers of nucleons. So getting an alpha out is reasonably probably, but getting a big fragment out is hard.

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I don't like the idea of "brain dead" examples; I think an example should have physical merit or not. But philosophy aside, I don't understand the physical model you're suggesting that's about "strong correlations is momentum space". Ohanian speaks of an average of 0.1-1 alpha particles that exist, as a unit, in a nucleus. What you have to say about ${}^{16}O$ in that sense? Do you have any textbook or article you're quoting or can refer me to? –  Yuval Apr 20 '13 at 21:05
Your answer implies that one proton and one neutron would be a better leaving group than an alpha particle. –  JoeHobbit Apr 20 '13 at 21:15
I can't answer @JoeHobbit's objection at all: that's been a puzzle to me too, though I do note that deuteron emission would flip the last term of the semi-empirical mass approximation and may be energetically disfavored in some cases. No solid references, I've only delved into nuclear theory once when I had the opportunity to take if from Walecka. The momenta of groups $\approx A/2$ of nucleons must be anti-correlated at some level to maintain the CoM, and the independent-pair model---i.e. 2-particle correlations---is a popular starting place for investigations of nuclear structure. –  dmckee Apr 21 '13 at 14:44
This question seems relevant. physics.stackexchange.com/questions/23615/… –  JoeHobbit Apr 22 '13 at 21:33
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Apparently what I'm suggesting is called cluster decay, and can happen. According to Wikipedia:

Theoretically any nucleus with Z > 40 for which the released energy (Q value) is a positive quantity, can be a cluster-emitter. In practice, observations are severely restricted to limitations imposed by currently available experimental techniques which require a sufficiently short half-life, $T_c$ < $10^{32}$ s, and a sufficiently large branching ratio B > $10^{ −17}$.

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You have to realize we are talking quantum mechanics here, and not just balancing energies. There exist quantum mechanical solutions of the strong potential binding nucleons ( protons and neutrons) that are more stable than others. In the shell model the so called magic numbers are well described .

Nuclei which have neutron number and proton (atomic) numbers each equal to one of the magic numbers are called "double magic", and are especially stable against decay. Examples of double magic isotopes include helium-4 (4He), oxygen-16 (16O), calcium-40 (40Ca), calcium-48 (48Ca), nickel-48 (48Ni) and lead-208 (208Pb).

Magic numbers are found experimentally before the shell model described them, so it is another success of the shell model.

Edit after comments:

In a heavy nucleus the nuclei cluster, as you point in your answer. They do not cluster like building bricks, but according to quantum mechanical rules. To start with, though, the combination possibilities have to be considered because they will enter in the calculation of the probability of decay. There are many more combinations for alpha than oxygen. But it is not only the problem of combination in numbers but also of matching quantum numbers and masses so that the split can happen, if it is allowed energetically. It is a many body quantum mechanical problem , but I think the combinatorial diminution of probabilities plays a large role qualitatively in not finding many heavy nuclei break ups into heavier than alpha subsets.

Here is an article for clustering in light nuclei, and one, again for alpha , in heavy nuclei. It is an active region of research in nuclear physics currently.

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Thanks for the answer. I am familiar with the shell model - I'm writing up a report just in these subjects. I still don't see how this has to do with my question. The magic numbers, as far as I understand, simply denote anomalies in the binding energy graph. The binding energy is what it is. And it's pretty high for 9Be. For that matter, why isn't 16O emitted? –  Yuval Apr 20 '13 at 14:41
16O emitted from what? You have to describe the reaction. If isotopes are stable, they are in shells . If they have a lifetime the decay products will break up into two or more lower nuclei and it is not only the binding energy that matters but also the quantum numbers, the numbers of neutrons to protons and the spin states of the resultant nuclei, as well as the total binding energy break needed. The anomalies in the graph are explained by the shell model . Here ie.lbl.gov/decay/parent.pdf you can see that the decays start with beta, and alpha come out after atomic number 153. –  anna v Apr 20 '13 at 15:07
So my question would be for example, why would any of the isotopes after ${}^{145}Pm$ decay via ${}^4H$ emission and not via ${}^{16}O$ emission. –  Yuval Apr 20 '13 at 16:38
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