Subsections


14.11 Alpha Decay

In alpha decay a nucleus emits an “alpha particle,” later identified to be simply a helium-4 nucleus. Since the escaping alpha particle consists of two protons plus two neutrons, the atomic number $Z$ of the nucleus decreases by two and the mass number $A$ by four. This section explains why alpha decay occurs.


14.11.1 Decay mechanism

Figure 14.8 gives decay data for the nuclei that decay exclusively through alpha decay. Nuclei are much like cherries: they have a variable size that depends mainly on their mass number $A$, and a charge $Z$ that can be shown as different shades of red. You can even define a stem for them, as explained later. Nuclei with the same atomic number $Z$ are joined by branches.

Figure 14.8: Half-life versus energy release for the atomic nuclei marked in NUBASE 2003 as showing pure alpha decay with unqualified energies. Top: only the even values of the mass and atomic numbers cherry-picked. Inset: really cherry-picking, only a few even mass numbers for thorium and uranium! Bottom: all the nuclei except one. [pdf]
\begin{figure}
\centering
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...,0)[b]{5}}
\put(-66,210){\makebox(0,0)[b]{6}}
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Not shown in figure 14.8 is the unstable beryllium isotope $\fourIdx{8}{4}{}{}{\rm {Be}}$, which has a half-life of only 67 as, (i.e. 67 10$\POW9,{-18}$ s), and a decay energy of only 0.092 MeV. As you can see from the graph, these numbers are wildly different from the other, much heavier, alpha-decay nuclei, and inclusion would make the graph very messy.

Note the tremendous range of half-lives in figure 14.8, from mere nanoseconds to quintillions of years. And that excludes beryllium’s attoseconds. In the early history of alpha decay, it seemed very hard to explain how nuclei that do not seem that different with respect to their numbers of protons and neutrons could have such dramatically different half-lives. The energy that is released in the decay process does not vary that much, as figure 14.8 also shows.

To add to the mystery in those early days of quantum mechanics, if an alpha particle was shot back at the nucleus with the same energy that it came out, it would not go back in! It was reflected by the electrostatic repulsion of the positively charged nucleus. So, it had not enough energy to pass through the region of high potential energy surrounding the nucleus, yet it did pass through it when it came out.

Gamow, and independently Gurney & Condon, recognized that the explanation was quantum tunneling. Tunneling allows a particle to get through a potential energy barrier even if classically it does not have enough energy to do so, chapter 7.12.2.

Figure 14.9: Schematic potential for an alpha particle that tunnels out of a nucleus.
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...ut(-86,64){\makebox(0,0)[r]{$V_{\rm Coulomb}$}}
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Figure 14.9 gives a rough model of the barrier.The horizontal line represents the total energy of the alpha particle. Far from the nucleus, the potential energy $V$ of the alpha particle can be defined to be zero. Closer to the nucleus, the potential energy of the alpha particle ramps up due to Coulomb repulsion. However, right at the outer edge $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $R$ of the nucleus itself, the strong but very short-range attractive nuclear force pops up, and the combined potential energy plummets almost vertically downwards to some low value $V_n$. In between the radial position $r_1$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $R$ and some larger radius $r_2$, the potential energy exceeds the total energy that the alpha particle has available. Classically, the alpha particle cannot penetrate into this region. However, in quantum mechanics it retains a very small probability of doing so.

The region in between $r_1$ and $r_2$ is called the “Coulomb barrier.” It is a poorly chosen name, because the barrier is only a Coulomb one for an alpha particle trying to get in the nucleus. For an alpha particle trying to get out, it is a nuclear force barrier; here the Coulomb force assists the tunneling particle to get through the barrier and escape. The term nuclear barrier would avoid this ambiguity. Therefore physicists do not use it.

Now, to get a rough picture of alpha decay, imagine an alpha particle wave packet rattling around inside the nucleus trying to escape. Each time it hits the barrier at $r_1$, it has a small chance of escaping. Eventually it gets lucky.

Assume that the alpha particle wave packet is small enough that the motion can be assumed to be one-di­men­sion­al. Then the small chance of escaping each time it hits the barrier is approximately given by the analysis of chapter 7.13 as

\begin{displaymath}
\fbox{$\displaystyle
T \approx e^{-2\gamma_{12}}
\qqua...
...bar} \int_{r_1}^{r_2} \sqrt{2m_\alpha(V-E)} {\,\rm d}r
$} %
\end{displaymath} (14.12)

The fact that this probability involves an exponential is the basic reason for the tremendous range in half-lives: exponentials can vary greatly in magnitude for relatively modest changes in their argument.


14.11.2 Comparison with data

The previous subsection explained alpha decay in terms of an imprisoned alpha particle tunneling out of the nucleus. To verify whether that is reasonable, the next step is obviously to put in some ballpark numbers and see whether the experimental data can be explained.

First, the energy $E$ of the alpha particle may be found from Einstein’s famous expression $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, section 14.7. Just find the difference between the rest mass of the original nucleus and the sum of that of the final nucleus and the alpha particle, and multiply by the square speed of light. That gives the energy release. It comes out primarily as kinetic energy of the alpha particle, ignoring any excitation energy of the final nucleus. (A reduced mass can be used to allow for recoil of the nucleus.) Note that alpha decay cannot occur if $E$ is negative; the kinetic energy of the alpha particle cannot be negative.

It may be noted that the energy release $E$ in a nuclear process is generally called the “$Q$-​value.” The reason is that one of the most common other quantities used in nuclear physics is the so-called quadrupole moment $Q$. Also, the total nuclear charge is indicated by $Q$, as is the quality factor of radiation, while projection and rotation operators, and second points are often also indicted by $Q$. The underlying idea is that when you are trying to figure out some technical explanation, then if almost the only mathematical symbol used is $Q$, it provides a pretty strong hint that you are probably reading a book on nuclear physics.

The nuclear radius $R$ approximately defines the start $r_1$ of the region that the alpha particle has to tunnel through, figure 14.9. It can be ballparked reasonably well from the number of nucleons $A$; according to section 14.10,

\begin{displaymath}
R \approx R_A \sqrt[3]{A} \qquad R_A = 1.23 \mbox{ fm}
\end{displaymath}

where f, femto, is 10$\POW9,{-15}$. That is a lot smaller than the typical Bohr radius over which electrons are spread out. Electrons are far away and are not really relevant.

It should be pointed out that the results are very sensitive to the assumed value of $r_1$. The simplest assumption would be that at $r_1$ the alpha particle would have its center at the nuclear radius of the remaining nucleus, computed from the above expression. But very noticeable improvements are obtained by assuming that at $r_1$ the center is already half the radius of the alpha particle outside. (In literature, it is often assumed that the alpha particle is a full radius outside, which means fully outside but still touching the remaining nucleus. However, half works better and is maybe somewhat less implausible.)

The good news about the sensitivity of the results on $r_1$ is that conversely it makes alpha decay a reasonably accurate way to deduce or verify nuclear radii, [30, p. 57]. You are hardly likely to get the nuclear radius noticeably wrong without getting into major trouble explaining alpha decay.

The number of escape attempts per unit time is also needed. If the alpha particle has a typical velocity $v_\alpha$ inside the original nucleus, it will take it a time of about $2r_0$$\raisebox{.5pt}{$/$}$$v_\alpha$ to travel the $2r_0$ diameter of the nucleus. So it will bounce against the barrier about $v_\alpha$$\raisebox{.5pt}{$/$}$$2r_0$ times per second. That is sure to be a very large number of times per second, the nucleus being so small, but each time it hits the perimeter, it only has a miniscule $e^{-2\gamma_{12}}$ chance of escaping. So it may well take trillions of years before it is successful anyway. Even so, among a very large number of nuclei a few will get out every time. Remember that a mol of atoms represents in the order of 10$\POW9,{23}$ nuclei; among that many nuclei, a few alpha particles are likely to succeed whatever the odds against. The relative fraction of successful escape attempts per unit time is by definition the reciprocal of the lifetime $\tau$;

\begin{displaymath}
\frac{v_\alpha}{2r_0} e^{-2\gamma_{12}} = \frac{1}{\tau}
\end{displaymath} (14.13)

Multiply the lifetime by $\ln2$ to get the half-life.

The velocity $v_\alpha$ of the alpha particle can be ballparked from its kinetic energy $E-V_n$ in the nucleus as $\sqrt{2(E-V_n)/m_\alpha}$. Unfortunately, finding an accurate value for the nuclear potential $V_n$ inside the nucleus is not trivial. But have another look at figure 14.8. Forget about engineering ideas about acceptable accuracy. A 50% error in half-life would be invisible seen on the tremendous range of figure 14.8. Being wrong by a factor 10, or even a factor 100, two orders of magnitude, is ho-hum on the scale that the half-life varies. So, the potential energy $V_n$ inside the nucleus can be ballparked. The current results use the typical value of $\vphantom0\raisebox{1.5pt}{$-$}$35 MeV given in [30, p. 252].

That leaves the value of $\gamma_{12}$ to be found from the integral over the barrier in (14.12). Because the nuclear forces are so short-range, they should be negligible over most of the integration range. So it seems reasonable to simply substitute the Coulomb potential everywhere for $V$. The Coulomb potential is inversely proportional to the radial position $r$, and it equals $E$ at $r_2$, so $V$ can be written as $V$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Er_2$$\raisebox{.5pt}{$/$}$$r$. Substituting this in, and doing the integral by making a change of integration variable to $u$ with $r$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r_2\sin^2u$, produces

\begin{displaymath}
\gamma_{12} = \frac{\sqrt{2m_\alpha E}}{\hbar} r_2
\left...
...{r_1}{r_2}\right)}
-\arcsin\sqrt{\frac{r_1}{r_2}}
\right]
\end{displaymath}

The last two terms within the square brackets are typically relatively small compared to the first one, because $r_1$ is usually fairly small compared to $r_2$. Then $\gamma_{12}$ is about proportional to $\sqrt{E}r_2$. But $r_2$ itself is inversely proportional to $E$, because the total energy of the alpha particle equals its potential energy at $r_2$;

\begin{displaymath}
E = \frac{(Z-Z_\alpha)e\,Z_\alpha e}{4\pi\epsilon_0 r_2}
\qquad Z_\alpha=2
\end{displaymath}

That makes $\gamma_{12}$ about proportional to 1$\raisebox{.5pt}{$/$}$$\sqrt{E}$ for a given atomic number $Z$.

So if you plot the half-life on a logarithmic scale, and the energy $E$ on an reciprocal square root scale, as done in figure 14.8, they should vary linearly with each other for a given atomic number. This does assume that the variations in number of escape attempts are also ignored. The predicted slope of linear variation is indicated by the stems on the cherries in figure 14.8. Ideally, all cherries connected by branches should fall on a single line with this slope. The figure shows that this is quite reasonable for even-even nuclei, considering the rough approximations made. For nuclei that are not even-even, the deviations from the predicted slope are more significant. The next subsection discusses the major sources of error.

Figure 14.10: Half-life predicted by the Gamow / Gurney & Condon theory versus the true value. Top: even-even nuclei only. Bottom: all the nuclei except one. [pdf]
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The bottom line question is whether the theory, rough as it may be, can produce meaningful values for the experimental half-lives, within reason. Figure 14.10 shows predicted half-lives versus the actual ones. Cherries on the black line indicate that the correct value is predicted. It is clear that there is no real accuracy to the predictions in any normal sense; they are easily off by several orders of magnitude. What can you expect without an accurate model of the nucleus itself? However, the predictions do successfully reproduce the tremendous range of half-lives and they do not deviate from the correct values that much compared to that tremendous range. It is hard to imagine any other theory besides tunneling that could do the same.

The worst performance of the theory is for the $\fourIdx{209}{83}{}{}{\rm {Bi}}$ bismuth isotope indicated by the rightmost dot in figure 14.10. Its true half-life of 19 Ey, 19 10$\POW9,{18}$ years, is grossly underestimated to be just 9 Py, 9 10$\POW9,{15}$ years. Then again, since the universe has only existed about 14 10$\POW9,{9}$ years, who is going to live long enough to complain about it? Essentially none of the bismuth-209 that has ever been created in the universe has decayed. It took until 2003 for physicists to observe that bismuth-209 actually did decay; it is still listed as stable in many references. For $\fourIdx{8}{4}{}{}{\rm {Be}}$, which is not shown in the figure, the predicted half-life is 55 as (55 10$\POW9,{-18}$ s), versus a true value of 67 as.


14.11.3 Forbidden decays

You may wonder why there is so much error in the theoretical predictions of the half life. Or why the theory seems to work so much better for even-even nuclei than for others. A deviation by a factor 2,000 like for bismuth-209 seems an awful lot, rough as the theory may be.

Some of the sources of inaccuracy are self-evident from the theoretical description as given. In particular, there is the already mentioned effect of the value of $r_1$. It is certainly possible to correct for deviations from the Coulomb potential near the nucleus by a suitable choice of the value of $r_1$. However, the precise value that kills off the error is unknown, and unfortunately the results strongly depend on that value. To fix this would require an accurate evaluation of the nuclear force potential, and that is very difficult. Also, the potential of the electrons would have to be included. The alpha particle does reach a distance of the order of a tenth of a Bohr radius from the nucleus at the end of tunneling. The Bohr radius is here taken to be based on the actual nuclear charge, not the hydrogen one.

Also, the picture of a relatively compact wave packet of the alpha particle rattling around assumes that that the size of that wave packet is small compared to the nucleus. That spatial localization is associated with increased uncertainty in momentum, which implies increased energy. And the kinetic energy of the alpha particle is not really known anyway, without an accurate value for the nuclear force potential.

A very major other problem is the assumption that the final alpha particle and nucleus end up in their ground states. If either ends up in an excited state, the energy that the alpha particle has available for escape will be correspondingly reduced. Now the alpha particle will most certainly come out in its ground state; it takes over 20 MeV to excite an alpha particle. But for most nuclei, the remaining nucleus cannot be in its ground state if the mechanism is as described.

The main reason is angular momentum conservation. The alpha particle has no net internal angular angular momentum. Also, it was assumed that the alpha particle comes out radially, which means that there is no orbital angular momentum either. So the angular momentum of the nucleus after emission must be the same as that of the nucleus before the emission. That is no problem for even-even nuclei, because it is the same; even-even nuclei all have zero internal angular momentum in their ground state. So even-even nuclei do not suffer from this problem.

However, almost all other nuclei do. All even-odd and odd-even nuclei and almost all odd-odd ones have nonzero angular momentum in their ground state. Usually the initial and final nuclei have different values. That means that alpha decay that leaves the final nucleus in its ground state violates conservation of angular momentum. The decay process is called “forbidden.” The final nucleus must be excited if the process is as described. That energy subtracts from that of the alpha particle. Therefore the alpha particle has less energy to tunnel through, and the true half-life is much longer than computed.

Note in the bottom half of figure 14.10 how many nuclei that are not even-even do indeed have half-lifes that are orders of magnitude larger than predicted by theory. Consider the example of bismuth-209, with a half-life 2,000 times longer than predicted. Bismuth-209 has a spin, i.e.  an azimuthal quantum number, of $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. However, the decay product thallium-205 has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ in its ground state. If you check out the excited states of thallium-205, there is an excited state with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, but its excitation energy would reduce the energy of the alpha particle from 3.2 MeV to 1.7 MeV, making the tunneling process very much slower.

And there is another problem with that. The decay to the mentioned exited state is not possible either, because it violates conservation of parity, chapter 7.3 and 7.4. Saying the alpha particle comes out radially, as done above is not really correct. The proper quantum way to say that the alpha particle comes out with no orbital angular momentum is to say that its wave function varies with angular location as the spherical harmonic $Y_0^0$, chapter 4.2.3. In spectroscopic terms, it comes out in an s-wave. Now the initial bismuth atom has odd parity; its complete wave function changes sign if you everywhere replace ${\skew0\vec r}$ by $\vphantom0\raisebox{1.5pt}{$-$}$${\skew0\vec r}$. But the alpha particle, the excited thallium state, and the $Y_0^0$ orbital motion all have even parity; there is no change of sign. That means that the total final parity is even too, so the final parity is not the same as the initial parity. That violates conservation of parity so the process cannot occur.

Thallium-205 does not have excited states below 3.2 MeV that have been solidly established to have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 9}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ and odd parity, so you may start to wonder whether alpha decay for bismuth-209 is possible at all. However, the alpha particle could of course come out with orbital angular momentum. In other words it could come out with a wave function that has an angular dependence according to $Y_l^m$ with the azimuthal quantum number $l$ equal to one or more. These states have even parity if $l$ is even and odd parity when $l$ is odd. Quantum mechanics then allows the thallium-205 excited state to have any spin $j$ in the range $\vert\frac92-l\vert$ $\raisebox{-.3pt}{$\leqslant$}$ $j$ $\raisebox{-.3pt}{$\leqslant$}$ $\frac92+l$ as long as its parity is odd or even whenever $l$ is even or odd.

For example, bismuth-209 could decay to the ground state of thallium-205 if the orbital angular momentum of the alpha particle is $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5. Or it could decay to an excited $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 7}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ state with an excitation energy of 0.9 MeV if $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. The problem is that the kinetic energy in the angular motion subtracts from that available for the radial motion, making the tunneling, once again, much slower. In terms of the radial motion, the angular momentum introduces an additional effective potential $l(l+1)\hbar^2$$\raisebox{.5pt}{$/$}$$2m_{\alpha}r^2$, compare the analysis of the hydrogen atom in chapter 4.3.2. Note that this effect increases rapidly with $l$. However, the decay of bismuth-209 appears to be to the ground state anyway; the measured energy of the alpha particle turns out to be 3.14 MeV. The predicted half-life including the effective potential is found to be 4.6 Ey, much better than the one computed in the previous section.

One final source of error should be mentioned. Often alpha decay can proceed in a number of ways and to different final excitation energies. In that case, the specific decay rates must be added together. This effect can make the true half-life shorter than the one computed in the previous subsection. But clearly, this effect should be minor on the scale of half-lifes of figure 14.10. Indeed, while the predicted half-lifes of many nuclei are way below the true value in the figure, few are significantly above it.


14.11.4 Why alpha decay?

The final question that begs an answer is why do so many nuclei so specifically want to eject an helium-4 nucleus? Why none of the other nuclei? Why not the less tightly bound, but lighter deuteron, or the more tightly bound, but heavier carbon-12 nucleus? The answer is subtle.

To understand the reason, reconsider the analysis of the previous subsection for a more general ejected nucleus. Assume that the ejected particle has an atomic number $Z_1$ and mass $m_1$. As mentioned, the precise number of escape attempts is not really that important for the half life; almost all the variation in half-life is through the quantity $\gamma_{12}$. Also, to a first approximation the ratio of start to end of the tunneling domain, $r_1$$\raisebox{.5pt}{$/$}$$r_2$, can be ignored. Under those conditions, $\gamma_{12}$ is proportional to

\begin{displaymath}
\gamma_{12} \propto \sqrt{\frac{m_1}{E}} Z_1(Z-Z_1)
\end{displaymath}

It is pretty much all in there.

As long as the ejected particle has about the usual 8 MeV binding energy per nucleon, the square root in the expression above does not vary that much. In such cases the energy release $E$ is about proportional to the amount of nucleons ejected. Table 14.3 gives some example numbers. That makes $\gamma_{12}$ about proportional to $Z_1$, and the greatest chance of tunneling out then occurs by far for the lightest nuclei. It explains why the alpha particle tunnels out instead of heavier nuclei. It is not that a heavier nucleus like carbon-14 cannot be emitted, it is just that an alpha particle has already done so long before carbon-14 gets the chance. In fact, for radium-223 it has been found that one carbon-14 nucleus is ejected for every billion alpha particles. That is about consistent with the computed half-lifes of the events as shown in table 14.3.


Table 14.3: Candidates for nuclei ejected by uranium-238, radium-223, and fermium-256.
\begin{table}\begin{displaymath}
\renewedcommand{arraystretch}{1.2}
\begin{a...
...\,10^{31} \\
\hline\hline
\end{array}
\end{displaymath}
\end{table}


But the argument that $Z_1$ should be as small as possible should make protons or neutrons, not the alpha particle, the ones that can escape most easily. However, these do not have any binding energy. While protons or neutrons are indeed ejected from nuclei that have a very large proton, respectively neutron excess, normally the energy release for such emissions is negative. Therefore the emission cannot occur. Beta decay occurs instead to adjust the ratio between protons and neutrons to the optimum value. Near the optimum value, you would still think it might be better to eject a deuteron than an alpha. However, because the binding energy of the deuteron is only a single MeV per nucleon, the energy release is again negative. Among the light nuclei, the alpha is unique in having almost the full 8 MeV of binding energy per nucleon. It is therefore the only one that produces a positive energy release.

The final problem is that the arguments above seem to show that spontaneous fission cannot occur. For, is the fission of say fermium-256 into two tin-128 nuclei not just ejection of a tin-128 nucleus, leaving a tin-128 nucleus? The arguments above say that alpha decay should occur much before this can happen.

The problem is that the analysis of alpha decay is inapplicable to fission. The numbers for fission-scale half-lifes in table 14.3 are all wrong. Fission is indeed a tunneling event. However, it is one in which the energy barrier is disintegrating due to a global instability of the nuclear shape. That instability mechanism strongly favors large scale division over short scale ones. The only hint of this in table 14.3 are the large values of $r_1$$\raisebox{.5pt}{$/$}$$r_2$ for fission-scale events. When $r_1$$\raisebox{.5pt}{$/$}$$r_2$ becomes one, the tunneling region is gone. But long before that happens, the region is so small compared to the size of the ejected nucleus that the basic ideas underlying the analysis have become meaningless. Even ignoring the fact that the nuclear shapes have been assumed spherical and they are not in fission.

Thus, unlike table 14.3 suggests, fermium-256 does fission. The two fragments are usually of different size, but not vastly so. About 92% of fermium-256 nuclei spontaneously fission, while the other 8% experience alpha decay. Uranium-238 decays for 99.999,95% through $\alpha$ decay, and for only 0.000,05% through spontaneous fission. Although the amount of fission is very small, it is not by far as small as the numbers in table 14.3 imply. Fission is not known to occur for radium-223; this nucleus does indeed show pure alpha decay except for the mentioned rare carbon-14 emission.