Subsections


14.19 Beta decay

Beta decay is the decay mechanism that affects the largest number of nuclei. It is important in a wide variety of applications, such as betavoltaics and PET imaging.


14.19.1 Energetics Data

In beta decay, or more specifically, beta-minus decay, a nucleus converts a neutron into a proton by emitting an electron. An electron antineutrino is also emitted. The number of neutrons $N$ decreases by one unit, and the number of protons $Z$ increases by one.

The new nucleus must be lighter than the original one. Classical mass conservation would say that the reduction in nuclear mass must equal the mass of the emitted electron plus the (negligible) mass of the antineutrino. However, Einstein’s mass-energy relation implies that that is not quite right. Mass is equivalent to energy, and the rest mass reduction of the nucleus must also provide the kinetic energies of the neutrino and electron and the (much smaller) one that the nucleus itself picks up during the decay.

Still, the bottom line is that the nuclear mass reduction must be at least the rest mass of the electron. In energy units, it must be at least 0.511 MeV, the rest mass energy of the electron.

Figure 14.45: Energy release in beta decay of even-odd nuclei. [pdf]
\begin{figure}
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...9,128.3,'147\,\,\,\,\ \ \ '
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Figure 14.46: Energy release in beta decay of odd-even nuclei. [pdf]
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...9,138.2,'146\,\,\,\,\ \ \ '
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Figure 14.47: Energy release in beta decay of odd-odd nuclei. [pdf]
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...9,130.3,'147\,\,\,\,\ \ \ '
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Figure 14.48: Energy release in beta decay of even-even nuclei. [pdf]
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...9,141.8,'148\,\,\,\,\ \ \ '
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Figures 14.45 through 14.48 show the nuclear mass reduction for beta decay as the vertical coordinate. The reduction exceeds the rest mass energy of the electron only above the horizontal center bands. The left half of each square indicates the nucleus before beta decay, the right half the one after the decay. The horizontal coordinate indicates the atomic numbers, with the values and element symbols as indicated. Neutron numbers are listed at the square itself. Lines connect pairs of nuclei with equal neutron excesses.

If the left half square is colored blue, beta decay is observed. Blue left half squares are only found above the center bands, so the mass reduction is indeed at least the mass of the electron. However, some nuclei are right on top of the band.

In beta-plus decay, the nucleus converts a proton into a neutron instead of the other way around. To find the energy release in that process, the figures may be read the other way around. The nucleus before the decay is now the right hand one, and the decay is observed when the right hand half square is red. The energy release is now positive downward, so it is now below the center bands that the mass reduction is sufficient to produce the rest mass of a positron that can carry the proton’ positive charge away. The positron, the anti-particle of the electron, has the same mass but opposite charge as the electron.

But note that red right-half squares extend to within the center bands. The reason is that instead of emitting a positron, the nucleus can capture an electron from the atomic electron cloud surrounding the nucleus. In that case, rather than having to come up with an electron mass worth of energy, the nucleus receives an infusion of that amount of energy.

It follows that the left-hand nucleus will suffer beta decay if the square is above the top of the band, while the right hand nucleus will suffer electron capture if the square is below the top of the band. Therefore at most one nucleus of each pair can be stable.

Electron capture is also called inverse beta decay. However, it is not really the inverse process. The reason is that these processes also emit neutrinos. In particular, beta decay emits an electron antineutrino in addition to the electron. Inverse beta decay does not capture an electron antineutrino, just an electron; instead it emits an electron neutrino. In fact, during the formation of neutron stars, inverse beta decay on steroids, tremendous amounts of these neutrinos are emitted, taking along a large amount of the available energy of the star.

More generally, in nuclear reactions absorption of a particle works out much the same as emission of the corresponding anti-particle, at least as far as conserved quantities other than energy are concerned. Capture of an electron adds one unit of negative charge, while emission of a positron removes one unit of positive charge. Either way, the nuclear charge becomes one unit more negative. In beta decay, a neutron with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ produces a proton and an electron, each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$, and a neutrino is needed to ensure that angular momentum is conserved. However, it can do that either by being emitted or being absorbed.

Below the center bands in figures 14.46 through 14.48, both electron capture and positron emission are possible. Electron capture has still an energy reduction advantage of two electron masses over positron emission. On the other hand, the wave functions of atomic electrons are so big compared to the size of the nucleus that the electron is very unlikely to be found inside the nucleus. A high-energy positron created by the nucleus itself can have a much shorter wave length.

Note that $\fourIdx{40}{19}{}{}{\rm {K}}$ potassium-40, with 21 neutrons, appears above the band in figure 14.47, indicating that it suffers beta decay. But it also appears below the band in figure 14.48, so that it also suffers electron capture and positron emission.

The magic neutron numbers are quite visible in the figures. For example, diagonal bands at neutron numbers 50, 82, and 126 are prominent in all four figures. Consider for example figure 14.46. For the 50/49 neutron nuclei, beta decay takes the tightly bound 50th neutron to turn into a proton. That requires relatively large energy, so the energy release is reduced. For the neighboring 52/51 nuclei, beta decay takes the much less tightly bound 52nd neutron, and the energy release is correspondingly higher.

The magic proton numbers tend to show up as step-downs in the curves. For example, consider the nuclei at the vertical $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 50 line also in figure 14.46. In the In/Sn (indium/tin) beta decay, the beta decay neutron becomes the tightly bound 50th proton, and the energy release is correspondingly high. In the Sb/Te (antimony/tellurium) decay, the neutron becomes the less tightly bound 52nd proton, and the energy release is lower.

When the neutron and proton magic number lines intersect, combined effects can be seen. One pointed out by Mayer in her Noble prize acceptance lecture [[10]] is the decay of argon-39. It has 18 protons and 21 neutrons. If you interpolate between the neighboring pairs of nuclei on the same neutron excess line in figure 14.45, you would expect argon-39 to be below the top of the center band, hence to be stable against beta decay. But the actual energy release for argon-39 is unusually high, and beta decay it does. Why is it unusually high? For the previous pairs of nuclei, beta decay converts a neutron in the 9-20 shell into a proton in the same shell. For the later pairs, beta decay converts a neutron in the 21-28 shell to a proton in the same shell. Only for argon-39, beta decay converts a neutron in the 21-28 shell into a proton in the lower energy 9-20 shell. The lowering of the major shell releases additional energy, and the decay has enough energy to proceed.

In figures 14.45 and 14.46, the lowest line for the lightest nuclei is unusually smooth. These lines correspond to a neutron excess of 1 or $\vphantom0\raisebox{1.5pt}{$-$}$1, depending on whether it is before or after the decay. The pairs of nuclei on these two lines are mirror nuclei. During beta decay the neutron that turns into a proton transfers from the neutron shells into the exact same position in the proton shells. Because of charge independence, the nuclear energy does not change. The Coulomb energy does change, but as a relatively small, long-range effect, it changes fairly gradually.

These lines also show that beta-plus decay and electron capture become energetically favored when the nuclei get heavier. That is to be expected since this are nuclei with almost no neutron excess. For them it is energetically favorable to convert protons into neutrons, rather than the other way around.


14.19.2 Von Weizsäcker approximation

Since the von Weizsäcker formula of section 14.10.2 predicts nuclear mass, it can be used to predict whether beta-minus or beta-plus/electron capture will occur.

The mathematics is relatively simple, because the mass number $A$ remains constant during beta decay. For a given mass number, the von Weizsäcker formula is just a quadratic in $Z$. Like in the previous subsection, consider again pairs of nuclei with the same $A$ and one unit difference in $Z$. Set the mass difference equal to the electron mass and solve the resulting equation for $Z$ using simple algebra.

It is then seen that beta-minus decay, respectively beta-plus decay / electron capture occurs for a pair of nuclei depending whether the average $Z$ value is less, respectively greater, than

\begin{displaymath}
\fbox{$\displaystyle
Z_{\rm bd} =
A \frac{4C_a +m_{\rm...
...{\rm p}-m_{\rm e}+C_cC_zA^{-1/3}}{8C_a + 2C_c A^{2/3}}
$} %
\end{displaymath} (14.50)

where the constants $C_{.}$ are as given in section 14.10.2. The nucleon pairing energy must be ignored in the derivation, so the result may be off by a pair of nuclei for even-even and odd-odd nuclei.

Figure 14.49: Examples of beta decay. [pdf]
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\centering
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\begin{picture}(...
...(0,0)[l]{$\fourIdx{40}{18}{}{}{\rm Ar}$}}
}
\end{picture}
\end{figure}

The result is plotted as the black curve in the decay graph figure 14.49. It gives the location where the change in nuclear mass is just enough for either beta-minus decay or electron capture to occur, with nothing to spare. The curve locates the stable nuclei fairly well. For light nuclei, the curve is about vertical, indicating there are equal numbers of protons and neutrons in stable nuclei. For heavier nuclei, there are more neutrons than protons, causing the curve to deviate to the right, the direction of increasing neutron excess.

Because of the pairing energy, stable even-even nuclei can be found well away from the curve. Conversely, stable odd-odd nuclei are hard to find at all. In fact, there are only four: hydrogen-2 (deuterium), lithium-6, boron-10, and nitrogen-14. For comparison, there are 150 stable even-even ones. For nuclei of odd mass number, it does not make much difference whether the number of protons is odd or the number of neutrons: there are 49 stable odd-even nuclei and 53 stable even-odd ones.

(There is also the bizarre excited $\fourIdx{180\rm {m}}{73}{}{}{\rm {Ta}}$ nucleus that is stable, and is odd-odd to boot. But that is an excited state and another story, which is discussed under gamma decay. The ground state $\fourIdx{180}{73}{}{}{\rm {Ta}}$ has a half life of only 8 hours, as a cooperative heavy odd-odd nucleus should.)

As an example of the instability of odd-odd nuclei, consider the curious case of potassium-40, $\fourIdx{40}{19}{}{}{K}$. It has both an odd number of protons, 19, and of neutrons, 21. Potassium-40 is pretty much on top of the stable line, as evident from the fact that both its neighbors, odd-even isotopes potassium-39 and potassium-41, are stable. But potassium-40 itself is unstable. It does have a lifetime comparable to the age of the universe; long enough for significant quantities to accumulate. About 0.01% of natural potassium is potassium-40.

But decay it does. Despite the two billion year average lifetime, there are so many potassium-40 nuclei in a human body that almost 5,000 decay per second anyway. About 90% do so through beta decay and end up as the doubly-magic calcium-40. The other 10% decay by electron capture or positron emission and end up as even-even argon-40, with 18 protons and 22 neutrons. So potassium-40 suffers all three beta decay modes, the only relatively common nucleus in nature that does.

Admittedly only 0.001% decays through positron emission. The nuclear mass difference of 0.99 MeV with argon-40 is enough to create a positron, but not by much. Before a positron can be created, potassium is almost sure to have captured an electron already. For a nucleus like xenon-119 the mass difference with iodine-119 is substantially larger, 4.5 MeV, and about 4 in 5 xenon-119 nuclei decay by positron emission, and the fifth by electron capture.

It is energetically possible for the potassium-40 decay product calcium-40 to decay further into argon-40, by capturing two electrons from the atom. Energetically possible means that this does not require addition of energy, it liberates energy, so it can occur spontaneously. Note that calcium-40 would have to capture two electrons at the same time; capturing just one electron would turn it into potassium-40, and that requires external energy addition. In other words, calcium-40 would have to skip over the intermediate odd-odd potassium 40. While it is possible, it is believed that calcium-40 is stable; if it decays at all, its half-life must be more than 5.9 zettayear (5.9 10$\POW9,{21}$ year).

But some even-even nuclei do decay through “double beta-minus” decay. For example, germanium-76 with 32 protons and 44 neutrons will in a couple of zettayear emit two electrons and so turn into even-even selenium-76, skipping over odd-odd arsenic-76 in the process. However, since the entire lifetime of the universe is much less than the blink of an eye compared to a zettayear, this does not get rid of much germanium-76. About 7.5% of natural germanium is germanium-76.

The reduced stability of odd-odd nuclei is the main reason that technetium (Tc) and promethium (Pm) can end up with no stable isotopes at all while their immediate neighbors have many. Both technetium and promethium have an odd-odd isotope sitting right on top of the separating line between beta-minus and beta-plus decay; technetium-98 respectively promethium-146. Because of the approximation errors in the von Weizsäcker formula, they are not quite on the theoretical curve in figure 14.49. However, examination of the experimental nuclear masses shows the excess mass reduction for beta-minus decay and electron capture to be virtually identical for these odd-odd nuclei. And in fact promethium-146 does indeed decay both ways. Technetium-98 could too, but does not; it finds it quicker to create an electron than to capture one from the surrounding atom.

Because the theoretical stable line slopes towards the right in figure 14.49, only one of the two odd-even isotopes next to technetium-98 should be unstable, and the same for the ones next to promethium-146. However, the energy liberated in the decay of these odd-even nuclei is only a few hundred keV in each case, far below the level for which the von Weizsäcker formula is anywhere meaningful. For technetium and promethium, neither neighboring isotope is stable. This is a qualitative failure of the von Weizsäcker model. But it is rare; it happens only for these two out of the lowest 82 elements. Few books even mention it is a fundamental failure of the formula.


14.19.3 Kinetic Energies

The kinetic energy of nuclear decay products is important to understand the correct nature of the decay.

Historically, one puzzling observation in beta decay was the kinetic energies with which the electrons came out. When the beta decay of a collection of nuclei of a given type is observed, the electrons come out with a range of kinetic energies. In contrast, in the alpha decay of a collection of nuclei of a given type, all alpha particles come out with pretty much the exact same kinetic energy.

Consider the reason. The total kinetic energy release in the decay of a given nucleus is called the $Q$ value. Following Einstein’s famous relation $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$, the $Q$ value in alpha decay is given by the reduction in the net rest mass energy during the decay:

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{\rm N1} c^2 - m_{\rm N2} c^2 - m_\alpha c^2
$} %
\end{displaymath} (14.51)

where 1 indicates the nucleus before the decay and 2 the nucleus after the decay.

Since energy must be conserved, the reduction in rest mass energy given by the $Q$-​value is converted into kinetic energy of the decay products. Classical analysis makes that:

\begin{displaymath}
Q = {\textstyle\frac{1}{2}} m_{\rm N2} v^2_{\rm N2} + {\textstyle\frac{1}{2}} m_\alpha v^2_\alpha
\end{displaymath}

This assumes that the initial nucleus is at rest, or more generally that the decay is observed in a coordinate system moving with the initial nucleus. Linear momentum must also be conserved:

\begin{displaymath}
m_{\rm N1} \vec v_{\rm N1}
= m_{\rm N2} \vec v_{\rm N2} + m_\alpha \vec v_\alpha
\end{displaymath}

but since the velocity of the initial nucleus is zero,

\begin{displaymath}
m_{\rm N2} \vec v_{\rm N2} = - m_\alpha \vec v_\alpha
\end{displaymath}

Square both sides and divide by $2m_{\rm {N2}}$ to get:

\begin{displaymath}
{\textstyle\frac{1}{2}} m_{\rm N2}v^2_{\rm N2}
= \frac {m_\alpha}{m_{\rm N2}}{\textstyle\frac{1}{2}} m_\alpha v^2_\alpha
\end{displaymath}

Now, excluding the special case of beryllium-8, the mass of the alpha particle is much smaller than that of the final nucleus. So the expression above shows that the kinetic energy of the final nucleus is much less than that of the alpha particle. The alpha particle runs away with almost all the kinetic energy. Its kinetic energy is almost equal to $Q$. Therefore it is always the same for a given initial nucleus, as claimed above. In the special case that the initial nucleus is beryllium-8, the final nucleus is also an alpha particle, and each alpha particle runs away with half the kinetic energy. But still, each alpha particle always comes out with a single value for its kinetic energy, in this case ${\textstyle\frac{1}{2}}Q$.

In beta decay, things would be pretty much the same if just an electron was emitted. The electron too would come out with a single kinetic energy. The fact that it did not led Pauli to propose that another small particle also comes out. That particle could carry away the rest of the kinetic energy. It had to be electrically neutral like a neutron, because the nuclear charge change is already accounted for by the charge taken away by the electron. The small neutral particle was called the neutrino by Fermi. The neutrino was also required for angular momentum conservation: a proton and an electron each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ have net spin 0 or 1, not $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ like the original neutron.

The neutrino that comes out in beta-minus decay is more accurately called an electron antineutrino and usually indicated by $\bar\nu$. The bar indicates that it is counted as an antiparticle.

The analysis of the kinetic energy of the decay products changes because of the presence of an additional particle. The $Q$-​value for beta decay is

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{{\rm{N}}1}c^2 - m_{{\rm{N}}2} c^2 - m_{\rm e}c^2 - m_{\bar\nu} c^2
$} %
\end{displaymath} (14.52)

However, the rest mass energy of the neutrino can safely be ignored. At the time of writing, numbers less than a single eV are bandied around. That is immeasurably small compared to the nuclear rest mass energies which are in terms of GeV. In fact, physicists would love the neutrino mass to be nonnegligible: then they could figure out what is was!

As an aside, it should be noted that the nuclear masses in the $Q$ values are nuclear masses. Tabulated values are invariably atomic masses. They are different by the mass of the electrons and their binding energy. Other books therefore typically convert the $Q$-​values to atomic masses, usually by ignoring the electronic binding energy. But using atomic masses in a description of nuclei, not atoms, is confusing. It is also a likely cause of mistakes. (For example, [30, fig. 11.5] seems to have mistakenly used atomic masses to relate isobaric nuclear energies.)

It should also be noted that if the initial and/or final nucleus is in an excited state, its mass can be computed from that of the ground state nucleus by adding the excitation energy, converted to mass units using $E$ $\vphantom0\raisebox{1.5pt}{$=$}$ $mc^2$. Actually, nuclear masses are usually given in energy units rather than mass units, so no conversion is needed.

Because the amount of kinetic energy that the neutrino takes away varies, so does the kinetic energy of the electron. One extreme case is that the neutrino comes out at rest. In that case, the given analysis for alpha decay applies pretty much the same way for beta decay if the alpha is replaced by the electron. This gives the maximum kinetic energy at which the electron can come out to be $Q$. (Unlike for the alpha particle, the mass of the electron is always small compared to the nucleus, and the nucleus always ends up with essentially none of the kinetic energy.) The other extreme is that the electron comes out at rest. In that case, it is the neutrino that pretty much takes all the kinetic energy. Normally, both electron and neutrino each take their fair share of kinetic energy. So usually the kinetic energy of the electron is somewhere in between zero and $Q$.

A further modification to the analysis for the alpha particle must be made. Because of the relatively small masses of the electron and neutrino, they come out moving at speeds close to the speed of light. Therefore the relativistic expressions for momentum and kinetic energy must be used, chapter 1.1.2.

Consider first the extreme case that the electron comes out at rest. The relativistic energy expression gives for the kinetic energy of the neutrino:

\begin{displaymath}
T_{\bar\nu} = \sqrt{(m_{\bar\nu} c^2)^2 + (pc)^2} - m_{\bar\nu} c^2
\end{displaymath} (14.53)

where $c$ is the speed of light and $p$ the momentum. The nucleus takes only a very small fraction of the kinetic energy, so $T_{\bar\nu}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $Q$. Also, whatever the neutrino rest mass energy $m_{\bar\nu}c^2$ may be exactly, it is certainly negligibly small. It follows that ${T}_{\bar\nu}\approx{Q}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $pc$.

The small fraction of the kinetic energy that does end up with the nucleus may now be estimated, because the nucleus has the same magnitude of momentum $p$. For the nucleus, the nonrelativistic expression may be used:

\begin{displaymath}
T_{\rm N2} = \frac{p^2}{2m_{\rm N2}} = pc \frac{pc}{2m_{\rm N2}c^2}
\end{displaymath} (14.54)

The final fraction is very small because the energy release $pc$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $Q$ is in MeV while the nuclear mass is in GeV. Therefore the kinetic energy of the nucleus is indeed very small compared to that of the neutrino. If higher accuracy is desired, the entire computation may now be repeated, starting from the more accurate value $T_{\bar\nu}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $Q-T_{\rm {N2}}$ for the kinetic energy of the neutrino.

The extreme case that the neutrino is at rest can be computed in much the same way, except that the rest mass energy of the electron is comparable to $Q$ and must be included in the computation of $pc$. If iteration is not desired, an exact expression for $pc$ can be derived using a bit of algebra:

\begin{displaymath}
pc = \sqrt{
\frac{[E^2 - (E_{\rm N2}+E_{\rm e})^2]
[E^...
...E_{\rm e})^2]}{4E^2}}
\qquad E = E_{\rm N2} + E_{\rm e} + Q
\end{displaymath} (14.55)

where $E_{\rm {N2}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $m_{\rm {N2}}c^2$ and $E_{\rm {e}}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${m_{\rm e}}c^2$ are the rest mass energies of the final nucleus and electron. The same formula may be used in the extreme case that the electron is at rest and the neutrino is not, by replacing $E_{\rm {e}}$ by the neutrino rest mass, which is to all practical purposes zero.

In the case of beta-plus decay, the electron becomes a positron and the electron antineutrino becomes an electron neutrino. However, antiparticles have the same mass as the normal particle, so there is no change in the energetics. (There is a difference if it is written in terms of atomic instead of nuclear masses.) In case of electron capture, it must be taken into account that the nucleus receives an infusion of mass equal to that of the captured electron. The $Q$-​value becomes

\begin{displaymath}
\fbox{$\displaystyle
Q = m_{{\rm{N}}1}c^2 + m_{\rm e}c^2 - m_{{\rm{N}}2} c^2 - m_{\bar\nu} c^2
- E_{\rm B,ce}
$} %
\end{displaymath} (14.56)

where $E_{\rm {B,ce}}$ is the electronic binding energy of the captured electron. Because this is an inner electron, normally a K or L shell one, it has quite a lot of binding energy, too large to be ignored. After the electron capture, an electron farther out will drop into the created hole, producing an X-ray. If that electron leaves a hole behind too, there will be more X-rays. The energy in these $X$-​rays subtracts from that available to the neutrino.

The binding energy may be ballparked from the hydrogen ground state energy, chapter 4.3, by simply replacing $e^2$ in it by $e^2Z$. That gives:

\begin{displaymath}
\fbox{$\displaystyle
E_{\rm B,ce} \sim 13.6\, Z^2\mbox{ eV}
$} %
\end{displaymath} (14.57)

The ballparks for electron capture in figure 14.52 use
\begin{displaymath}
E_{\rm B,ce} \sim {\textstyle\frac{1}{2}} (\alpha Z)^2 m_{\rm e}c^2
\left(1+ {\textstyle\frac{1}{4}}(\alpha Z)^2\right)
\end{displaymath} (14.58)

in an attempt to partially correct for relativistic effects, which are significant for heavier nuclei. Here $\alpha$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1/137 is the so-called fine structure constant. The second term in the parentheses is the relativistic correction. Without that term, the result is the same as (14.57). See addendum {A.38} for a justification.


14.19.4 Forbidden decays

Energetics is not all there is to beta decay. Some decays are energetically fine but occur extremely slowly or not at all. Consider calcium-48 in figure fig:betdec2e. The square is well above the center band, so energy-wise there is no problem at all for the decay to scandium-48. But it just does not happen. The half life of calcium-48 is 53 10$\POW9,{18}$ years, more than a billion times the entire lifetime of the universe. And when decay does happen, physicists are not quite sure anyway how much of it is beta decay rather than double beta decay.

The big problem is angular momentum conservation. As an even-even nucleus, calcium-48 has zero spin, while scandium-48 has spin 6 in its ground state. To conserve angular momentum during the decay, the electron and the antineutrino must therefore take six units of spin along. But to the extend that the nuclear size can be ignored, the electron and antineutrino come out of a mathematical point. That means that they come out with zero orbital angular momentum. They have half a unit of spin each, and there is no way to produce six units of net spin from that. The decay is forbidden by angular momentum conservation.

Of course, calcium-48 could decay to an excited state of scandium-48. Unfortunately, only the lowest two excited states are energetically possible, and these have spins 5 and 4. They too are forbidden.


14.19.4.1 Allowed decays

To understand what beta decays are forbidden, the first step is to examine what decays are allowed.

Consider the spins of the electron and antineutrino. They could combine into a net spin of zero. If they do, it is called a “Fermi decay.” Since the electron and antineutrino take no spin away, in Fermi decays the nuclear spin cannot change.

The only other possibility allowed by quantum mechanics is that the spins of electron and antineutrino combine into a net spin of one; that is called a “Gamow-Teller decay.” The rules of quantum mechanics for the addition of angular momentum vectors imply:

\begin{displaymath}
\fbox{$\displaystyle
\vert j_{\rm N1} - j_{{\rm e}\bar\n...
...ox{-.7pt}{$\leqslant$}}j_{\rm N1} + j_{{\rm e}\bar\nu}
$} %
\end{displaymath} (14.59)

where $j_{\rm {N1}}$ indicates the spin of the nucleus before the decay, $j_{\rm {N2}}$ the one after it, and $j_{{\rm {e}}\bar\nu}$ is the combined angular momentum of electron and antineutrino. Since $j_{{\rm {e}}\bar\nu}$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 for allowed Gamow-Teller decays, spin can change one unit or stay the same. There is one exception; if the initial nuclear spin is zero, the final spin cannot be zero but must be one. Transitions from spin zero to zero are only allowed if they are Fermi ones. But they are allowed.

Putting it together, the angular momentum can change by up to one unit in an allowed beta decay. Also, if there is no orbital angular momentum, the parities of the electron and antineutrino are even, so the nuclear parity cannot change. In short

\begin{displaymath}
\fbox{$\displaystyle
\mbox{allowed:}\qquad
\vert\Delta...
...}{$\leqslant$}}1 \qquad \Delta \pi_{\rm N} = \mbox{no}
$} %
\end{displaymath} (14.60)

where $\Delta$ indicates the nuclear change during the decay, $j_{\rm {N}}$ the spin of the nucleus, and $\pi_{\rm {N}}$ its parity.

One simple example of an allowed decay is that of a single neutron into a proton. Since this is a $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ to $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ decay, both Fermi and Gamow-Teller decays occur. The neutron has a half-life of about ten minutes. It can be estimated that the decay is 18% Fermi and 82% Gamow-Teller, [30, p. 290].

Some disclaimers are in order. Both the discussion above and the following one for forbidden decays are nonrelativistic. But neutrinos are very light particles that travel at very close to the speed of light. For such relativistic particles, orbital angular momentum and spin get mixed-up. That is much like they get mixed-up for the photon. That was such a headache in describing electromagnetic transitions in chapter 7.4.3. Fortunately, neutrinos turn out to have some mass. So the given arguments apply at least under some conditions, even if such conditions are never observed.

A much bigger problem is that neutrinos and antineutrinos do not conserve parity. That is discussed in more detail a later subsection, Above, this book simply told you a blatant lie when it said that the electron-antineutrino system, (or the positron-neutrino system in beta-plus decay), comes off with zero parity. A system involving a single neutrino or antineutrino does not have definite parity. And parity is not conserved in the decay process anyway. But the initial and final nuclear states do have definite parity (to within very high accuracy). Fortunately, it turns out that you get the right answers for the change in nuclear parity if you simply assume that the electron and antineutrino come off with the parity given by their orbital angular momentum.

No you cannot have your money back. You did not pay any.

A relativistic description of neutrinos can be found in {A.43}.


14.19.4.2 Forbidden decays allowed

As noted at the start of this subsection, beta decay of calcium-48 requires a spin change of at least 4 and that is solidly forbidden. But forbidden is not quite the same as impossible. There is a small loophole. A nucleus is not really a mathematical point, it has a nonzero size.

Classically that would not make a difference, because the orbital angular momentum would be much too small to make up the deficit in spin. A rough ballpark of the angular momentum of, say, the electron would be $pR$, with $p$ its linear momentum and $R$ the nuclear radius. Compare this with the quantum unit of angular momentum, which is $\hbar$. The ratio is

\begin{displaymath}
\frac{pR}{\hbar} = \frac{pc\, R}{\hbar c} = \frac{pc\, R}{197\mbox{ MeV fm}}
\end{displaymath}

with $c$ the speed of light. The product $pc$ is comparable to the energy release in the beta decay and can be ballparked as on the order of 1 MeV. The nuclear radius ballparks to 5 fm. As a result, the classical orbital momentum is just a few percent of $\hbar$.

But quantum mechanics says that the orbital momentum cannot be a small fraction of $\hbar$. Angular momentum is quantized to values $\sqrt{l(l+1)}\hbar$ where $l$ must be an integer. For $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 the angular momentum is zero, for $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 the angular momentum is $\sqrt{2}\hbar$. There is nothing in between. An angular momentum that is a small fraction of $\hbar$ is not possible. Instead, what is small in quantum mechanics is the probability that the electron has angular momentum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. If you try long enough, it may happen.

In particular, $pR$$\raisebox{.5pt}{$/$}$$\hbar$ gives a rough ballpark for the quantum amplitude of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state. (The so-called Fermi theory of beta decay, {A.44}, can be used to justify this and other assumptions in this section.) The probability is the square magnitude of the quantum amplitude, so the probability of getting $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 is roughly $(pR/\hbar)^2$ smaller than getting $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0. That is about 3 or 4 orders of magnitude less. It makes decays that have $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 that many orders of magnitude slower than allowed decays, all else being the same. But if the decay is energetically possible, and allowed decays are not, it will eventually happen. (Assuming of course that some completely different decay like alpha decay does not happen first.)

Decays with $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 are called “first-forbidden decays.” The electron and neutrino can then take up to 2 units of angular momentum away through their combined orbital angular momentum and spin. So the nuclear spin can change up to two units. Orbital angular momentum has negative parity if $l$ is odd, so the parity of the nucleus must change during the decay. Therefore the possible changes in nuclear spin and parity are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{first-forbidden:}\qquad
\ve...
...$\leqslant$}}2 \qquad \Delta \pi_{\rm{N}} = \mbox{yes}
$} %
\end{displaymath} (14.61)

That will not do for calcium-48, because at least 4 units of spin change is needed. In second-forbidden decays, the electron and neutrino come out with a net orbital angular momentum $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2. Second forbidden decays are another 3 or 4 order of magnitude slower still than first forbidden ones. The nuclear parity remains unchanged like in allowed decays. Where both allowed and second forbidden decays are possible, the allowed decay should be expected to have occurred long before the second forbidden one has a chance. Therefore, the interesting second-forbidden cases cases are those that are not allowed ones:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{second-forbidden:}\qquad
\v...
... 2\mbox{ or } 3 \qquad \Delta \pi_{\rm{N}} = \mbox{no}
$} %
\end{displaymath} (14.62)

In third forbidden decays, $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3. The transitions that become possible that were not in first forbidden ones are:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{third-forbidden:}\qquad
\ve...
...mbox{ or } 4
\qquad \Delta \pi_{\rm{N}} = \mbox{yes}
$} %
\end{displaymath} (14.63)

These transitions are still another 3 or 4 orders of magnitude slower than second forbidden ones. And they do not work for calcium-48, as both the calcium-48 ground state and the three reachable scandium-48 states all have equal, positive, parity.

Beta decay of calcium-48 is possible through fourth-forbidden transitions:

\begin{displaymath}
\fbox{$\displaystyle
\mbox{fourth-forbidden:}\qquad
\v...
... 4\mbox{ or } 5 \qquad \Delta \pi_{\rm{N}} = \mbox{no}
$} %
\end{displaymath} (14.64)

This allows decay to either the 5$\POW9,{+}$ and 4$\POW9,{+}$ excited states of scandium-48. However, fourth forbidden decays are generally impossibly slow.


14.19.4.3 The energy effect

There is an additional effect slowing down the beta decay of the 0$\POW9,{+}$ calcium-48 ground state to the 5$\POW9,{+}$ excited scandium-48 state. The energy release, or $Q$-​value, of the decay is only about 0.15 MeV.

One reason that is bad news, (or good news, if you like calcium-48), is because it makes the momentum of the electron and neutrino correspondingly small. The ratio $pR$$\raisebox{.5pt}{$/$}$$\hbar$ is therefore quite small at about 0.01. And because this is a fourth forbidden decay, the transition is slowed down by a ballpark $((pR/\hbar)^{-2})^4$; that means a humongous factor 10$\POW9,{16}$ for $pR$$\raisebox{.5pt}{$/$}$$\hbar$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0.01. If a 1 MeV allowed beta decay may take on the order of a day, you can see why calcium-48 is effectively stable against beta decay.

Figure 14.50: The Fermi integral. It shows the effects of energy release and nuclear charge on the beta decay rate of allowed transitions. Other effects exists. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...$}}
\put(-50,92){\makebox(0,0)[l]{$\beta^+$}}
\end{picture}
\end{figure}

There is another, smaller, effect. Even if the final nucleus is the 5$\POW9,{+}$ excited scandium-48 state, with a single value for the magnetic quantum number, there is still more than one final state to decay to. The reason is that the relative amounts of energy taken by the electron and neutrino can vary. Additionally, their directions of motion can also vary. The actual net decay rate is an integral of the individual decay rates to all these different states. If the $Q$-​value is low, there are relatively few states available, and this reduces the total decay rate too. The amount of reduction is given by the so-called “Fermi integral” shown in figure 14.50. A decay with a $Q$ value of about 0.15 MeV is slowed down by roughly a factor thousand compared to one with a $Q$ value of 1 MeV.

The Fermi integral shows beta plus decay is additionally slowed down, because it is more difficult to create a positron at a strongly repelling positively charged nucleus. The relativistic Fermi integral also depends on the nuclear radius, hence a bit on the mass number. Figure 14.50 used a ballpark value of the mass number for each $Z$ value, {A.44}.

The Fermi integral applies to allowed decays, but the general idea is the same for forbidden decays. In fact, half-lives $\tau_{1/2}$ are commonly multiplied by the Fermi integral $f$ to produce a “comparative half-life,” or “$ft$-​value” that is relatively insensitive to the details of the decay besides the degree to which it is forbidden. The $ft$-​value of a given decay can therefore be used to ballpark to what extent the decay is forbidden.

You see how calcium-48 can resist beta-decay for 53 10$\POW9,{18}$ years.


14.19.5 Data and Fermi theory

Figure 14.51 shows nuclei that decay primarily through beta-minus decay in blue. Nuclei that decay primarily through electron capture and beta-plus decay are shown in red. The sizes of the squares indicate the decay rates. Note the tremendous range of decay rates. It corresponds to half-lives ranging from milliseconds to 10$\POW9,{17}$ years. This is much like the tremendous range of half-lives in alpha decay. Decays lasting more than about twenty years are shown as a minimum-size dot in figure 14.51; many would be invisible shown on the true scale.

Figure 14.51: Beta decay rates. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...20,0){\makebox(0,0)[b]{?}}
}
\end{picture}}
\end{picture}
\end{figure}

The decay rates in figure 14.51 are color coded according to a guesstimated value for how forbidden the decay is. Darker red or blue indicate more forbidden decays. Note that more forbidden decays tend to have much lower decay rates. (Yellow respectively green squares indicate nuclei for which the degree to which the decay is forbidden could not be found by the automated procedures used.)

Figure 14.52: Beta decay rates as fraction of a ballparked value. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...0,0)[r]{$\fourIdx{236}{93}{}{}{\rm Np}$}}
}
\end{picture}
\end{figure}

Figure 14.52 shows the decay rates normalized with a theoretical guesstimate for them. Note the greatly reduced range of variation that the guesstimate achieves, crude as it may be. One major success story is for forbidden decays. These are often so slow that they must be shown as minimum-size dots in figure 14.51 to be visible. However, in figure 14.52 they join the allowed decays as full-size squares. Consider in particular the three slowest decays among the data set. The slowest of all is vanadium-50, with a half-life of 150 10$\POW9,{15}$ year, followed by cadmium-113 with 8 10$\POW9,{15}$ year, followed by indium-115 with 441 10$\POW9,{12}$ year. (Tellurium-123 has only a lower bound on its half life listed and is not included.) These decay times are long enough that all three isotopes occur naturally. In fact, almost all naturally occurring indium is the unstable isotope indium-115. Their dots in figure 14.51 become full squares in figure 14.52.

Another eye-catching success story is $\fourIdx{3}{1}{}{}{\rm {H}}$, the triton, which suffers beta decay into $\fourIdx{3}{2}{}{}{\rm {He}}$, the stable helion. The decay is allowed, but because of its miniscule energy release, or $Q$-​value, it takes 12 years anyway. Scaled with the ballpark, this slow decay too becomes a full size square.

The ballparks were obtained from the Fermi theory of beta decay, as discussed in detail in addendum {A.44}. Unlike the relatively simple theory of alpha decay, the Fermi theory is elaborate even in a crude form. Taking beta-minus decay as an example, the Fermi theory assumes a pointwise interaction between the wave functions of the neutron that turns into a proton and those of the electron/antineutrino pair produced by the decay. (Quantum mechanics allows the neutron before the decay to interact with the electron and neutrino that would exist if it had already decayed. That is a twilight effect, as discussed in chapter 5.3 and more specifically in addendum {A.24} for gamma decay.) The strength of the interaction is given by empirical constants.

Note that for many nuclei no ballparks were found. One major reason is that the primary decay mechanism is not necessarily to the ground state of the final nucleus. If decay to the ground state is forbidden, decay to a less-forbidden excited state may dominate. Therefore, to correctly estimate the decay rate for a given nucleus requires detailed knowledge about the excited energy states of the final nucleus. The energies of these excited states must be sufficiently accurately known, and they may not be. In particular, for a few nuclei, the energy release of the decay, or $Q$-​value, was computed to be negative even for the ground state. This occurred for the electron capture of $\fourIdx{163}{67}{}{}{\rm {Ho}}$, $\fourIdx{193}{78}{}{}{\rm {Pt}}$, $\fourIdx{194}{80}{}{}{\rm {Hg}}$, $\fourIdx{202}{82}{}{}{\rm {Pb}}$, and $\fourIdx{205}{82}{}{}{\rm {Pb}}$, and for the beta decay of $\fourIdx{187}{75}{}{}{\rm {Re}}$ and $\fourIdx{241}{94}{}{}{\rm {Pu}}$. According to the normal Fermi theory, the decay cannot occur if the $Q$-​value is negative. To be sure, there is some energy slop, and decays with slightly negative $Q$-​values could theoretically still happen, {A.44}. But that is irrelevant here because the $Q$-​values in question are much smaller than the estimated electronic binding energy (14.8). In fact they are comparable to the difference in electronic binding energy between initial and final nucleus or less. Since the binding energy is just an estimate, the computed $Q$-​values should not be trusted.

In addition to the energy of the excited states, their spins and parities must also be accurately known. The reason is that they determine to what level the decay is forbidden, hence slowed down. The computer program that produced figures 14.51 and 14.52 assumed conservatively that if no unique value for spin and/or parity was given, it might be anything. Also, while there was obviously no way for the program to account for any excited states whose existence is not known, the program did allow for the possibility that there might be additional excited states above the highest energy level known. This is especially important well away from the stable line where the excited data are often sparse or missing altogether. All together, for about one third of the nuclei processed, the uncertainty in the ballparked decay rate was judged too large to be accepted. For the remaining nuclei, the level to which the decay was forbidden was taken from the excited state that gave the largest contribution to the decay rate.

The Fermi ballparks were constructed such that the true decay rate should not be significantly more than the ballparked one. In general they met that requirement, although for about 1% of the nuclei, the true decay rate was more ten times the ballparked ones, reaching up to 370 times for $\fourIdx{253}{100}{}{}{\rm {Fm}}$. All these cases were for first-forbidden decays with relatively low $Q$-​values. Since they included both beta minus and electron capture decays, a plausible explanation may be poor $Q$-​values. However, for forbidden decays, the correction of the electron/positron wave function for the effect of the nuclear charge is also suspect.

Note that while the true decay rate should not be much more than the ballparked one, it is very possible for it to be much less. The ballpark does not consider the details of the nuclear wave function, because that is in general prohibitively difficult. The ballpark simply hopes that if a decay is not strictly forbidden by spin or parity at level $l$, the nuclear wave function change will not for some other reason make it almost forbidden. But in fact, even if the decay is theoretically possible, the part of the Hamiltonian that gives rise to the decay may produce a nuclear wave function that has little probability of being the right one. In that case the decay is slowed down proportional to that probability.

As an example, compare the decay processes of scandium-41 and calcium-47. Scandium-41, with 21 protons and 20 neutrons, decays into its mirror twin calcium-41, with 20 protons and 21 neutrons. The decay is almost all due to beta-plus decay to the ground state of calcium-41. According to the shell model, the lone proton in the 4f$_{7/2}$ proton shell turns into a lone neutron in the 4f$_{7/2}$ neutron shell. That means that the nucleon that changes type is already in the right state. The only thing that beta decay has to do is turn it from a proton into a neutron. And that is in fact all that the decay Hamiltonian does in the case of Fermi decay. Gamow-Teller decays also change the spin. The nucleon does not have to be moved around spatially. Decays of this type are called “superallowed.” (More generally, superallowed decays are defined as decays between isobaric analog states, or isospin multiplets. Such states differ only in nucleon type. In other words, they differ only in the net isospin component $T_3$.) Superallowed decays proceed at the maximum rate possible. Indeed the decay of scandium-41 is at 1.6 times the ballparked value.

All the electron capture / beta-plus decays of the nuclei immediately to the left of the vertical $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $N$ line in figures 14.51 and 14.52 are between mirror nuclei, and all are superallowed. They are full-size squares in figure 14.52. Superallowed beta-minus decays occur for the triton mentioned earlier, as well as for a lone neutron.

But now consider the beta-minus decay process of calcium-47 to scandium-47. Calcium-47 has no protons in the 4f$_{7/2}$ proton shell, but it has 7 neutrons in the 4f$_{7/2}$ neutron shell. That means that it has a 1-neutron hole in the 4f$_{7/2}$ neutron shell. Beta decay to scandium-47 will turn one of the 7 neutrons into a lone proton in the 4f$_{7/2}$ proton shell.

At least one source claims that in the odd-particle shell model “all odd particles are treated equivalently,” so that we might expect that the calcium-47 decay is superallowed just like the scandium-41 one. That is of course not true. The odd-particle shell model does emphatically not treat all odd particles equivalently. It only says that, effectively, an even number of nucleons in the shell pair up into a state of zero net spin, leaving the odd particle to provide the net spin and electromagnetic moments. That does not mean that the seventh 4f$_{7/2}$ neutron can be in the same state as the lone proton after the decay. In fact, if the seventh neutron was in the same state as the lone proton, it would blatantly violate the antisymmetrization requirements, chapter 5.7. Whatever the state of the lone proton might be, 7 neutrons require 6 more independent states. And each of the 7 neutrons must occupy all these 7 states equally. It shows. The nuclear wave function of calcium-47 produced by the decay Hamiltonian matches up very poorly with the correct final wave function of scandium-47. The true decay rate of calcium-47 is therefore about 10,000 times smaller than the ballpark.

As another example, consider the beta-plus decay of oxygen-14 to nitrogen-14. Their isobaric analog states were identified in figure 14.44. Decay to the ground state is allowed by spin and parity, at a ballparked decay rate of 0.23/s. However, the true decay proceeds at a rate 0.01/s, which just happens to be 1.6 times the ballparked decay rate to the 0$\POW9,{+}$ excited isobaric analog state. One source notes additionally that over 99% of the decay is to the analog state. So decay to the ground state must be contributing less than a percent to the total decay. And that is despite the fact that decay to the ground state is allowed too and has the greater $Q$-​value. The effect gets even clearer if you look at the carbon-14 to nitrogen-14 beta-minus decay. Here the decay to the isobaric analog state violates energy conservation. The decay to the ground state is allowed, but it is more than 10,000 times slower than ballpark.

Superallowed decays like the one of oxygen-14 to the corresponding isobaric analog state of nitrogen-14 are particularly interesting because they are 0$\POW9,{+}$ to 0$\POW9,{+}$ decays. Such decays cannot occur through the Gamow-Teller mechanism, because in Gamow-Teller decays the electron and neutrino take away one unit of angular momentum. That means that decays of this type can be used to study the Fermi mechanism in isolation.

The horror story of a poor match up between the nuclear wave function produced by the decay Hamiltonian and the final nuclear wave function is lutetium-176. Lutetium-176 has a 7$\POW9,{-}$ ground state, and that solidly forbids decay to the 0$\POW9,{+}$ hafnium-176 ground state. However, hafnium has energetically allowed 6$\POW9,{+}$ and 8$\POW9,{+}$ excited states that are only first-forbidden. Therefore you would not really expect the decay of lutetium-176 to be particularly slow. But the spin of the excited states of hafnium is due to collective nuclear rotation, and these states match up extremely poorly with the ground state of lutetium-176 in which the spin is intrinsic. The decay rate is a stunning 12 orders of magnitude slower than ballpark. While technically the decay is only first-forbidden, lutetium is among the slowest decaying unstable nuclei, with a half-life of almost 40 10$\POW9,{12}$ year. As a result, it occurs in significant quantities naturally. It is commonly used to determine the age of meteorites. No other ground state nucleus gets anywhere close to that much below ballpark. The runner up is neptunium-236, which is 8 orders of magnitude below ballpark. Its circumstances are similar to those of lutetium-176.

The discussed examples show that the Fermi theory does an excellent job of predicting decay rates if the differences in nuclear wave functions are taken into account. In fact, if the nuclear wave function can be accurately accounted for, like in 0$\POW9,{+}$ to 0$\POW9,{+}$ superallowed decays, the theory will produce decay rates to 3 digits accurate, [30, table 9.2]. The theory is also able to give accurate predictions for the distribution of velocities with which the electrons or positrons come out. Data on the velocity distributions can in fact be used to solidly determine the level to which the decay is forbidden by plotting them in so-called “Fermi-Kurie plots.” These and many other details are outside the scope of this book.


14.19.6 Parity violation

For a long time, physicists believed that the fundamental laws of nature behaved the same when seen in the mirror. The strong nuclear force, electromagnetism, and gravity all do behave the same when seen in the mirror. However, in 1956 Lee and Yang pointed out that the claim had not been tested for the weak force. If it was untrue there, it could explain why what seemed to be a single type of K-meson could decay into end products of different parity. The symmetry of nature under mirroring leads to the law of conservation of parity, chapter 7.3. However, if the weak force is not the same under mirroring, parity can change in weak processes, and therefore, the decay products could have any net parity, not just that of the original K-meson.

Figure 14.53: Parity violation. In the beta decay of cobalt-60, left, the electron preferentially comes out in the direction that a left-handed screw rotating with the nuclear spin would move. Seen in the mirror, right, that becomes the direction of a right-handed screw.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...akebox(0,0)[b]{$\fourIdx{60}{27}{}{}{\rm Co}$}}
\end{picture}
\end{figure}

Wu and her coworkers therefore tested parity conservation for the beta decay of cobalt-60 nuclei. These nuclei were cooled down to extremely low temperatures to cut down on their thermal motion. That allowed their spins to be aligned with a magnetic field, as in the left of figure 14.53. It was then observed that the electrons preferentially came out in the direction of motion of a left-handed screw rotating with the nuclear spin. Since a left-handed screw turns into a right-handed one seen in the mirror, it followed that indeed the weak force is not the same seen in the mirror. The physics in the mirror is not the correct physics that is observed.

Since the weak force is weak, this does not affect parity conservation in other circumstances too much. Formally it means that eigenfunctions of the Hamiltonian are not eigenfunctions of the parity operator. However, nuclear wave functions still have a single parity to very good approximation; the amplitude of the state of opposite parity mixed in is of the order of 10$\POW9,{-7}$, [30, p. 313]. The probability of measuring the opposite parity is the square of that, much smaller still. Still, if a decay is absolutely forbidden when parity is strictly preserved, then it might barely be possible to observe the rare decays allowed by the component of the wave function of opposite parity.

An additional operation can be applied to the mirror image in 14.53 to turn it back into a physically correct decay. All particles can be replaced by their antiparticles. This operation is called “charge conjugation,” because among other things it changes the sign of the charge for each charged particle. In physics, you are always lucky if a name gets some of it right. Some of the particles involved may actually be charged, and conjugation is a sophisticated-​sounding term to some people. It is also a vague term that quite conceivably could be taken to mean reversal of sign by people naive enough to consider conjugation sophisticated. Charge conjugation turns the electrons going around in the loops of the electromagnet in figure 14.53 into positrons, so the current reverses direction. That must reverse the sign of the magnetic field if the physics is right. But so will the magnetic moment of anticobalt-60 nucleus change sign, so it stays aligned with the magnetic field. And physicist believe the positrons will preferentially come out of anticobalt-60 nuclei along the motion of a right-handed screw.

Besides this combined charge conjugation plus parity (CP) symmetry of nature, time symmetry is also of interest here. Physical processes should remain physically correct when run backwards in time, the same way you can run a movie backwards. It turns out that time symmetry too is not completely absolute, and neither is CP symmetry for that matter. However, if all three operations, charge conjugation (C), mirroring (P), and time inversion (T), together are applied to a physical process, the resulting process is believed to always be physically correct. There is a theorem, the CPT theorem, that says so under relatively mild assumptions.