Subsections


14.20 Gamma Decay

Nuclear reactions and decays often leave the final nucleus in an quantum state of elevated energy. Such an excited state may lower its energy by emitting a photon of electromagnetic radiation. That is called gamma decay. It is a common way to evolve to the ground state.

Gamma decay is in many respect similar to alpha and beta decay discussed in earlier sections. However, the type of nucleus does not change in gamma decay. Both the atomic and mass number of the nucleus stay the same. (Of course, an excited nuclear state can suffer alpha or beta decay instead of gamma decay. That however is not the subject of this section.)

Gamma decay of excited nuclei is the direct equivalent of the decay of excited electron states in atoms. The big difference between gamma decay and the radiation emitted by the electrons in atoms is energy. The energy of the photons emitted by nuclei is typically even higher than that of the X-ray photons emitted by inner electrons. Therefore the radiation emitted by nuclei is generally referred to as “gamma rays.”

Both atomic radiation and nuclear gamma decay were analyzed in considerable detail in chapter 7.4 through 7.8 and addenda {A.20} through {A.25}. There is no point in repeating all that here. Instead this section will merely summarize the key points and discuss some actual observations.

However, the existing data on gamma decay is enormous. Consider NuDat 2, a standard data base. At the time of writing, it contains over 3,100 nuclei. Almost every nucleus but the deuteron has many excited energy levels; there are over 160,000 in NuDat 2. Gamma decays can proceed between different states, and NuDat 2 contains over 240,000 of them. There is no way that this book can cover all that data. The coverage given in this section will therefore be anecdotal or random rather than comprehensive.

However, based on a simple model, at least ballpark transition rates will be established. These are called the Weisskopf units. They are commonly used as reference values, to give some context to the measured transition rates.

One big limitation of gamma decay is for nuclear states of zero spin. A state of zero spin cannot transition to another state of zero spin by emitting a photon. As discussed in chapter 7.4, this violates conservation of angular momentum.

But there are other ways that a nucleus can get rid of excess energy besides emitting an electromagnetic photon. One way is by kicking an atomic electron out of the surrounding atom. This process is called “internal conversion” because the electron is outside the nucleus. It allows transitions between states of zero spin.

For atoms, two-photon emission is a common way to achieve decays between states of zero angular momentum. However, for nuclei this process is less important because internal conversion usually works so well.

Internal conversion is also important for other transitions. Gamma decay is slow between states that have little difference in energy and/or a big difference in spin. For such decays, internal conversion can provide a faster alternative.

If the excitation energy is high, it is also possible for the nucleus to create an electron and positron pair from scratch. Since the quantum uncertainty in position of the pair is far too large for them to be confined within the small nucleus, this is called “internal pair creation.”


14.20.1 Energetics

The reduction in nuclear energy during gamma decay is called the $Q$-​value. This energy comes out primarily as the energy of the photon, though the nucleus will also pick up a bit of kinetic energy, called the recoil energy.

Recoil energy will usually be ignored, so that $Q$ gives the energy of the photon. The photon energy is related to its momentum and frequency through the relativistic mass-energy and Planck-Einstein relations:

\begin{displaymath}
Q = E_{\rm N1} - E_{\rm N2} = p c = \hbar \omega
\end{displaymath} (14.65)

Typical tabulations list nuclear excitation energies as energies, rather than as nuclear masses. Unfortunately, the energies are usually in eV instead of SI units.

In internal conversion, the nucleus does not emit a photon, but kicks an electron out of the surrounding atomic electron cloud. The nuclear energy reduction goes into kinetic energy of the electron, plus the binding energy required to remove the electron from its orbit:

\begin{displaymath}
Q = E_{\rm N1} - E_{\rm N2} = T_{\rm e} + E_{\rm B,e}
\end{displaymath} (14.66)


14.20.2 Forbidden decays

The decay rate in gamma decay is to a large extent dictated by what is allowed by conservation of angular momentum and parity. The nucleus is almost a mathematical point compared to the wave length of a typical photon emitted in gamma decay. Therefore, it is difficult for the nucleus to give the photon additional orbital angular momentum. That is much like what happens in alpha and beta decay.

The photon has one unit of spin. If the nucleus does not give it additional orbital angular momentum, the total angular momentum that the photon carries off is one unit. That means that the nuclear spin cannot change by more than one unit.

(While this is true, the issue is actually somewhat more subtle than in the decay types discussed previously. For a photon, spin and orbital angular momentum are intrinsically linked. Because of that, a photon always has some orbital angular momentum. That was discussed in chapter 7.4.3 and in detail in various addenda such as {A.21}. However, the inherent orbital angular momentum does not really change the story. The bottom line remains that it is unlikely for the photon to be emitted with more than one unit of net angular momentum.)

The nuclear spin can also stay the same, instead of change by one unit, even if a photon with one unit of angular momentum is emitted, In classical terms the one unit of angular momentum can go into changing the direction of the nuclear spin instead of its magnitude, chapter 7.4.2. However, this only works if the nuclear spin is nonzero.

Parity must also be preserved, chapter 7.4. Parity is even, or 1, if the wave function stays the same when the positive direction of all three Cartesian axes is reversed. Parity is odd, or $\vphantom0\raisebox{1.5pt}{$-$}$1, if the wave function changes sign. Parities of separate sources are multiplied together to combine them. That is unlike for angular momentum, where separate angular momenta are added together.

In the normal, or allowed, decays the photon is emitted with odd parity. Therefore, the nuclear parity must reverse during the transition, chapter 7.4.2.

(To be picky, the so-called weak force does not preserve parity. This creates a very small uncertainty in nuclear parities. That then allows a very small probability for transitions in which the apparent parity is not conserved. But the probability for this is so small that it can almost always be ignored.)

Allowed transitions are called electric transitions because the nucleus interacts mainly with the electric field of the photon. More specifically, they are called “electric dipole transitions” for reasons originating in classical electromagnetics, chapter 7.7.2. For practical purposes, a dipole transition is one in which the photon is emitted with one unit of net angular momentum.

Transitions in which the nuclear spin change is greater than one unit, or in which the nuclear parity does not change, or in which the spin stays zero, are called forbidden. Despite the name, most such decays will usually occur given enough time. However they are generally much slower.

One way that forbidden transitions can occur is that the nucleus interacts with the magnetic field instead of the electric field. This produces what are called magnetic transitions. Magnetic transitions tend to be noticeably slower than corresponding electric ones. In magnetic dipole transitions, the photon has one unit of net angular momentum just like in electric dipole transitions. However, the photon now has even parity. Therefore magnetic dipole transitions allow the nuclear parity to stay the same.

Transitions in which the nuclear spin changes by more than one unit are possible through emission of a photon with additional orbital angular momentum. That allows a net angular momentum of the photon greater than one. But at a price. Each unit of additional net angular momentum slows down the typical decay rate by roughly 5 orders of magnitude.

The horror story is tantalum-180m. There are at the time of writing 256 ground state nuclei that are classified as stable. And then there is the excited nucleus tantalum-180m. Stable nuclei should be in their ground state, because states of higher energy decay into lower energy ones. But tantalum-180m has never been observed to decay. If it decays at all, it has been established that its half life cannot be less than 10$\POW9,{15}$ year. The universe has only existed for less than 10$\POW9,{10}$ years, and so tantalum-180m occurs naturally.

Figure 14.54: Energy levels of tantalum-180. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
% vertical spaci...
....2,0){\makebox(0,0)[l]{1}}
}
\end{picture}}
\end{picture}
\end{figure}

The tantalum-180 ground state shows no such idiocy. It is unstable as any self-respecting heavy odd-odd nucleus should be. In fact it disintegrates within about 8 hours through both electron capture and beta-minus decay at comparable rates. But tantalum-180m is an excited state with a humongous spin of 9. Figure 14.54 shows the excited energy levels of tantalum-180; tantalum-180m is the second excited energy level. It can only decay to the 1$\POW9,{+}$ ground state and to an 2$\POW9,{+}$ excited state. It has very little energy available to do either. The decay would require the emission of a photon with at least seven units of orbital angular momentum, and that just does not happen in a thousand years. Nor in a petayear.

You might think that tantalum-180m could just disintegrate directly through electron capture or beta decay. But those processes have the same problem. There is just no way for tantalum-180m to get rid of all that spin without emitting particles with unlikely large orbital angular momentum. So tantalum-180m will live forever, spinning too fast to reach the sweet oblivion of the quick death that waits below.

Electric transitions are often generically indicated as ${\rm {E}}\ell$ and magnetic ones as ${\rm {M}}\ell$. Here $\ell$ indicates the net angular momentum of the photon. That is the maximum nuclear spin change that the transition can achieve. So electric dipole transitions are ${\rm {E1}}$, and magnetic dipole transitions are ${\rm {M1}}$. Names have also been given to the higher multipole orders. For example, $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 transitions are quadrupole ones, $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 3 octupole, $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 4 hexadecapole, $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 5 triakontadipole, $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 6 hexacontatetrapole, etcetera. (If you are wondering, the prefixes in these names are powers of two, expressed in a mixture of Latin and Greek.)

For electric transitions, the nuclear parity changes when $\ell$ is odd. For magnetic transitions, it changes when $\ell$ is even. The transition rules are summarized in table 14.4.


Table 14.4: Nuclear spin and parity changes in electromagnetic multipole transitions.
\begin{table}\begin{displaymath}
\begin{array}{rcc}\hline\hline
& \mbox{maxi...
...n 0 transitions.}} \ \hline
\end{array}
\end{displaymath}
\end{table}


That leaves transitions from nuclear spin 0 to nuclear spin 0. Such transitions cannot occur through emission of a photon, period. For such transitions, conservation of angular momentum would require that the photon is emitted without angular momentum. But a photon cannot have zero net angular momentum. You might think that the spin of the photon could be canceled through one unit of orbital angular momentum. However, because the spin and orbital angular momentum of a photon are linked, it turns out that this is not possible, {A.21}.

Decay from an excited state with spin zero to another state that also has spin zero is possible through internal conversion or internal pair production. In principle, it could also be achieved through two-photon emission, but that is a very slow process that has trouble competing with the other two.

One other approximate conservation law might be mentioned here, isospin. Isospin is conserved by nuclear forces, and its charge component is conserved by electromagnetic forces, section 14.18. It can be shown that to the extent that isospin is conserved, certain additional selection rules apply. These involve the quantum number of square isospin $t_T$, which is the isospin equivalent of the azimuthal quantum number for the spin angular momentum of systems of fermions. Warburton & Weneser [47] give the following rules:

1.
Electromagnetic transitions are forbidden unless $\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0, or $\pm$1. (Here $\Delta$ means the difference between the initial and final nuclear states).
2.
Corresponding $\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\pm$1 transitions in conjugate nuclei are identical in all properties. (Conjugate means that the two nuclei have the numbers of protons and neutrons swapped. Corresponding transitions means transitions between equivalent levels, as discussed in section 14.18.)
3.
Corresponding ${\rm {E1}}$ transitions in conjugate nuclei -- whether $\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 or $\pm$1 -- have equal strengths.
4.
$\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ${\rm {E1}}$ transitions in self-conjugate nuclei are forbidden. (Self-​conjugate nuclei have the same number of protons as neutrons.)
5.
Corresponding $\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ${\rm {M1}}$ transitions in conjugate nuclei are expected to be of approximately equal strength, within, say, a factor of two if the transitions are of average strength or stronger.
6.
$\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ${\rm {M1}}$ transitions in self-conjugate nuclei are expected to be weaker by a factor of 100 than the average ${\rm {M1}}$ transition strength.
7.
$\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ${\rm {M}}\ell$ transitions in conjugate nuclei are expected to be of approximately equal strength if the transitions are of average strength or stronger.
8.
$\Delta{t}_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ${\rm {M}}\ell$ transitions in self-conjugate nuclei are expected to be appreciably weaker than average.
The last four rules involve an additional approximation besides the assumption that isospin is conserved.

In a nutshell, expect that the transitions will be unexpectedly slow if the isospin changes by more than one unit. Expect the same for nuclei with equal numbers of protons and neutrons if the isospin does not change at all and it is an ${\rm {E1}}$ or magnetic transition.

As an example, [30, p. 390], consider the decay of the 1$\POW9,{-}$ isobaric analog state common to carbon-14, nitrogen-14, and oxygen-14 in figure 14.44. This state has $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1. For oxygen-14, it is the lowest excited state. Its decay to the $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1, 0$\POW9,{+}$, ground state is an ${\rm {E1}}$ transition that is allowed by the spin, parity, and isospin selection rules. And indeed, the 1$\POW9,{-}$ excited state decays rapidly to the ground state; the half-life is about 0.000,012 fs (femtoseconds). That is even faster than the Weisskopf ballpark for a fully allowed decay, subsection 14.20.4, which gives about 0.009 fs here. But for nitrogen-14, the equivalent transition is not allowed because of rule 4 above. Nitrogen-14 has 7 protons and 7 neutrons. And indeed, the partial half life of this transition is 2.7 fs. That is very much longer. Based on rule 3 above, it is expected that the decay rate of the 1$\POW9,{-}$ state in carbon-14 is similar to the one in oxygen-14. Unfortunately, experimentally it has only been established that its half-life is less than 7 fs.

Some disclaimers are appropriate for this example. As far as the oxygen transition is concerned, the NuDat 2 [[12]] data do not say what the dominant decay process for the oxygen-14 state is. Nor what the final state is. So it might be another decay process that dominates. The next higher excited state, with 0.75 MeV more energy, decays 100% through proton emission. And two orders of magnitude faster than Weisskopf does seem a lot, figure 14.61.

As far as the nitrogen transition is concerned, the decay processes are listed in NuDat 2. The decay is almost totally due to proton emission, not gamma decay. The actual half-life of this state is 0.000,02 fs; the 2.7 fs mentioned above is computed using the given decay branching ratios. The 2.7 fs is way above the 0.006 fs Weisskopf estimate, but that is quite common for ${\rm {E1}}$ transitions.

It may be more reasonable to compare the forbidden nitrogen 8 MeV to 2.3 MeV transition to the allowed 8 MeV to ground state, and 8 MeV to 4 MeV transitions. They are all three ${\rm {E1}}$ transitions. Corrected for the differences in energy release, the forbidden transition is 20 times slower than the one to the ground state, and 25 times slower than the one to the 4 MeV state. So apparently, being forbidden seems to slow down this transition by a factor of roughly 20. It is significant, though it is not that big on the scale of figure 14.61.

As another example, the nitrogen transition from the 1$\POW9,{-}$ 5.7 MeV $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 state to the $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 ground state is also forbidden, while the transition to the 0$\POW9,{+}$ $t_T$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 state is now permitted. And indeed, the decay to the ground state is about ten times slower, when corrected for energy release, [30, p. 391].

More comprehensive data may be found in [47].


14.20.3 Isomers

An “isomer” is a long lasting excited state of a nucleus. Usually, an excited nucleus that does not disintegrate through other means will drop down to lower energies through the emission of photons in the gamma ray range. It will then end up back in the ground state within a typical time in terms of fs, or about 10$\POW9,{-15}$ second.

But sometimes a nucleus gets stuck in a metastable state that takes far longer to decay. Such a state is called an isomeric state. Krane [30, p. 174] ballparks the minimum lifetime to be considered a true isomeric state at roughly 10$\POW9,{-9}$ s, Bertulani [5, p. 244] gives 10$\POW9,{-15}$ s, and NuDat 2 [[12]] uses 10$\POW9,{-1}$ s with qualification in their policies and 10$\POW9,{-9}$ s in their glossary. Don’t you love standardization? In any case, this book will not take isomers serious unless they have a lifetime comparable to 10$\POW9,{-3}$ second. Why would an excited state that cannot survive for a millisecond be given the same respect as tantalum-180m, which shows no sign of kicking the bucket after 10$\POW9,{15}$ years?

But then, why would any excited state be able to last very much more than the typical 10$\POW9,{-15}$ s gamma decay time in the first place? The main reason is angular momentum conservation. It is very difficult for a tiny object like a nucleus to give a photon much angular momentum. Therefore, transitions between states of very different angular momentum will be extremely slow, if they occur at all. Such transitions are highly forbidden, or using a better term, hindered.

If an excited state has a very different spin than the ground state, and there are no states in between the two that are more compatible, then that excited state is stuck. But why would low spin states be right next to high spin states? The main reason is found in the shell model, and in particular figure 14.13. According to the shell model, just below the magic numbers of 50, 82, and 126, high spin states are pushed into regions of low spin states by the so-called spin-orbit interaction. That is a recipe for isomerism if there ever was one.

Therefore, it should be expected that there will be many isomers below the magic numbers of 50, 82, and 126. And that these isomers will have the opposite parity of the ground state, because the high spin states are pushed into low spin states of opposite parity.

Figure 14.55: Half-life of the longest-lived even-odd isomers. [pdf]
\begin{figure}
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...akebox(0,0)[b]{uncertain}}
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And so it is. Figure 14.55 shows the half-lifes of the longest-lasting exited states of even $Z$ and odd $N$ nuclei. The groups of isomers below the magic neutron numbers are called the “islands of isomerism.” The difference in spin from the ground state is indicated by the color. A difference in parity is indicated by a minus sign. Half-lives over 10$\POW9,{14}$ s are shown as full-size squares.

Figure 14.56: Half-life of the longest-lived odd-even isomers. [pdf]
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...akebox(0,0)[b]{uncertain}}
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Figure 14.56 shows the islands for odd $Z$, even $N$ nuclei.

Figure 14.57: Half-life of the longest-lived odd-odd isomers. [pdf]
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...
{$\fourIdx{180\rm m}{73}{}{}{\rm Ta}$}}
}
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\end{figure}

For odd-odd nuclei, figure 14.57, the effects of proton and neutron magic numbers get mixed up. Proton and neutron excitations may combine into larger spin changes, providing one possible explanation for the isomers of light nuclei without parity change.

Figure 14.58: Half-life of the longest-lived even-even isomers. [pdf]
\begin{figure}
\centering
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...akebox(0,0)[b]{uncertain}}
}
\end{picture}}
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For even-even nuclei, figure 14.58, there is very little isomeric activity.


14.20.4 Weisskopf estimates

Gamma decay rates can be ballparked using the so-called Weisskopf estimates:

\begin{displaymath}
\fbox{$\displaystyle
\lambda^{{\rm{E}}\ell} = C_{{\rm{E}...
...M}}\ell} = C_{{\rm{M}}\ell} A^{(2\ell-2)/3}Q^{2\ell+1}
$} %
\end{displaymath} (14.67)


\begin{displaymath}
\begin{array}{rccccc}\hline\hline
\ell: & 1 & 2 & \quad3...
...& 10 & 3.3 10^{-6} & 7.4 10^{-13}
\ \hline
\end{array}
\end{displaymath}

Here the decay rates are per second, $A$ is the mass number, and $Q$ is the energy release of the decay in MeV. Also $\ell$ is the maximum nuclear spin change possible for that transition. As discussed in subsection 14.20.2, electric transitions require that the nuclear parity flips over when $\ell$ is odd, and magnetic ones that it flips over when $\ell$ is even. In the opposite cases, the nuclear parity must stay the same. If there is more than one decay process involved, add the individual decay rates.

Figure 14.59: Weisskopf ballpark half-lifes for electromagnetic transitions versus energy release. Broken lines include ballparked internal conversion. [pdf]
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...5,206){\makebox(0,0)[rb]{227}}
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Figure 14.60: Moszkowski ballpark half-lifes for magnetic transitions versus energy release. Broken lines include ballparked internal conversion. [pdf]
\begin{figure}
\centering
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\begin{picture}(...
...b]{80}}
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The estimates are plotted in figure 14.59. For magnetic transitions the better Moszkowski estimates are shown in figure 14.60. (Internal conversion is discussed in subsection 14.20.7, where the ballparks are given.)

A complete derivation and discussion of these estimates can be found {A.25}. Note that many sources have errors in their formulae and/or graphs or use non-SI units, {A.25.9}. The correct formulae in SI units are in {A.25.8}.

These estimates are derived under the assumption that only a single proton changes states in the transition. They also assume that the multipole order is the lowest possible, given by the change in nuclear spin. And that the final state of the proton has angular momentum $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Some correction factors are available to allow for different multipole orders and different final angular momenta, {A.25.8}. There are also correction factors to allow for the fact that really the proton and the rest of the nucleus move around their common center of gravity. Similar correction factors can allow for the case that a single neutron instead of a proton makes the transition. See {A.25.8} for more.

The initial and final proton states assumed in the estimates are further very simple, {A.25.8}. They are like a shell model state with a simplified radial dependence. Corrections exist for that radial dependence too. But the way the estimates are mostly used in practice is as a reference. The actual decay rate in a transition is compared to the Weisskopf or Moszkowski estimates. These estimates are therefore used as units to express decay rates in.

If there is a big difference, it gives hints about the nature of the transition process. For example, the actual decay rate is often orders of magnitude smaller than the estimate. That can indicate that the state produced by the decay Hamiltonian has only a small probability of being the correct final nuclear state. In other words, there may be a poor match-up or “little overlap” between the initial and final nuclear states. It is implicit in the simple proton states used in the estimates that the state produced by the decay Hamiltonian has a good chance of being right. But actual ${\rm {E}}1$ transitions can easily be three or more orders of magnitude slower than estimate, as shown in the next subsection. That is similar to what was observed for the ballparks for beta decays given in section 14.19.5. One reason may be that some of these transitions are approximately forbidden by isospin conservation.

Conversely, the observed transition rate may be several orders of magnitude more rapid than the estimate. That may indicate that a lot of nucleons are involved in the transition. Their contributions can add up. This is frequently related to shape changes in nonspherical nuclei. For example, ${\rm {E}}2$ transitions, which are particularly relevant to deformed nuclei, may easily be orders of magnitude faster than estimate.

Interestingly, ${\rm {M}}4$ transitions tend to be quite close to the mark. Recall that the shell model puts the highest spin states of one harmonic oscillator shell right among the lowest spin states of the next lower shell, 14.13. Transitions between these states involve a parity change and a large change in spin, leading to E3 and M4 transitions. They resemble single-particle transitions as the Weisskopf and Moszkowski estimates assume. The estimates tend to work well for them. One possible reason that they do not end up that much below ballpark as E1 transitions may be that these are heavy nuclei. For heavy nuclei the restrictions put on by isospin may be less confining.

Finally, what other books do not point out is that there is a problem with electrical transitions in the islands of isomerism. There is serious concern about the correctness of the very Hamiltonian used in such transitions, {N.14}. This problem does not seem to affect magnetic multipole transitions in the nonrelativistic approximation.

Another problem not pointed out in various other books is for magnetic transitions. Consider the shell model states, figure 14.13. They allow many transitions inside the bands that by their unit change in angular momentum and unchanged parity are ${\rm {M}}1$ transitions. However, these states have a change in orbital angular momentum equal to two units. The single-particle model on which the Weisskopf and Moszkowski estimates are based predicts zero transition rate for such ${\rm {M}}1$ transitions. It does not predict the Moszkowski or Weisskopf values given above and in the figures. In general, the predicted single-particle transition rates are zero unless the multipole order $\ell$ satisfies, {A.25.8}

\begin{displaymath}
\vert l_1-l_2\vert \mathrel{\raisebox{-.7pt}{$\leqslant$}}\ell \mathrel{\raisebox{-.7pt}{$\leqslant$}}l_1+l_2
\end{displaymath}

where $l_1$ and $l_2$ are the initial and final orbital azimuthal quantum numbers. Fortunately this is only an issue for magnetic transitions, {A.25.8}.

Note that the single-particle model does give a nontrivial prediction for say an 2p$_{1/2}$ to 2p$_{3/2}$ ${\rm {M}}1$ transition. That is despite the fact that the simplified Hamiltonian on which it is based would predict zero transition rate for the model system. For say a 4p$_{3/2}$ to 2p transition, the Weisskopf and Moszkowski units also give a nontrivial prediction. That, however, is due to the incorrect radial estimate (A.187). The correct single-particle model on which they are based would give this transition rate as zero. Fortunately, transitions like 4p$_{3/2}$ to 2p are not likely to be much of a concern.


14.20.5 Comparison with data

This subsection compares the theoretical Weisskopf and Moszkowski estimates of the previous section with actual data. The data are from NuDat 2, [[12]]. The plotted values are a broad but further quite random selection of data of apparently good quality. A more precise description of the data selection procedure is in {N.34}. Internal conversion effects, as discussed in subsection 14.20.7, have been mathematically removed using the conversion constants given by NuDat 2. Computed decay rates were checked against the decay rates in W.u. as given by NuDat 2.

Figure 14.61: Comparison of electric gamma decay rates with theory. [pdf]
\begin{figure}
\centering
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...}
\put(353,375.8){\makebox(0,0)[bl]{odd-odd}}
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Figure 14.62: Comparison of magnetic gamma decay rates with theory. [pdf]
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Figures 14.61 and 14.62 show the results. What is plotted is the half life, scaled to an (harmonic) average nucleus size. In particular,

\begin{displaymath}
\mbox{E$\ell$:}\quad
\tau_{1/2,\rm red} = \tau_{1/2} \le...
...\rm red} = \tau_{1/2} \left(\frac{A}{32}\right)^{2(\ell-1)/3}
\end{displaymath}

The horizontal coordinate in the figures indicates the energy release $Q$.

The multipole levels are color coded. The Weisskopf values are shown as broken lines. The solid lines are an attempt at a “best guess” based on the single-particle model. For the electric transitions, they simply use the empirical radial factor {A.25.8} (A.187). For the magnetic transitions, the best guess was based on the more accurate Moszkowski estimates. The empirical radial factor was again used. The momentum factor {A.25.8} (A.189) at minimum multipole order was averaged between the proton and neutron values. The $g$ values in those factors were in turn averaged between their free space and theoretical values.

Symbol size indicates the nucleus size. Symbol shape indicates whether the numbers of protons and neutrons are even or odd. A minus sign indicates that the initial parity is odd; otherwise it is even. The final parity follows from the multipole order and type. An open symbol center indicates that the multipole level $\ell$ is higher than needed in the transition. More precisely, it indicates that it is higher than the change in nuclear spin.

Please, your mouth is hanging open. It makes you look very goofy. You can almost pretend that the magnetic data are not really as bad as they look, if you cover up those numbers along the vertical axis with your arm.

There is no doubt that if engineers got data like that, they would conclude that something is terribly and fundamentally wrong. Physicists however pooh-pooh the problems.

First of all, physics textbooks typically only present the ${\rm {M4}}$ data graphically like this. Yes, the ${\rm {M4}}$ transitions are typically only an order of magnitude or so off. According to the figures here, this good agreement happens only for the ${\rm {M4}}$ data. Have a look at the ${\rm {E1}}$ and ${\rm {E2}}$ data. They end up pretty much in the same humongous cloud of scattered data. In physics textbooks you do not really see it, as these data are presented in separate histograms. And for some reason, in those histograms the ${\rm {E2}}$ transitions are typically only half an order of magnitude above estimate, rather than 2.5 orders. (The ${\rm {E1}}$ transitions in those histograms are similar to the data presented here.)

Consider now a typical basic nuclear textbook for physicist. According to the book, disagreements of several orders of magnitude from theory can happen. The difference between “can happen” and are normal is not defined. The book further explains: “In particular, experimental disintegration rates smaller than the ones predicted by [the Weisskopf estimates] can mean that [the Weisskopf radial factor {A.25.8} (A.187)] is not very reasonable and that the small overlap of the [initial and final nuclear wave functions] decreases the values of $\lambda$.”

However, the best guess ${\rm {E1}}$ line in figure 14.61 uses a better radial estimate. It is not exactly enough to get anywhere near the typical data.

And poor overlap is an easy cop-out since nuclear wave functions are not known. For example, it does not explain why some multipole orders like ${\rm {E1}}$ have a very poor overlap of wave functions, and others do not.

Transition rates many orders of magnitude smaller than theory must have a good reason. Random deviations from theory are not a reasonable explanation. Having a transition rate typically four orders of magnitude smaller than a reasonable theoretical estimate is like routinely hitting the bull’s eye of a 10 cm target to within a mm. There must be something causing this.

But what might that be? Conservation of angular momentum and parity are already fully accounted for. To be sure, conservation of isospin is not. However, isospin is an approximate symmetry. It is not accurate enough to explain reductions by 4 or 5 orders of magnitude. The two examples mentioned in subsection 14.20.2 managed just 1 order of magnitude slow down. And not all ${\rm {E1}}$ transitions are forbidden anyway. And light nuclei, for which isospin conservation is presumably more accurate, seem no worse than heavier ones in figure 14.61. Actually, the two best data are small nuclei.

To be sure, the above arguments implicitly assume that a bull’s eye is hit by an incredibly accurate cancelation of opposite terms in the so-called matrix element that describes transitions. There is an alternate possibility. The final nuclear wave function could be zero where the initial one is nonzero and vice versa. In that case, the integrand in the matrix element is everywhere zero, and no accurate cancellations are needed.

Figure 14.63: Comparisons of decay rates between the same initial and final states.
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...7){\makebox(0,0)[bl]{odd-odd}}
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But consider now the top half of figure 14.63. Here mixed ${\rm {E1+M2}}$ transitions are plotted. These transitions take sometimes place through the electric dipole mechanism and sometimes through the magnetic quadrupole one. Note that there are three very fast ${\rm {M2}}$ transitions, the first, fourth, and tenth. These transitions occur at rates of 49, 58, respectively 21 times faster than the best guess based on the single-particle model. So the initial and final wave functions must unavoidably overlap very well. But how then to explain that the corresponding electric rates are 31,000, 7,800, and 200 times slower than best guess? The initial and final wave functions are the same. While there is a different operator sandwiched in between, {A.25}, the picture still becomes that one operator apparently achieves a bull’s eye of perfect cancellation. There should be an explanation.

The example textbook also notes: “Experimental values higher than predicted by [the Weisskopf estimates] can mean, on the other hand, that the transition involves the participation of more than one nucleon or even a collective participation of the whole nucleus.” The textbook notes in particular that the reason that most ${\rm {E2}}$ transitions are faster than theory is due to the fact that these transitions are common among collective bands, especially rotational bands in deformed nuclei.

This is a well established argument. In principle 50 protons transitioning could indeed explain why many ${\rm {E2}}$ transitions in figure 14.61 end up on the order of a rough factor 50$\POW9,{2}$ faster than theory.

But there are again some problems. For one, the mixed ${\rm {E1+M2}}$ transitions discussed earlier are not among states in the same rotational bands. The nuclear parity flips over in them. But the mentioned examples showed that several ${\rm {M2}}$ transitions were also much faster than single-particle theory. While that might still be due to collective motion, it does not explain why the ${\rm {E1}}$ transitions then were so slow.

Consider also the bottom of figure 14.63. Here mixed ${\rm {M1+E2}}$ transitions are plotted. Note that the ${\rm {E2}}$ transitions are again much faster than theory, with few exceptions. But how then to explain that the ${\rm {M1}}$ transitions between the same initial and final states are much slower than theory? More of these miraculously accurate cancellations? There are quite a few transitions at the higher energies where the ${\rm {M1}}$ transition proceeds slower than the ${\rm {E2}}$ one, despite the difference in multipole order.

It is true that the orbital effect is relatively minor in magnetic transitions of minimal multipole order, {A.25.8}. But the transitions given by black symbols with open centers in the bottom of figure 14.63 are not of minimal multipole order. And in any case, relatively minor gets nowhere near to explaining differences of four or five orders of magnitude in relative decay rates.

The same problem exists for the idea that the special nature of transitions within rotational bands might somehow be responsible. Surely the idea of rotational bands is not be far accurate enough to explain the humongous differences? And some of the worst offenders in figure 14.63 are definitely not between states in the same rotational band. Those are again the ones where the black symbol has an open center; the nuclear spin does not change in those transitions.

Also, why do ${\rm {E3}}$ transitions act much like ${\rm {E1}}$ transitions in the first half of the energy range and like ${\rm {E2}}$ ones over the second half? It seems weird.

The example textbook concludes: “In figure [...] one notes very good agreement between theoretical values and experimental ones for ${\rm {M4}}$. This behavior is typical for transitions of high multipolarity.” Based on figures 14.61 and 14.62, it seems very optimistic to call ${\rm {M4}}$ transitions typical for high multipolarity.

You will find similar discussions in other textbooks. They also tend to give the same old graphs that do not really show the problems. Only Quantum Mechanics for Engineers tells it like it is. It shows you comprehensive data clearly. That allows you to make up your own mind about them based on solid information.


14.20.6 Cage-of-Faraday proposal

The biggest problem with the data of the previous subsection is to explain why transitions end up so far below ballpark. Consider in particular the ${\rm {E1}}$ transitions in figures 14.61 and 14.63. How come that they are not just occasionally, but typically slower than theory by four orders of magnitude? There are 65 randomly chosen ${\rm {E1}}$ transitions plotted in figure 14.61. Out of these 65, only one manages to achieve the best guess theoretical transition rate. That is a boron-10 transition. (You may have to strain your eyes to see it, it is such a small nucleus. It is right on top of the best-guess line, just before 1 MeV.) On the other hand, one transition is slower than best guess by more than 8 orders of magnitude, and another three are slower by more than 7 orders of magnitude.

Compare that with the 67 ${\rm {E2}}$ transitions. Only one manages the three orders of magnitude slower than best guess that is so ho-hum for ${\rm {E1}}$ transitions. That transition is a 2,340 keV $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 29}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ to 2,063 keV $\leavevmode\kern.03em
\raise.7ex\hbox{\the\scriptfont0 25}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em\kern-.05em\rule{0pt}{8pt}^{+}$ $\fourIdx{205}{85}{}{}{\rm {At}}$ one. The amount of spin that is involved here is not exactly run-of-the mill. Note also that the three runners-up for being far above the ${\rm {E2}}$ line are ${\rm {E1}}$ transitions, rather than ${\rm {E2}}$ ones...

This systematic lack of agreement with theory needs a decent explanation, not just some pooh-poohing. Generally speaking, it would be relatively easy to explain transition rates much larger than theory. “Apparently we did not include something whose effect is much larger than the effects considered.” But hitting the virtually-zero transition rate pretty much every time is not so easy. The something else now has to achieve a near-perfect elimination of an effect that is already there.

In particular, it seems hard to explain why effects this large and this systematic occur within the mental picture of a bunch of individual nucleons doing their own thing. So maybe the reason is that the nucleons do not do their own thing; that they respond in some way collectively to the electromagnetic field.

Unfortunately, modeling that would make the quantum mechanics even more difficult than it is already. So maybe it is again time to do what has been done before for nuclei; look for macroscopic models. And surely the macroscopic model that stands out in killing off electromagnetic effects is the cage of Faraday.

Maybe then the nuclear surface acts as such a cage. Assume at first that macroscopically speaking the nuclear surface is spherical and conducting. Then electric changes in the interior of the nucleus would not leak out. That would kill off the capability of transitions to emit radiation. Mathematically, if no radiation is emitted, it implies that the electric matrix elements sum to zero, compare chapter 13.3.3 (13.27) and say {A.25.7}. Charges can still move around inside the nucleus, but that does not produce radiation outside it. Note that unlike a macroscopic cage, here the cage shields the outside from the inside.

Also unlike macroscopic cages of Faraday, the nuclear surface must stay spherical to achieve this. The reason is that the nucleus contains a net charge. A conducting surface makes the field tangential to the surface zero. But if there is a net charge inside, then there is always a nonzero field outside. In that case the electric field outside will not stay unperturbed if the conducting surface changes shape.

At least for relatively light nuclei with small excitations, surface tension would promote a spherical surface. And “surface roughness” would not necessarily make much of a difference. That is just like small holes in a macroscopic cage do not make a difference. The field outside the nucleus is governed by the so-called Laplace equation. This equation is known to kill off short-scale perturbations quickly. Another way to say this is that random surface roughness would not produce much of a contribution to radiation that varies slowly with angular position, in particular ${\rm {E1}}$ radiation. (For a very small nucleus like boron-10, surface roughness might be relatively more important.)

On the other hand, changes in a deformed nuclear surface shape would definitely produce nontrivial long-range electric field perturbations. Now nuclei are often modeled as spheroids or ellipsoids. Changes in such a shape would produce quadrupole and hexadecapole perturbations in the electric field outside the nuclei. So they would produce ${\rm {E2}}$ radiation, but not ${\rm {E1}}$ radiation. That is exactly what is needed to make some sense out of the electric transitions.

While macroscopic cages of Faraday do not block static magnetic fields, they do block changes in magnetic fields. So conceptually the model could also explain why magnetic transitions of low multipole order are often so slow. Note that there is no net magnetic charge inside the cage. Magnetic monopoles do not exist. So surface shape would not necessarily affect magnetic transitions much.


14.20.7 Internal conversion

In internal conversion, a nucleus gets rid of excitation energy by kicking an atomic electron out of the atom. This is most important for transitions between states of zero spin. For such transitions, the normal gamma decay process of emitting a photon is not possible since a photon cannot be emitted with zero angular momentum. However, the ejected electron, called the “conversion electron,” can keep whatever angular momentum it has. (For practical purposes, that is zero. Electrons that are not in s states have negligible probabilility of being found inside the nucleus.) Transitions in which no angular momentum is emitted by the nucleus are called ${\rm {E}}0$ transitions.

A ballpark decay rate for ${\rm {E}}0$ internal conversion can be found in Blatt & Weisskopf [7,8, p. 621]. Where else. Converted to look similar to the gamma decay Weisskopf estimate (A.190), it reads

\begin{displaymath}
\fbox{$\displaystyle
\lambda^{{\rm{E}}0}_{\rm Blatt \& W...
...ga\equiv\frac{Q}{\hbar} \quad k\equiv\frac{Q}{\hbar c}
$} %
\end{displaymath} (14.68)

Here $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0{\hbar}c$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1/137 is the fine structure constant and $Q$ is the energy release. Further ${m_{\rm e}}c^2$ is the rest mass energy of the electron, which is about half an MeV. The initial and final parities need to be the same.

Note that the first three factors in the expression above look much like an ${\rm {E}}2$ electric transition. However, the next three factors are very small, though less so for heavy nuclei. On the other hand the final factor can be very large if the energy release $Q$ is much less than an MeV. So ${\rm {E}}0$ internal conversion is relatively speaking most effective for low-energy transitions in heavy nuclei.

Putting in the numbers gives the equivalent of (14.67) as

\begin{displaymath}
\fbox{$\displaystyle
\lambda^{{\rm{E}}0} = 3.8 Z^3 A^{4/3} Q^{1/2}
$} %
\end{displaymath} (14.69)

Once again, the energy release should be in MeV and then the decay rate will be per second. Note that absolutely speaking the decay rate does in fact increase with the energy release, but very weakly.


Table 14.5: Half lifes for E0 transitions.
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Table 14.5 shows how the estimate stands up to scrutiny. The listed ${\rm {E}}0$ transitions are all those for which NuDat 2, [[12]], gives useful and unambiguous data. All these turn out to be 0$\POW9,{+}$ to 0$\POW9,{+}$ transitions.

The second-last column in the table shows a scaled half life. It is scaled to some (harmonic) average nucleus size $Z$ $\vphantom0\raisebox{1.5pt}{$=$}$ 16, $A$ $\vphantom0\raisebox{1.5pt}{$=$}$ 32. In particular

\begin{displaymath}
\tau_{1/2,\rm red} \equiv
\tau_{1/2} \frac{Z^3}{16^3} \frac{A^{4/3}}{32^{4/3}}
\end{displaymath}

The final column shows what the scaled half life should be according to the theoretical estimate above. Note that, excluding the final three nuclei, the agreement is not too bad, as they come. What is an order of magnitude or so between friends? After the previous subsections everything would look accurate.

However, the final three nuclei decay much more rapidly than internal conversion predicts. A second decay process occurs here: electron-positron pair creation. This requires that the nuclear energy release $Q$ is at least large enough to provide the rest mass energy of the electron and positron. That is a bit over a MeV. However, as the table suggests, to get a significant effect, more energy is needed. There should be enough additional energy to give the electron and positron relativistically nontrivial kinetic energies.

In transitions other than between states of zero spin, normal gamma decay is possible. But even in those decays, internal conversion and pair production may compete with gamma decay. They are especially important in highly forbidden gamma decays.

The so-called “internal conversion coefficient” $\alpha_\ell$ gives the internal conversion rate of a transition as a fraction of its gamma decay rate:

\begin{displaymath}
\fbox{$\displaystyle
\alpha_\ell = \frac{\lambda_{\rm IC}}{\lambda_\gamma}
$} %
\end{displaymath} (14.70)

The following ballpark values for the internal conversion coefficient in electric and magnetic transitions can be derived ignoring relativistic effects and electron binding energy:

\begin{displaymath}
\fbox{$\displaystyle
\alpha_{{\rm{E}}\ell} = \frac{1}{n^...
... Z)^3
\left(\frac{2 m_{\rm e}c^2}{Q}\right)^{\ell+3/2}
$}
\end{displaymath} (14.71)

Here, once again, $\ell$ is the multipole order of the decay, $Q$ the nuclear energy release, and $\alpha$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1/137 the fine structure constant. Further $n$ is the principal quantum number of the atomic shell that the conversion electron comes from, Note the brilliance of using the same symbol for the internal conversion coefficients as for the fine structure constant. This book will use subscripts to keep them apart.

The above estimates are very rough. They are routinely off by a couple of orders of magnitude. However, they do predict a few correct trends. Internal conversion is relatively more important compared to gamma decay if the energy release $Q$ of the decay is low, if the multipolarity $\ell$ is high, and if the nucleus is heavy. Ejection from the $n$ $\vphantom0\raisebox{1.5pt}{$=$}$ 1 K shell tends to dominate ejection from the other shells, but not to a dramatic amount.

(You might wonder why the earlier ballpark for ${\rm {E}}0$ transitions looks mathematically like an $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 rate, instead of some $\ell$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 one. The reason is that ${\rm {E}}0$ transitions do not create an electromagnetic field outside the nucleus, compare for example chapter 7.4.3. So the interaction with the electron is limited to the interior of the nucleus. That reduces the magnitude of the interaction greatly.)

Internal conversion is especially useful for investigating nuclei because the conversion coefficients are different for electric and magnetic transitions. Therefore, detailed decay measurements can shed light on the question whether a given transition is an electric or a magnetic one. Since they also depend strongly on the multipole order, they also help establish that. To be sure, the estimates above are not by far accurate enough to do these things. But much more accurate values have been computed using relativistic theories and tabulated.

Internal pair formation supplements internal conversion, [7,8, p. 622]. The pair formation rate is largest where the internal conversion rate is smallest. That is in the region of low atomic number and high transition energies.

One very old reference incorrectly states that internal conversion happens when a gamma ray emitted by the nucleus knocks a surrounding electron out of the atom. Such a process, the photoelectric effect, is in principle possible, but its probability would be negligibly small. Note in particular that in many decays, almost no gamma rays are emitted but lots of conversion electrons. (While the interaction between the nucleus and the conversion electron is of course caused by photons, these are virtual photons. They would not come out of the nucleus even if you stripped away the atomic electrons.)

It may be noted that internal conversion is not unique to nuclei. Energetic atomic electron transitions can also get rid of their energy by ejection of another electron. The ejected electrons are called “Auger electrons.” They are named after the physicist Auger, who was the first man to discover the process. (Some unscrupulous woman, Lise Meitner, had discovered and published it earlier, selfishly attempting to steal Auger’s credit, {N.35}). In fact, internal conversion can give rise to additional Auger electrons as other electrons rush in to fill the internal converion electron hole. And so can electron capture.