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I'm trying to figure out the energy balance for the $\beta^+$ and electron capture (ec) decay modes of Al-26. According to the decay data at http://www.nndc.bnl.gov/nudat2/decaysearchdirect.jsp?nuc=26AL&unc=nds and http://www.nucleide.org/DDEP_WG/Nuclides/Al-26_tables.pdf, the endpoint energy for the $\beta^+$ mode is 1173.42 keV, which, if I understand it correctly, would correspond to a neutrino-free decay as all the energy and momentum are balanced by the positron. However, there seems to be no energy level of the daughter nuclide Mg-26 that corresponds to a gamma ray of 2830.72 keV, which would be necessary to arrive at the ground-level energy difference of 4004.19 keV between Al-26 and Mg-26.

By contrast, the energies tabulated for ec, 1065.78 and 2195.47 keV, correspond exactly to the observed gammas with 2938.41 and 1808.72 keV, respectively, that are required to bring the excited daughter Mg-26 into its ground state.

What am I missing? I would have expected that the $\beta^+$ and the ec feed the same energy levels of Mg-26, especially as I don't see separate gamma energies tabulated for both modes.

Tom

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  • $\begingroup$ Not neutrino-free because leptin number still needs to be conserved. But very low neutrino momentum. $\endgroup$ – dmckee♦ Mar 20 '17 at 15:02
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[...] to arrive at the ground-level energy difference of $4004.19~\text{keV}$ between Al-26 and Mg-26.

Electron capture decay of an Al-26 atom (including 13 electrons) leaves an (excited) Mg-26 nucleus together with 12 electrons and a neutrino (of nearly negligible energy), while $\beta^+$ decay of an Al-26 atom results in an (excited) Mg-26 nucleus together with 13 electrons, a positron, and a neutrino (whose kinetic energy may be significant in general; but is negligible by definition if we consider the "end-point"-energy of emitted positrons).

Assuming then that the tabulated "ground state difference" refers to the entire atoms, and that the $\beta^+$ decay of an Al-26 leads to the excited Mg-26 nucleus $1808.72~\text{keV}$ "above ground level" (i.e. the lower of the two excited levels mentioned in the question), and neglecting kinetic energies except for the positron, then

$$E_{endpoint}[ \, e^+ \, ] + m_{e^+}~c^2 + m_{e^-}~c^2 + \Delta_E[ \, \text{Mg}^*, \text{Mg} \, ] \approx \Delta_E[ \, \text{Al}, \text{Mg} \, ], $$ $$1173.42~\text{keV} + (2 \times 511.00~\text{keV}) + 1808.72~\text{keV} = 4004.14~\text{keV} \approx 4004.19~\text{keV}.$$

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  • $\begingroup$ Of course, I forgot the annihilation energy of the positron with an electron! Thanks, this largely settles the problem. One more subtle thing I still wonder about concerns the gammas, however: their energies, which correspond to different excitation levels, have to be weighted according to the occupation of those levels, but does one also have to multiply them with the gamma intensities tabulated there? I would think so, but here (and in other nuclides) the intensities do not add up to 100%, which looks like a violation of energy conservation to me. $\endgroup$ – TomR Mar 21 '17 at 0:38
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    $\begingroup$ As a general rule the neutrino can carry away kinetic energy on the same order as the electron can. It's not a good idea to assume that it can be neglected. Now, when the electron has the endpoint energy then the neutrino energy is very small, but that's a conditional statement. $\endgroup$ – dmckee♦ Mar 21 '17 at 2:05
  • $\begingroup$ @TomR: "Thanks, this largely settles the problem." -- Terrific! (You're welcome.) "the gammas, however: their energies, which correspond to different excitation levels, have to be weighted according to the occupation of those levels, but does one also have to multiply them with the gamma intensities tabulated there?" -- Surely there's a (nuclear-physics-complicated) relation between energies and probabilities (branching fractions, relative intensities) of various possible processes. But, hopefully, that's another question to be answered. (Yours above was rather about "plain kinematics".) $\endgroup$ – user12262 Mar 21 '17 at 3:52
  • $\begingroup$ @dmckee: "As a general rule the neutrino can carry away kinetic energy on the same order as the electron can. It's not a good idea to assume that it can be neglected." -- Right; this seems to be related to the distinction in the tables referenced in the OP between "Energy" (which may denote "mean kinetic energy of emitted positrons") and "End-point energy". "Now, when the electron has the endpoint energy then the neutrino energy is very small, but that's a conditional statement." -- Correct; and this condition was stipulated in the OP. $\endgroup$ – user12262 Mar 21 '17 at 3:53
  • $\begingroup$ @dmckee: Only after posting the comment above I realized that the formulation in my answer had been imprecise. So: thanks for pointing that out. I hope that my answer is now corrected accordingly. $\endgroup$ – user12262 Mar 21 '17 at 4:03

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