beta+/ec decay of Al-26: confused about energy balance 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
 A: 
[...] 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 
$$E_{endpoint}[ \, e^+ \, ] + m_{e^+}~c^2 + m_{e^-}~c^2 + \Delta_E[ \, \text{Mg}^*, \text{Mg} \, ] \approx \Delta_E[ \, \text{Al}, \text{Mg} \, ], $$
$$1173.42~\text{keV} + (2 \times 511.00~\text{keV}) + 1808.72~\text{keV} = 4004.14~\text{keV} \approx 4004.19~\text{keV}.$$
