Electron Capture and B+ decay

[binbagsss]

When electron capture and β+ decay occur in proton-rich nuclide, in order to obtain a lower E state, is the lower E state of the atom reached by 1 electron capture of the same energy as the state reached due to one β+ decay?
- I ask because I know that in order for EC or B+ to occur, the daughter nuclues must be of a greater BE than the mother BE, and the ΔBE must be sufficient to provide enough energy for the reaction to occur. ( In B+ decay to account for the fact that the neutron is heavier, and in EC is it to remove the electron from its orbital shell?).

And so the fact that when BE is not sufficient for B+ emission to occur, EC is the sole decay mode, must be due to the fact that removing a electron from its orbital shell requires less energy than B+ decay - assuming the E of the daughter nucleus is the same in both cases, i.e. same Δ BE)

Sincere apologoies if this is hard to read and a little long-winded. :shy:

p.s - my thoughts as to why this may the case: in both cases a proton is turned into a neutron, in EC with one less electron - removed from the lower E state- and in B+ with the emission of a positron - which I thought would be most likely to annhilate with a electron, as close aas possible, in a lower E state again removing it.

 

[jbsteele]

No, the lower energy state of the atom reached by one electron capture is not the same as the state reached due to one β+ decay. The energy states of these two processes are different because they involve different nuclear reactions. In electron capture, a proton is converted into a neutron by capturing an inner shell electron, which requires less energy than in β+ decay, which involves the emission of a positron. Therefore, the lower energy state reached by electron capture is lower than that reached by β+ decay.