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 t
he 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.
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.