I was studying nuclear physics this day and I read about radioactive decays. $\beta$-plus decay turns one proton in the nucleus into one neutron, one positron and one neutrino.
I was wondering about the electron count in the process of $\beta$-plus decay and also in any other decays. Let's suppose if we have a $\beta$-plus decay process: $^{33}_{17}Cl \longrightarrow ^{33}_{16}S + e^+ + v_e$.
I would calculate the mass difference $\Delta m$ of this process like this (ignoring the mass of the neutrino):
$\Delta m = m_{nucleus}(^{33}_{17}Cl) - (m_{nucleus}(^{33}_{16}S) + m_e) \\ \Delta m = m_{atom}(^{33}_{17}Cl) - 17m_e - (m_{atom}(^{33}_{16}S) - 16m_e + m_e)$
I know that $^{33}_{17}Cl$ has 17 protons and thus 17 electrons. But why does $^{33}_{16}S$ have 16 electrons? Shouldn't it still have 17 electrons? Where does the 17th electron go? Same question arises in $\beta$-minus decay; one extra electron seems to appear into the daughter nucleus.