Electron count in radioactive decay

Question

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.

Answer 1

The full equation is as follows

$$\Large\rm ^{33}_{17}Cl \to {}^{33}_{16}S^- + {}^{\;\;0}_{+1}\beta^+ + {}^{0}_{0} \nu_{\rm e}$$

The daughter therefore has an extra electron compared with the neutral atom and the extra electron is eventually ejected by the daughter ion.

For this reaction to occur two electron masses have to be created and because of this such a reaction rarely occurs naturally.