# Electron count in radioactive decay

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.

• All explained here. as you are left with a daughter which is ionized. – Farcher May 17 '20 at 15:01
• If the daughter nucleus is ionized, it should have 17 electrons? Why are we still only subtracting mass of 16 electrons in the above equations... @Farcher – zu238 May 17 '20 at 16:53

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