# Beta Decay Energy

Consider $\beta ^-$ decay. \begin{align*} ^{198} Au \rightarrow ^{198} Hg + e^- + \bar{v_e} \end{align*}

The decay energy is given by the difference in mass between multiplied by the speed of light. So, you'd take the mass of the Au nucleus - the mass of the Hg nucleus - the mass of the electron to find the change in mass (I think we leave out the mass of the antineutrino since its really small, therefore the equation for the decay energy is actually an upper limit).

My textbook goes on to say that the decay energy can be written in terms of the mass of the atom, and not the nuclear mass.

\begin{align} \frac{Q}{c^2} = M_p - M_D \end{align} where $Q$ is the decay energy, $M_p$ is the mass of the parent atom, and $M_D$ is the mass of the daughter atom. My book derived this by adding $Z$ electrons to both the parent nucleus and the decay products. If we take the initial example above,

\begin{align} ^{198} Au + 198e^- \rightarrow ^{198} Hg + 198e^- + e^- + \bar{v_e} \end{align}

which becomes, Mass of Au atom $\rightarrow$ Mass of Hg atom + $e^- + \bar{v_e}$

or

\begin{align} M_p \rightarrow M_D + e^-1 \end{align} (I leave out the antineutrino here). So my question is, why is the decay energy just $M_p - M_D$ and not $M_p - M_D -e^-$? For $\beta ^+$ decay, my textbook says that the decay energy is

\begin{align} \frac{Q}{c^2} = M_p - (M_D -2m_e) \end{align} They derived this in the same fashion. So my question also applies to this decay process as well.

\begin{align} ^{198} Au + 198e^- \rightarrow ^{198} Hg + 198e^- + e^- + \bar{v_e} \end{align}
\begin{align} ^{198} Au + 79e^- \rightarrow ^{198} Hg + 80e^- + \bar{v_e} \end{align}
• I understand the reasoning but not the formalism because $^{198} Au$ is the nuclide symbol and it includes the electrons, protons and neutrons, imo. Why add 79 electron masses again? – Helder Velez Mar 14 '15 at 3:31