# $\beta^-$ and $\beta^+$ $Q$ value issue

It's unclear to me why the following doesn't work when I calculate the $$Q$$ value for $$\beta^-$$ and $$\beta^+$$ decay. If we start with a parent nucleus $$P_z$$ with $$Z$$ electrons decaying through $$\beta^-$$ producing a daughter nucleus $$D_{z+1}$$ with $$Z+1$$ electrons:

$$Q = (P_z + Z*m_e) - (D_{Z+1} + (Z+1)*m_e + m_e).$$

Assuming that the Daughter is charge neutral and $$m_e$$ is the mass of an electron I find:

$$Q = P_z - D_{Z+1} -2m_e$$

I was expecting to get: $$Q = P_z - D_{Z+1}$$

For $$\beta^+$$ decay I have a similar problem where I get: $$Q = P_z - D_{Z-1}$$ when I expect to get $$Q = P_z - D_{Z-1} -2m_e$$

Can someone please point out where I have gone wrong?

• Does this answer your question? When to involve electron mass in energy calculation of beta decay Commented May 20, 2020 at 22:20
• Not really in my case $P_Z$ and $D_Z$ are atomic nuclei (I shall update my question to reflect this). Hence I don't think the electrons are included in this. I guess my question is more related to this en.wikipedia.org/wiki/Beta_decay#Energy_release. Why do we need to convert from the mass of the nucleus to atomic masses.. Commented May 20, 2020 at 22:29
• But you have electrons included in your calculations. Compare what is in that question with yours - it shows how to include the creation of the relevant beta. Commented May 20, 2020 at 22:31
• Yeah but the question suggests that I have double-counted electrons and I don't think I have. Commented May 20, 2020 at 22:32
• If you prefer, try physics.stackexchange.com/q/186897, or search for 'beta decay q value' and find other Q/A. Commented May 20, 2020 at 23:25

## 1 Answer

When you calculate Q values of nuclear reactions you should use the nuclear masses of the nuclei involved. The tables of masses available to us are tables of neutral atomic masses.

Therefore, to get the nuclear mass you should subtract the electron masses from the tabulated atomic masses. In your case for $$\beta^-$$ decay you should have $$Q = (P_Z - Z*m_e) - (D_{Z+1} - (Z+1)*m_e + m_e),$$ where $$P_Z$$ and $$D_{Z+1}$$ will be the tabulated neutral atomic masses. I think you'll get what you expected, both in this case and the $$\beta^+$$ case.

You should also try it for the electron capture and alpha decay cases.

• This is sound advice for decays of bare nuclei. But for heavy nuclei, the electron binding energies are also significant.
– rob
Commented Mar 16, 2022 at 20:17