In analyzing mass-energy calculations involving beta decay, it is important to avoid simple bookkeeping errors. You look at the balanced nuclear reaction, and use tables of isotope masses to calculate the loss of mass (and its conversion into energy)
The problem comes from the two terms "mass of the final atom" and "tabulated mass of the final atom"
Consider this example:
$$
_{12}^{23}\mathrm{Mg} \longrightarrow \, _{11}^{23}\mathrm{Na} + e^{+} + \nu
$$
It is basic in all nuclear reactions that the sum of all the mass and energy on the LHS must equal the sum of all mass and energy on the RHS. The mass of the parent isotope must be greater (perhaps only slightly) than the combined mass of the daughter isotope plus the mass of the positron, with the missing mass appearing as energy split between the positron and neutrino.
BUT
When you sit down to do the calculation, there's a problem: a bookkeeping problem.
You started with a normal, run-of-the-mill magnesium atom, with a surrounding cloud of 12 electrons. That's the mass you find listed for that isotope of magnesium, and that's what you write down to start your calculation.
Now you look at the daughter sodium isotope. It's sitting in the reaction vessel, still with the 12 orbital electrons it had a picosecond ago, when it was a magnesium atom.. But when you look up the mass of the sodium isotope, the tabulated value only includes the mass of the 11 orbital electrons that all normal sodium atoms have.
So to replicate the masses in the actual reaction, you need to add an electron mass to the tabulated sodium isotope mass.
Finally, you can look at the positron produced and look up its tabulated mass for the calculation. There's your required mass difference of two positron masses...
You are modeling the following reaction:
Nitrogen-13 -> Carbon-13 + Positron + neutrino
This reaction is in the nucleus, so you should use the mass of the nucleus in the mass balance, not the atomic mass. However, we generally don't have the nucleus mass available, so we use the atomic masses. The atomic masses include the mass of the electrons.
The equation you should be using:
Δm=nucleus(N13)-(nucleus(C13) + positron mass)
Add 7 electrons to each side to use atomic mass:
Δm=nucleus(N13)+ 7(electron mass) -(nucleus(C13) + 7(electron mass) + positron mass)
Combine the nucleus mass and electron mass to get atomic mass. Note that the carbon atom only has 6 electrons, so you are left with an extra electron.
Δm=atomic(N13) -(atomic(C13) + electron mass + positron mass)
This is where the extra electron mass comes from.
Best Answer
When they show a decay with a certian amount of energy, this energy is net of the masses of the particles. So you get Co = $\beta$ + Ni + 0.31 MeV, the energy is attached to the ejected beta particle.