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In a beta minus decay, a neutron is converted to a proton, electron and antineutrino.

Why then for calculation of the rest mass of the products is (Z+1)) (mass of electron) being used? The number of electrons in the atom should remain the same. It should be Zme

Besides, since an electron is being created, the number of electrons on the rhs should be greater than the lhs but in this calculation they're the same.

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The notation $m\left(^A_ZX\right)$ is referring to the mass of the whole atom, not just the nucleus. Since only the nucleus is involved in beta decay, we want to know the mass of the nucleus, which is given (at least to a good approximation) by the mass of the atom minus the mass of all of the electrons in the atom. So: $$ m_{\rm nuc}=m\left(^A_ZX\right)-Zm_e$$

Similarly, $m\left(^A_{Z+1}Y\right)-(Z+1)m_e$ is really just a long way of writing $m_{\rm nuc,Y}$.

If you already had the masses of the nuclei directly, you would only need to include the mass of the single electron in the final state.

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  • $\begingroup$ Why wouldn't it be -Zme is my question. Atomic number is changing to z+1 but the number of electrons around the nucleus are not. $\endgroup$ – xasthor Jan 19 '18 at 15:13
  • $\begingroup$ @xasthor Because the electrons are irrelevant. What is in tables is the mass of an atom with $z+1$ protons and $z+1$ electrons. If you want to know the mass of the nucleus without those electrons, simply subtract of the electron masses. It doesn't matter if the nucleus itself has $z$ electrons or no electrons, as that doesn't change its mass. $\endgroup$ – Chris Jan 19 '18 at 22:18
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The masses in the formulas of your question, such as $$m\left(^A_Z X\right)$$ are atomic masses of neutral atoms. Those are the masses which are actually experimentally determined and tabulated. In order to evaluate nuclear decay energies, however, we need to use nuclear masses which are approximated by subtracting the masses of the electrons attached to the neutral atom. It is the mathematics of this procedure which make the number of electron masses the same on each side for $\beta^-$ decay.

On the other hand, for positron decay ($\beta^+$), the number of electron masses don't balance. There is an excess of 2 electron masses which means that the atomic mass of the parent must exceed the atomic mass of the daughter by 1022 keV for that process to occur.

Obviously, this procedure ignores the binding energies of these electrons, but the error introduced is small because the binding energies are on the order of a few eVs to a few keVs (for more massive atoms). The mass energy of an electron is 511 keV.

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  • $\begingroup$ Why wouldn't it be -Zme is my question. Atomic number is changing to z+1 but the number of electrons around the nucleus are not. $\endgroup$ – xasthor Jan 19 '18 at 15:15
  • $\begingroup$ To calculate the nuclear mass which you need, you take the atomic mass of the nuclide in question and subtract all the electron masses for that nuclide. It doesn't matter what the decay mode is. You need the nuclear mass, and that means you must take away all the electrons from the tabulated atomic mass. $\endgroup$ – Bill N Jan 19 '18 at 16:41

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