<P> where m N (X Z A) (\ displaystyle m_ (N) \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right)) is the mass of the nucleus of the X atom, m e (\ displaystyle m_ (e)) is the mass of the electron, and m ν _̄ e (\ displaystyle m_ ((\ overline (\ nu)) _ (e))) is the mass of the electron antineutrino . In other words, the total energy released is the mass energy of the initial nucleus, minus the mass energy of the final nucleus, electron, and antineutrino . The mass of the nucleus m is related to the standard atomic mass m by </P> <Dl> <Dd> m (X Z A) c 2 = m N (X Z A) c 2 + Z m e c 2 − ∑ i = 1 Z B i (\ displaystyle m \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right) c ^ (2) = m_ (N) \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right) c ^ (2) + Zm_ (e) c ^ (2) - \ sum _ (i = 1) ^ (Z) B_ (i)). </Dd> </Dl> <Dd> m (X Z A) c 2 = m N (X Z A) c 2 + Z m e c 2 − ∑ i = 1 Z B i (\ displaystyle m \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right) c ^ (2) = m_ (N) \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right) c ^ (2) + Zm_ (e) c ^ (2) - \ sum _ (i = 1) ^ (Z) B_ (i)). </Dd> <P> That is, the total atomic mass is the mass of the nucleus, plus the mass of the electrons, minus the sum of all electron binding energies B for the atom . This equation is rearranged to find m N (X Z A) (\ displaystyle m_ (N) \ left ((\ ce (^ (\ mathit (A)) _ (\ mathit (Z)) X)) \ right)), and m N (X Z + 1 A ′) (\ displaystyle m_ (N) \ left ((\ ce (^ (\ mathit (A)) _ ((\ mathit (Z)) + 1) X')) \ right)) is found similarly . Substituting these nuclear masses into the Q - value equation, while neglecting the nearly - zero antineutrino mass and the difference in electron binding energies, which is very small for high - Z atoms, we have </P>

Which of the following is not a part of beta decay