1. The problem statement, all variables and given/known data 1) Calculate the Q-value for the electron emission beta decay of Co-60 * 2) Calculate the Q-value for the positron emission beta decay of Mg-23 * * (both correct to 6 significant figures) Atomic Masses (u) electron = 0.0005485803 Co-60 = 59.9338222 Ni-60 = 59.9307906 Mg-23 = 22.9941249 Na-23 = 22.98976967 Atomic Number Co-27 Ni-28 Mg-12 Na-11 2. Relevant equations Q = Σmic2-Σmfc2 Beta decay (electron emission) = X ---> Y(Z+1) + -1e + v_ where v_ = anti-neutrino Beta decay (positron emission) = X ---> Y(Z-1) + +1e + v where v = neutrino 3. The attempt at a solution 1 (Co-60) Q = (59.9338222 - (59.9307906+0.0005485803))*c2 = 59.9338222 - 59.9313392 = 0.0024830 * 931.5020 = 2.3129 = 2.31290 MeV (correct to 6 significant figures) 2 (Mg-23) Q = (22.9941249 - (22.98976967+0.0005485803))*c2 = 22.9941249-22.99031825 = 0.0038066*931.5020 = 3.54580 MeV (correct to 6 significant figures) Q is positive in both cases. Q is the energy that is liberated in the reaction. Am i correct in neglecting and mass that the anti-neutrino and neutrino have?