Nuclear Reactions and Mass

Quantum Theory & Nuclear Physics The lastest spaceship for interplanatary travel uses the following equation for propulsion... (3,1)H --> (3,2)H + e- + antinutrino where (a,b) represent (atomic mass, atomic number) Based on the equation, what is the "total energy" produced by this reaction...? -first of all my guess is that the (3,2)H is not correct and should possibly be (3,2)He...would you tend to agree with that. -then based on that, we know that in order to calculate the E that the mass of the products has to be less that the mass of the reactants. I have used the following: m (3,1)H = 3.016049 u m (3,2)He = 3.016029 u m e- = 0.00054858 u m antineutrino = 0 (?) Using that information, my products weigh more than the intitial weight of the tritium. Any suggestion(s) how I am to resolve this weight dilema? THks, Ken
- Ken Baratko (age 65)
Houston, Tx, USA

Those atomic masses already include the electrons. So including the electron mass in the products is double-counting.

Mike W.

(published on 06/09/2011)