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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?
- 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.
(published on 06/09/2011)
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