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Mass/charge balance of Beta decay/nuclear reactions

  1. Apr 20, 2009 #1
    1. The problem statement, all variables and given/known data
    There are three different reactions we did in class.
    B-: 6C14 --> 7N14 + e-
    B+: 6C11 --> 5B11 + e+
    Nuclear rxn: alpha particle + 7N14 --> 8O17 + e+

    2. Relevant equations
    Q tot = (mass inital - mass final) * 931.5 MeV
    Charge balance


    3. The attempt at a solution
    In class, for B- the teacher wrote down:
    Q= (Mass of C nucleus) - (mass of N nucleus) - 1 e-
    = (mass of C atom - 6 e-) - (mass of N atom - 7 e-) - (mass of 1 e-)
    so -6 -(-7) - 1 = 0 so Q= (mass of C atom) - (mass of N atom)

    B+
    Q = (mass of C nuc) - (Mass of B nuc) - (mass of e+)
    Q= (mass of C atom - 6 e-) - (mass of B atom - 5 e-) - mass of 1 e-
    Q= -6 -(-5) -1 = -2, so he said Q total = (mass of C) - (mass of B + 2e-)

    Nuclear rxn
    Q = (mass of alpha nuc) + (mass of N nuc) - (mass of O nuc) - (mass of e+)
    Q = (mass of He - 2 e-) + (mass of N atom - 7 e-) - (mass of O atom - 8 e-) - (mass of e+)
    Q = -2 + -7 - (-8) -1 = -2 e-
    However, he said in class that the Q = (mass of alpha particle + mass of N atom) - (mass of O atom)
    Can someone please clear this up for me? I really don't get the charge/mass balance. Thanks so much
     
  2. jcsd
  3. Apr 21, 2009 #2
    so basically you have to make sure that the mass and charge is the same before the reaction as it is after.
    remember energy cant be created, it simply changes from one form to another =]
    and also for B- emission you would have an anti-electron-neutrino and for B+ emission you would have an electron-neutrino!
     
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