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'simple' deacy problem

  1. Oct 16, 2004 #1
    I think I am doing this correctly, but getting a strange result. Here's the original problem dealing with decay modes:

    The isobars 37Ar and 37Cl have binding energies/nucleon of 8.279 Mev and 8.336 Mev, respectively. Show which nuclide decay to the other spontaneously and by what decay mode. (p=1.007277 u, n=1.008665u, 1u=931.5 Mev). Neglect contribution of electrons

    Now, keep in mind that in reality, there is EC decay from 37Ar->37Cl.

    First thing I did was to calculate Q for the reaction, which meant calculating the masses. I am neglecting electrons like the problem says and I think this is what is putting me off.

    37Cl Mass=17*1.007277+20*1.008665=37.297009u
    37Ar Mass=18*1.007277+19*1.008665=37.295621u

    Oh no!! Q must be positive, so now I have 37Cl->37Ar by beta decay...

    Q=37Cl M - 37Ar M = .001988u * 931.5MeV/u=1.292922MeV

    So did I do this wrong or does my professor want me to give him the fantasy world mode of decay? Am I supposed to use the energy/nucleon somehow? Help, I need all the points I can get. Thanks.
    Last edited: Oct 17, 2004
  2. jcsd
  3. Oct 17, 2004 #2
    Ok, I figured it out. I need to take the binding energy and subtract it from the total mass of the separate nucleons to calculate the true mass to find Q. Hooray!
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