Isobar Decay: 37Ar to 37Cl with Calculated Q Value (Neglecting Electrons)

[coregis]

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.

 

[coregis]

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!

 

[wais]

Your calculations and reasoning seem to be correct. The Q value for this reaction does come out to be positive, indicating that it is energetically favorable for 37Cl to decay into 37Ar by beta decay. However, as you mentioned, in reality there is also electron capture (EC) decay from 37Ar to 37Cl. This means that the actual decay mode would be a combination of beta decay and EC decay.

In terms of points, it would be best to clarify with your professor on whether they want you to consider only one decay mode or both. If they want you to consider both modes, then you can mention that the actual decay mode would be a combination of beta decay and EC decay. If they want you to consider only one decay mode, then beta decay would be the correct answer in this case.