When Cu-64 undergoes positron emission, we have $$_{29}^{64}\text{Cu}_{35}\rightarrow_{28}^{64}\text{Ni}_{36}^-+\beta^++\nu + Q_{\beta^+}.$$ The $Q$ value could be calculated as 0.653 MeV. I wonder why this emission is possible? Shouldn't the requirement of $\beta^+$ decay be $Q_{\beta^+}\geq1.022$ MeV (2 electron masses)? I also noticed the typical values of $Q_{\beta^+}$ should be 2-4 MeV, so I'm confused.
Nuclear Physics – Why ?+ Decay is Possible for Cu-64
nuclear-physicsradiation
Best Answer
Here is a level scheme for the decay from the Evaluated Nuclear Structure Data File (click to embiggen):
The ground-state to ground-state $Q$-value is $\rm 1.675\,MeV$ for electron capture, so your $Q_{\beta^+} = \rm 0.653\,MeV$ already accounts for the mass of the extra electron-positron pair.
Note that the decay to the excited state, where $\rm 1.3\,MeV$ of the $Q$-value is carried away by a photon, can proceed only by electron capture.