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.
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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.
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$\begingroup$ Thanks so much for the answer, so in this case, is it true that the positron emission could happen because the ground state to ground state Q value is greater than that limit? If so, do we just need to verify $Q_\beta^+\geq0$ to decide if a positron emission is possible? $\endgroup$– IGYFeb 21, 2022 at 5:41
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2$\begingroup$ Rather $Q_{\beta^+}$>$0$, or equivalently $Q_{\beta^-}$>$1.022$ MeV for ground state to ground state. A process with zero energy has no phase space to go to. And if $Q_{\beta^+}$>$0$ but very small, the probability of ${\beta^+}$ decay will be extremely small, theoretically possible, but since electron capture is possible with a large phase space to go to, it will dominate and ${\beta^+}$ decay will never be observed in practice. $\endgroup$– AlfredFeb 21, 2022 at 6:22
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1$\begingroup$ @Alfred is correct, except he meant $Q_{\epsilon}>1.022\rm\,MeV$ for electron capture. The existence of a $\beta^-$ decay is mostly irrelevant to the competition between positron decay versus electron capture. $\endgroup$– rob ♦Feb 21, 2022 at 9:37
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$\begingroup$ @rob Yes, of course, this is what I meant. Sorry ! $\endgroup$– AlfredMar 2, 2022 at 8:17