# Why is $\beta^+$ decay possible for $\text{Cu}$-64?

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.

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.

• 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?
– IGY
Feb 21, 2022 at 5:41
• 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. Feb 21, 2022 at 6:22
• @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.
– rob
Feb 21, 2022 at 9:37
• @rob Yes, of course, this is what I meant. Sorry ! Mar 2, 2022 at 8:17