2
$\begingroup$

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.

$\endgroup$

1 Answer 1

6
$\begingroup$

Here is a level scheme for the decay from the Evaluated Nuclear Structure Data File (click to embiggen):

Level scheme

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – IGY
    Feb 21, 2022 at 5:41
  • 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. $\endgroup$
    – Alfred
    Feb 21, 2022 at 6:22
  • 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
  • $\begingroup$ @rob Yes, of course, this is what I meant. Sorry ! $\endgroup$
    – Alfred
    Mar 2, 2022 at 8:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.