This'll be back to basics for many of you, but here's something I still don't get.
How can nuclear decay of an unstable atom both create and annihilate positron-electron pairs?
You have an unstable atom, say Mg-22 (nucleon: 22, proton: 12), which decays and emits positrons and gamma rays. I get that the positrons impacting the electrons in the atom's electron shells cause annihilation of the pair. But the decay also creates positron-electron pairs. How?
Gamma decay is the emission of photons. You are thinking of $\beta$ decay.
When the particle is decaying, if it emits a $W^{-}$boson, it will subsequently decay and create an electron ($e^-$), and an electron antineutrino ($\overline{v}_e$), the antimatter particle to an electron neutrino. It will also flip a neutron into a proton. This is known as $\beta^-$decay. In contrast, if it emits a $W^{+}$boson, it will decay into a positron ($e^+$), and an electron neutrino ($v_e$). It changes a proton into a neutron. This is $\beta^+$decay.
If a positron were to interact with an electon in $W^+$ decay, it could annihilate and produce photon(s).
So to answer your question, no. It does not create electron/positron pairs. It creates one electron/positron and an electron antineutrino/neutrino respectively.
