Beta-plus: too many protons

The mirror image of beta-minus. Again, start with why: the ladders force the weak force into play.

1. The ladders are imbalanced the other way

Fluorine-18 has 9 protons and 9 neutrons. Its top protons sit high while lower neutron rungs sit empty. Dropping a proton onto a neutron rung releases enough energy to pay for making a heavier neutron plus a positron — but only because the whole nucleus becomes much more tightly bound.

A free proton is stable (lucky, since every hydrogen atom is one). A neutron is heavier than a proton, so a free proton turning into one would be making mass from nothing.

2. The reaction

Step through it for fluorine-18. The weak force turns an up quark into a down quark, emitting a $W^+$ that immediately becomes the positron + neutrino:

Step through a β⁺ decay (fluorine-18)

Step 1Fluorine-18: 9 protons, 9 neutrons. The only stable fluorine has 10 neutrons, so this one is proton-heavy.

Step 2Zoom into one proton: two up quarks and a down quark.

Step 3The weak force turns an up quark into a down quark, emitting a W⁺. The charge balances: +⅔ = −⅓ + (+1).

Step 4The W⁺ immediately becomes a positron (e⁺) and a neutrino (ν), both created on the spot from released energy.

Step 5The positron and neutrino fly off, sharing up to 0.63 MeV. The proton is now a neutron (u d d). With 8 protons the nucleus is oxygen-18, which is stable.

Step 6The positron is antimatter. Within about a millimeter of tissue it slows down and meets an ordinary electron, and both are destroyed. Their combined mass becomes two gamma rays of exactly 511 keV flying in opposite directions. PET scanners detect these back-to-back pairs to pinpoint where F-18 (attached to sugar) collects in the body.

$u \to d + e^+ + \nu \qquad p \to n + e^+ + \nu$ ${}^{18}_{9}\mathrm{F} \to {}^{18}_{8}\mathrm{O} + e^+ + \nu$

F-18 has $1.656\ \mathrm{MeV}$ to spare, leaving up to $0.634\ \mathrm{MeV}$ of motion after paying the $1.022\ \mathrm{MeV}$ positron cost. The quark diagram reads the same as beta-minus, mirrored:

Rule	Before ( $u$ )	After ( $d + e^+ + \nu$ )
Charge	$+\tfrac{2}{3}$	$-\tfrac{1}{3} + 1 + 0$
Lepton count	$0$	$-1 + 1 = 0$ (anti-electron forces a regular neutrino)

3. What happens to the positron

The positron is antimatter. Within about a millimeter of tissue it slows down, meets an ordinary electron, and both are destroyed:

$e^+ + e^- \to \gamma + \gamma \quad (511\ \mathrm{keV\ each,\ opposite\ directions})$

PET scanners detect those back-to-back pairs to pinpoint where F-18 (attached to sugar) collects in the body.

A nucleus with less than $1.022\ \mathrm{MeV}$ to spare cannot emit a positron. It uses electron capture instead.

Next: electron capture — the low-energy route.