Classification of Particles

A $\text{hadron}$ is any particle built from quarks, and so it can feel the $\text{strong nuclear force}$. Protons and neutrons — the particles that make up every nucleus — are hadrons, but so are many short-lived particles produced in accelerators, such as pions and kaons.

A $\text{lepton}$ is a genuinely fundamental particle (not built from anything smaller) that does $\text{not}$ feel the strong force at all. The electron is a lepton; so is its heavier, unstable cousin the $\text{muon}$ (${\mu }^{-}$), and the (anti)neutrinos that accompany beta decay. Charged leptons feel the electromagnetic force, and all leptons feel the weak force — but none of them ever feel the strong force, which is exactly why an electron can sit right next to a nucleus without being dragged in by it.

Hadrons split into two subclasses depending on how many quarks they contain. A $\text{baryon}$ is built from three quarks (the proton is $uud$, the neutron $udd$); an $\text{antibaryon}$ is built from three antiquarks. A $\text{meson}$ is built from a quark $\text{and}$ an antiquark — the lightest examples are the pions (${\pi }^{+}$, ${\pi }^{-}$, ${\pi }^0$), which you’ll meet again as the exchange particles that hold nuclei together at short range in older models of the strong force.

This gives every hadron a $\text{baryon number}$, $B$: $+1$ for a baryon, $-1$ for an antibaryon, and $0$ for a meson (and for every lepton, which isn’t a hadron at all). Baryon number is found experimentally to be conserved in every known interaction.

Because baryon number is conserved, the total $B$ before an interaction must exactly equal the total $B$ after. This has a striking consequence: the proton, the $\text{lightest}$ baryon, cannot decay at all, because there is no lighter baryon (or combination with total $B=+1$) for it to decay into without violating either baryon number conservation or energy conservation. Every observation to date is consistent with the proton being stable.

The free neutron, by contrast, is not the lightest baryon — the proton is slightly lighter — so it $\text{can}$ decay: $n\to p+e^{-}+{\bar{\nu }}_e$ (ordinary ${\beta }^{-}$ decay). Check baryon number: left-hand side $B=+1$ (the neutron); right-hand side $B=+1$ (the proton) $+0$ (electron) $+0$ (antineutrino) $=+1$. Balanced, so the decay is allowed.

Leptons are assigned a $\text{lepton number}$, $L$: $+1$ for a lepton (electron, muon, their neutrinos), $-1$ for an antilepton (positron, antimuon, antineutrinos), and $0$ for every hadron. Like baryon number, total lepton number must be conserved across any interaction or decay.

This is exactly the check that first justified the antineutrino in ${\beta }^{-}$ decay: without it, $n\to p+e^{-}$ would have lepton number $0$ on the left but $+1$ on the right — not conserved. Adding the antineutrino ($L=-1$) restores balance: $0=0+(+1)+(-1)$.