Hadrons and leptonsAQA A-Level Physics: Revision notes

Hadrons are particles that are subject to the strong interaction. There are two classes of hadron:

• baryons: the proton and neutron, and the antibaryons antiproton and antineutron
• mesons: the pion (π⁺, π⁰, π⁻) and the kaon (K⁺, K⁰, K⁻)

The proton is the only stable baryon. All other baryons eventually decay into a proton, for example the neutron: n → p + e⁻ + $\bar{\nu}_e$.

Baryon number B is a quantum number that is always conserved.

• baryons: B = +1
• antibaryons: B = −1
• mesons and leptons: B = 0

This explains why the proton is stable: it is the lightest baryon, so it cannot decay without breaking conservation of baryon number.

Worked example. p + p → p + π⁺ would change B from +2 to +1, so it cannot happen. n → p + e⁻ + $\bar{\nu}_e$ changes B from +1 to +1 + 0 + 0, so it can.

The pion is the exchange particle of the strong nuclear force between nucleons. Kaons can decay into pions, for example K⁺ → π⁺ + π⁰ and K⁰ → π⁺ + π⁻.

In these decays charge and baryon number are conserved (B = 0 on both sides).

Leptons are not affected by the strong interaction. You need the electron, the muon and their neutrinos (electron-type and muon-type only), and their antiparticles.

• e⁻ and $\nu_e$ have electron lepton number Le = +1; e⁺ and $\bar{\nu}_e$ have Le = −1
• μ⁻ and $\nu_\mu$ have muon lepton number Lμ = +1; μ⁺ and $\bar{\nu}_\mu$ have Lμ = −1

Both lepton numbers are conserved separately. The muon decays into an electron: μ⁻ → e⁻ + $\bar{\nu}_e$ + $\nu_\mu$. Check: Lμ is +1 before and after (from $\nu_\mu$), and Le is 0 before and +1 − 1 = 0 after.

Particle physics depends on the collaborative efforts of large teams of scientists and engineers. Accelerators and detectors are huge and costly, they produce vast amounts of data, and results must be validated by independent teams, peer review and repetition before the new knowledge is accepted.