The Hadrons — Question 8

Question 8

Baryon-antibaryon annihilation bounds the multiplicity of massive products

A proton-antiproton pair annihilates at rest. Use rest energy 938 MeV for each and assume all final particles are identical neutral mesons of rest energy 140 MeV. Ignore spin and other selection rules for this kinematic bound.

1. Find the largest integer meson multiplicity permitted by rest-energy conservation.

2. For ten mesons, find total available kinetic energy and mean kinetic energy per meson.

Question 8 - Solution

Baryon-antibaryon annihilation bounds the multiplicity of massive products

(a) Method and calculation

$$140N\le2(938)=1876\Rightarrow N\le13.4.$$$$\boxed{N_{\max}=13}\quad\text{as an energy-only upper bound}.$$

(b) Interpretation and further calculation

$$\boxed{K_{\mathrm{total}}=1876-10(140)=476\,\mathrm{MeV}}.$$$$\boxed{\langle K\rangle=476/10=47.6\,\mathrm{MeV}}.$$

Physical check

The mean does not imply equal particle energies. Charge, angular momentum and dynamics can exclude channels within an energy-only bound.