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Elementary Particles

Problem 6.294

There is a narrow beam of negative pions with kinetic energy TT equal to the rest energy of these particles. Find the ratio of fluxes at the sections of the beam separated by a distance l=l= =20 m.=20 \mathrm{~m} . The proper mean lifetime of these pions is τ0=25.5 ns\tau_{0}=25.5 \mathrm{~ns}

Reveal Answer
J/J0=exp⁡[−lmc/τ0T(T+2mc2)]=0.22, where m is the rest mass of a negative pionJ / J_{0}=\exp \left[-l m c / \tau_{0} \sqrt{T\left(T+2 m c^{2}\right)}\right]=0.22, \text { where } m \text { is the rest mass of a negative pion}
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