There is a narrow beam of negative pions with kinetic energy T equal to the rest energy particles. Find the ratio of fluxes at the sections of the beam separated by a distance l=20m. The proper mean lifetime of these pions is τ0=25.5ns.
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Updated on : 2022-09-05
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Here η=mc2T=1 so the life time of the pion in the laboratory frame is η=(1+η)τ0=2τ0
The law of radioactive decay implies that the flux decreases by the factor. J0J=e−t/τ=e−l/ντ=e−l/cτ0η(2+η)