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There is a narrow beam of negative pions with kinetic energy T equal to the rest energy of these particles. Find the ratio of fluxes at the sections of the beam separated by a distance l=20 m. The proper mean lifetime of these poins is tau_(0)=25.5ns. |
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Answer» Solution :Here `eta=(T)/(mc^(2))=1` so the life time of the PION in the laboratory frame is `eta=(1+eta)tau_(0)=2tau_(0)` The law of RADIOACTIVE decay implies that the flux DECREASE by the factor. `(J)/(J_(0))=e^(-t//tau)=e^(-l//vt)=e^(-l//ctau_(0))sqrt(eta(2+eta))` `=exp(-(m c l)/(tau_(0)sqrt(T(T+2mc^(2)))))=0.221` |
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