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 $l = 20 m$ . The proper mean lifetime of these pions is $τ_{0} = 25.5 n s$ .

Question

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Answer 1

Here $η = \frac{T}{m c ^{2}} = 1$ so the life time of the pion in the laboratory frame is

$η = (1 + η) τ_{0} = 2 τ_{0}$

The law of radioactive decay implies that the flux decreases by the factor.
$\frac{J}{J _{0}} = e^{- t / τ} = e^{- l / ν τ} = e^{- l / c τ_{0}} η (2 + η)$

$= e x p (- \frac{m c l}{τ _{0} T ( T + 2 m c ^{2} )}) = 0.221$

Answer 2

Here

η = \frac{T}{m c ^{2}} = 1

so the life time of the pion in the laboratory frame is

η = (1 + η) τ_{0} = 2 τ_{0}

The law of radioactive decay implies that the flux decreases by the factor.

\frac{J}{J _{0}} = e^{- t / τ} = e^{- l / ν τ} = e^{- l / c τ_{0}} η (2 + η)

= e x p (- \frac{m c l}{τ _{0} T ( T + 2 m c ^{2} )}) = 0.221