Unstable pions are produced as a beam in a nuclear reaction experiment . The pions leave the target at a speed of 0.995c. The intensity of the beam reduces to half its original value as the beam travels a distance of 39 m . Find the half - life of pions (a) in the laboratory frame, (b) in their rest frame.

Explanation

(a) The intensity of the pion beam reduces to half its original value in one half - life. The half - life of the pions as measured in the laboratory is
` t_(1/2) = 39 m / 0995c = 39 m / 0.995 xx 3 xx 10 ^(8) m s^(-1)`
`= 1.3 xx 10^(-7).` (b) The events - a pion leaving the target and its decaying - occur at the same place in the pion - frame . thus, the time measured in the pion - frame is the proper time and is the smallest. It is equal to
` t'_(1/2) = t_(1/2) (sqrt 1 - v^(2) / c^(2)) = (1.3 xx 10^(-7) s ) (sqrt 1- (0.995)^(2)`
`= 1.3 xx 10^(-8) s.`