A particular particle created in a nuclear reactor leaves a 1cm track before decaying. Assuming that the particle moved at `0.995c` , calculate the life of the particle (a) in the lab frame and (b) in the frame of the particle .

Explanation

Given `d = 1 cm , v = 0.995 c`
(a) Time in Lab frame
` = d / v = 1 xx 10^(-2) / 0.995 c`
` 1 xx 10^(-2) / 0.995 xx 3 xx 10^(8)`
` = 33.5 xx 10^(-12) = 33.5 ps`
(b) In the frame of the particle
` t ' = 1 (sqrt 1-v^(2) / c^(2))`
` = 33.5 xx 10^(12) / (sqrt 1 -(0.995)^(2))`
` = 335.41 ps` .