A certain particle is unstable and has a velocity of 085 c r

A certain particle is unstable and has a velocity of 0.85 c, relative to the lab. If it exists for 25 ns, in its own rest frame, how long does it exist in the lab? how far does it travel relative to the lab? According to an observer in the particle\'s rest frame, how far apart are the points in the lab where the particle is created and where it decays?

Solution

Y = 1 / sqrt [ 1- (v/c)^2]

= 1 / sqrt[1 - (0.85c/c)^2] = 1.898

(a) t = Y t0 = 1.898 x 25ns = 47.46 ns

(B) L = v t = (0.85c) (47.46 ns)

= (0.85 x 3 x 10^8) ( 47.46 x 10^-9 ) = 12.1 m

(C) L0 = L / Y = 12.1 / 1.898 = 6.38 m