Question

A particular particle created in a nuclear reactor leaves a 1 cm 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.

Answer 1

Given:-
Length of the track, d = 1 cm
Velocity of the particle, v = 0.995c

(a) Life of the particle in the lab frame is given by

\[t = \frac{d}{v} = \frac{0 . 01}{0 . 995c}\]

\[ = \frac{0 . 01}{0 . 995 \times 3 \times {10}^8}\]

\[ = 33 . 5 \times {10}^{- 12} s = 33 . 5\text{ ps}\]

(b) Let the life of the particle in the frame of the particle be t'. Thus,

\[t' = \frac{t}{\sqrt{1 - v^2 / c^2}}\]

\[t' = \frac{33 . 5 \times {10}^{- 12}}{\sqrt{1 - \left( 0 . 995 \right)^2}}\]

\[t' = 3 . 3541 \times {10}^{- 12} s = 3 . 3541\text{ ps}\]