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
$$\begin{gathered} t=\frac{d}{v}=\frac{0.01}{0.995c} \\ =\frac{0.01}{0.995\times 3\times {10}^8} \\ =33.5\times {10}^{-12}\mathrm{s}=33.5\mathrm{ps} \end{gathered}$$

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

$t\text{'}=\frac{t}{\sqrt{1-v^2/c^2}}$
$t\text{'}=\frac{33.5\times {10}^{-12}}{\sqrt{1-{\left(0.995\right)}^2}}$

$t\text{'}=3.3541\times {10}^{-12}\mathrm{s}=3.3541\mathrm{ps}$