Question
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:
  1. In the lab frame and.
  2. In the frame of the particle.

Answer

$\text{d}=1\text{cm}, \ \text{v}=0.995\text{c}$
  1. Time in Laboratory frame $=\frac{\text{d}}{\text{v}}=\frac{1\times10^{-2}}{0.995\text{c}}$
$=\frac{1\times10^{-2}}{0.995\times3\times10^{8}}=33.5\times10^{-12}=33.5\text{PS}$
  1. In the frame of the particle
$\text{t}'=\frac{\text{t}}{\sqrt{1-\frac{\text{v}^2}{\text{c}^2}}}=\frac{33.5\times10^{-12}}{\sqrt{1-(0.995)^2}}=335.41\text{PS}$

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