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प्रश्न
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.
उत्तर
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,
\[t' = \frac{t}{\sqrt{1 - v^2 / c^2}}\]
\[t' = \frac{33 . 5 \times {10}^{- 12}}{\sqrt{1 - \left( 0 . 995 \right)^2}}\]
\[t' = 3 . 3541 \times {10}^{- 12} s = 3 . 3541\text{ ps}\]
