Answer to Question #338938 in Mechanics | Relativity for ayushi

Question

Explain the following observation on the basis of theory of relativity, the half-life of a

particular particle as measured in the laboratory comes out to be 4.0 × 10-8 s when its speed is 0.8c

and 3.0 × 10-8 s when its speed is 0.6c.

Answer 1

$t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}=9\cdot 10^{-8}~s.$