Question #338938

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.


Expert's answer

t=t01−v2c2=9⋅10−8 s.t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}=9\cdot 10^{-8}~s.


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