A particle called the pion lives on average only about 2.6 ｘ10⁻⁸ s when at rest in the laboratory. It then changes to another form. How long would such a particle live when shooting through space at 0.95c?[Ans. 8.3 x 10⁻⁸ s)





Given:


Proper time at rest = t₀ = 2.6 ｘ10⁻⁸ s

Observed Speed in space = v = 0.95 c 


To Find:

Measured time of the pion to live in space = t = ?


Solution:


The time dilation formula in special theory of relativity is given by: 

t = t₀√1-v2c2

by putting values

t = 2.6ｘ10⁻⁸s√1-(0.95c)2c2

t = 2.6ｘ10⁻⁸s√1-0.9035c2c2

c² will be canceled so,

t = 2.6ｘ10⁻⁸s√1-0.9025

t = 2.6ｘ10⁻⁸s√0.0875

t = 2.6ｘ10⁻⁸s0.312

t = 8.333 ｘ10⁻⁸ s    -----------------Ans.

Thus, shooting a pion at speed v = 0.95 c will change to another form after an interval of  8.333 ｘ10⁻⁸ s. 


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