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 = `\frac {t₀}{sqrt {1 - frac {v^2}{c^2}}}`

by putting values

t = `\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - frac {(0.95c)^2}{c^2}}}`

t = `\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - frac {0.9035 \cancel {c^2}}{\cancel {c^2}}}}`

c² will be canceled so,

t = `\frac {2.6 ｘ10⁻⁸ s}{sqrt {1 - 0.9025}}`

t = `\frac {2.6 ｘ10⁻⁸ s}{sqrt {0.0875}}`

t = `\frac {2.6 ｘ10⁻⁸ s}{0.312}`

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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