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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