Answer to Question #152883 in Mechanics | Relativity for Abhay

Question

The lifetime of π+-meson is 2.5x10-8 sec in its own frame of reference. If a beam of these mesons of velocity 2.4x108 m/s is produced, calculate the distance, the beam can travel before the flux of meson is reduced to 1/e2 times the initial value

Answer 1

Answer

Lifetime is given by

$\tau=\frac{\tau}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{2.5\times10^{-8}}{\sqrt{1-\frac{v^2}{c^2}}}$

$=5.60\times10^{-8}s$

the distance, the beam can travel before the flux of meson is reduced to 1/e^2 times the initial value

$Speed =\frac{distance }{time}$

So

Distance=$2.4\times10^8\times5.60\times10^{-8}$

=13.44m

But according to question meson is reduced to 1/e2 times the initial value so it's become

Distance

$=13.44/e^2=10m$