#### Here $$ \eta = \dfrac{T}{mc^{2}} = 1 $$ so the life time of the pion in the laboratory frame is

$$ \eta = (1 + \eta) \tau_{0} = 2 \tau_{0} $$

The law of radioactive decay implies that the flux decreases by the factor.

$$ \dfrac{J}{J_{0}} = e^{-t/\tau} = e^{-l/ \nu\ \tau} = e^{-l/c \tau_{0}} \sqrt{\eta (2 + \eta)} $$

$$ = exp \left (- \dfrac{m\ c\ l}{\tau_{0} \sqrt{T (T + 2m\ c^{2})}} \right ) = 0.221 $$