A pion is created in the higher reaches of Earth’s atmosphere when an incoming high-energy cosmic-ray particle collides with an atomic nucleus. A pion so formed descends toward Earth with a speed of 0.99c. In a reference frame in which they are at rest, pion decay with an average life of 26 ns. As measured in a frame fixed with respect to Earth, how far (on the average) will such a pion move through the atmosphere before it decays?

A pion is a part of the meson family where a meson particle combines a quark and an anti-quark. A pion is denoted by $\pi$and is of three types, namely ${\pi }^{+},{\pi }^o\&{\pi }^{-}$.

Suppose two consecutive events occur at the same location; the time interval measured in the same inertial reference frame is called proper time. And the time interval measured in any other reference frame relative to this frame will be longer than the proper time. The formula given below is used to determine the time interval in another frame.

$\Delta t=\frac{\Delta t_o}{\sqrt{1-u^2/c^2}}$

Where $\Delta t_o$is the proper time, and $\Delta t$is the interval measured by an observer moving with a relative speed $u$.

Here the pion is moving at 0.99c, and the proper time of decay is 26 ns; substituting these values in the above equation,

$$\begin{gathered} \Delta t=\frac{26\times {10}^{-9}s}{\sqrt{1-\frac{{\left(0.99c\right)}^2}{c^2}}} \\ =\frac{26\times {10}^{-9}s}{\sqrt{0.0199}} \\ =184ns \end{gathered}$$

The time interval for the pion to decay as measured from the earth’s frame is 184 ns. In this period, the pion must have travelled an average distance of

$$\begin{gathered} u=\frac{d}{\Delta t} \\ d=u\Delta t \\ =(0.99\times 3\times {10}^{8}m/s)(184\times {10}^{-9}s) \\ \approx 55m \end{gathered}$$

Thus. the time interval for the pion to decay as measured from the earth’s frame is 184 ns. In this period, the pion must have travelled an average distance of 55 m.