91Ó°ÊÓ

The height for collision above Earth’s surface is H=120 km

The total energy of pion is $E=1.35×{10}^5\text{ M±ð³Õ}$

The duration for pion decay is $t=35\text{ n²õ}$

The rest energy of pion is $E_0=139.6\text{ M±ð³Õ}$

The rest energy of pion is the energy of pion before collision due to mass of the pion. If kinetic energy of pion is taken with rest energy then it is called total energy.

The Lorentz factor for pion is given as:

$\mathrm{γ}=\frac{E}{E_0}$

Substitute all the values in the above equation.

$\begin{array}{l}\mathrm{γ}=\frac{1.35×{10}^5\text{ M±ð³Õ}}{139.6\text{ M±ð³Õ}}\\ \mathrm{γ}=967.05\end{array}$

The distance of pion from Earth’s surface is given as:

$h=\mathrm{γ}â‹\dots câ‹\dots t$

Here, $\text{c}$ is the speed of light and its value is $3×{10}^8\text{″¾}/\text{s}$.

Substitute all the values in the above equation.

$\begin{array}{l}h=\left(967.05\right)(3×{10}^8\text{″¾}/\text{s})\left(35\text{ n²õ}\right)\left(\frac{{10}^{−9}\text{ s}}{1\text{ n²õ}}\right)\\ h=10154.03\text{″¾}\\ h=\left(10154.03\text{″¾}\right)\left(\frac{1\text{ k³¾}}{1000\text{″¾}}\right)\\ h≈10\text{ k³¾}\end{array}$

The distance of the position of collision for pion from sea level is given as:

$d=H−h$

Substitute all the values in the above equation.

$\begin{array}{l}d=120\text{ k³¾}−10\text{ k³¾}\\ d=110\text{ k³¾}\end{array}$

Therefore, the distance of the position of collision for pion from sea level is $110\text{ k³¾}$.