Question

In a high-energy collision between a cosmic-ray particle and a particle near the top of Earth’s atmosphere, 120 km above sea level, a pion is created. The pion has a total energy E of $\mathbf{1}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathbf{MeV}$and is travelling vertically downward. In the pion’s rest frame, the pion decays 35.0 ns after its creation. At what altitude above sea level, as measured from Earth’s reference frame, does the decay occur? The rest energy of a pion is 139.6 MeV.

Short answer

The distance of the position of collision for pion from sea level is 110 km.

Step-by-step solution

Identification of given data

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

The total energy of pion is $E=1.35\times {10}^5\text{\hspace{0.17em}MeV}$

The duration for pion decay is $t=35\text{\hspace{0.17em}ns}$

The rest energy of pion is $E_0=139.6\text{\hspace{0.17em}MeV}$

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.

Determination of distance from the Earth’s surface for energy collision

The Lorentz factor for pion is given as:

$\gamma =\frac{E}{E_0}$

Substitute all the values in the above equation.

$\begin{array}{l}\gamma =\frac{1.35\times {10}^5\text{\hspace{0.17em}MeV}}{139.6\text{\hspace{0.17em}MeV}}\\ \gamma =967.05\end{array}$

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

$h=\gamma \cdot c\cdot t$

Here, $\text{c}$ is the speed of light and its value is $3\times {10}^8\text{\hspace{0.17em}m}/\text{s}$.

Substitute all the values in the above equation.

$\begin{array}{l}h=\left(967.05\right)(3\times {10}^8\text{\hspace{0.17em}m}/\text{s})\left(35\text{\hspace{0.17em}ns}\right)\left(\frac{{10}^{-9}\text{\hspace{0.17em}s}}{1\text{\hspace{0.17em}ns}}\right)\\ h=10154.03\text{\hspace{0.17em}m}\\ h=\left(10154.03\text{\hspace{0.17em}m}\right)\left(\frac{1\text{\hspace{0.17em}km}}{1000\text{\hspace{0.17em}m}}\right)\\ h\approx 10\text{\hspace{0.17em}km}\end{array}$

Determination of distance of collision from sea level

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.

$\begin{array}{l}d=120\text{\hspace{0.17em}km}-10\text{\hspace{0.17em}km}\\ d=110\text{\hspace{0.17em}km}\end{array}$

Therefore, the distance of the position of collision for pion from sea level is $110\text{\hspace{0.17em}km}$.