A famous experiment detected 527 muons per hour at the top of Mt. Washington, New Hemisphere, elevation 1910 m . At sea level, the same equipment detected 395 muons per hour. A discriminator selected for muons whose speed was between 0.9950 c and 0.9954c . Given that the mean lifetime T of a muon in a frame in which it is at rest is $\mathbf{2}\mathbf{.}\mathbf{2}\mathbf{}\mathit{\mu }\mathit{s}$and that in this frame the number of muons decays exponentially with time according to $\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\mathbf{i}\mathbf{/}\mathbf{T}}$, show that the results obtained in the experiment are sensible.

Consider a lifetime of muons at rest is $T=2.2\mu s$.

Consider the elevation is ${\iota }_0=1910\mathrm{m}$.

Consider a minimum speed of muons is$v_{min}=0.9950c$.

Consider a maximum speed of muons is $v_{max}=0.9954c$.

Consider a number of muons detected at sea level is N = 395.

Consider a number of muons detected at mountain is$N_0=395$ .

Write the formula of number of muons decays exponentially with time.

$\mathbf{N}\mathbf{=}{\mathbf{N}}_{\mathbf{0}}{\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{T}}}$ …… (1)

Here, ${\mathbf{e}}^{\mathbf{-}\frac{\mathbf{t}}{\mathbf{T}}}$is muons decay exponentially.

The following equation may be used to show how a relativistic effect causes an object's length to contract:

$l=l_0\sqrt{1-\frac{v^2}{c^2}}$

According to the following connection, the quantity of muons decreases exponentially with time:

$N=N_0{e}^{-\frac{t}{T}}$

Since the muons' speed ranges between 0.9950c and 0.9954c , we may use this range to calculate their average speed as follows:

$$\begin{gathered} v_{ave}=\frac{v_{max}+v_{min}}{2} \\ =\frac{0.9954c+0.9950c}{2} \\ =0.9952c \end{gathered}$$

The distance from the peak to sea level is also impacted by relativistic effects since muons are moving so quickly. Consequently, the separation would decrease. Eq. (1) may be used to compute the distance:

$$\begin{gathered} l=l_0\sqrt{1-\frac{v_{ave}^2}{c^2}} \\ =\left(1910\right)\sqrt{1-\frac{{\left(0.9952c\right)}^2}{c^2}} \\ =1910\sqrt{1-\frac{{\left(0.9952c\right)}^2}{c^2}} \\ =1910\sqrt{1-{\left(0.9952\right)}^2} \end{gathered}$$

Solve further as:

l= 187 m

Hence, it would take a distance of 187 m for the muons to travel the distance.

Next, since we know that the speed, time, and distance have the following relationship:$v=\frac{d}{t}$.We can obtain the time it takes for the muons to travel a distance of 187 m as follows:

$$\begin{gathered} v_{ave}=\frac{d}{t} \\ t=\frac{d}{v_{ave}} \\ =\frac{187}{0.9952c} \\ =\frac{187}{0.9952.3.{10}^8} \end{gathered}$$

Solve further as:

$t=0.63\mu s$

We were able to determine the muons' journey time, therefore we used Eq. (2) to get the connection when the muons' number decays exponentially with time:

Determine the number of muons decays exponentially with time.

Substitute -0.63 for t and 2.2 for T into equation (2).

$$\begin{gathered} \frac{N}{N_0}=e^{-\frac{t}{T}} \\ =e^{-\frac{0.63}{2.2}} \\ \approx 0.75 \end{gathered}$$

Experimentally, it was obtained that $N_0=527$and $N=395$, we verify the result by obtaining the ratio:

$$\begin{gathered} \frac{N}{N_0}=\frac{395}{527} \\ \approx 0.75 \end{gathered}$$

The experiment's outcomes are logical since they are based on the results and Eq (2).