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Chapter 26: Problem 23

The mean (average) lifetime of a muon in its rest frame is $2.2 \mu \mathrm{s}\(. A beam of muons is moving through the lab with speed \)0.994 c .$ How far on average does a muon travel through the lab before it decays?

Short Answer

Expert verified
Answer: The muon travels on average about 15.47 x 10^6 meters in the lab frame before it decays.

Step by step solution

01

Find the time dilation factor

Use the relativistic time dilation formula to find the time dilation factor, which is given by \(\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\), where \(v\) is the speed of the muon, and \(c\) is the speed of light. We are given \(v = 0.994c\). Plugging the values, we get: \(\gamma = \frac{1}{\sqrt{1 - \frac{(0.994c)^2}{c^2}}} = \frac{1}{\sqrt{1 - 0.994^2}}\)
02

Calculate the time dilation factor

Now, calculate the value of \(\gamma\): \(\gamma = \frac{1}{\sqrt{1 - 0.994^2}} \approx 7.089\)
03

Convert the mean lifetime to the lab frame

Using the time dilation factor, convert the mean lifetime of the muon in its rest frame to the lab frame by multiplying the rest frame mean lifetime by the time dilation factor, \(\gamma\). The mean lifetime in the rest frame, \(t_r\), is given as \(2.2 \mu s\). So, the mean lifetime in the lab frame, \(t_l\), will be: \(t_l = \gamma \cdot t_r \approx 7.089 \times 2.2 \mu s\)
04

Calculate the mean lifetime in the lab frame

Now, calculate the value of the mean lifetime in lab frame: \(t_l \approx 7.089 \times 2.2 \mu s = 15.596 \mu s\)
05

Find the average distance traveled in the lab frame

To find the average distance traveled by a muon in the lab frame before it decays, multiply the mean lifetime in the lab frame by the speed of the muon: \(distance = t_l \cdot v \approx 15.596 \mu s \times 0.994c\)
06

Calculate the average distance traveled

Finally, calculate the value of the distance: \(distance \approx 15.596 \mu s \times 0.994c \approx 15.47 \times 10^6 m\) The muon travels on average about \(15.47 \times 10^6 m\) in the lab frame before it decays.

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A neutron (mass 1.00866 u) disintegrates into a proton (mass $1.00728 \mathrm{u}\( ), an electron (mass \)0.00055 \mathrm{u}$ ), and an antineutrino (mass 0 ). What is the sum of the kinetic energies of the particles produced, if the neutron is at rest? $\left(1 \mathrm{u}=931.5 \mathrm{MeV} / \mathrm{c}^{2} .\right)$
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