ne of the many fundamental particles in nature is the muon mu. This particle acts very much like a "heavy electron." It has a mass of 106 MeV/c^2, compared to the electron's mass of just 0.511 MeV/c^2. (We are using E = mc^2 to obtain the mass in units of energy and the speed of light c).
Unlike the electron, though, the muon has a finite lifetime, after which it decays into an electron and two very light particles called neutrinos (nu). We'll ignore the neutrinos throughout this problem.
If the muon is at rest, the characteristic time that it takes it to decay is about 2.2 microseconds. Most of the time, though, particles such as muons are not at rest and, if they are moving relativistically, their lifetimes are increased by time dilation. In this problem we will explore some of these relativistic effects.
Cosmic rays are constantly raining down on the earth from outer space. (Figure 1)The majority of these cosmic rays are protons, and when upper atmosphere, they can convert into particles called pions (), which subsequently decay into muons with a characteristic lifetime of T.
their rest frame). These muons can then continuo down toward the earth until they too decay into electrons, which are so light that they stop atmosphere). Let's look at how time dilation affects those cosmic rays.
Suppose that a cosmic-ray proton crashes into a nitrogen molecule in the upper atmosphere. 45 km above the earth's surface, producing a pia a muon. Assume that the pion has a velocity of 99.99% the speed of light and decays into a muon, which, owing to kinematics, has a downward velocity of 99.9943% the speed of light.
let’s consider the effect of time dilation how far would the pion actually travel before decaying?
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Answer:
To get the number of half-lives passed, we divide the time of flight from part a by the length of a half-life, as
given in the statement of the problem:
# of half lives = 2×10−4
1.5×10−6 ≈ 133
The fraction of remaining muons will decrease by 1
2
every half life, so the fraction after a given number N of
half-lives will be:
1
2 ×
1
2 ×
1
2
· · · N times · · · × 1
2 = ( 1
2
)
N .
This makes the fraction remaining equal to 1/2 to the power 133:
fraction remaining = ( 1
2
)
133 ≈ 10−40. That’s not very many muons
Explanation:
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