/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 74 A pion is created in the higher ... [FREE SOLUTION] | 91Ó°ÊÓ

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A pion is created in the higher reaches of Earth's atmosphere when an incoming high-energy cosmic-ray particle collides with an atomic nucleus. A pion so formed descends toward Earth with a speed of \(0.99 c .\) In a reference frame in which they are at rest, pionsdecay with an average life of \(26 \mathrm{~ns}\). As measured in a frame fixed with respect to Earth, how far (on the average) will such a pion move through the atmosphere before it decays?

Short Answer

Expert verified
The pion travels approximately 54.4 meters before decaying.

Step by step solution

01

Understand the Time Dilation Concept

Since the pion is moving at a significant fraction of the speed of light, time dilation must be considered. In the pion's own frame, its average lifetime is 26 ns. However, in the Earth's reference frame, its lifetime appears longer due to time dilation effects.
02

Apply Time Dilation Formula

Use the time dilation formula to find the dilated time (\( t' \)) experienced by the pion in the Earth's reference frame. The formula is \( t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \), where \( t = 26 \) ns (the proper time), \( v = 0.99c \), and \( c \) is the speed of light.
03

Calculate the Lorentz Factor

Calculate the Lorentz factor (\( \gamma \)), which is \( \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \). With \( v = 0.99c \), \( \gamma \approx \frac{1}{\sqrt{1 - 0.99^2}} = \frac{1}{\sqrt{1 - 0.9801}} = \frac{1}{\sqrt{0.0199}} \).
04

Evaluate the Dilated Lifetime

Substitute \( t = 26 \) ns and the calculated value of \( \gamma \) into the time dilation formula. \( t' = 26 \times \gamma \) ns. Calculate \( \gamma \approx 7.0888 \), so \( t' = 26 \times 7.0888 \approx 184.3 \) ns.
05

Calculate the Distance the Pion Travels

Use the dilated time to calculate the average distance the pion travels before decaying, using the formula \( d = v \times t' \). Substituting \( v = 0.99c \) and \( t' \approx 184.3 \) ns, we find \( d = 0.99c \times 184.3 \times 10^{-9} \) seconds. Convert speed to meters per second: \( c \approx 3 \times 10^8 \) m/s, hence \( d \approx 0.99 \times 3 \times 10^8 \times 184.3 \times 10^{-9} \approx 54.4 \) meters.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Special Relativity
Special relativity is a groundbreaking theory developed by Albert Einstein in 1905. It revolutionizes our understanding of space and time. One of its central ideas is that the laws of physics are the same for all non-accelerating observers. This means that the speed of light is always constant, no matter how fast someone is moving relative to the light source.

A key aspect of special relativity is that time and space are relative rather than absolute. This means that time can actually "stretch" depending on how fast an object is moving. This concept is often dramatically illustrated using something called time dilation, where time appears to pass at different rates for different observers based on their velocity.

This principle is crucial when examining particles like pions, which can travel close to the speed of light. Understanding special relativity helps us predict how such particles behave from different points of view, exemplified by pions reaching much farther than they would otherwise be expected to before decaying.
Lorentz Factor
When discussing high-speed particles, the Lorentz factor (\(\gamma\)) plays a central role in special relativity phenomena such as time dilation and length contraction. The Lorentz factor is calculated using the formula:\[\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}\]where \(v\) is the velocity of the moving object and \(c\) is the speed of light.

For instance, if a pion moves at a speed of \(0.99c\), the Lorentz factor becomes significant, magnifying the effects of time dilation. In this case, using the values provided, the Lorentz factor is approximately \(\gamma \approx 7.0888\).

This means that from an observer on Earth, the pion's lifetime appears almost seven times longer than its actual lifetime from its rest frame. This extended lifetime translates to the pion traveling greater distances before decaying.
Cosmic Rays
Cosmic rays are high-energy particles originating from outer space that reach Earth. They consist largely of protons and atomic nuclei. When these cosmic rays collide with atmospheric molecules, they can yield a variety of new particles, including pions.

The study of cosmic rays helps scientists understand fundamental processes occurring in the universe. Notably, the interactions these rays have with Earth's atmosphere provide insight into particle physics, such as the behavior of particles traveling at significant fractions of the speed of light.

Pions arise from such high-energy collisions and are vital biological markers in special relativity for observing time dilation effects. By examining how far pions travel before decaying, scientists can effectively illustrate how time dilation makes them travel farther than they would in a static frame of reference.
Pion Decay
Pions are subatomic particles that play a significant role in the field of particle physics. These particles are usually generated in high-energy environments, such as those provided by cosmic ray collisions in the atmosphere.

In their rest frame, pions have a relatively short lifespan, around 26 nanoseconds on average before they decay. However, when they move at speeds close to the speed of light, their lifetime appears extended to an observer in a stationary reference frame, such as an observer on Earth. This is due to the effects of time dilation.

As a result of this extended lifetime, pions can travel surprising distances before decaying. In the given problem, this translates to a pion traveling about 54.4 meters before decay, illustrating how special relativity affects the perceived rates of fast-moving particles. Understanding pion decay in the context of relativity not only helps us understand particle behavior but also provides a practical demonstration of time dilation in action.

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Most popular questions from this chapter

(a) How much energy is released in the explosion of a fission bomb containing \(3.0 \mathrm{~kg}\) of fissionable material? Assume that \(0.10 \%\) of the mass is converted to released energy. (b) What mass of TNT would have to explode to provide the same energy release? Assume that each mole of TNT liberates 3.4 MJ of energy on exploding. The molecular mass of TNT is \(0.227 \mathrm{~kg} / \mathrm{mol}\). (c) For the same mass of explosive, what is the ratio of the energy released in a nuclear explosion to that released in a TNT explosion?

How much work must be done to increase the speed of an electron (a) from \(0.18 c\) to \(0.19 c\) and (b) from \(0.98 c\) to \(0.99 c\) ? Note that the speed increase is \(0.01 c\) in both cases.

A spaceship whose rest length is \(350 \mathrm{~m}\) has a speed of \(0.82 c\) with respect to a certain reference frame. \(A\) micrometeorite, also with a speed of \(0.82 c\) in this frame, passes the spaceship on an antiparallel track. How long does it take this object to pass the ship as measured on the ship?

Space cruisers \(A\) and \(B\) are moving parallel to the positive direction of an \(x\) axis. Cruiser \(A\) is faster, with a relative speed of \(v=0.900 c\), and has a proper length of \(L=200 \mathrm{~m}\). According to the pilot of \(A\), at the instant \((t=0)\) the tails of the cruisers are aligned, the noses are also. According to the pilot of \(B\), how much later are the noses aligned?

To four significant figures, find the following when the kinetic energy is \(10.00 \mathrm{MeV}:\) (a) \(\gamma\) and (b) \(\beta\) for an electron \(\left(E_{0}=\right.\) \(0.510998 \mathrm{MeV}\) ), (c) \(\gamma\) and (d) \(\beta\) for a proton \(\left(E_{0}=938.272 \mathrm{MeV}\right)\) and (e) \(\gamma\) and (f) \(\beta\) for an \(\alpha\) particle \(\left(E_{0}=3727.40 \mathrm{MeV}\right)\).

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