/*! 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} Q7PE Suppose a \[{{\rm{W}}^{\rm{ - }}... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 33: Q7PE (page 1212)

Suppose a \[{{\rm{W}}^{\rm{ - }}}\]created in a bubble chamber lives for \[{\rm{5}}{\rm{.00 \times 1}}{{\rm{0}}^{{\rm{ - 25}}}}{\rm{\;s}}\]. What distance does it move in this time if it is traveling at \[{\rm{0}}{\rm{.900c}}\]? Since this distance is too short to make a track, the presence of the \[{{\rm{W}}^{\rm{ - }}}\]must be inferred from its decay products. Note that the time is longer than the given \[{{\rm{W}}^{\rm{ - }}}\]lifetime, which can be due to the statistical nature of decay or time dilation.

Short Answer

Expert verified

The distance the particle travels is\[1.35 \times {10^{ - 16}}\;\;{\rm{m}}\].

Step by step solution

01

Definition of work done

The length of the path traveled by the particle gives the distance traveled. It can

Be described as the product of speed and time taken to travel the given distance.

02

Given Data

Lifespan of the particle is-\(\Delta t = 5.00 \times {10^{ - 25}}\;{\rm{s}}\)

The speed of the particle is-\(v = 0.900c\)

03

Finding the distance traveled by particle

We apply the equation to calculate the particle's journey distance.\[{\rm{d = vt}}\]and solve it for\[d\]

\[\begin{array}{c}D = vt\\ = \left( {0.900 \times 3 \times {{10}^8}\;\;{\rm{m/s}}} \right) \times \left( {5.00 \times {{10}^{ - 25}}\;{\rm{s}}} \right)\\ = 1.35 \times {10^{ - 16}}\;\;{\rm{m}}\end{array}\]

Therefore, the distance the particle travels is\[1.35 \times {10^{ - 16}}\;\;{\rm{m}}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The decay mode of the positive tau is\({{\bf{\tau }}^ + } \to {\rm{ }}{{\bf{\mu }}^ + }{\rm{ }} + {\rm{ }}{{\bf{\nu }}_{\bf{\mu }}}{\rm{ }} + {\rm{ }}{{\bf{\bar \nu }}_{\bf{\tau }}}\).

(a) What energy is released?

(b) Verify that charge and lepton family numbers are conserved.

(c) The \({\tau ^ + }\)is the antiparticle of the \({\tau ^ - }\). Verify that all the decay products of the \({\tau ^ + }\)are the antiparticles of those in the decay of the \({\tau ^ - }\) given in the text.

Accelerators such as the Triangle Universities Meson Facility (TRIUMF) in British Columbia produce secondary beams of pions by having an intense primary proton beam strike a target. Such "meson factories" have been used for many years to study the interaction of pions with nuclei and, hence, the strong nuclear force. One reaction that occurs is\({{\rm{\pi }}^{\rm{ + }}}{\rm{ + p}} \to {{\rm{\Delta }}^{{\rm{ + + }}}} \to {{\rm{\pi }}^{\rm{ + }}}{\rm{ + p}}\), where the \({{\rm{\Delta }}^{{\rm{ + + }}}}\)is a very short-lived particle. The graph in Figure \({\rm{33}}{\rm{.26}}\)shows the probability of this reaction as a function of energy. The width of the bump is the uncertainty in energy due to the short lifetime of the\({{\rm{\Delta }}^{{\rm{ + + }}}}\).

(a) Find this lifetime.

(b) Verify from the quark composition of the particles that this reaction annihilates and then re-creates a d quark and a \({\rm{\bar d}}\)antiquark by writing the reaction and decay in terms of quarks.

(c) Draw a Feynman diagram of the production and decay of the \({{\rm{\Delta }}^{{\rm{ + + }}}}\)showing the individual quarks involved.

(a) Is the decay \({\sum ^{\rm{ - }}} \to {\rm{n + }}{\pi ^{\rm{ - }}}\) possible considering the appropriate conservation laws? State why or why not.

(b) Write the decay in terms of the quark constituents of the particles.

Why does the \({\eta ^0}\) meson have such a short lifetime compared to most other mesons?

There are particles called bottom mesons or \({\rm{B}}\)-mesons. One of them is the \({{\rm{B}}^{\rm{ - }}}\)meson, which has a single negative charge; its baryon number is zero, as are its strangeness, charm, and topness. It has a bottomness of \({\rm{ - 1}}\). What is its quark configuration?
See all solutions

Recommended explanations on Physics Textbooks

Scientific Method Physics

Read Explanation

Mechanics and Materials

Read Explanation

Atoms and Radioactivity

Read Explanation

Medical Physics

Read Explanation

Turning Points in Physics

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.