/*! 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} Q32E If a muon is traveling at 0.999c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

If a muon is traveling at 0.999c, what are its momentum and kinetic energy? (The mass of such a muon at rest in the laboratory is 207 times the electron mass.)

Short Answer

Expert verified

The momentumP=1.27*10-18kg/s and kinetic energy K=3.63*10-10J

Step by step solution

01

Step 1:Formula of relativistic momentum and kinetic energy(K).

K=mc21-v2/c2-γmv

Where,

P→momentumγ→lorentzfactorm→restmassv→velocity

Kinetic energy is,

width="179" height="74" role="math">P=mv2-v2/c2=γmvK=(γ-1)mc2

02

Step 2:Calculating the momentum.

Given a muon is travelling at 0.999c

The mass of muon is207me=1.89*10-28kg

Therefore the momentum is,

role="math" localid="1664101880227" P=mv1-v2/c2=(1.89*10-28kg)(0.999*3*108m/s)1-(0.999c)2/c2=1.27*10-18

03

Step 3:Calculating the kinetic energy.

The kinetic energy is,

P=mv1-v2/c2-mc2=(1.89*10-28kg)(3*108m/s)21-(0.999c)2/c2-(1.89*1028kg)(3*108m/s)2=3.63*10-10J

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 91Ó°ÊÓ!

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

A 4.78 - MeV alpha particle from aR226a decay makes a head-on collision with a uranium nucleus. A uranium nucleus has 92 protons. (a) What is the distance of closest approach of the alpha particle to the center of the nucleus? Assume that the uranium nucleus remains at rest and that the approach is much greater than the radius of the uranium nucleus. (b) What is the force on the alpha particle at the instant when it is at the distance of closest approach?

Can Compton scattering occur with protons as well as electrons? For example, suppose a beam of x rays is directed at a target of liquid hydrogen. (Recall that the nucleus of hydrogen consists of a single proton.) Compared to Compton scattering with electrons, what similarities and differences would you expect? Explain.

(a) Calculate the minimum energy required to remove one proton from the nucleusC612. This is called the proton-removal energy. (Hint: Find the difference between the mass of a C612nucleus and the mass of a proton plus the mass of the nucleus formed when a proton is removed from C612. (b) How does the proton-removal energy for C612compare to the binding energy per nucleon for C612, calculated using Eq. (43.10)?

An electron is moving past the square well shown in Fig. 40.13. The electron has energy E=3U0. What is the ratio of the de Broglie wavelength of the electron in the regionx>Lto the wavelength for 0<x>L?

Creating a Particle. Two protons (each with rest massM=1.67×10-27kg ) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces ann0 particle (see Chapter 44). The rest mass of then0 is m=9.75×10-28kg. (a) If the two protons and then0 are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in . (c) What is the rest energy of the n0, expressed in ? (d) Discuss the relationship between the answers to parts (b) and (c).

See all solutions

Recommended explanations on Physics Textbooks

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.