Physics for Scientists and Engineers with Modern Physics, Technology Update
Physics for Scientists and Engineers with Modern Physics, Technology Update
9th Edition
ISBN: 9781305401969
Author: SERWAY, Raymond A.; Jewett, John W.
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 39, Problem 66AP

(a)

To determine

The speed of a proton that has the same kinetic energy as the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 66AP

The speed of the proton that has the same kinetic energy as the electron is 0.0236c_.

Explanation of Solution

Write the expression for the kinetic energy of the electron.

  Ke=mec2(γe1)                                                                                                      (I)

Here, Ke is the kinetic energy of the electron, me is the mass of the electron, c is the speed of light, and γe is the Lorentz factor for the electron.

Write the expression for the kinetic energy of the proton.

  Kp=mpc2(γp1)                                                                                                    (II)

Here, Kp is the kinetic energy of the proton, mp is the mass of the proton, and γp is the Lorentz factor of the proton.

Given that the kinetic energy of electron and proton are equal.

  Kp=Ke                                                                                                                   (III)

Use expressions (I) and (II) in expression (III).

  mec2(γe1)=mpc2(γp1)                                                                                    (IV)

Solve expression (IV) for γp.

  γp=1+mec2(γe1)mpc2                                                                                                (V)

Write the expression for Lorentz factor of electron.

  γe=11(uec)2                                                                                                      (VI)

Here, ue is the speed of electron.

Write the expression for Lorentz factor of proton.

  γp=11(upc)2                                                                                                    (VII)

Here, up is the speed of proton.

Use expression (VII) for up.

  up=c1γp2                                                                                                      (VIII)

Conclusion:

Substitute 0.750c for ue in equation (VI) to find γe.

  γe=11(0.750c)2c2=1.5119

Substitute 9.1×1031kg for me, 3.00×108m/s for c, 1.67×1027kg for mp, and 1.5119 for γe in equation (V) to find γp.

  γp=1+(9.1×1031kg)(3.00×108m/s)2(1.51191)1.6×1019kgm2/s2(1.67×1027kg)(3.00×108m/s)21.6×1019kgm2/s2=1.000279

Substitute 1.000279 for γp in equation (VIII) to find up.

  up=c1(1.000279)2=0.0236c

Therefore, the speed of the proton that has the same kinetic energy as the electron is 0.0236c_.

(b)

To determine

The speed of the proton that has the same momentum as the electron.

(b)

Expert Solution
Check Mark

Answer to Problem 66AP

The speed of the proton that has the same momentum as the electron is 6.18×104c_.

Explanation of Solution

Write the expression for the relativistic momentum of electron.

  pe=γemeue                                                                                                             (IX)

Here, pe is the relativistic momentum of electron.

Write the expression for the relativistic momentum of proton.

  pp=γpmpup                                                                                                             (X)

Here, pp is the relativistic momentum of proton.

Momentum of electron and proton are equal.

  pe=pp                                                                                                                   (XI)

Use expressions (IX) and (X) in (XI).

  γemeue=γpmpup                                                                                                    (XII)

Solve equation (XII) for γp.

  up=γemeuempγp                                                                                                          (XIII)

Conclusion:

Substitute 1.5119 for γe, 9.1×1031kg for me, 0.750c for ue, 1.67×1027kg for mp, and 1.000279 for γp in equation (XIII) to find up.

  up=(1.5119)(9.1×1031kg)(0.750c)(1.67×1027kg)(1.000279)=6.18×104c

Therefore, the speed of the proton that has the same momentum as the electron is 6.18×104c_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 39 Solutions

Physics for Scientists and Engineers with Modern Physics, Technology Update

Ch. 39 - A spacecraft zooms past the Earth with a constant...Ch. 39 - Prob. 3OQCh. 39 - Prob. 4OQCh. 39 - Prob. 5OQCh. 39 - Prob. 6OQCh. 39 - Prob. 7OQCh. 39 - Prob. 8OQCh. 39 - Prob. 9OQCh. 39 - Prob. 10OQCh. 39 - Prob. 1CQCh. 39 - Prob. 2CQCh. 39 - Prob. 3CQCh. 39 - Prob. 4CQCh. 39 - Prob. 5CQCh. 39 - Prob. 6CQCh. 39 - Prob. 7CQCh. 39 - Prob. 8CQCh. 39 - Prob. 9CQCh. 39 - Prob. 10CQCh. 39 - Prob. 11CQCh. 39 - Prob. 12CQCh. 39 - Prob. 13CQCh. 39 - Prob. 14CQCh. 39 - Prob. 1PCh. 39 - In a laboratory frame of reference, an observer...Ch. 39 - The speed of the Earth in its orbit is 29.8 km/s....Ch. 39 - Prob. 4PCh. 39 - A star is 5.00 ly from the Earth. At what speed...Ch. 39 - Prob. 6PCh. 39 - Prob. 7PCh. 39 - Prob. 8PCh. 39 - Prob. 9PCh. 39 - An astronaut is traveling in a space vehicle...Ch. 39 - Prob. 11PCh. 39 - Prob. 12PCh. 39 - Prob. 13PCh. 39 - Prob. 14PCh. 39 - Prob. 15PCh. 39 - Prob. 16PCh. 39 - Prob. 17PCh. 39 - A cube of steel has a volume of 1.00 cm3 and mass...Ch. 39 - Prob. 19PCh. 39 - Prob. 20PCh. 39 - Prob. 21PCh. 39 - Review. In 1963, astronaut Gordon Cooper orbited...Ch. 39 - Prob. 23PCh. 39 - Prob. 24PCh. 39 - Prob. 25PCh. 39 - Prob. 26PCh. 39 - Prob. 27PCh. 39 - Prob. 28PCh. 39 - Prob. 29PCh. 39 - Prob. 30PCh. 39 - Prob. 31PCh. 39 - Prob. 32PCh. 39 - Prob. 33PCh. 39 - Prob. 34PCh. 39 - Prob. 35PCh. 39 - Prob. 36PCh. 39 - Prob. 37PCh. 39 - Prob. 38PCh. 39 - Prob. 39PCh. 39 - Prob. 40PCh. 39 - Prob. 41PCh. 39 - Prob. 42PCh. 39 - Prob. 43PCh. 39 - Prob. 44PCh. 39 - Prob. 45PCh. 39 - Prob. 46PCh. 39 - Prob. 47PCh. 39 - (a) Find the kinetic energy of a 78.0-kg...Ch. 39 - Prob. 49PCh. 39 - Prob. 50PCh. 39 - Prob. 51PCh. 39 - Consider electrons accelerated to a total energy...Ch. 39 - Prob. 53PCh. 39 - Prob. 54PCh. 39 - Prob. 55PCh. 39 - Prob. 56PCh. 39 - Prob. 57PCh. 39 - Prob. 58PCh. 39 - Prob. 59PCh. 39 - Prob. 60PCh. 39 - Prob. 61PCh. 39 - An unstable particle with mass m = 3.34 1027 kg...Ch. 39 - Prob. 63PCh. 39 - Prob. 64PCh. 39 - Prob. 65PCh. 39 - Prob. 66APCh. 39 - Prob. 67APCh. 39 - Prob. 68APCh. 39 - Prob. 69APCh. 39 - Prob. 70APCh. 39 - Prob. 71APCh. 39 - Prob. 72APCh. 39 - Prob. 73APCh. 39 - Prob. 74APCh. 39 - Prob. 75APCh. 39 - Prob. 76APCh. 39 - Prob. 77APCh. 39 - Prob. 78APCh. 39 - Prob. 79APCh. 39 - Prob. 80APCh. 39 - Prob. 81APCh. 39 - Prob. 82APCh. 39 - An alien spaceship traveling at 0.600c toward the...Ch. 39 - Prob. 84APCh. 39 - Prob. 85APCh. 39 - Prob. 86APCh. 39 - Prob. 87APCh. 39 - Prob. 88CPCh. 39 - The creation and study of new and very massive...Ch. 39 - Prob. 90CPCh. 39 - Owen and Dina are at rest in frame S, which is...
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
University Physics Volume 3
Physics
ISBN:9781938168185
Author:William Moebs, Jeff Sanny
Publisher:OpenStax
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Length contraction: the real explanation; Author: Fermilab;https://www.youtube.com/watch?v=-Poz_95_0RA;License: Standard YouTube License, CC-BY