EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC
1st Edition
ISBN: 9781337684668
Author: Katz
Publisher: VST
bartleby

Concept explainers

Question
Book Icon
Chapter 39, Problem 51PQ

(a)

To determine

The rest mass energy of the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 51PQ

The rest mass energy of an electron is 512keV or 8.20×1014J.

Explanation of Solution

Write the expression for the rest mass energy.

    Erest=mrestc2                        (I)

Here, mrest is the rest mass,c is the speed of the light and Erest is the total energy of the electron at rest.

Conclusion:

Substitute 9.11×1031kg for mrest and 3.00×108m/s for c in equation (I).

    Erest=(9.11×1031kg)(3.00×108m/s)2=8.20×1014J                 (II)

Convert J to keV by multiplying with (1keV1.602×1016J) in equation (II).

Here, keV is the kilo electron volt, and J is joule.

    Erest=8.20×1014J(1keV1.602×1016J)=8.20×10141.602×1016(keV×JJ)=512keV

Therefore, the rest mass energy of an electron is 512keV or 8.20×1014J.

(b)

To determine

The total energy of an electron.

(b)

Expert Solution
Check Mark

Answer to Problem 51PQ

The total energy of an electron is 1.30×1013J or 812keV.

Explanation of Solution

Write the expression for total energy of an electron.

    E=K+Erest                  (III)

Here, E is the total energy of the electron, K is the kinetic energy and Erest is the total energy of the electron at rest.

Conclusion:

Substitute 300.0keV for K and 512keV for Erest in equation (III).

    E=300.0keV+512keV=812keV                     (IV)

Convert keV to J by multiplying with (1.602×1016J1keV) in equation (IV).

    E=812keV(1.602×1016J1keV)=812(1.602×1016)(keV×JkeV)=1.30×1013J

Therefore, the total energy of an electron is 1.30×1013J or 812keV.

(c)

To determine

The speed of the electron.

(c)

Expert Solution
Check Mark

Answer to Problem 51PQ

The speed of the electron is 0.7762c.

Explanation of Solution

Write the expression of the relativistic kinetic energy of the particle.

    K(γ1)mrestc2                                          (V)

Here, mrest is the rest mass of the particle, v is the moving velocity of the particle velocity, γ=11(vc)2 is the Lorentz factor, c is the speed of light and K is the relativistic energy.

Write the expression for Lorentz factor.

    γ=11(vc)2

Substitute 11(vc)2 for γ in the Equation (V)

    K=(11(vc)21)mrestc2                               (VI)

Convert the unit of K from keV to J by multiplying (1.602×1016J1keV) with 300.0keV.

    K=300.0keV=300.0keV(1.602×1016J1keV)=4.806×1014J                             (VII)

Conclusion:

Substitute 4.806×1014J for K, 9.11×1031kg for mrest and 3.00×108m/s for c in equation (VI).

    4.806×1014J=(11(vc)21)(9.11×1031kg)(3.00×108m/s)24.806×1014J(9.11×1031kg)(3.00×108m/s)2=(11(vc)21)1(vc)2=11+(4.806×1014J(9.11×1031kg)(3.00×108m/s)2)1(vc)2=(11+0.586)2

Solve the equation further to find v.

    1(vc)2=0.3975(vc)2=(10.3975)(vc)=0.6025=0.7762c

Therefore, the speed of the electron is 0.7762c.

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!
Students have asked these similar questions
2-3. Consider the situation of the reflection of a pulse at the interface of two string described in the previous problem. In addition to the net disturbances being equal at the junction, the slope of the net disturbances must also be equal at the junction at all times. Given that p1 = 4.0 g/m, H2 = 9.0 g/m and Aj = 0.50 cm find 2. A, (Answer: -0.10 cm) and 3. Ay. (Answer: 0.40 cm)please I need to show all work step by step problems 2 and 3
From number 2 and 3 I just want to show all problems step by step please do not short cut look for formula
Plz don't use chatgpt pls will upvote

Chapter 39 Solutions

EBK WEBASSIGN FOR KATZ'S PHYSICS FOR SC

Ch. 39 - Prob. 6PQCh. 39 - Prob. 7PQCh. 39 - Prob. 8PQCh. 39 - Prob. 9PQCh. 39 - Prob. 10PQCh. 39 - Prob. 11PQCh. 39 - Prob. 12PQCh. 39 - Prob. 13PQCh. 39 - Prob. 14PQCh. 39 - Prob. 15PQCh. 39 - Prob. 16PQCh. 39 - Prob. 17PQCh. 39 - Prob. 18PQCh. 39 - Prob. 19PQCh. 39 - Prob. 20PQCh. 39 - Prob. 21PQCh. 39 - Prob. 22PQCh. 39 - Prob. 23PQCh. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - A starship is 1025 ly from the Earth when measured...Ch. 39 - Prob. 26PQCh. 39 - Prob. 27PQCh. 39 - Prob. 28PQCh. 39 - Prob. 29PQCh. 39 - Prob. 30PQCh. 39 - Prob. 31PQCh. 39 - Prob. 32PQCh. 39 - Prob. 33PQCh. 39 - Prob. 34PQCh. 39 - Prob. 35PQCh. 39 - Prob. 36PQCh. 39 - Prob. 37PQCh. 39 - Prob. 38PQCh. 39 - As measured in a laboratory reference frame, a...Ch. 39 - Prob. 40PQCh. 39 - Prob. 41PQCh. 39 - Prob. 42PQCh. 39 - Prob. 43PQCh. 39 - Prob. 44PQCh. 39 - Prob. 45PQCh. 39 - Prob. 46PQCh. 39 - Prob. 47PQCh. 39 - Prob. 48PQCh. 39 - Prob. 49PQCh. 39 - Prob. 50PQCh. 39 - Prob. 51PQCh. 39 - Prob. 52PQCh. 39 - Prob. 53PQCh. 39 - Prob. 54PQCh. 39 - Prob. 55PQCh. 39 - Prob. 56PQCh. 39 - Consider an electron moving with speed 0.980c. a....Ch. 39 - Prob. 58PQCh. 39 - Prob. 59PQCh. 39 - Prob. 60PQCh. 39 - Prob. 61PQCh. 39 - Prob. 62PQCh. 39 - Prob. 63PQCh. 39 - Prob. 64PQCh. 39 - Prob. 65PQCh. 39 - Prob. 66PQCh. 39 - Prob. 67PQCh. 39 - Prob. 68PQCh. 39 - Prob. 69PQCh. 39 - Prob. 70PQCh. 39 - Joe and Moe are twins. In the laboratory frame at...Ch. 39 - Prob. 72PQCh. 39 - Prob. 73PQCh. 39 - Prob. 74PQCh. 39 - Prob. 75PQCh. 39 - Prob. 76PQCh. 39 - Prob. 77PQCh. 39 - In December 2012, researchers announced the...Ch. 39 - Prob. 79PQCh. 39 - Prob. 80PQCh. 39 - How much work is required to increase the speed of...Ch. 39 - Prob. 82PQCh. 39 - Prob. 83PQCh. 39 - Prob. 84PQCh. 39 - Prob. 85PQ
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: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning