Physics for Scientists and Engineers: Foundations and Connections
Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN: 9781337026345
Author: Katz
Publisher: Cengage
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Chapter 39, Problem 50PQ

(a)

To determine

The observed mass of electron in the lab.

(a)

Expert Solution
Check Mark

Answer to Problem 50PQ

The observed mass of electron is 1.444×1030kg.

Explanation of Solution

Write relativistic kinetic energy equation of an electron.

    K=mc2mrestc2                                                             (I)

Here, K is the kinetic energy, m is the observed mass of electron, mrest is the rest mass of electron and c is the speed of light.

Write the equation of mass transformation.

    m=γmrest                                                             (II)

Here, mrest is the rest mass of baseball and γ is the Lorentz factor.

Write the equation of Lorentz factor.

    γ=11(vc)2

Here, v is the speed of electron.

Substitute, γ=11(vc)2 in the equation (II).

    m=11(vc)2mrest                                          (III)

Substitute γmrest for m in equation (I).

    K=γmrestc2mrestc2=(γ1)mrestc2                              (IV)

Substitute, γ=11(vc)2 in the above equation.

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

Write the equation for relativistic momentum of equation.

    p=11(vc)2mrestv                                     (VI)

Here, p is the momentum of electron.

Convert the unit of K from keV to J.

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

Substitute 4.806×1014J for K, 9.11×1031kg for mrest and 3×108m/s for c in equation (V).

    4.806×1014J=(11(vc)21)(9.11×1031kg)(3×108m/s)24.806×1014J(9.11×1031kg)×(9×1016m2/s2)=(11(vc)21)

Rearrange the above equation.

    1(vc)2=11+(4.806×1014J(9.11×1031kg)×(9×1016m2/s2))1(vc)2=(11+0.586J1/kg1/m2/s2)(vc)2=1(0.63)2v2=0.602c2

Take square roots on both sides.

    v=0.776c

Conclusion:

Substitute 0.776c for v, 9.11×1031kg for mrest and 3×108m/s for c in equation (III).

    m=11(0.776cc)2(9.11×1031kg)=11(0.776×3×108m/s3×108m/s)2(9.11×1031kg)=1.58545(9.11×1031kg)=1.444×1030kg

Therefore, the observed mass of electron is 1.444×1030kg.

(b)

To determine

The momentum of electron.

(b)

Expert Solution
Check Mark

Answer to Problem 50PQ

The momentum of electron is 3.362×1022kgm/s.

Explanation of Solution

Substitute 0.776c for v, 9.11×1031kg for mrest and 3×108m/s for c in equation (VI).

    p=11(0.776cc)2(9.11×1031kg)(0.776c)=11((0.776)(3×108m/s)3×108m/s)2(9.11×1031kg)(0.776×3×108m/s)=1.58545(9.11×1031kg)(0.776×3×108m/s)=3.362×1022kgm/s

Conclusion:

Therefore, the momentum of electron is 3.362×1022kgm/s.

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Chapter 39 Solutions

Physics for Scientists and Engineers: Foundations and Connections

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