Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 39, Problem 39.66AP

(a)

To determine

The speed of a proton when its kinetic energy is equal to the kinetic energy of the electron.

(a)

Expert Solution
Check Mark

Answer to Problem 39.66AP

The speed of a proton is 0.0236c .

Explanation of Solution

Given info: The speed of an electron is 0.750c .

Write the equation of kinetic energy of proton.

Kp=(γp1)mpc2

Here,

γp is a constant.

mp is the mass of proton.

c is the speed of light.

The value of γp is 11(up/c)2 , where up is the speed of a proton.

Substitute 11(up/c)2 for γp in above equation to find Kp .

Kp=(11(up/c)21)mpc2

Write the equation of kinetic energy of electron.

Ke=(γe1)mec2

Here,

γe is a constant.

me is the mass of an electron.

The value of γe is,

γe=11(ue/c)2

Here,

ue is the speed of an electron.

Substitute 0.750c for ue in above equation to find γe .

γe=11(0.750c/c)2γe=1.5119

Thus, the value of γe is 1.5119 .

Substitute 1.5119 for γe in above equation to find Ke .

Ke=(1.51191)mec2Ke=0.5119mec2

The kinetic energy of proton is equal to the kinetic energy of electron.

Kp=Ke

Substitute (11(up/c)21)mpc2 for Kp and 0.5119mec2 for Ke in above equation.

(11(up/c)21)mpc2=0.5119mec211(up/c)2=0.5119memp+1up=c11(0.5119memp+1)2

The mass of proton is 1.67×1027kg and the mass of an electron is 9.11×1031kg .

Substitute 1.67×1027kg for mp and 9.11×1031kg for me in above equation to find up .

up=c11(0.5119(9.11×1031kg)(1.67×1027kg)+1)2=0.0236c

Conclusion:

Therefore, the speed of a proton is 0.0236c .

(b)

To determine

The speed of a proton when its momentum is equal to the momentum of an electron.

(b)

Expert Solution
Check Mark

Answer to Problem 39.66AP

The speed of a proton is 6.18×104c .

Explanation of Solution

Given info: The speed of an electron is 0.750c .

Write the equation of momentum of proton.

pp=γpmpup

Substitute 11(up/c)2 for γp in above equation to find pp .

pp=11(up/c)2mpup

Write the equation of momentum of electron.

pe=γemeue

Substitute 1.5119 for γe , 9.11×1031kg for me and 0.750c for ue in above equation to find pe .

pe=(1.5119)(9.11×1031kg)(0.750c)pe=(1.033×1030kg)c

The momentum of proton and electron are same.

pp=pe

Substitute 11(up/c)2mpup for pp and (1.033×1030kg)c for pe in above equation.

11(up/c)2mpup=(1.033×1030kg)c

Substitute 1.67×1027kg for mp in above equation to find up .

11(up/c)2(1.67×1027kg)up=(1.033×1030kg)cupc=(6.18×104)1(upc)2up=c3.819×1071+(3.819×107)up=6.18×104c

Conclusion:

Therefore, the speed of a proton is 6.18×104c .

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Chapter 39 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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