EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100663987
Author: Jewett
Publisher: Cengage Learning US
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Chapter 39, Problem 39.59P

The rest energy of an electron is 0.511 MeV. The rest energy of a proton is 938 MeV. Assume both particles have kinetic energies of 2.00 MeV. Find the speed of (a) the electron and (b) the proton. (c) By what factor does the speed of the electron exceed that of the proton? (d) Repeat the calculations in parts (a) through (c) assuming both particles have kinetic energies of 2 000 MeV.

(a)

Expert Solution
Check Mark
To determine
The speed of the electron.

Answer to Problem 39.59P

The speed of the electron is 0.979c .

Explanation of Solution

Given info: The kinetic energy of the electron and proton is 2.00MeV , the rest energy of the electron and proton are 0.511MeV and 938MeV respectively.

Formula to calculate the total energy of particle is,

E=γmc2 (1)

Formula to calculate the total energy of the particle is,

E=K+mc2 (2)

Here,

K is the kinetic energy.

m is the mass of particle.

c is the speed of the light in the vacuum.

E is the total energy of the particle.

Equate the equation (1) and equation (2).

γmc2=K+mc2γ=Kmc2+1 (3)

Formula to calculate the Lorentz factor is,

γ=11v2c2

Substitute 11v2c2 for γ in equation (3).

11v2c2=Kmc2+1

Rearrange the above equation for v ,

v=c[1(mc2K+mc2)2]1/2 (4)

Formula to calculate the rest energy of particle is,

ER=mc2

Substitute mc2 for ER in equation (4).

v=c[1(ERK+ER)2]1/2 (5)

The speed of the electron is,

velectron=c[1(ERK+ER)2]1/2 (6)

Here,

velectron is the speed of the electron.

Substitute 0.511MeV for ER , and 2.00MeV for K in equation (6) to find the velectron .

velectron=c[1(0.511MeV2.00MeV+0.511MeV)2]1/2=0.979c

Thus, the speed of the electron is 0.979c .

Conclusion:

Therefore, the speed of the electron is 0.979c .

(b)

Expert Solution
Check Mark
To determine
The speed of the proton.

Answer to Problem 39.59P

The speed of the proton is 0.0652c .

Explanation of Solution

Given info: The kinetic energy of the electron and proton is 2.00MeV , the rest energy of the electron and proton are 0.511MeV and 938MeV .

From equation (5), the speed of the particle is given as,

v=c[1(ERK+ER)2]1/2

The speed of the proton is,

vproton=c[1(ERK+ER)2]1/2

Here,

vproton is the speed of the proton.

Substitute 938MeV for ER , and 2.00MeV for K to find the vproton .

v=c[1(938MeV2.00MeV+938MeV)2]1/2=0.0652c

Thus, the speed of the proton is 0.0652c .

Conclusion:

Therefore, the speed of the proton is 0.0652c .

(c)

Expert Solution
Check Mark
To determine
The factor by which speed of electron exceed that of the proton.

Answer to Problem 39.59P

Answer The factor by which speed of electron exceed that of the proton is 15.0 .

Explanation of Solution

Given info: The kinetic energy of the electron and proton is 2.00MeV , the rest energy of the electron and proton are 0.511MeV and 938MeV .

The ratio of the speed of the electron and proton is,

ratio=velectronvproton

Substitute 0.0625c for vproton and 0.979c for velectron .

velectronvproton=0.979c0.0652cvelectronvproton=15.01velectron=15.01vprotonvelectron15vproton

Thus, the factor by which speed of electron exceed that of the proton is 15.0 .

Conclusion:

Therefore, the factor by which speed of electron exceed that of the proton is 15.0 .

(d)

Expert Solution
Check Mark
To determine
The speeds of the electron and proton and the factor by which speed of electron exceed that of the proton.

Answer to Problem 39.59P

The speeds of the electron and proton are 0.99999997c and 0.948c respectively and the factor by which speed of electron exceed that of the proton is 1.06 .

Explanation of Solution

Given info: The kinetic energy of the electron and proton is 2000MeV , the rest energy of the electron and proton are 0.511MeV and 938MeV .

The speed of the electron is,

velectron=c[1(ERK+ER)2]1/2 (7)

Substitute 0.511MeV for ER and 2000MeV for K in equation (7) to find the velectron .

velectron=c[1(0.511MeV2000MeV+0.511MeV)2]1/2=0.99999997c

The speed of the electron is,

vproton=c[1(ERK+ER)2] (8)

Substitute 938MeV for ER and 2000MeV for K in equation (8) to find the vproton .

vproton=c[1(938MeV2000MeV+938MeV)2]1/2=0.948c

The ratio of the speed of the electron and proton is,

ratio=velectronvproton (9)

Substitute 0.948c for vproton and 0.99999997c for velectron in equation (9).

velectronvproton=0.99999997c0.948cvelectronvproton=1.054velectron=1.0548vprotonvelectron1.06vproton

Thus, the speeds of the electron and proton are 0.99999997c and 0.948c respectively and the factor by which speed of electron exceed that of the proton is 1.06 .

Conclusion:

Therefore, the speeds of the electron and proton are 0.99999997c and 0.948c respectively and the factor by which speed of electron exceed that of the proton is 1.06 .

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Chapter 39 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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