(a) The kinetic energy of the proton

Question

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Answer 1

Question

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

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Answer 2

Question

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

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Answer 3

Question

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

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Answer 4

Question

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

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Answer 5

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

Answer 6

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

Answer 7

Chapter 28, Problem 33QAP

To determine

(a)

The kinetic energy of the proton

Answer 8

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+Kp+Kp−E=2mpc2+2Kp

Kp=E−2mpc22

=2.5×109−2( 9.38× 108 c2)c22 =3.12×108eV =0.312GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

Answer 13

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

Answer 14

To determine

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Answer 15

Expert Solution

Answer to Problem 33QAP

Kinetic energy of the proton and antiproton =0.312GeV

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

Total energy is =2.5GeV=2.5×109eV

Mass of the proton and antiproton are equal that is mp=1.67×10−27Kg

Formula used:

E=mc2

Calculation:

E=mpc2 =(1.67×10−27)(3×108) =1.503×10−10J

Convert Joules into eV

1.503×10−10J=1.503×10−10J×1eV1.6×10−19J=9.38×108eV

E=mpc2+mp−c2+1.25Kp+Kp−E=2mpc2+2.25Kp

Kp=E−2mpc22.25

=2.5×109−2( 9.38× 108 c2)c22.25 =2.77×108eV =0.277GeV

Conclusion:

Kinetic energy of the proton and antiproton =0.312GeV

Answer 20

Ch. 28 - Prob. 1QAP Ch. 28 - Prob. 2QAP Ch. 28 - Prob. 3QAP Ch. 28 - Prob. 4QAP Ch. 28 - Prob. 5QAP Ch. 28 - Prob. 6QAP Ch. 28 - Prob. 7QAP Ch. 28 - Prob. 8QAP Ch. 28 - Prob. 9QAP Ch. 28 - Prob. 10QAP

(a) The kinetic energy of the proton

Concept explainers

(a)

The kinetic energy of the proton

Answer to Problem 33QAP

Explanation of Solution

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Answer to Problem 33QAP

Explanation of Solution

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Chapter 28 Solutions

(a) The kinetic energy of the proton

Concept explainers

(a) The kinetic energy of the proton

Answer to Problem 33QAP

Explanation of Solution

(b) The kinetic energy of the proton with 1.25 more kinetic energy than antiproton

Answer to Problem 33QAP

Explanation of Solution

Want to see more full solutions like this?

Chapter 28 Solutions

(a)

The kinetic energy of the proton

(b)

The kinetic energy of the proton with 1.25 more kinetic energy than antiproton