(a) The kinetic energy of the proton

Question

Want to see more full solutions like this?

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

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

Fresnel lens: You would like to design a 25 mm diameter blazed Fresnel zone plate with a first-order power of +1.5 diopters. What is the lithography requirement (resolution required) for making this lens that is designed for 550 nm? Express your answer in units of μm to one decimal point. Fresnel lens: What would the power of the first diffracted order of this lens be at wavelength of 400 nm? Express your answer in diopters to one decimal point. Eye: A person with myopic eyes has a far point of 15 cm. What power contact lenses does she need to correct her version to a standard far point at infinity? Give your answer in diopter to one decimal point.

Paraxial design of a field flattener. Imagine your optical system has Petzal curvature of the field with radius p. In Module 1 of Course 1, a homework problem asked you to derive the paraxial focus shift along the axis when a slab of glass was inserted in a converging cone of rays. Find or re-derive that result, then use it to calculate the paraxial radius of curvature of a field flattener of refractive index n that will correct the observed Petzval. Assume that the side of the flattener facing the image plane is plano. What is the required radius of the plano-convex field flattener? (p written as rho )

3.37(a) Five free electrons exist in a three-dimensional infinite potential well with all three widths equal to \( a = 12 \, \text{Å} \). Determine the Fermi energy level at \( T = 0 \, \text{K} \). (b) Repeat part (a) for 13 electrons. Book: Semiconductor Physics and Devices 4th ed, NeamanChapter-3Please expert answer only. don't give gpt-generated answers, & please clear the concept of quantum states for determining nx, ny, nz to determine E, as I don't have much idea about that topic.

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

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!

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

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!

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

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!

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

Want to see more full solutions like this?

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