The momentum of the electron.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

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 2

Question

Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

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 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

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 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

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 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

Answer 6

Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

Answer 7

Chapter 27, Problem 27P

(a)

To determine

The momentum of the electron.

Answer 8

(a)

Expert Solution

Answer to Problem 27P

The momentum of the electron is 1.18×10−23 kg.m/s .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

College Physics, Chapter 27, Problem 27P

The figure shows the situation before and after the scattering process.

The Compton shift formula is,

λ'−λ=hmec(1−cos180°)=2hmec

- λ0 is the wavelength of the x-ray
- h is the Plank’s constant
- c is the speed of the light
- me is the mass of the electron

Substitute 0.110 nm for λ , 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 3.00×108 m/s for c .

λ'=(0.110 nm)(10−9 m1 nm)+2(6.63×10−34 J.s)(9.11×10−31 kg)(3.00×108 m/s)=0.115 nm

The momentum of the recoiling electron is,

pe=p+p'

- p is the momentum of the incident photon
- p' is the momentum of the scattered photon

pe=hλ+hλ'

Substitute 0.110 nm for λ , 0.115 nm for λ' , 6.63×10−34 J.s for h and 3.00×108 m/s for c .

pe=(6.63×10−34 J.s)(1(0.110 nm)(10−9 m1 nm)+1(0.115 nm)(10−9 m1 nm))=1.18×10−23 kg.m/s

Conclusion:

Thus, the momentum of the electron is 1.18×10−23 kg.m/s .

Answer 13

(b)

To determine

The kinetic energy of the electron.

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

Answer 14

(b)

To determine

The kinetic energy of the electron.

Answer 15

(b)

Expert Solution

Answer to Problem 27P

The kinetic energy of the electron is 478 eV .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The kinetic energy of the electron is,

KE=pe22me

Substitute 1.18×10−23 kg.m/s for pe and 9.11×10−31 kg for me .

KE=(1.18×10−23 kg.m/s)22(9.11×10−31 kg)(1 eV1.60×10−19 J)=478 eV

Conclusion:

Thus, the kinetic energy of the electron is 478 eV

Answer 20

Ch. 27.5 - Prob. 27.1QQ Ch. 27.5 - Prob. 27.2QQ Ch. 27.5 - Prob. 27.3QQ Ch. 27.6 - Prob. 27.4QQ Ch. 27.6 - Prob. 27.5QQ Ch. 27 - Prob. 1CQ Ch. 27 - Prob. 2CQ Ch. 27 - Prob. 3CQ Ch. 27 - Prob. 4CQ Ch. 27 - Prob. 5CQ

The momentum of the electron.

Concept explainers

The momentum of the electron.

Answer to Problem 27P

Explanation of Solution

The kinetic energy of the electron.

Answer to Problem 27P

Explanation of Solution

Want to see more full solutions like this?

Chapter 27 Solutions