The de Broglie wavelength of the electron.

Question

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Answer 1

Question

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

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Answer 2

Question

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

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Answer 3

Question

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

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Answer 4

Question

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

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Answer 5

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

Answer 6

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

Answer 7

Chapter 27, Problem 32P

(a)

To determine

The de Broglie wavelength of the electron.

Answer 8

(a)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the electron is 5.79×10−10 m .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The equation for de Broglie wavelength of the electron in terms of kinetic energy is,

λe=h2me(KE)

- h is the Plank’s constant
- me is the mass of the electron
- KE is the kinetic energy

Substitute 6.63×10−34 J.s for h , 9.11×10−31 kg for me and 4.50 eV for KE .

λe=6.63×10−34 J.s2(9.11×10−31 kg)(4.50 eV)(1.60×10−19 J1 eV)=5.79×10−10 m

Conclusion:

Thus, the de Broglie wavelength of the electron is 5.79×10−10 m .

Answer 13

(b)

To determine

The de Broglie wavelength of the bowling ball.

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Answer 14

(b)

To determine

The de Broglie wavelength of the bowling ball.

Answer 15

(b)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The equation for de Broglie wavelength of the bowling ball in terms of kinetic energy is,

λb=h2mb(KE)

- mb is the mass of the bowling ball

Substitute 6.63×10−34 J.s for h , 6.00 kg for mb and 4.50 eV for KE .

λb=6.63×10−34 J.s2(6.00 kg)(4.50 eV)(1.60×10−19 J1 eV)=2.26×10−25 m

Conclusion:

Thus, the de Broglie wavelength of the bowling ball is 2.26×10−25 m .

Answer 20

(c)

To determine

The de Broglie wavelength of the photon.

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

Answer 21

(c)

To determine

The de Broglie wavelength of the photon.

Answer 22

(c)

Expert Solution

Answer to Problem 32P

The de Broglie wavelength of the photon is 2.76×10−7 m .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The equation for de Broglie wavelength of the photon is,

λp=hcE

- c is the speed of light
- E is the energy of the photon

Substitute 6.63×10−34 J.s for h , 3.00×108 m/s for c and 4.50 eV for E .

λp=(6.63×10−34 J.s)(3.00×108 m/s)(4.50 eV)(1.60×10−19 J1 eV)=2.76×10−7 m

Conclusion:

Thus, the de Broglie wavelength of the photon is 2.76×10−7 m

Answer 27

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 de Broglie wavelength of the electron.

Concept explainers

The de Broglie wavelength of the electron.

Answer to Problem 32P

Explanation of Solution

The de Broglie wavelength of the bowling ball.

Answer to Problem 32P

Explanation of Solution

The de Broglie wavelength of the photon.

Answer to Problem 32P

Explanation of Solution

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Chapter 27 Solutions