The magnitude of de Broglie wavelength.

Question

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Answer 1

Question

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

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Answer 2

Question

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

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Answer 3

Question

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

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Answer 4

Question

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

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Answer 5

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

Answer 6

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

Answer 7

Chapter 40, Problem 1P

(a)

To determine

The magnitude of de Broglie wavelength.

Answer 8

(a)

Expert Solution

Answer to Problem 1P

The de Broglie wavelength is 1.26×10−10 m.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The angular wave number from the wave function is 5.00×1010 m−1.

Write the expression to calculate the wavelength.

λ=2πk

Here, λ is the wavelength and k is the angular wave number.

Write the expression to calculate 5.00×1010 m−1 for k in the above equation to calculate λ.

λ=2π5.00×1010 m−1=1.26×10−10 m

Conclusion:

Therefore, the de Broglie wavelength is 1.26×10−10 m.

Answer 13

(b)

To determine

The magnitude of momentum.

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

Answer 14

(b)

To determine

The magnitude of momentum.

Answer 15

(b)

Expert Solution

Answer to Problem 1P

The magnitude of momentum is 5.26×10−24 kg⋅m/s.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Write the expression to calculate the momentum.

p=hλ

Here, p is the momentum and h is the Planck’s constant.

Substitute 6.626×10−34 J⋅s for h and 1.26×10−10 m for λ in the above equation to calculate p.

p=6.626×10−34 J⋅s1.26×10−10 m=5.26×10−24 kg⋅m/s

Conclusion:

Therefore, the magnitude of momentum is 5.26×10−24 kg⋅m/s.

Answer 20

(c)

To determine

The kinetic energy in electron volts.

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

Answer 21

(c)

To determine

The kinetic energy in electron volts.

Answer 22

(c)

Expert Solution

Answer to Problem 1P

The kinetic energy in electron volts is 95.0 eV.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Write the expression to calculate the kinetic energy.

K=p22m

Here, K is the kinetic energy and m is the mass of electron.

Substitute 5.26×10−24 kg⋅m/s for p and 9.11×10−31 kg for m in the above equation to calculate K.

K=(5.26×10−24 kg⋅m/s)22(9.11×10−31 kg)=1.52×10−17 J(1 eV1.602×10−19 J)=95.0 eV

Conclusion:

Therefore, the kinetic energy in electron volts is 95.0 eV.

Answer 27

Ch. 40.1 - Prob. 40.1QQ Ch. 40.2 - Prob. 40.2QQ Ch. 40.2 - Prob. 40.3QQ Ch. 40.5 - Prob. 40.4QQ Ch. 40 - Prob. 1P Ch. 40 - Prob. 2P Ch. 40 - Prob. 3P Ch. 40 - Prob. 4P Ch. 40 - Prob. 5P Ch. 40 - Prob. 6P

The magnitude of de Broglie wavelength.

Concept explainers

The magnitude of de Broglie wavelength.

Answer to Problem 1P

Explanation of Solution

The magnitude of momentum.

Answer to Problem 1P

Explanation of Solution

The kinetic energy in electron volts.

Answer to Problem 1P

Explanation of Solution

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Chapter 40 Solutions