The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined. Concept Introduction: de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength. The expression given by the de Broglie is as follows: λ = h m v (1) Here, λ is a wavelength of particle. h is a Plank’s constant. m is a mass of particle. v is a velocity of mass. The expression to calculate the root-mean-square speed of particle is as follows: v rms = 3 k T m (2) Here, k is a Boltzmann constant and its value is 1.38 × 10 − 23 m 2 ⋅ kg ⋅ s − 2 ⋅ K − 1 . T is a temperature. m is a mass of a particle.

Question

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Answer 1

Question

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

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Answer 2

Question

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

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Answer 3

Question

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

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Answer 4

Question

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

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Answer 5

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

Answer 6

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

Answer 7

Chapter 7, Problem 7.103QE

(a)

Interpretation Introduction

Interpretation:

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm has to be determined.

Concept Introduction:

de Broglie established a relation for particles of matter that they also can behave as a wave and has a wavelength. The wavelength of the particle is inversely proportional to its mass. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

λ=hmv (1)

Here,

λ is a wavelength of particle.

h is a Plank’s constant.

m is a mass of particle.

v is a velocity of mass.

The expression to calculate the root-mean-square speed of particle is as follows:

vrms=3kTm (2)

Here,

k is a Boltzmann constant and its value is 1.38×10−23 m2⋅kg⋅s−2⋅K−1.

T is a temperature.

m is a mass of a particle.

Answer 8

(a)

Expert Solution

Answer to Problem 7.103QE

The velocity of electron required for an electron diffraction experiment if the wavelength is 100 pm is 7.27×106 m/s.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Rearrange equation (1) to calculate the value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 9.11×10−31 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(9.11×10−31 kg)(100 pm))(1 pm10−12 m)=7.27×106 m/s

Answer 13

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

Answer 14

(b)

Interpretation Introduction

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 7.103QE

The velocity of neutron with wavelength 100 pm is 3.97×103 m/s. The root-mean-square velocity at 300 K is 2.73×103 m/s.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Rearrange equation (1) to calculate value of velocity as follows:

v=hmλ (3)

Substitute 6.626×10−34 J⋅s for h, 1.67×10−27 kg for m and 100 pm for λ in equation (3).

v=(6.626×10−34 J⋅s(1.67×10−27 kg)(100 pm))(1 pm10−12 m)=3.97×103 m/s

Substitute 1.38×10−23 m2⋅kg⋅s−2⋅K−1 for k, 1.67×10−27 kg for m and 300 K for T in equation (2).

vrms=3(1.38×10−23 m2⋅kg⋅s−2⋅K−1)(300 K)(1.67×10−27 kg)=2.73×103 m/s

The value of the velocity of neutron with wavelength 100 pm is higher than the root mean square velocity at 300 K.

Answer 20

Ch. 7 - Prob. 7.1QE Ch. 7 - Prob. 7.2QE Ch. 7 - Prob. 7.3QE Ch. 7 - Prob. 7.4QE Ch. 7 - Prob. 7.5QE Ch. 7 - Prob. 7.6QE Ch. 7 - Prob. 7.7QE Ch. 7 - Prob. 7.8QE Ch. 7 - Prob. 7.9QE Ch. 7 - Prob. 7.10QE

Concept explainers

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Answer to Problem 7.103QE

Explanation of Solution

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).

Answer to Problem 7.103QE

Explanation of Solution

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Chapter 7 Solutions

Concept explainers

Videos

Answer to Problem 7.103QE

Explanation of Solution

Interpretation: The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared. Concept Introduction: Refer to part (a).

Answer to Problem 7.103QE

Explanation of Solution

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Chapter 7 Solutions

Interpretation:

The velocity of neutron with wavelength 100 pm has to be determined. Also, the velocity of neutron and root-mean-square speed of neutron at 300 K has to be compared.

Concept Introduction:

Refer to part (a).