Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 11, Problem 100AP
Interpretation Introduction

(a)

Interpretation:

The de-Broglie wavelength is to be calculated.

Concept Introduction:

In 1924, de-Broglie suggested the dual nature of microscopic particles and compared the energy value of both particle and wave nature particles. According to him, the wavelength of an electron (λ) moving with a speed of v is:

λ=hmv

Where,

  • h is Plank’s constant.
  • m is the mass of electron.

Expert Solution
Check Mark

Answer to Problem 100AP

The de-Broglie wavelength is calculated as 2.7×1012 m1.

Explanation of Solution

Mass of electron =9.10×1031kg

Speed of light =3×108ms1

Given: The speed of electron is 0.9 times the speed of light.

So the speed of electron, v=0.9×3×108ms-1

The formula for the calculation of de-Broglie wavelength is:

λ=hmv

The value of Plank’s constant, h=6.63×1034Js

Applying the formula for calculating wavelength:

λ=hmvλ=6.63×1034Js(9.10×1031kg)(0.9×3×108ms1)λ=2.7×1012 m1.

Interpretation Introduction

(b)

Interpretation:

The de-Broglie wavelength is to be calculated and the reason as to why macroscopic particles cannot be treated as wave particle is to be stated.

Concept Introduction:

In 1924, de-Broglie suggested the dual nature of microscopic particles and compared the energy value of both particle and wave nature particles. According to him, the wavelength of an electron (λ) moving with a speed of v is:

λ=hmv

Where,

  • h is Plank’s constant.
  • m is the mass of electron.

Expert Solution
Check Mark

Answer to Problem 100AP

The de-Broglie wavelength is calculated as 4.42×1034m1 and such magnitude is practically impossible to achieve.

Explanation of Solution

Given:

Mass of ball =150g

Conversion of grams to kilograms can be done as:

1000g=1kg

So,

150g=1501000kg=0.15kg

Speed of ball =10ms1

The formula for the calculation of de-Broglie wavelength is:

λ=hmv

The value of Plank’s constant, h=6.63×1034Js

Applying the formula for calculating wavelength:

λ=hmvλ=6.63×1034Js(0.150kg)(10ms1)λ=4.42×1034m1

For a ball of mass 0.150kg, it is practically impossible to have a wavelength of such magnitude and therefore, the wave theory of particles cannot be applied to the macroscopic particles.

Interpretation Introduction

(b)

Interpretation:

The de-Broglie wavelength is to be calculated and the reason as to why macroscopic particles cannot be treated as wave particle is to be stated.

Concept Introduction:

In 1924, de-Broglie suggested the dual nature of microscopic particles and compared the energy value of both particle and wave nature particles. According to him, the wavelength of an electron (λ) moving with a speed of v is:

λ=hmv

Where,

  • h is Plank’s constant.
  • m is the mass of electron.

Expert Solution
Check Mark

Answer to Problem 100AP

The de-Broglie wavelength is calculated as 4.42×1034m1 and such magnitude is practically impossible to achieve.

Explanation of Solution

Given:

Mass of ball =150g

Conversion of grams to kilograms can be done as:

1000g=1kg

So,

150g=1501000kg=0.15kg

Speed of ball =10ms1

The formula for the calculation of de-Broglie wavelength is:

λ=hmv

The value of Plank’s constant, h=6.63×1034Js

Applying the formula for calculating wavelength:

λ=hmvλ=6.63×1034Js(0.150kg)(10ms1)λ=4.42×1034m1

For a ball of mass 0.150kg, it is practically impossible to have a wavelength of such magnitude and therefore, the wave theory of particles cannot be applied to the macroscopic particles.

Interpretation Introduction

(c)

Interpretation:

The de-Broglie wavelength is to be calculated and the reason as to why macroscopic particles cannot be treated as wave particle is to be stated.

Concept Introduction:

In 1924, de-Broglie suggested the dual nature of microscopic particles and compared the energy value of both particle and wave nature particles. According to him, the wavelength of an electron (λ) moving with a speed of v is:

λ=hmv

Where,

  • h is Plank’s constant.
  • m is the mass of electron.

Expert Solution
Check Mark

Answer to Problem 100AP

The de-Broglie wavelength is calculated to be as 1.59×1035m1 and such magnitude is practically impossible to achieve.

Explanation of Solution

Given:

Mass of person =75kg

Speed of person =2kmh1

Conversion of speed in kilometer per hour to speed in meter per second can be done as:

1km=1000m1hour=3600s

So,

2kmh1=2×1000m3600s=1018ms1

The formula for the calculation of de-Broglie wavelength is:

λ=hmv

The value of Plank’s constant, h=6.63×1034Js

Applying the formula for calculating wavelength:

λ=hmvλ=6.63×1034Js(75kg)(1018ms1)λ=1.59×1035m1

For a person of mass 75kg, it is practically impossible to have a wavelength of such magnitude and therefore, the wave theory of particles cannot be applied to the macroscopic particles.

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Chapter 11 Solutions

Introductory Chemistry: A Foundation

Ch. 11 - Prob. 6ALQCh. 11 - rue or false? The hydrogen atom has a 3 orbital....Ch. 11 - Prob. 8ALQCh. 11 - ake sense of the fact that metals tend to lose...Ch. 11 - Show how using the periodic table helps you find...Ch. 11 - r Questions 11—13, you will need to consider...Ch. 11 - Prob. 12ALQCh. 11 - Prob. 13ALQCh. 11 - Prob. 14ALQCh. 11 - Prob. 15ALQCh. 11 - What evidence do we have that energy levels in an...Ch. 11 - Explain the hydrogen emission spectrum. Why is it...Ch. 11 - There am an infinite number of allowed transitions...Ch. 11 - You have learned that each orbital is allowed two...Ch. 11 - Atom A has valence electrons that are lower in...Ch. 11 - Prob. 21ALQCh. 11 - Prob. 1QAPCh. 11 - hat questions were left unanswered by Rutherford’s...Ch. 11 - Prob. 3QAPCh. 11 - Prob. 4QAPCh. 11 - Prob. 5QAPCh. 11 - Prob. 6QAPCh. 11 - he “Chemistry in Focus" segment Light as a Sex...Ch. 11 - Prob. 8QAPCh. 11 - hen lithium salts are heated in a flame, they emit...Ch. 11 - The energy of a photon of visible light emitted by...Ch. 11 - Prob. 11QAPCh. 11 - An excited atom can release some or all of its...Ch. 11 - How is the energy carried per photon of light...Ch. 11 - When an atom energy from outside, the atom goes...Ch. 11 - Describe briefly why the study of electromagnetic...Ch. 11 - What does it mean to say that the hydrogen atom...Ch. 11 - Because a given element’s atoms emit only certain...Ch. 11 - How does the energy possessed by an emitted photon...Ch. 11 - 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