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 - Prob. 19QAPCh. 11 - When a tube containing hydrogen atoms is energized...Ch. 11 - What are the essential points of Bohr‘s theory of...Ch. 11 - According to Bohr, what happens to the electron...Ch. 11 - How does the Bohr theory account for the observed...Ch. 11 - Why was Bohr's theory for the hydrogen atom...Ch. 11 - What major assumption (that was analogous to what...Ch. 11 - Discuss briefly the difference between an orbit...Ch. 11 - Discuss briefly the difference between an orbit...Ch. 11 - Section 11.6 uses a "firefly" analogy to...Ch. 11 - Your text describes the probability map for an s...Ch. 11 - Consider the following representation of a set of...Ch. 11 - What are the differences between the :math>2s...Ch. 11 - What overall shape do the 2p and 3p orbitals have?...Ch. 11 - Prob. 33QAPCh. 11 - When the electron in hydrogen is in the n=3...Ch. 11 - Although a hydrogen atom has only one electron,...Ch. 11 - Complete the following table. trong>Value of n...Ch. 11 - When describing the electrons in an orbital, we...Ch. 11 - Why can only two electrons occupy a particular...Ch. 11 - How does the energy of a principal energy level...Ch. 11 - The number of sublevels in a principal energy...Ch. 11 - According to the Pauli exclusion principle, a...Ch. 11 - Prob. 42QAPCh. 11 - Which of the following orbital designations...Ch. 11 - Which of the following orbital designations is...Ch. 11 - Which orbital is the first be filled in any atom?...Ch. 11 - When a hydrogen atom is in its ground state, in...Ch. 11 - Prob. 47QAPCh. 11 - How are the electron arrangements in a given group...Ch. 11 - Write the full electron configuration ( 1s22s2...Ch. 11 - To which element does each of the following...Ch. 11 - Write the full electron configuration...Ch. 11 - To which element does each of the following...Ch. 11 - Write the complete orbital diagram for each of the...Ch. 11 - Prob. 54QAPCh. 11 - Prob. 55QAPCh. 11 - For each of the following, give an atom and its...Ch. 11 - Why do we believe that the valence electrons of...Ch. 11 - Would you expect the Valence electrons of rubidium...Ch. 11 - Using the symbol of the previous noble gas in...Ch. 11 - Prob. 60QAPCh. 11 - Prob. 61QAPCh. 11 - How many valence electrons does each of the...Ch. 11 - How many 3d electrons are found in each of the...Ch. 11 - Based on the elements’ locations on the periodic...Ch. 11 - For each of the following elements, indicate which...Ch. 11 - Write the valence-electron configuration of each...Ch. 11 - Prob. 67QAPCh. 11 - The “Chemistry in Focus" segment The Chemistry of...Ch. 11 - What are some of the physical properties that...Ch. 11 - Prob. 70QAPCh. 11 - Give some similarities than exist among the...Ch. 11 - Give some similarities that exist among the...Ch. 11 - Which of the following elements most easily gives...Ch. 11 - Which elements in a given period (horizontal row)...Ch. 11 - Where are the most nonmetallic elements located on...Ch. 11 - Why do the metallic elements of a given period...Ch. 11 - Prob. 77QAPCh. 11 - The “Chemistry in Focus" segment Fireworks...Ch. 11 - Prob. 79QAPCh. 11 - In each of the following sets of elements, which...Ch. 11 - Arrange the following sets of elements in order of...Ch. 11 - In each of the following sets of elements,...Ch. 11 - Consider the bright line spectrum of hydrogen...Ch. 11 - Prob. 84APCh. 11 - The portion of the electromagnetic spectrum...Ch. 11 - A beam of light can be thought of as consisting of...Ch. 11 - Prob. 87APCh. 11 - The Energy levels of hydrogen (and other atoms)...Ch. 11 - According to Bohr, the electron in the hydrogen...Ch. 11 - Which of the following statements is false...Ch. 11 - Electrons found in the outemost principal energy...Ch. 11 - An unknown element is a nonmetal and has a...Ch. 11 - Prob. 93APCh. 11 - The current model of the atom in which essentially...Ch. 11 - Prob. 95APCh. 11 - Prob. 96APCh. 11 - Without referring to your textbook or a periodic...Ch. 11 - Prob. 98APCh. 11 - Prob. 99APCh. 11 - Prob. 100APCh. 11 - ight waves move through space at a speed of ters...Ch. 11 - Prob. 102APCh. 11 - ow does the attractive force that the nucleus...Ch. 11 - Prob. 104APCh. 11 - ased on the ground-state electron configuration of...Ch. 11 - lement X, which has a valence shell configuration...Ch. 11 - hy do we believe that the three electrons in the...Ch. 11 - Prob. 108APCh. 11 - Prob. 109APCh. 11 - Prob. 110APCh. 11 - Prob. 111APCh. 11 - ank the following elements in order of increasing...Ch. 11 - Prob. 113APCh. 11 - Prob. 114APCh. 11 - Prob. 115APCh. 11 - Prob. 116APCh. 11 - n each of the following sets of elements, indicate...Ch. 11 - Prob. 118APCh. 11 - Determine the maximum number of electrons that can...Ch. 11 - hich of the following statements is(are) true? The...Ch. 11 - ive the electron configurations for the following...Ch. 11 - Prob. 122CPCh. 11 - Prob. 123CPCh. 11 - Prob. 124CPCh. 11 - Prob. 125CPCh. 11 - Prob. 126CP
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