Sample Solution from
Inorganic Chemistry
5th Edition
ISBN: 9780321811059
Chapter 2
Problem 2.1P
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Textbook Problem

Determine the de Brogue wavelength of
a. an electron moving at 1/10 the speed of light.
b. a 400 g Frisbee moving at 10 km/h.
c. an 8.0-pound bowling ball rolling down the lane with a velocity of 2.0 meters per second.
d. a 13.7 g hummingbird flying at a speed of 30.0 miles per hour.

Expert Solution

a)

Interpretation Introduction

Interpretation:The de Broglie wavelength of electron that has 1/10th speed of light should be determined.

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

The expression given by the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of particle.

Answer

The de Broglie wavelength of electron that has 1/10th speed of light is 0.0242 nm .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of mass.

Value of h is 6.625×10−34 J⋅sec .

Value of m is 9.11×10−31 kg .

Value of v is 3.0×107 m/s .

Substitute the value in above equation.

  λ=hmv=6.625× 10 −34 J⋅sec( 9.11× 10 −31  kg)( 3.0× 10 7  m/s)=(2.42× 10 −11 m)( 1 nm 10 −9  m)=0.0242 nm

Expert Solution

b)

Interpretation Introduction

Interpretation:The de Broglie wavelength of 400 g Frisbee thathas 10 km/h speed should 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 particle is inversely related to mass of particle. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of particle.

Answer

The de Broglie wavelength of 400 g Frisbee that has 10 km/h speed is 5.96×10−34 m .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of mass.

Value of h is 6.625×10−34 J⋅sec .

Value of m is 400 g .

Value of v is 10 km/h .

Substitute the values in above equation.

  λ=hmv=( 6.625× 10 −34  J⋅sec ( 400 g )( 10 km/h ))( 10 3  g 1 kg)( 1 km 10 3  m)( 3600 sec 1 h)=5.96×10−34 m

Expert Solution

c)

Interpretation Introduction

Interpretation:The de Broglie wavelength of 8.0 lb bowling ballthat has 2.0 m/sec speed should 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 particle is inversely related to mass of particle. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of particle.

Answer

The de Broglie wavelength of 8.0 lb bowling ball that has 2.0 m/sec speed is 9.13×10−35 m .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of mass.

Value of h is 6.625×10−34 J⋅sec .

Value of m is 8.0 lb .

Value of v is 2.0 m/sec .

Substitute the value in above equation.

  λ=hmv=( 6.625× 10 −34  J⋅sec ( 8.0 lb )( 2.0 m/sec ))( 1 lb 453.592 g)( 10 3  g 1 kg)=9.13×10−35 m

Expert Solution

d)

Interpretation Introduction

Interpretation:The de Broglie wavelength of 13.7 g birdthat has 30.0 mi/hr speed should 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 particle is inversely related to mass of particle. The smaller the mass, the larger its wavelength.

The expression given by the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of particle.

Answer

The de Broglie wavelength of 13.7 g bird that has 30.0 mi/hr speed is 3.6×10−36 m .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  λ=hmv

Where,

  • λ is a wavelength of particle.
  • h is a Plank’s constant.
  • m is a mass of particle.
  • v is a velocity of mass.

Value of h is 6.625×10−34 J⋅sec .

Value of m is 13.7 g .

Value of v is 30.0 mi/hr .

Substitute the value in above equation.

  λ=hmv=( 6.625× 10 −34  J⋅sec ( 13.7 g )( 30.0 mi/hr ))( 1 mi 1609.34 m)( 3600 sec 1 h)=3.6×10−36 m

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