Chapter 2, Problem 2.1P

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.

Interpretation Introduction

Interpretation:The de Broglie wavelength of electron that has $\text{1/10th}$ 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:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ is a wavelength of particle.
• $h$ is a Plank’s constant.
• $m$ is a mass of particle.
• $v$ is a velocity of particle.

Answer to Problem 2.1P

The de Broglie wavelength of electron that has $\text{1/10th}$ speed of light is $0.0242\text{ nm}$ .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ 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 $6.625\times {10}^{-34}\text{ J}\cdot \text{sec}$ .

Value of $m$ is $9.11\times {10}^{-31}\text{ kg}$ .

Value of $v$ is $3.0\times {10}^7\text{ m/s}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{λ}=\frac{h}{mv}\\ =\frac{6.625\times {10}^{-34}\text{ J}\cdot \text{sec}}{\left(9.11\times {10}^{-31}\text{ kg}\right)\left(3.0\times {10}^7\text{ m/s}\right)}\\ =\left(2.42\times {10}^{-11}\text{ m}\right)\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =0.0242\text{ nm}\end{array}$

Interpretation Introduction

Interpretation:The de Broglie wavelength of $400\text{ g}$ Frisbee thathas $\text{10 km/h}$ 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:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ is a wavelength of particle.
• $h$ is a Plank’s constant.
• $m$ is a mass of particle.
• $v$ is a velocity of particle.

Answer to Problem 2.1P

The de Broglie wavelength of $400\text{ g}$ Frisbee that has $\text{10 km/h}$ speed is $5.96\times {10}^{-34}\text{ m}$ .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ 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 $6.625\times {10}^{-34}\text{ J}\cdot \text{sec}$ .

Value of $m$ is $400\text{ g}$ .

Value of $v$ is $\text{10 km/h}$ .

Substitute the values in above equation.

  $\begin{array}{c}\text{λ}=\frac{h}{mv}\\ =\left(\frac{6.625\times {10}^{-34}\text{ J}\cdot \text{sec}}{\left(400\text{ g}\right)\left(\text{10 km/h}\right)}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)\left(\frac{1\text{ km}}{{10}^3\text{ m}}\right)\left(\frac{3600\text{ sec}}{1\text{ h}}\right)\\ =5.96\times {10}^{-34}\text{ m}\end{array}$

Interpretation Introduction

Interpretation:The de Broglie wavelength of $8.0\text{ lb}$ bowling ballthat has $\text{2}\text{.0 m/sec}$ 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:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ is a wavelength of particle.
• $h$ is a Plank’s constant.
• $m$ is a mass of particle.
• $v$ is a velocity of particle.

Answer to Problem 2.1P

The de Broglie wavelength of $8.0\text{ lb}$ bowling ball that has $\text{2}\text{.0 m/sec}$ speed is $9.13\times {10}^{-35}\text{ m}$ .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ 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 $6.625\times {10}^{-34}\text{ J}\cdot \text{sec}$ .

Value of $m$ is $8.0\text{ lb}$ .

Value of $v$ is $\text{2}\text{.0 m/sec}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{λ}=\frac{h}{mv}\\ =\left(\frac{6.625\times {10}^{-34}\text{ J}\cdot \text{sec}}{\left(8.0\text{ lb}\right)\left(\text{2}\text{.0 m/sec}\right)}\right)\left(\frac{1\text{ lb}}{\text{453}\text{.592 g}}\right)\left(\frac{{10}^3\text{ g}}{1\text{ kg}}\right)\\ =9.13\times {10}^{-35}\text{ m}\end{array}$

Interpretation Introduction

Interpretation:The de Broglie wavelength of $13.7\text{ g}$ birdthat has $\text{30}\text{.0 mi/hr}$ 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:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ is a wavelength of particle.
• $h$ is a Plank’s constant.
• $m$ is a mass of particle.
• $v$ is a velocity of particle.

Answer to Problem 2.1P

The de Broglie wavelength of $13.7\text{ g}$ bird that has $\text{30}\text{.0 mi/hr}$ speed is $3.6\times {10}^{-36}\text{ m}$ .

Explanation of Solution

The expression to calculate the de Broglie is as follows:

  $\text{λ}=\frac{h}{mv}$

Where,

• $\text{λ}$ 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 $6.625\times {10}^{-34}\text{ J}\cdot \text{sec}$ .

Value of $m$ is $13.7\text{ g}$ .

Value of $v$ is $\text{30}\text{.0 mi/hr}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{λ}=\frac{h}{mv}\\ =\left(\frac{6.625\times {10}^{-34}\text{ J}\cdot \text{sec}}{\left(13.7\text{ g}\right)\left(\text{30}\text{.0 mi/hr}\right)}\right)\left(\frac{1\text{ mi}}{\text{1609}\text{.34 m}}\right)\left(\frac{3600\text{ sec}}{1\text{ h}}\right)\\ =3.6\times {10}^{-36}\text{ m}\end{array}$