Chapter 5, Problem 72A

Interpretation Introduction

Interpretation : The energy of one photon of infrared radiation is to be calculated.

Concept Introduction : Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v). The formula of energy is given as:

  $\begin{array}{l}E=h\nu \\ E=h\frac{c}{\lambda }\end{array}$

Where $h$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

Answer to Problem 72A

The energy of one photon of infrared radiation is $1.7\times {10}^{-21}\text{J}$ .

Explanation of Solution

Given information:

The wavelength is $1.2\times {10}^{-4}\text{m}$ .

It is known that the value of Planck’s constant is $\text{h}\text{=}6.626\times {10}^{-34}\text{J}\text{s}$ and the velocity of light is $c=\text{2}\text{.998}\times {\text{10}}^8\text{m/s}$ .

Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v).

The formula of energy is given as:

  $E=h\frac{c}{\lambda }$

Where $\text{h}$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

To calculate the energy of a photon, substitute the values in the above formula.

  $\begin{array}{c}E=h\frac{c}{\lambda }\\ =\left(6.626\times {10}^{-34}\text{J}\cancel{\text{s}}\right)\times \frac{\left(2.998\times {10}^8\cancel{\text{m}}\text{/}\cancel{\text{s}}\right)}{1.2\times {10}^{-4}\cancel{\text{m}}}\\ E=1.7\times {10}^{-21}\text{J}\end{array}$

The energy of one photon of infrared radiation is $1.7\times {10}^{-21}\text{J}$ .

Interpretation Introduction

Interpretation : The energy of one photon of ultraviolet radiation is to be calculated.

Concept Introduction : Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v). The formula of energy is given as

  $\begin{array}{l}E=h\nu \\ E=h\frac{c}{\lambda }\end{array}$

Where $h$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

Answer to Problem 72A

The energy of one photon of ultraviolet radiation is $1.7\times {10}^{-21}\text{J}$ .

Explanation of Solution

Given information:

The wavelength is $5.1\times {10}^{-7}\text{m}$ .

It is known that value of Planck’s constant is $h\text{=}6.626\times {10}^{-34}\text{J}\text{s}$ and velocity of light is $c=\text{2}\text{.998}\times {\text{10}}^8\text{m/s}$ .

Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v).

The formula of energy is given as

  $E=h\frac{c}{\lambda }$

Where $\text{h}$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

To calculate the energy of a photon, substitute the values in the above formula.

  $\begin{array}{c}E=h\frac{c}{\lambda }\\ =\left(6.626\times {10}^{-34}\text{J}\cancel{\text{s}}\right)\times \frac{\left(2.998\times {10}^8\cancel{\text{m}}\text{/}\cancel{\text{s}}\right)}{5.1\times {10}^{-7}\cancel{\text{m}}}\\ E=3.9\times {10}^{-19}\text{J}\end{array}$

The energy of one photon of visible radiation is $3.9\times {10}^{-19}\text{J}$ .

Interpretation Introduction

Interpretation : The energy of one photon of infrared radiation is to be calculated.

Concept Introduction : Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v). The formula of energy is given as:

  $\begin{array}{l}E=h\nu \\ E=h\frac{c}{\lambda }\end{array}$

Where $h$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

Answer to Problem 72A

The energy of one photon of ultraviolet radiation is $1.7\times {10}^{-17}\text{J}$ .

As the wavelength value decreases, the energy of the photon increases.

Explanation of Solution

Given information:

The wavelength is $1.4\times {10}^{-8}\text{m}$ .

It is known that the value of Planck’s constant is $h\text{=}6.626\times {10}^{-34}\text{J}\text{s}$ and the velocity of light is $c=\text{2}\text{.998}\times {\text{10}}^8\text{m/s}$ .

Planck demonstrated mathematically that a body's ability to absorb or emit radiation is inversely related to the frequency of the radiation (v).

The formula of energy is given as:

  $E=h\frac{c}{\lambda }$

Where $\text{h}$ is Planck’s constant, $c$ is the velocity of light, $\lambda$ is the wavelength, and $\nu$ is the frequency of the photon.

To calculate the energy of a photon, substitute the values in the above formula.

  $\begin{array}{c}E=h\frac{c}{\lambda }\\ =\left(6.626\times {10}^{-34}\text{J}\cancel{\text{s}}\right)\times \frac{\left(2.998\times {10}^8\cancel{\text{m}}\text{/}\cancel{\text{s}}\right)}{1.4\times {10}^{-8}\cancel{\text{m}}}\\ E=1.7\times {10}^{-17}\text{J}\end{array}$

The energy of one photon of ultraviolet radiation is $1.7\times {10}^{-17}\text{J}$ .

From the results of the energy of different radiation, the energy of one photon is inversely proportional to the wavelength. As the wavelength value decreases, the energy of the photon increases.