Chapter 5, Problem 88A

Interpretation Introduction

Interpretation : The energy of the photons absorbed by the electrons in glow-in-the-dark objects 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:

  $E=h\nu$

Where $\text{h}$ is Planck’s constant and $\nu$ is the frequency of the photon.

Answer to Problem 88A

The energy of the photons absorbed by the electrons in glow-in-the-dark objects will be in the range from $3\times {10}^{-19}\text{J}$ to $7\times {10}^{-19}\text{J}$ .

Explanation of Solution

Given information:

The wavelengths are $300\text{nm}$ and $700\text{nm}$ .

The formula of energy is given as

  $E=h\nu$

Where $\text{h}$ is Planck’s constant and $\nu$ is the frequency of the photon.

The frequency is how often waves cycle through a specific point in a unit of time.

The formula of frequency $\left(\nu \right)$ is given as:

  $\nu =\frac{c}{\lambda }$

Where $c$ is the velocity of light and $\lambda$ is the wavelength of the photon.

In order to calculate the energy first calculate the frequencies for the wavelength.

To calculate the frequency at wavelength $300\text{nm}$, substitute the values in the frequency formula:

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{300\cancel{\text{nm}}}\times \frac{{10}^9\cancel{\text{nm}}}{1\cancel{\text{m}}}\\ =4\times {10}^{14}{\text{s}}^{-1}\end{array}$

To calculate the frequency at wavelength $700\text{nm}$, substitute the values in the frequency formula:

  $\begin{array}{c}\nu =\frac{c}{\lambda }\\ =\frac{2.998\times {10}^8\cancel{\text{m}}\text{/s}}{700\cancel{\text{nm}}}\times \frac{{10}^9\cancel{\text{nm}}}{1\cancel{\text{m}}}\\ =1\times {10}^{15}{\text{s}}^{-1}\end{array}$

To calculate the energy at wavelength $300\text{nm}$, substitute the values in the energy formula:

  $\begin{array}{c}E=h\nu \\ =6.626\times {10}^{-34}\text{J }\cancel{\text{s}}\times 4\times {10}^{14}\cancel{{\text{s}}^{-1}}\\ =3\times {10}^{-19}\text{J}\end{array}$

To calculate the energy at wavelength $700\text{nm}$, substitute the values in the energy formula:

  $\begin{array}{c}E=h\nu \\ =6.626\times {10}^{-34}\text{J }\cancel{\text{s}}\times 1\times {10}^{15}\cancel{{\text{s}}^{-1}}\\ =7\times {10}^{-19}\text{J}\end{array}$

The range of the energy will be from $3\times {10}^{-19}\text{J}$ to $7\times {10}^{-19}\text{J}$ .