Chapter 5.3, Problem 15SP

Interpretation Introduction

Interpretation: The wavelength of radiation of a given frequency needs to be determined. Whether this radiation has a longer or shorter wavelength than red light needs to be explained.

Concept Introduction: The relation between wavelength and frequency of radiation is as follows:

  $v=\frac{c}{\lambda }$

Here, $v$ is the frequency of radiation, $\lambda$ is the wavelength of radiation, and c is the speed of light. The value of the speed of light is constant that is $3\times {10}^8\text{ m/s}$ .

Answer to Problem 15SP

20000 nm, a longer wavelength than red light.

Explanation of Solution

From the given frequency of radiation, the wavelength can be calculated as follows:

  $v=\frac{c}{\lambda }$

Rearranging the above formula,

  $\lambda =\frac{c}{v}$

Here, the value of the speed of light is constant that is $3\times {10}^8\text{ m/s}$ and the frequency of radiation is given $1.50\times {10}^{13}\text{ Hz}$ .

Substitute the values in the above formula,

  $\lambda =\frac{3\times {10}^8\text{ m/s}}{1.50\times {10}^{13}\text{ Hz}}$

Here,

  $1\text{ Hz}=1{\text{ s}}^{-1}$

Thus,

  $\begin{array}{c}\lambda =\frac{3\times {10}^8\text{ m/s}}{1.50\times {10}^{13}\text{ Hz}\left(\frac{1{\text{ s}}^{-1}}{1\text{ Hz}}\right)}\\ =2\times {10}^{-5}\text{ m}\end{array}$

This can be converted into nm using the below conversion:

  $1\text{ nm}={10}^{-9}\text{ m}$

Thus,

  $\begin{array}{l}\lambda =2\times {10}^{-5}\text{ m}\left(\frac{1\text{ nm}}{{10}^{-9}\text{ m}}\right)\\ =2\times {10}^4\text{ nm}\\ =20000\text{ nm}\end{array}$

Now, the wavelength of red-light ranges from 620-720 nm thus, the given radiation has a longer wavelength than red light.

Conclusion

Thus, the wavelength of radiation of a given frequency is 20000 nm. Also, this radiation has a longer wavelength than red light.