Chapter U5.115, Problem 4E

Interpretation Introduction

Interpretation:

The type of radiation that has a wavelength of $\text{6}\text{.12}\times {\text{10}}^2\text{ m}$ needs to be determined.

Concept introduction:

The formula which relates frequency of the light and the wavelength of the light is:

$\nu \text{ = }\frac{\text{c}}{\text{λ}}$

Where v is the frequency of the light, c is velocity of light ( ${\text{3×10}}^{\text{8}}\text{ m/sec}$ ), and λ is wavelength of the light.

Answer to Problem 4E

The type of radiation that has a wavelength of $\text{6}\text{.12}\times {\text{10}}^2\text{ m}$ is radio-wave.

Explanation of Solution

The formula which relates frequency of the light and the wavelength of the light is:

$\nu \text{ = }\frac{\text{c}}{\text{λ}}$ - (1)

Where v is the frequency of the light, c is velocity of light ( ${\text{3×10}}^{\text{8}}\text{ m/sec}$ ), and λ is wavelength of the light.

Wavelength, $\lambda =\text{ 6}\text{.12}\times {\text{10}}^2\text{ m}$ (given)

Substituting the values in formula (1):

$\begin{array}{l}\nu \text{ = }\frac{{\text{3×10}}^{\text{8}}\text{ m/sec}}{\text{6}\text{.12}\times {\text{10}}^2\text{ m}}\\ \nu \text{ = 0}{\text{.4902×10}}^6{\text{ sec}}^{-1}\\ \nu \text{ = 4}{\text{.902×10}}^5{\text{ sec}}^{-1}\end{array}$

Since, $1\text{ Hz = }\frac{1}{\sec }$ so,

$\nu \text{ = 4}{\text{.902×10}}^5\text{ Hz}$

Since, the frequency range for radio wave radiation is ${\text{< 3×10}}^{\text{9}}\text{ Hz}$ so, the type of radiation that has a wavelength of $\text{6}\text{.12}\times {\text{10}}^2\text{ m}$ is radio wave.