Chapter 7, Problem 36E

Interpretation Introduction

Interpretation:

The diagrams of waves are to be drawn, the wavelength that X-rays and radio waves have on this scale is to be calculated and the range of wavelength visible to our eyes are to be determined.

Concept introduction:

The distance between two consecutive crests of a wave of light is called the wavelength of light.

The arrangement of lights according to the wavelength associated with it, is called a spectrum.

Electromagnetic radiation with a certain wavelength is called light.

To convert the m to nm, the conversion factor is as: $\frac{1\text{nm}}{1{\text{0}}^{-\text{9}}\text{m}}$

To convert nm to mm, the conversion factor is as: $\frac{0.1\text{mm}}{1\text{nm}}$

Answer to Problem 36E

Solution: The diagram of a wave of the violet edge $\lambda =40\text{mm}$ and the diagram of the wave of the red edge $\lambda =78\text{mm}$ are as:

The wavelength that X-rays and radio waves have on this scale are $\lambda =1\times {10}^{-2}\text{mm}$ and $\lambda =1\times {10}^{12}\text{mm}$, respectively. The range of wavelength visible to our eyes is covering X-rays to radio waves.

Explanation of Solution

Given information:

The violet edge of spectrum appears at a wavelength of $400\text{nm}$. The red edge of spectrum appears at a wavelength of $780\text{nm}$ and the given scale is $1\text{}\text{nm}\text{=}\text{0}\text{.1}\text{mm}$.

The wavelength of X-rays, radio waves and visible range in mm is calculated as follows:

For X-rays, $\lambda =1\times {10}^{-10}\text{m}$, convert the wavelength from meter to nm, and then in mm.

$\begin{array}{c}\lambda =1\times {10}^{-10}\cancel{\text{m}}\times \frac{1\cancel{\text{nm}}}{{10}^{-9}\cancel{\text{m}}}\times \frac{0.1\text{mm}}{1\cancel{\text{nm}}}\\ =1\times {10}^{-2}\text{mm}\end{array}$

For radio waves, $\lambda =1\times {10}^4\text{m}$, convert the wavelength from meter to nm and then in mm,

$\begin{array}{c}\lambda =1\times {10}^4\cancel{\text{m}}\times \frac{1\cancel{\text{nm}}}{{10}^{-9}\cancel{\text{m}}}\times \frac{0.1\text{mm}}{1\cancel{\text{nm}}}\\ =1\times {10}^{12}\text{mm}\end{array}$

For violet and red edged visible range, $\lambda =400\text{nm}$ and $\lambda =780\text{nm}$, respectively, convert the wavelength from nm to mm,

For Violet edge: $\begin{array}{c}\lambda =400\cancel{\text{nm}}\times \frac{0.1\text{mm}}{1\cancel{\text{nm}}}\\ =40\text{mm}\end{array}$

For Red edge: $\begin{array}{c}\lambda =780\cancel{\text{nm}}\times \frac{0.1\text{mm}}{1\cancel{\text{nm}}}\\ =78\text{mm}\end{array}$

The sketches of the waves associated with wave lengths 40mm and 78 mm are as:

Conclusion

Diagrams of waves are drawn as follows;

The X-rays and radio waves on this scale are $1\times {10}^{-2}\text{mm}$ and $1\times {10}^{12}\text{mm}$. The range of wavelength visible to our eyes is covering X-ray to radio waves.