Chapter 3, Problem 21Q

Calculate the wavelength, in nanometers, of the following wave frequencies:

1. a. 6.79 × 10¹⁴ s⁻¹
2. b. 4.44 × 10¹² Hz

Interpretation Introduction

Interpretation:

Using the given frequency value, the wavelength in nanometers has to be calculated.

Concept-Introduction:

Wavelength: The distance between successive peaks in a wave and measured in units of length.

Frequency: The number of waves passed a point in a certain amount of time.

Relation between wavelength and frequency:

$\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$ and represents the maximum velocity that light is able to travel through air.

Explanation of Solution

Given: Frequency $\text{= 6}\text{.79}\times {10}^{14}{\text{ s}}^{-1}$, c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$

Using the relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$

$\begin{array}{l}\left(\text{6}{\text{.79×10}}^{\text{14}}{\text{s}}^{\text{-1}}\right)\text{ =}\frac{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\text{λ}}\\ \\ \left(\text{λ}\right)\text{ =}\frac{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\left(\text{6}{\text{.79×10}}^{\text{14}}{\text{s}}^{\text{-1}}\right)}\text{=}\text{0}{\text{.442×10}}^{8-14}\text{ m }\left(\text{or}\right)442\times 10^{-9}m\end{array}$

Therefore, the wavelength in nanometer is $442\times 10^{-9}m\left(or\right)442nm$.

Interpretation Introduction

Interpretation:

Using the given frequency value, the wavelength in nanometers has to be calculated.

Concept-Introduction:

Wavelength: The distance between successive peaks in a wave and measured in units of length.

Frequency: The number of waves passed a point in a certain amount of time.

Relation between wavelength and frequency:

$\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$

Where, c is constant, the value of c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$ and represents the maximum velocity that light is able to travel through air.

Explanation of Solution

Given: Frequency $\text{= 4}{\text{.44×10}}^{\text{12}}\text{ Hz}$, c is $\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)$

Using the relation $\text{Frequency}\left(\text{ν}\right)\text{=}\frac{\text{speed}\text{of}\text{light}\left(\text{c}\right)}{\text{wavelength}\left(\text{λ}\right)}$

$\begin{array}{l}\left(\text{4}{\text{.44 ×10}}^{\text{12}}{\text{s}}^{\text{-1}}\right)\text{ =}\frac{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\text{λ}}\\ \\ \left(\text{λ}\right)\text{ =}\frac{\left(\text{3}{\text{.00×10}}^{\text{8}}\text{m/s}\right)}{\left(\text{4}{\text{.44 ×10}}^{\text{12}}{\text{s}}^{\text{-1}}\right)}\text{=}\text{0}{\text{.6756756×10}}^{8-12}\text{ m }\left(\text{or}\right)67567.56\times 10^{-9}m\end{array}$

Therefore, the wavelength in nanometer is $67567.56\times 10^{-9}m\left(or\right)67567nm$.