Chapter 34, Problem 54AP

To determine

The classification of waves of frequencies $2\text{Hz}$, $2\text{kHz}$, $2\text{MHz}$, $2\text{GHz}$, $2\text{THz}$, $2\text{PHz}$, $2\text{EHz}$, $2\text{ZHz}$ and $2\text{YHz}$ on the electromagnetic spectrum and the classification of waves of wavelengths $2\text{km}$, $2\text{m}$, $2\text{mm}$, $2\text{μm}$, $2\text{nm}$, $2\text{pm}$, $2\text{fm}$, and $2\text{am}$

Answer to Problem 54AP

The classification of waves on the basis of frequency is shown below.

Frequency, $f$	Wavelength, $\lambda =\frac{c}{f}$		Classification
$2\text{Hz}$	$150\text{Mm}$		Radio
$2\text{kHz}$= $2\times {10}^3\text{Hz}$		$150\text{km}$		Radio
$2\text{MHz}$= $2\times {10}^6\text{Hz}$		$150\text{m}$		Radio
$2\text{GHz}$= $2\times {10}^9\text{Hz}$		$150\text{cm}$		Microwave
$2\text{THz}$= $2\times {10}^{12}\text{Hz}$		$150\text{μm}$		Infrared
$2\text{PHz}$= $2\times {10}^{15}\text{Hz}$		$150\text{nm}$		Ultraviolet
$2\text{EHz}$= $2\times {10}^{18}\text{Hz}$		$150\text{pm}$		X-ray
$2\text{ZHz}$= $2\times {10}^{21}\text{Hz}$		$150\text{fm}$		Gamma ray
$2\text{YHz}$= $2\times {10}^{24}\text{Hz}$		$150\text{am}$		Gamma ray

The classification of waves on the basis of wavelength is shown below.

Wavelength, $\lambda$	Frequency, $f=\frac{c}{\lambda }$		Classification
$2\text{km}$= $2\times {10}^3\text{m}$	$1.5\times {10}^5\text{Hz}$		Radio
$2\text{m}$= $2\times {10}^0\text{m}$		$1.5\times {10}^8\text{Hz}$		Radio
$2\text{mm}$= $2\times {10}^{-3}\text{m}$		$1.5\times {10}^{11}\text{Hz}$		Microwave
$2\text{μm}$= $2\times {10}^{-6}\text{m}$		$1.5\times {10}^{14}\text{Hz}$		Infrared
$2\text{nm}$= $2\times {10}^{-9}\text{m}$		$1.5\times {10}^{17}\text{Hz}$		Ultraviolet/ X-ray
$2\text{pm}$= $2\times {10}^{-12}\text{m}$		$1.5\times {10}^{20}\text{Hz}$		X-ray/Gamma ray
$2\text{fm}$= $2\times {10}^{-15}\text{m}$		$1.5\times {10}^{23}\text{Hz}$		Gamma ray
$2\text{am}$= $2\times {10}^{-18}\text{m}$		$1.5\times {10}^{26}\text{Hz}$		Gamma ray

Explanation of Solution

The formula to calculate the wavelength is,

$\lambda =\frac{c}{f}$ (1)

Here,

$c$ is the speed of light.

$f$ is the frequency of the wave.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{Hz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{Hz}\right)}\\ =15\times {10}^7\text{m}\times \frac{{10}^{-6}\text{Mm}}{1\text{m}}\\ =150\text{Mm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{kHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{kHz}\times \frac{{10}^3\text{Hz}}{1\text{kHz}}\right)}\\ =15\times {10}^4\text{m}\times \frac{{10}^{-3}\text{km}}{1\text{m}}\\ =150\text{km}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{MHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{MHz}\times \frac{{10}^6\text{Hz}}{1\text{MHz}}\right)}\\ =1.5\times {10}^2\text{m}\\ =150\text{m}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{GHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{GHz}\times \frac{{10}^9\text{Hz}}{1\text{GHz}}\right)}\\ =1.5\times {10}^{-1}\text{m}\times \frac{{10}^3\text{cm}}{1\text{m}}\\ =150\text{cm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{THz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{THz}\times \frac{{10}^{12}\text{Hz}}{1\text{THz}}\right)}\\ =1.5\times {10}^{-4}\text{m}\times \frac{{10}^6\text{μm}}{1\text{m}}\\ =150\text{μm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{PHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{PHz}\times \frac{{10}^{15}\text{Hz}}{1\text{PHz}}\right)}\\ =1.5\times {10}^{-7}\text{m}\times \frac{{10}^9\text{nm}}{1\text{m}}\\ =150\text{nm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{EHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{EHz}\times \frac{{10}^{18}\text{Hz}}{1\text{EHz}}\right)}\\ =1.5\times {10}^{-10}\text{m}\times \frac{{10}^{12}\text{pm}}{1\text{m}}\\ =150\text{pm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{ZHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{ZHz}\times \frac{{10}^{21}\text{Hz}}{1\text{ZHz}}\right)}\\ =1.5\times {10}^{-13}\text{m}\times \frac{{10}^{15}\text{fm}}{1\text{m}}\\ =150\text{fm}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{YHz}$ for $f$ in equation (1) to find the value of $\lambda$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{YHz}\times \frac{{10}^{24}\text{Hz}}{1\text{YHz}}\right)}\\ =1.5\times {10}^{-16}\text{m}\times \frac{{10}^{18}\text{am}}{1\text{m}}\\ =150\text{am}\end{array}$

From the above calculation, the table for the wavelength can be deduces and the respective classification of waves is shown.

Frequency, $f$	Wavelength, $\lambda =\frac{c}{f}$		Classification
$2\text{Hz}$	$150\text{Mm}$		Radio
$2\text{kHz}$= $2\times {10}^3\text{Hz}$		$150\text{km}$		Radio
$2\text{MHz}$= $2\times {10}^6\text{Hz}$		$150\text{m}$		Radio
$2\text{GHz}$= $2\times {10}^9\text{Hz}$		$150\text{cm}$		Microwave
$2\text{THz}$= $2\times {10}^{12}\text{Hz}$		$150\text{μm}$		Infrared
$2\text{PHz}$= $2\times {10}^{15}\text{Hz}$		$150\text{nm}$		Ultraviolet
$2\text{EHz}$= $2\times {10}^{18}\text{Hz}$		$150\text{pm}$		X-ray
$2\text{ZHz}$= $2\times {10}^{21}\text{Hz}$		$150\text{fm}$		Gamma ray
$2\text{YHz}$= $2\times {10}^{24}\text{Hz}$		$150\text{am}$		Gamma ray

The formula to calculate the frequency is,

$f=\frac{c}{\lambda }$ (2)

Here,

$c$ is the speed of light.

$\lambda$ is the wavelength of the wave.

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{km}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}\\ =1.5\times {10}^5\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{m}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{m}\right)}\\ =1.5\times {10}^8\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{mm}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{mm}\times \frac{{10}^{-3}\text{m}}{1\text{mm}}\right)}\\ =1.5\times {10}^{11}\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{μm}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{μm}\times \frac{{10}^{-6}\text{m}}{1\text{μm}}\right)}\\ =1.5\times {10}^{14}\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{nm}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{nm}\times \frac{{10}^{-9}\text{m}}{1\text{nm}}\right)}\\ =1.5\times {10}^{17}\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{pm}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{pm}\times \frac{{10}^{-12}\text{m}}{1\text{nm}}\right)}\\ =1.5\times {10}^{20}\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{fm}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{am}\times \frac{{10}^{-15}\text{m}}{1\text{am}}\right)}\\ =1.5\times {10}^{23}\text{Hz}\end{array}$

Substitute $3\times {10}^8\text{m}/\text{s}$ for $c$ and $2\text{am}$ for $\lambda$ in equation (2) to find the value of $f$.

$\begin{array}{c}\lambda =\frac{\left(3\times {10}^8\text{m}/\text{s}\right)}{\left(2\text{am}\times \frac{{10}^{-18}\text{m}}{1\text{am}}\right)}\\ =1.5\times {10}^{26}\text{Hz}\end{array}$

From the above calculation the table for the frequency can be deduces and the respective classification of waves is shown below.

Wavelength, $\lambda$	Frequency, $f=\frac{c}{\lambda }$		Classification
$2\text{km}$= $2\times {10}^3\text{m}$	$1.5\times {10}^5\text{Hz}$		Radio
$2\text{m}$= $2\times {10}^0\text{m}$		$1.5\times {10}^8\text{Hz}$		Radio
$2\text{mm}$= $2\times {10}^{-3}\text{m}$		$1.5\times {10}^{11}\text{Hz}$		Microwave
$2\text{μm}$= $2\times {10}^{-6}\text{m}$		$1.5\times {10}^{14}\text{Hz}$		Infrared
$2\text{nm}$= $2\times {10}^{-9}\text{m}$		$1.5\times {10}^{17}\text{Hz}$		Ultraviolet/ X-ray
$2\text{pm}$= $2\times {10}^{-12}\text{m}$		$1.5\times {10}^{20}\text{Hz}$		X-ray/Gamma ray
$2\text{fm}$= $2\times {10}^{-15}\text{m}$		$1.5\times {10}^{23}\text{Hz}$		Gamma ray
$2\text{am}$= $2\times {10}^{-18}\text{m}$		$1.5\times {10}^{26}\text{Hz}$		Gamma ray

Conclusion:

Therefore, the table of classification of waves with the frequencies is shown below.

Frequency, $f$	Wavelength, $\lambda =\frac{c}{f}$		Classification
$2\text{Hz}$	$150\text{Mm}$		Radio
$2\text{kHz}$= $2\times {10}^3\text{Hz}$		$150\text{km}$		Radio
$2\text{MHz}$= $2\times {10}^6\text{Hz}$		$150\text{m}$		Radio
$2\text{GHz}$= $2\times {10}^9\text{Hz}$		$150\text{cm}$		Microwave
$2\text{THz}$= $2\times {10}^{12}\text{Hz}$		$150\text{μm}$		Infrared
$2\text{PHz}$= $2\times {10}^{15}\text{Hz}$		$150\text{nm}$		Ultraviolet
$2\text{EHz}$= $2\times {10}^{18}\text{Hz}$		$150\text{pm}$		X-ray
$2\text{ZHz}$= $2\times {10}^{21}\text{Hz}$		$150\text{fm}$		Gamma ray
$2\text{YHz}$= $2\times {10}^{24}\text{Hz}$		$150\text{am}$		Gamma ray

Therefore, the table of classification of waves with the wavelengths is shown below.

Wavelength, $\lambda$	Frequency, $f=\frac{c}{\lambda }$		Classification
$2\text{km}$= $2\times {10}^3\text{m}$	$1.5\times {10}^5\text{Hz}$		Radio
$2\text{m}$= $2\times {10}^0\text{m}$		$1.5\times {10}^8\text{Hz}$		Radio
$2\text{mm}$= $2\times {10}^{-3}\text{m}$		$1.5\times {10}^{11}\text{Hz}$		Microwave
$2\text{μm}$= $2\times {10}^{-6}\text{m}$		$1.5\times {10}^{14}\text{Hz}$		Infrared
$2\text{nm}$= $2\times {10}^{-9}\text{m}$		$1.5\times {10}^{17}\text{Hz}$		Ultraviolet/ X-ray
$2\text{pm}$= $2\times {10}^{-12}\text{m}$		$1.5\times {10}^{20}\text{Hz}$		X-ray/Gamma ray
$2\text{fm}$= $2\times {10}^{-15}\text{m}$		$1.5\times {10}^{23}\text{Hz}$		Gamma ray
$2\text{am}$= $2\times {10}^{-18}\text{m}$		$1.5\times {10}^{26}\text{Hz}$		Gamma ray