Chapter 22, Problem 63P

(a)

To determine

The frequency of EM wave with wavelength as the size of the thickness of a piece of paper.

Answer to Problem 63P

The frequency of EM wave with wavelength as the size of the thickness of a piece of paper is $5.0\times {10}^{12}\text{Hz}$

Explanation of Solution

Write an expression for the frequency of EM wave with wavelength as the size of the thickness of a piece of paper.

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

Here, $f$ is the frequency, $c$ is the speed of light and $\lambda$ is the wavelength.

Conclusion:

Substitute $3.0\times {10}^8\text{m/s}$ for $c$ and $60\text{μm}$ for $\lambda$ in equation (I) to find $f$.

    $\begin{array}{c}f=\frac{3.0\times {10}^8\text{m/s}}{\left(60\text{μm}\right)\left(\frac{1\text{m}}{{10}^6\text{μm}}\right)}\\ =\frac{3.0\times {10}^8\text{m/s}}{60\times {10}^{-6}\text{m}}\\ =5.0\times {10}^{12}\text{Hz}\end{array}$

Thus, the frequency of EM wave with wavelength as size of the thickness of a piece of paper is $5.0\times {10}^{12}\text{Hz}$

(b)

To determine

The frequency of EM wave with wavelength as the size of the length of a soccer field.

Answer to Problem 63P

The frequency of EM wave with wavelength as the size of the length of a soccer field is $3.3\times {10}^6\text{Hz}$

Explanation of Solution

Write an expression for the frequency of EM wave with wavelength as the size of the length of a soccer field.

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

Conclusion:

Substitute $3.0\times {10}^8\text{m/s}$ for $c$ and $91\text{m}$ for $\lambda$ in equation (II) to find $f$.

    $\begin{array}{c}f=\frac{3.0\times {10}^8\text{m/s}}{91\text{m}}\\ =3.3\times {10}^6\text{Hz}\end{array}$

Thus, the frequency of EM wave with wavelength the size of as the length of a soccer field is $3.3\times {10}^6\text{Hz}$.

(c)

To determine

The frequency of EM wave with wavelength as the size of diameter of Earth.

Answer to Problem 63P

The frequency of EM wave with wavelength as the size of diameter of Earth is $23.5\text{Hz}$

Explanation of Solution

Write an expression for the frequency of EM wave with wavelength as the size of diameter of Earth.

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

Conclusion:

Substitute $3.0\times {10}^8\text{m/s}$ for $c$ and $2\left(6.371\times {10}^6\text{m}\right)$ for $\lambda$ in equation (III) to find $f$.

    $\begin{array}{c}f=\frac{3.0\times {10}^8\text{m/s}}{2\left(6.371\times {10}^6\text{m}\right)}\\ =\frac{3.0\times {10}^8\text{m/s}}{12.742\times {10}^6\text{m}}\\ =23.5\text{Hz}\end{array}$

Thus, the frequency of EM wave with wavelength as the size of diameter of Earth is $23.5\text{Hz}$

(d)

To determine

The frequency of EM wave with wavelength the size of the distance from Earth to Sun.

Answer to Problem 63P

The frequency of EM wave with wavelength the size of the distance from Earth to Sun is $2.00\times {10}^{-3}\text{Hz}$

Explanation of Solution

Write an expression for the frequency of EM wave with wavelength the size of the thickness of distance from Earth to Sun.

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

Conclusion:

Substitute $3.0\times {10}^8\text{m/s}$ for $c$ and $1.50\times {10}^{11}\text{m}$ for $\lambda$ in equation (IV) to find $f$.

    $\begin{array}{c}f=\frac{3.0\times {10}^8\text{m/s}}{1.50\times {10}^{11}\text{m}}\\ =2.00\times {10}^{-3}\text{Hz}\end{array}$

Thus, the frequency of EM wave with wavelength the size of the distance from Earth to Sun is $2.00\times {10}^{-3}\text{Hz}$