Chapter 5, Problem 31QP

(a)

To determine

Wavelengths of AM and FM signals.

Answer to Problem 31QP

Wavelength of AM is $\underset{\_}{379.74\text{ m}}$ and wavelength of FM is $\underset{\_}{3.05\text{ m}}$.

Explanation of Solution

Write the relation between speed of signal, frequency of signal and wavelength of signal.

    $c=\nu \lambda$        (I)

Here, $c$ is the speed of light,     $\nu$ is the frequency of the signal, $\lambda$ is the wavelength of signal.

For AM, the frequency of the signal received by radio is given as

    ${\nu }_{\text{AM}}=7.90\times {10}^5\text{Hz}$        (II)

For AM, the frequency of the signal received by radio is given as

    ${\nu }_{\text{FM}}=9.83\times {10}^7\text{Hz}$        (III)

Conclusion:

Substituting $7.90\times {10}^5\text{Hz}$ for ${\nu }_{\text{AM}}$ , $9.83\times {10}^7\text{Hz}$ for ${\nu }_{\text{FM}}$ and $3\times {10}^8\text{m/s}$ for $c$ in equation (I), For AM,

    ${\lambda }_{\text{AM}}=\frac{3\times {10}^8}{7.90\times {10}^5}\text{m}=379.74\text{ m}$

For FM,

    ${\lambda }_{\text{FM}}=\frac{3\times {10}^8}{9.83\times {10}^7}\text{m}=3.05\text{ m}$

Therefore, wavelength of AM is $\underset{\_}{379.74\text{ m}}$ and wavelength of FM is $\underset{\_}{3.05\text{ m}}$.

(b)

To determine

Whether AM or FM broadcast in higher frequencies.

Answer to Problem 31QP

The FM broadcast in higher frequencies.

Explanation of Solution

The frequency of the AM radio waves produced by the radio station is given as ${\nu }_{\text{AM}}=7.90\times {10}^5\text{Hz}$. The frequency of the FM radio waves produced by the radio station is given as ${\nu }_{\text{FM}}=9.83\times {10}^7\text{Hz}$.

By comparing the value of frequency of AM and FM, the FM broadcast in higher frequency than the AM. Due to this reason FM is better than the AM. FM can travel more distances than AM due to less dissipation than AM.

Conclusion:

Therefore, the FM radio signal has more frequency.

(c)

To determine

The photon energies of AM and FM.

Answer to Problem 31QP

The photon energy of AM is and photon energy of FM is $\underset{\_}{5.2\times {10}^{-28}\text{ J}}$ and $\underset{\_}{6.51\times {10}^{-26}\text{ J}}$.

Explanation of Solution

Write the relation between photon energy and frequency.

    $E=h\nu$        (IV)

Here, $E$ is the energy of photon, $h$ is the Planck’s constant, $\nu$ is the frequency of photon.

For AM, the frequency of the signal received by radio is given as

    ${\nu }_{\text{AM}}=7.90\times {10}^5\text{Hz}$

For AM, the frequency of the signal received by radio is given as 

  ${\nu }_{\text{FM}}=9.83\times {10}^7\text{Hz}$

Conclusion:

Substituting $7.90\times {10}^5\text{Hz}$ for ${\nu }_{\text{AM}}$ , $9.83\times {10}^7\text{Hz}$ for ${\nu }_{\text{FM}}$ and $6.626\times {10}^{-34}\text{J s}$ for $h$ in equation (IV), For AM,

  $\begin{array}{c}{E}_{\text{AM}}=6.626\times {10}^{-34}\times \text{7}\text{.9}\times {\text{10}}^5\text{ J}\\ =5.2\times {10}^{-28}\text{ J}\end{array}$

For FM,

  $\begin{array}{c}{E}_{\text{AM}}=6.626\times {10}^{-34}\times \text{9}\text{.83}\times {\text{10}}^7\text{ J}\\ =6.51\times {10}^{-26}\text{ J}\end{array}$

Therefore, photon energies of AM is $\underset{\_}{5.2\times {10}^{-28}\text{ J}}$ and wavelength of FM is $\underset{\_}{6.51\times {10}^{-26}\text{ J}}$.