21st Century Astronomy
21st Century Astronomy
6th Edition
ISBN: 9780393428063
Author: Kay
Publisher: NORTON
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Chapter 5, Problem 31QP

(a)

To determine

Wavelengths of AM and FM signals.

(a)

Expert Solution
Check Mark

Answer to Problem 31QP

Wavelength of AM is 379.74 m_ and wavelength of FM is 3.05 m_.

Explanation of Solution

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

    c=νλ        (I)

Here, c is the speed of light,     ν is the frequency of the signal, λ is the wavelength of signal.

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

    νAM=7.90×105Hz        (II)

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

    νFM=9.83×107Hz        (III)

Conclusion:

Substituting 7.90×105Hz for νAM , 9.83×107Hz for νFM and 3×108m/s for c in equation (I), For AM,

    λAM=3×1087.90×105m=379.74 m

For FM,

    λFM=3×1089.83×107m=3.05 m

Therefore, wavelength of AM is 379.74 m_ and wavelength of FM is 3.05 m_.

(b)

To determine

Whether AM or FM broadcast in higher frequencies.

(b)

Expert Solution
Check Mark

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 νAM=7.90×105Hz. The frequency of the FM radio waves produced by the radio station is given as νFM=9.83×107Hz.

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.

(c)

Expert Solution
Check Mark

Answer to Problem 31QP

The photon energy of AM is and photon energy of FM is 5.2×1028 J_ and 6.51×1026 J_.

Explanation of Solution

Write the relation between photon energy and frequency.

    E=hν        (IV)

Here, E is the energy of photon, h is the Planck’s constant, ν is the frequency of photon.

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

    νAM=7.90×105Hz

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

  νFM=9.83×107Hz

Conclusion:

Substituting 7.90×105Hz for νAM , 9.83×107Hz for νFM and 6.626×1034J s for h in equation (IV), For AM,

  EAM=6.626×1034×7.9×105 J=5.2×1028 J

For FM,

  EAM=6.626×1034×9.83×107 J=6.51×1026 J

Therefore, photon energies of AM is 5.2×1028 J_ and wavelength of FM is 6.51×1026 J_.

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