An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.2 MHz. Calculate the energy of the photons emitted by the AM radio station? Calculate the energy of the photons emitted by the FM radio station? Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.
An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.2 MHz. Calculate the energy of the photons emitted by the AM radio station? Calculate the energy of the photons emitted by the FM radio station? Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.2 MHz. Calculate the energy of the photons emitted by the AM radio station? Calculate the energy of the photons emitted by the FM radio station? Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.
![### Problem Statement
An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.2 MHz.
### Part C
**Question:**
Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.
**Input Fields:**
- **\( E_2 = \)** [J]
- **\( \frac{E_1}{E_2} = \)** [ratio]
(Submit buttons and request answer links are available for both input sections.)
### Explanation
The problem involves comparing the energies of photons from two different radio frequencies: AM (Amplitude Modulation) and FM (Frequency Modulation).
- **AM Frequency**: 1010 kHz (or 1.01 x 10^6 Hz)
- **FM Frequency**: 98.2 MHz (or 9.82 x 10^7 Hz)
**Key Concept:**
The energy of a photon is directly proportional to its frequency and can be calculated using the equation:
\[ E = h \cdot f \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \( f \) is the frequency of the photon.
The task involves calculating \( E_2 \) (the energy of FM photons) and the ratio \( \frac{E_1}{E_2} \) (the energy of AM photons to FM photons) to understand the comparison between these two electromagnetic waves.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Facd26fc6-fbad-4ec2-a17c-1465df39071f%2Ff70c413b-8bed-462c-9271-71095d6e7373%2F5mpfgmj_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.2 MHz.
### Part C
**Question:**
Compare the energy of the photons emitted by the AM radio station with the energy of the photons emitted by the FM radio station.
**Input Fields:**
- **\( E_2 = \)** [J]
- **\( \frac{E_1}{E_2} = \)** [ratio]
(Submit buttons and request answer links are available for both input sections.)
### Explanation
The problem involves comparing the energies of photons from two different radio frequencies: AM (Amplitude Modulation) and FM (Frequency Modulation).
- **AM Frequency**: 1010 kHz (or 1.01 x 10^6 Hz)
- **FM Frequency**: 98.2 MHz (or 9.82 x 10^7 Hz)
**Key Concept:**
The energy of a photon is directly proportional to its frequency and can be calculated using the equation:
\[ E = h \cdot f \]
where:
- \( E \) is the energy of the photon,
- \( h \) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J s}\)),
- \( f \) is the frequency of the photon.
The task involves calculating \( E_2 \) (the energy of FM photons) and the ratio \( \frac{E_1}{E_2} \) (the energy of AM photons to FM photons) to understand the comparison between these two electromagnetic waves.
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