To determine the frequency and energy of AM and FM waves, we can use the relationship between wavelength (λ), frequency (f), and the speed of light (c), given by the formula:
\[ c = f \cdot \lambda \]
Since the speed of light \( c \) is approximately \( 3 \times 10^8 \) meters per second, we can rearrange the formula to find frequency:
\[ f = \frac{c}{\lambda} \]
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AM Waves: Wavelength (λ) is approximately 100 meters. \[ f_{AM} = \frac{3 \times 10^8 , \text{m/s}}{100 , \text{m}} = 3 \times 10^6 , \text{Hz} , (or , 3 , \text{MHz}) \]
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FM Waves: Wavelength (λ) is approximately 1 meter. \[ f_{FM} = \frac{3 \times 10^8 , \text{m/s}}{1 , \text{m}} = 3 \times 10^8 , \text{Hz} , (or , 300 , \text{MHz}) \]
From these calculations, we see that:
- FM has a higher frequency than AM (300 MHz vs. 3 MHz).
Next, regarding energy, since energy (E) of a wave is directly proportional to its frequency, the relationship can be expressed as:
\[ E = h \cdot f \]
Where \( h \) is Planck's constant. Since FM has a higher frequency than AM, it will also have higher energy.
Therefore, based on these calculations, the correct response is:
FM has a higher frequency and higher energy.
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