To determine which type of wave has a higher frequency, we can use the relationship between wavelength (λ), frequency (f), and the speed of light (c), given by the formula:
\[ c = f \cdot \lambda \]
where:
- \( c \) is the speed of light (approximately 3 x 10^8 meters per second),
- \( f \) is the frequency,
- \( \lambda \) is the wavelength.
Given that AM waves have a wavelength of approximately 100 meters and FM waves have a wavelength of approximately 1 meter, we can express their frequencies:
-
For AM waves:
- \( λ_{AM} = 100 , \text{m} \)
- \( f_{AM} = \frac{c}{λ_{AM}} = \frac{3 \times 10^8}{100} = 3 \times 10^6 , \text{Hz} \) or 3 MHz.
-
For FM waves:
- \( λ_{FM} = 1 , \text{m} \)
- \( f_{FM} = \frac{c}{λ_{FM}} = \frac{3 \times 10^8}{1} = 3 \times 10^8 , \text{Hz} \) or 300 MHz.
From this, we see that:
- FM has a higher frequency (300 MHz) compared to AM (3 MHz).
As for energy, the energy \( E \) of a photon can be expressed by the equation:
\[ E = h \cdot f \]
where:
- \( E \) is energy,
- \( h \) is Planck's constant (approximately 6.626 x 10^-34 Joule seconds),
- \( f \) is frequency.
Since energy is directly proportional to frequency, we can conclude that FM waves, having a higher frequency, also possess higher energy.
So, the correct response is:
FM has a higher frequency and higher energy.
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