Answer:
Signal A has a longer wavelength than signal B.
Explanation:
To know the the correct answer to the question given above, we shall determine the energy and the wavelength of both radio signals.
This is illustrated below:
1. Determination of the energy of both radio signals
i. Energy of signal A
Frequency (f) = 98.7 MHz = 98.7×10⁶ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =.?
E = hf
E = 6.63×10¯³⁴ × 98.7×10⁶
E = 6.54×10¯²⁶ J
ii. Energy of signal B
Frequency (f) = 100.3 MHz = 100.3×10⁶ Hz
Planck's constant (h) = 6.63×10¯³⁴ Js
Energy (E) =.?
E = hf
E = 6.63×10¯³⁴ × 100.3×10⁶
E = 6.65×10¯²⁶ J
2. Determination of the wavelength of both radio signals.
i. Wavelength of signal A:
Frequency (f) = 98.7 MHz = 98.7×10⁶ Hz
Velocity (v) = 3×10⁸ m/s
Wavelength (λ) =?
v = λf
3×10⁸ = λ × 98.7×10⁶
Divide both side by 98.7×10⁶
λ = 3×10⁸ / 98.7×10⁶
λ = 3.04 m
ii. Wavelength of signal B
Frequency (f) = 100.3 MHz = 100.3×10⁶ Hz
Velocity (v) = 3×10⁸ m/s
Wavelength (λ) =?
v = λf
3×10⁸ = λ × 100.3×10⁶
Divide both side by 100.3×10⁶
λ = 3×10⁸ / 100.3×10⁶
λ = 2.99 m
The above calculation is summarised in the table below:
Signal >> Energy >>>>> Wavelength
A >>>>>> 6.54×10¯²⁶ >> 3.04
B >>>>>> 6.65×10¯²⁶ >> 2.99
From the above illustration,
1. Signal A has lesser energy than signal B.
2. Signal A has a longer wavelength than signal B.
