Question

ADD NOTE O QUESTION GUIDE EXIT TEST

Answer 1

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.