High School

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------------------------------------------------ QUESTION 3:

3.1) Define the term "Electromagnetic Waves."
(2 points)

3.2) A certain EM wave has a period of [tex]2 \times 10 \text{ s}[/tex]. Calculate the energy of a photon of this EM wave.
(6 points)

3.3) Radio waves have a speed of [tex]3 \times 10 \text{ m/s}[/tex]. A radio station tests which wavelength between two radios, with frequencies 97.6 MHz and 92.4 MHz respectively, will be the largest?
(5 points)

3.4) Will a gamma-ray have a higher, lower, or the same frequency as a radio wave? Justify your answer.
(2 points)

Answer :

Sure! Let's break down each part of the question step by step.

### 3.1) Define the term Electromagnetic Waves

Electromagnetic waves are waves that are formed by the oscillation of electric and magnetic fields. These waves can travel through a vacuum and carry energy from one place to another. They include a wide range of waves such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.

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### 3.2) Calculate the energy of a photon with a period [tex]\( T = 2 \times 10 \)[/tex] seconds

To find the energy of a photon, we use the formula:

[tex]\[ E = h \times f \][/tex]

where:
- [tex]\( E \)[/tex] is the energy of the photon,
- [tex]\( h = 6.626 \times 10^{-34} \)[/tex] Joule-seconds (Planck's constant),
- [tex]\( f \)[/tex] is the frequency of the wave.

First, we need to find the frequency [tex]\( f \)[/tex]. The frequency is the inverse of the period:

[tex]\[ f = \frac{1}{T} = \frac{1}{2 \times 10} = 0.05 \, \text{Hz} \][/tex]

Now, we can calculate the energy [tex]\( E \)[/tex]:

[tex]\[ E = 6.626 \times 10^{-34} \times 0.05 = 3.313 \times 10^{-35} \, \text{Joules} \][/tex]

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### 3.3) Compare wavelengths for frequencies 97.6 MHz and 92.4 MHz.

The wavelength [tex]\( \lambda \)[/tex] of a wave is given by the formula:

[tex]\[ \lambda = \frac{c}{f} \][/tex]

where:
- [tex]\( c = 3 \times 10^8 \, \text{m/s} \)[/tex] is the speed of light,
- [tex]\( f \)[/tex] is the frequency of the wave.

For a frequency of 97.6 MHz:

[tex]\[ f_1 = 97.6 \times 10^6 \, \text{Hz} \][/tex]

[tex]\[ \lambda_1 = \frac{3 \times 10^8}{97.6 \times 10^6} \approx 3.07 \, \text{meters} \][/tex]

For a frequency of 92.4 MHz:

[tex]\[ f_2 = 92.4 \times 10^6 \, \text{Hz} \][/tex]

[tex]\[ \lambda_2 = \frac{3 \times 10^8}{92.4 \times 10^6} \approx 3.25 \, \text{meters} \][/tex]

The wavelength for 92.4 MHz is larger.

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### 3.4) Compare frequencies: Gamma-rays vs radio waves.

Gamma rays have a much higher frequency compared to radio waves. This is because gamma rays are found on the extreme high-frequency end of the electromagnetic spectrum, while radio waves are on the low-frequency end. Therefore, gamma rays contain a higher frequency than radio waves.