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.
---
### 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]
---
### 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.
---
### 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.
### 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.
---
### 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]
---
### 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.
---
### 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.
