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Waves - The Electromagnetic Spectrum

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Electromagnetic (EM) waves are transverse waves consisting of oscillating electric and magnetic fields that are perpendicular to each other and the direction of energy transfer.

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Unlike mechanical waves, EM waves do not require a medium to travel and can propagate through a vacuum at the constant speed of light, c≈3.0×108 m/sc \approx 3.0 \times 10^8 \text{ m/s}.

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The Electromagnetic Spectrum is the range of all types of EM radiation, ordered by frequency or wavelength: Radio waves (longest λ\lambda, lowest ff), Microwaves, Infrared, Visible Light, Ultraviolet, X-rays, and Gamma rays (shortest λ\lambda, highest ff).

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The energy of an EM wave is directly proportional to its frequency (E∝fE \propto f). High-frequency waves like X-rays and Gamma rays are 'ionizing radiation', meaning they carry enough energy to remove electrons from atoms.

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The wave equation relates speed, frequency, and wavelength: v=fλv = f\lambda. For all EM waves in a vacuum, this is expressed as c=fλc = f\lambda.

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Visible light ranges from approximately 400 nm400 \text{ nm} (violet) to 700 nm700 \text{ nm} (red) in wavelength.

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Applications include: Radio waves for telecommunications, Microwaves for satellite transmissions and cooking, Infrared for night vision and thermal imaging, and X-rays for medical diagnostics.

📐Formulae

v=fλv = f \lambda

c=fλc = f \lambda

f=1Tf = \frac{1}{T}

E=hfE = hf

c≈3.0×108 m/sc \approx 3.0 \times 10^8 \text{ m/s}

💡Examples

Problem 1:

A specific green light has a wavelength of 5.2×10−7 m5.2 \times 10^{-7} \text{ m}. Calculate its frequency in a vacuum.

Solution:

f=cλ=3.0×108 m/s5.2×10−7 m≈5.77×1014 Hzf = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{5.2 \times 10^{-7} \text{ m}} \approx 5.77 \times 10^{14} \text{ Hz}

Explanation:

We use the speed of light cc as the velocity and rearrange the wave equation c=fλc = f\lambda to solve for frequency ff.

Problem 2:

A radio station broadcasts at a frequency of 98.0 MHz98.0 \text{ MHz}. What is the wavelength of these radio waves?

Solution:

λ=cf=3.0×108 m/s98.0×106 Hz≈3.06 m\lambda = \frac{c}{f} = \frac{3.0 \times 10^8 \text{ m/s}}{98.0 \times 10^6 \text{ Hz}} \approx 3.06 \text{ m}

Explanation:

First, convert frequency from megahertz to hertz (1 MHz=106 Hz1 \text{ MHz} = 10^6 \text{ Hz}). Then divide the speed of light by the frequency to find the wavelength.

Problem 3:

Determine the period TT of an X-ray with a frequency of 3.0×1018 Hz3.0 \times 10^{18} \text{ Hz}.

Solution:

T=1f=13.0×1018 Hz≈3.33×10−19 sT = \frac{1}{f} = \frac{1}{3.0 \times 10^{18} \text{ Hz}} \approx 3.33 \times 10^{-19} \text{ s}

Explanation:

The period is the reciprocal of the frequency, representing the time taken for one complete oscillation.

The Electromagnetic Spectrum - Revision Notes & Key Formulas | IB Grade 9 Science