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Waves - Electromagnetic spectrum

Grade 12IGCSEPhysics

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. They do not require a medium for propagation and can travel through a vacuum.

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All electromagnetic waves travel at the same speed in a vacuum, which is approximately c=3.0imes108 m/sc = 3.0 imes 10^8 \text{ m/s}.

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The electromagnetic spectrum is arranged in order of increasing frequency (ff) and decreasing wavelength (λ\lambda): Radio waves, Microwaves, Infrared (IR), Visible light, Ultraviolet (UV), X-rays, and Gamma rays.

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The energy of an EM wave is directly proportional to its frequency, expressed as E=hfE = hf. Therefore, high-frequency waves like X-rays and Gamma rays are more energetic and ionizing.

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Ionizing radiation (UV, X-rays, Gamma rays) has enough energy to remove electrons from atoms, which can damage biological molecules like DNA and cause mutations or cancer.

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Radio waves are used for long-range communication; Microwaves for satellite communication and cooking; Infrared for remote controls and thermal imaging; Visible light for photography; Ultraviolet for sun tanning and sterilization; X-rays for medical imaging; and Gamma rays for radiotherapy.

📐Formulae

v=fλv = f \lambda

c=fλc = f \lambda

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

E=hfE = hf

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

💡Examples

Problem 1:

A local radio station broadcasts at a frequency of 98.0 MHz98.0 \text{ MHz}. Calculate the wavelength of these radio waves in a vacuum.

Solution:

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

Explanation:

Since all EM waves travel at the speed of light (cc) in a vacuum, we rearrange the wave equation to solve for wavelength. Note that 1 MHz=106 Hz1 \text{ MHz} = 10^6 \text{ Hz}.

Problem 2:

Calculate the frequency of a green light wave with a wavelength of 530 nm530 \text{ nm}.

Solution:

f=cλf = \frac{c}{\lambda} f=3.0×108 m/s530×10−9 mf = \frac{3.0 \times 10^8 \text{ m/s}}{530 \times 10^{-9} \text{ m}} f≈5.66×1014 Hzf \approx 5.66 \times 10^{14} \text{ Hz}

Explanation:

To find the frequency, we divide the speed of light by the wavelength. The unit 'nm' (nanometers) must be converted to meters (1 nm=10−9 m1 \text{ nm} = 10^{-9} \text{ m}) for consistency with the SI units of the speed of light.

Electromagnetic spectrum - Revision Notes & Key Formulas | IGCSE Grade 12 Physics