krit.club logo

Waves - Electromagnetic spectrum

Grade 11IGCSEPhysics

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

🔑Concepts

•

Electromagnetic (EM) waves are transverse waves consisting of oscillating electric and magnetic fields. They do not require a medium to travel and can move through a vacuum.

•

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

•

The EM spectrum is ordered by increasing frequency and decreasing wavelength: Radio waves, Microwaves, Infrared, Visible Light (Red to Violet), Ultraviolet, X-rays, and Gamma rays.

•

The energy of an EM wave is directly proportional to its frequency ff. Therefore, Gamma rays carry the highest energy while Radio waves carry the lowest.

•

Higher frequency waves such as Ultraviolet, X-rays, and Gamma rays are ionizing radiation, meaning they have enough energy to remove electrons from atoms, which can damage DNADNA and cause cancer.

•

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

•

Applications include: Radio waves (broadcasting), Microwaves (satellite communication and cooking), Infrared (thermal imaging and remote controls), X-rays (medical imaging and security).

📐Formulae

v=fλv = f \lambda

c=fλc = f \lambda

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

💡Examples

Problem 1:

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

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:

To find the wavelength, rearrange the wave equation c=fλc = f \lambda to solve for λ\lambda. Ensure the frequency is converted from megahertz (MHzMHz) to hertz (HzHz) by multiplying by 10610^6.

Problem 2:

An X-ray has a wavelength of 2.0×10−10 m2.0 \times 10^{-10} \text{ m}. Determine its frequency.

Solution:

f=cλ=3.0×108 m/s2.0×10−10 m=1.5×1018 Hzf = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m/s}}{2.0 \times 10^{-10} \text{ m}} = 1.5 \times 10^{18} \text{ Hz}

Explanation:

Using the wave speed formula for light, divide the speed of light cc by the given wavelength λ\lambda to find the frequency ff.

Electromagnetic spectrum - Revision Notes & Key Formulas | IGCSE Grade 11 Physics