From radio to gamma rays: wave properties, photon energy, and how to master graphs for IB Physics SL.
📡 What is the electromagnetic spectrum?
The electromagnetic (EM) spectrum is the continuous range of all possible frequencies of electromagnetic radiation. It spans from very long wavelength radio waves (km scale) to extremely short gamma rays (smaller than an atomic nucleus).
All EM waves travel at the same speed in a vacuum: c = 3.00 × 10⁸ m s⁻¹. They differ only in frequency (f) and wavelength (λ) — and therefore in the energy they carry.
Fig 1 — The electromagnetic spectrum. Note the inverse relationship: long wavelength ↔ low frequency ↔ low energy (left) and short wavelength ↔ high frequency ↔ high energy (right).
💡 Examiner insight: You must be able to order any set of EM radiations by frequency (or wavelength). Mnemonic: Radio, Microwave, Infrared, Visible, Ultraviolet, X-ray, Gamma.
⚡ How are EM waves created?
An oscillating electric charge (for example, electrons moving up and down in an antenna) produces a changing electric field, which in turn generates a changing magnetic field. These two fields regenerate each other and propagate outward as a transverse wave — no medium required.
Fig 2 — Propagation of an EM wave. The electric field and magnetic field oscillate perpendicular to each other and to the direction of travel.
📐 The wave equation — linking f and λ
For any EM wave in a vacuum, the speed \(c\) is constant. Therefore frequency and wavelength are inversely proportional:
$$c = f\lambda \quad\Rightarrow\quad f = \dfrac{c}{\lambda} \ \text{ and }\ \lambda = \dfrac{c}{f}$$
If frequency doubles, wavelength halves — and vice versa.
c = 3.00×10⁸ m/sf ∝ 1/λλ ∝ 1/f
⚠️ Common mistake: Thinking that different EM waves travel at different speeds in vacuum. They all travel at \(c\). Speed only changes in a medium (e.g., glass, water).
🔋 Photon model — quantised energy
EM waves carry energy in discrete packets called photons. The energy of a single photon depends only on its frequency (not on its intensity):
$$E = hf$$
where \(h = 6.63 \times 10^{-34}\) J s (Planck's constant). Using the wave equation, we also get:
$$E = \dfrac{hc}{\lambda}$$
A useful shortcut for the IB data booklet: \(hc = 1240\) eV nm. That means if λ is in nanometres, \(E\) (eV) \(= 1240 / \lambda\) (nm).
✓ Exam tip: You will often be asked to convert between joules and electronvolts. Remember 1 eV = 1.60 × 10⁻¹⁹ J.
📈 Graphical mastery for IB exams
Two essential graphs appear frequently in Papers 1, 2 and 3:
Fig 3 — Photon energy vs. frequency: straight line through origin, slope = h = 6.63×10⁻³⁴ J s. Higher frequency = higher photon energy.
Fig 4 — Left: E ∝ 1/λ (hyperbola). Right: linearised plot E vs. 1/λ gives a straight line with slope = hc.
What examiners ask: Identify the slope, calculate h or hc from a graph, predict the energy for a given frequency, or determine which region of the spectrum a data point belongs to.
🔧 Worked example — calculating wavelength and photon energy
1
State the known values: A microwave oven operates at f = 2.45 GHz. \(c = 3.00\times10^8\) m s⁻¹, \(h = 6.63\times10^{-34}\) J s.
2
Find the wavelength: Convert GHz → Hz: 2.45 GHz = 2.45×10⁹ Hz. \(\lambda = c/f = (3.00\times10^8)/(2.45\times10^9) = 0.122\) m.
3
Find photon energy in joules: \(E = hf = (6.63\times10^{-34})(2.45\times10^9) = 1.62\times10^{-24}\) J.
4
Convert to eV: \(E = (1.62\times10^{-24})/(1.60\times10^{-19}) = 1.01\times10^{-5}\) eV (very low energy — non-ionising).
💡 Examiner note: Always include units and show conversion steps. Marks are often awarded for the correct use of scientific notation.
🌍 Real-world connections
Fig 5 — Applications across the spectrum: X-rays (medical imaging), microwaves (cooking & radar), infrared (thermal cameras), radio (astronomy & communication).
Each region interacts with matter differently because of photon energy: radio waves pass through walls, UV causes sunburn (ionisation starts near 3.1 eV), and gamma rays sterilise medical equipment.
Practice check — multiple choice
1. Which of the following types of electromagnetic radiation has the longest wavelength?
A Ultraviolet
B Radio waves
C Gamma rays
D Visible light
2. The energy of a photon is directly proportional to:
A its wavelength
B its frequency
C the speed of light
D the amplitude of the wave
3. A photon of red light has wavelength 700 nm. What is its approximate energy in eV? (Use \(hc = 1240\) eV nm)
A 0.56 eV
B 0.89 eV
C 1.77 eV
D 3.54 eV
4. Which statement about X-rays and radio waves is true?
A Radio waves travel faster in vacuum than X-rays.
B X-rays have a lower frequency than radio waves.
C Radio waves carry more energy per photon than X-rays.
D X-rays have a shorter wavelength and higher photon energy.
Practice check — fill in the blanks
Complete the statements:
1. All electromagnetic waves travel at the same speed in a vacuum, which is m s⁻¹.
2. The energy of a photon is given by \(E = \) (in terms of frequency) or \(E = \) (in terms of wavelength).
3. The region of the EM spectrum immediately below visible light (lower frequency) is ; the region immediately above visible light is .
Exam-style questions
Exam question · Calculate
(a) A gamma ray photon has a wavelength of \(5.0 \times 10^{-13}\) m. Calculate its frequency. [2 marks]
(b) Hence calculate the energy of this photon in joules and in eV. [3 marks]
Model answer (5 marks total):
(a) \(f = \dfrac{c}{\lambda} = \dfrac{3.00\times10^8}{5.0\times10^{-13}} = 6.0\times10^{20}\) Hz ✓
(b) \(E = hf = (6.63\times10^{-34})(6.0\times10^{20}) = 3.98\times10^{-13}\) J ✓
Convert to eV: \(E = \dfrac{3.98\times10^{-13}}{1.60\times10^{-19}} = 2.49\times10^{6}\) eV ✓ (or 2.49 MeV)
✓ Unit conversions correct. ✓ Final answers to 2 significant figures.
Exam question · Explain
Ultraviolet (UV) radiation is more harmful to living tissue than visible light, even at the same low intensity. Using the photon model, explain why. [3 marks]
Model answer (3 marks):
✓ UV has a higher frequency / shorter wavelength than visible light.
✓ From \(E = hf\), each UV photon carries more energy than a visible photon.
✓ Higher-energy photons can ionise atoms / break chemical bonds in DNA (even if the total intensity — number of photons per second — is low).
(Award any two of the above for full marks, third is bonus clarity.)
