Physics • Year 12 • Module 7 • Lesson 1

The Electromagnetic Spectrum

Apply your understanding of c = fλ and E = hf to real data, graph interpretation, and scenario reasoning.

Apply · Data & Reasoning

1. Calculate and complete — EM wave data table

The table below gives partial data for six electromagnetic wave sources. Use c = fλ and E = hf (h = 6.63 × 10−34 J·s, c = 3.00 × 108 m/s) to complete the missing values. 12 marks (2 per row)

Source / region Wavelength (λ) Frequency (f) Photon energy (J) Photon energy (eV)
FM radio (103 MHz) 103 × 106 Hz
Microwave oven 12.2 cm
Red visible light 650 nm
Blue visible light 6.67 × 1014 Hz
UV (from sun) 290 nm
Medical X-ray 3.0 × 1018 Hz
Stuck? Use λ = c/f or f = c/λ first, then E = hf. To convert J to eV: divide by 1.60 × 10−19. Revisit the Worked Example card in the lesson.

2. Interpret the graph — photon energy vs frequency

The graph below shows photon energy (E) plotted against frequency (f) for a range of EM waves. 7 marks

0 1 2 3 4 5 0 1 2 3 4 5 6 7 Frequency f (× 10¹&sup4; Hz) Photon energy E (× 10−¹&sup9; J) Δf = 2 × 10¹&sup4; Hz ΔE

Figure 2. Photon energy vs frequency for EM radiation. Gradient = Planck's constant h. Illustrative data.

2.1 Describe the relationship shown by the graph. 1 mark

2.2 Use the graph to calculate the gradient. State what physical quantity the gradient represents and give its unit. 3 marks

2.3 From the graph, read the photon energy for blue visible light at f = 6.67 × 1014 Hz. Convert your reading to electronvolts. 2 marks

2.4 A student claims the y-intercept is non-zero because photons must have “base energy” even at zero frequency. Evaluate this claim using the graph and the equation E = hf. 1 mark

Stuck? Gradient = rise/run = ΔE / Δf. At f = 0, E = hf = h(0) = 0, so the line passes through the origin.

3. Scenario — EM waves at the CSIRO Parkes Radio Telescope (Murriyang)

The Murriyang (Parkes) telescope in NSW detects radio waves from pulsars. One pulsar signal has a frequency of 2.50 GHz. 8 marks

3.1 Calculate the wavelength of this radio signal. Show your working. 2 marks

3.2 Calculate the energy of one photon of this signal in joules and in electronvolts. 3 marks

3.3 The telescope also detects X-rays with λ = 0.10 nm using a different detector. Calculate the frequency of these X-rays and show that their photon energy is approximately 12.4 keV. 3 marks

Stuck? 1 GHz = 109 Hz; 1 nm = 10−9 m. Revisit the Worked Example card in the lesson.

4. Predict and justify — UV sterilisation in Australian hospitals

A UV sterilisation lamp emits light at 254 nm. This wavelength is used in Australian hospitals to kill bacteria on surfaces without chemical disinfectants. 7 marks

4.1 Calculate the photon energy of 254 nm UV light in both joules and electronvolts. 3 marks

4.2 Compare this photon energy to that of visible red light (650 nm) and explain, using E = hf, why UV is biologically effective at killing bacteria when red light is not. 2 marks

4.3 Predict whether infrared light at 1000 nm would be more or less effective than 254 nm UV at bacterial sterilisation. Justify using photon energy reasoning. 2 marks

Stuck? Shorter wavelength means higher frequency means more energy per photon (E = hf = hc/λ). Compare photon energies numerically.

5. Compare the regions of the EM spectrum

Complete the table by filling in wavelength, frequency, photon energy (relative), and one harm/benefit for each region. 6 marks — any 3 rows fully completed

RegionWavelength orderFrequency orderPhoton energy (high/med/low)One biological or technological effect
Radio
Infrared
Visible
UV
X-ray
Gamma
Stuck? Remember: wavelength and frequency are inversely proportional. Photon energy increases with frequency (E = hf). Revisit Card 1 in the lesson.
Answers — Do not peek before attempting

Q1 — Data table (12 marks, 2 per row)

FM radio (103 MHz): λ = c/f = 3.00×108 / 103×106 = 2.91 m; E = hf = 6.63×10−34 × 103×106 = 6.83×10−26 J = 4.27×10−7 eV.

Microwave oven (λ = 0.122 m): f = c/λ = 3.00×108/0.122 = 2.46×109 Hz; E = 6.63×10−34×2.46×109 = 1.63×10−24 J = 1.02×10−5 eV.

Red light (λ = 650 nm): f = 3.00×108/650×10−9 = 4.62×1014 Hz; E = 6.63×10−34×4.62×1014 = 3.06×10−19 J = 1.91 eV.

Blue light (f = 6.67×1014 Hz): λ = 3.00×108/6.67×1014 = 4.50×10−7 m = 450 nm; E = 6.63×10−34×6.67×1014 = 4.42×10−19 J = 2.76 eV.

UV (λ = 290 nm): f = 3.00×108/290×10−9 = 1.03×1015 Hz; E = 6.63×10−34×1.03×1015 = 6.83×10−19 J = 4.27 eV.

X-ray (f = 3.0×1018 Hz): λ = 3.00×108/3.0×1018 = 1.0×10−10 m = 0.10 nm; E = 6.63×10−34×3.0×1018 = 1.99×10−15 J = 12.4 keV.

Q2.1 — Graph relationship

The graph shows a linear (directly proportional) relationship passing through the origin: photon energy E is directly proportional to frequency f, consistent with E = hf.

Q2.2 — Gradient (3 marks)

Using two well-separated points, e.g. (2, 1.326) and (6, 3.978): gradient = (3.978 − 1.326) × 10−19 / (6 − 2) × 1014 = 2.652×10−19 / 4×1014 = 6.63×10−34 J·s [1]. The gradient represents Planck's constant h [1]. Unit: J·s (joule-seconds) [1].

Q2.3 — Read blue light energy (2 marks)

At f = 6.67×1014 Hz, read E ≈ 4.4×10−19 J [1]. In eV: E = 4.4×10−19 / 1.60×10−19 = 2.75 eV [1]. (Accept 4.3–4.5 × 10−19 J from graph reading.)

Q2.4 — Evaluate y-intercept claim (1 mark)

The claim is incorrect. The graph passes through the origin (E = 0 when f = 0), which is consistent with E = hf: when f = 0 Hz, E = h × 0 = 0 J. There is no “base energy” for a photon at zero frequency.

Q3 — Murriyang scenario

3.1 λ = c/f = 3.00×108 / 2.50×109 = 0.120 m = 12.0 cm [2].

3.2 E = hf = 6.63×10−34 × 2.50×109 = 1.66×10−24 J [1]; in eV: 1.66×10−24 / 1.60×10−19 = 1.04×10−5 eV [1]. (Accept 1.0×10−5 eV.) Award 1 mark for correct method even if arithmetic error.

3.3 f = c/λ = 3.00×108/0.10×10−9 = 3.0×1018 Hz [1]; E = hf = 6.63×10−34×3.0×1018 = 1.99×10−15 J [1]; in keV: 1.99×10−15/1.60×10−19 = 12 440 eV ≈ 12.4 keV [1]. QED.

Q4 — UV sterilisation

4.1 f = c/λ = 3.00×108/254×10−9 = 1.18×1015 Hz [1]; E = hf = 6.63×10−34×1.18×1015 = 7.82×10−19 J [1]; in eV: 7.82×10−19/1.60×10−19 = 4.89 eV [1].

4.2 Red light at 650 nm has E = 1.91 eV; UV at 254 nm has E = 4.89 eV — approximately 2.6 times more energetic per photon [1]. UV photons have sufficient energy to break covalent bonds in bacterial DNA (bond energies typically 3–6 eV), disrupting replication and killing cells. Red photons do not carry enough energy to cause ionisation or bond cleavage [1].

4.3 Infrared at 1000 nm has f = 3.0×1014 Hz, so E = 6.63×10−34×3.0×1014 = 1.99×10−19 J = 1.24 eV — much less than the 4.89 eV UV photon [1]. Infrared photons lack the energy to break DNA bonds, so infrared would be far less effective at sterilisation than 254 nm UV [1].

Q5 — Compare and contrast table (sample answers)

Radio: longest λ (~103–10−1 m); lowest f; low photon energy; used in communication (AM/FM), no biological harm.

Infrared: λ ~10−3–7×10−7 m; low-medium f; medium-low; felt as heat; high intensity can cause skin burns.

Visible: λ ~400–700 nm; medium f; medium; enables human vision; no ionisation risk.

UV: λ ~10−8 m; medium-high f; medium-high; causes sunburn and DNA damage; also used for sterilisation.

X-ray: λ ~10−10 m; high f; high; medical imaging of bone; ionising — dangerous in high doses.

Gamma: shortest λ (~10−12 m); highest f; highest; cancer radiotherapy; highly ionising; produced by nuclear decay.

Award 2 marks per fully completed row (wavelength + frequency + energy + effect). Accept any valid biological or technological effect.

HSCScience • Physics • Year 12 • Module 7 • Lesson 1 • Worksheet 2 (Apply)