Year 12 PhysicsModule 7⏱ ~25 min5 MC · 2 Short AnswerLesson 1 of 14
The Electromagnetic Spectrum
You'll be able to calculate frequency from wavelength and vice versa.
Maxwell's electromagnetic theory predicted waves travelling at the measured speed of light. Hertz later generated and detected radio waves in the laboratory, providing experimental support for that prediction. For a representative wavelength of 66 cm, $f=c/\lambda\approx4.54\times10^8$ Hz, or 454 MHz; every region of the spectrum obeys the same vacuum relation $c=f\lambda$.
Today's hook: Take a radio wave with wavelength 66 cm, so $f=c/\lambda\approx454$ MHz, and compare it with a gamma ray of wavelength $10^{-12}$ m. Both travel at $c$ in vacuum. Which carries more energy per photon, and by what factor?
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You’re here
Orient and predict
Predict how a radio wave and a gamma ray differ, then set your goals and vocabulary for the electromagnetic spectrum.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Consider a radio wave and a gamma ray. A radio wave might have a wavelength of 100 m, while a gamma ray has a wavelength of $10^{-12}$ m.
Before reading on, answer:
Which wave has the higher frequency? Explain your reasoning.
Which wave carries more energy per photon? Why?
Both waves travel at the same speed in vacuum. Does this mean they have the same effect on matter? Explain.
Warm-up, in the electromagnetic spectrum, which type of radiation has the shortest wavelength?
Learning Intentions
goals
Know, The EM Spectrum
Seven regions: radio, microwave, IR, visible, UV, X-ray, gamma
All EM waves are transverse and travel at $c = 3.0\times10^8$ m/s in vacuum
Relationship: $c = f\lambda$
Understand, Photon Energy
Photon energy $E = hf = hc/\lambda$
Higher frequency = higher energy per photon
Why gamma photons can ionise atoms while radiofrequency exposure is non-ionising but can still heat tissue at sufficient intensity
Can Do, Calculate & Compare
Calculate frequency from wavelength and vice versa
Calculate photon energy in joules and electronvolts
Compare wavelength, frequency and energy across regions
Scan these before reading
vocab
Electromagnetic waveA transverse wave consisting of oscillating electric and magnetic fields perpendicular to each other and to the direction of propagation.
PhotonA quantum of electromagnetic energy with $E = hf$.
Wavelength ($\lambda$)The distance between successive crests of a wave, measured in metres.
Frequency ($f$)The number of wave cycles per second, measured in hertz (Hz).
Planck's constant ($h$)$h = 6.63 \times 10^{-34}$ J·s. Relates photon energy to frequency.
Cross-lesson links: M7 opens by mapping the full electromagnetic spectrum. You already know from M6 (Faraday/Maxwell) that changing magnetic flux induces electric fields and vice versa, L01 shows this is the physical origin of self-propagating electromagnetic waves. Every wavelength in the spectrum obeys the same Maxwell equations.
Core Content
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Map the seven regions of the spectrum
See what every electromagnetic wave has in common, and what changes as you move from radio to gamma.
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The Electromagnetic Spectrum
+5 XP
One phenomenon, seven regions
Tune a radio to an FM station and music reaches you through nothing but empty space. Stand outside in the dark and warmth from the sun still finds you. Undergo an X-ray and the image forms without the beam touching you. In each case, energy has arrived as an oscillating electric and magnetic field, a self-propagating electromagnetic wave requiring no medium, travelling at exactly $c = 3.00 \times 10^8$ m/s. All electromagnetic waves share three fundamental properties: they are transverse waves, they travel at the speed of light in vacuum, and they do not require a medium.
The electromagnetic spectrum is traditionally divided into seven regions, ordered by increasing frequency (and decreasing wavelength):
Figure 1, The electromagnetic spectrum: all regions travel at speed $c$ in vacuum, differing only in wavelength and frequencyText description
The seven regions in order of increasing frequency and photon energy, which is decreasing wavelength:
Radio: typical wavelength ~10³ m.
Microwave: typical wavelength ~10⁻² m.
Infrared: typical wavelength ~10⁻⁵ m.
Visible: typical wavelength 400–700 nm.
Ultraviolet: typical wavelength ~10⁻⁸ m.
X-ray: typical wavelength ~10⁻¹⁰ m.
Gamma: typical wavelength ~10⁻¹² m.
All regions are electromagnetic waves travelling at c in vacuum; they differ only in wavelength and frequency.
Radio waves ($\lambda \sim 10^3$ m): Used for communication, AM/FM radio, television and mobile phones. Their photons are non-ionising; sufficiently intense radiofrequency fields can still deposit energy and heat tissue.
Microwaves ($\lambda \sim 10^{-2}$ m): Used in microwave ovens (heating water molecules), radar, and Wi-Fi. Can cause heating of tissue.
Infrared ($\lambda \sim 10^{-5}$ m): Felt as heat. Used in remote controls, thermal imaging, and night vision. Can cause burns at high intensity.
Visible light ($\lambda \sim 400$–$700$ nm): The only region our eyes can detect. Red has the longest wavelength ($\approx 700$ nm); violet the shortest ($\approx 400$ nm).
Ultraviolet ($\lambda \sim 10^{-8}$ m): Causes sunburn and skin damage. Used in sterilisation and forensics. Can damage DNA.
X-rays ($\lambda \sim 10^{-10}$ m): Pass through soft tissue but are absorbed by bone. Used in medical imaging and airport security. Ionising, dangerous in high doses.
Gamma rays ($\lambda \sim 10^{-12}$ m): Highest energy, produced by nuclear decay and cosmic events. Used in cancer treatment (radiotherapy). Highly ionising and dangerous.
Stop & Check
Arrange the following in order of increasing photon energy: red light, blue light, infrared, ultraviolet, X-rays. Explain your reasoning using the relationship $E = hf$.
The electromagnetic spectrum has seven regions ordered by increasing frequency: radio, microwave, infrared, visible, UV, X-ray, gamma. All EM waves are transverse and travel at $c = 3.00 \times 10^8$ m/s in vacuum, no medium required. Higher frequency means shorter wavelength and greater photon energy ($E = hf$).
Pause, copy the highlighted definition into your book before moving on.
Which of the following correctly orders three regions of the EM spectrum from lowest to highest photon energy?
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Link speed, frequency and wavelength
Use $c = f\lambda$ and $E = hf$, then track what changes when a wave enters a medium.
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Wave Properties and the Speed of Light
+5 XP
The universal constant that connects wavelength and frequency
We just saw that the EM spectrum spans seven regions all travelling at $c$. That raises a question: how do frequency and wavelength relate mathematically, and what determines a photon's energy? This card answers it → the wave equation $c = f\lambda$ and the photon energy formula $E = hf$.
All electromagnetic waves in vacuum obey the fundamental wave equation $c = f\lambda$, where $c = 3.00 \times 10^8$ m/s is the speed of light in vacuum, $f$ is frequency in hertz (Hz), and $\lambda$ is wavelength in metres (m). Since $c$ is constant, wavelength and frequency are inversely proportional: high frequency means short wavelength, and vice versa.
Each photon of electromagnetic radiation carries energy:
Essential Equations
$c = f\lambda$ ($c = 3.00\times10^8$ m/s in vacuum)
These equations reveal why gamma radiation can ionise atoms whereas radiofrequency radiation cannot: a gamma photon carries $\sim10^{15}$ times more energy than a radio photon. Biological risk also depends on absorbed dose and exposure conditions; non-ionising radiation can still cause heating at sufficient intensity.
When EM waves enter a medium such as glass or water, they slow down. The speed in a medium is $v = c/n$, where $n$ is the refractive index. The frequency stays the same, but the wavelength decreases: $\lambda_{\text{medium}} = \lambda_{\text{vacuum}}/n$.
Figure 2, Wave parameters: wavelength is the distance between successive crests; amplitude is the maximum displacement from equilibriumText description
A wave oscillates about a dashed equilibrium line.
The wavelength λ is marked between two successive crests.
The amplitude is marked from the equilibrium line up to a crest: the maximum displacement.
Frequency is the number of cycles per second; for EM waves c = fλ.
Stop & Check
A microwave oven operates at 2.45 GHz. Calculate the wavelength of these microwaves. If these waves enter water ($n = 1.33$), what is their new wavelength and speed?
For all EM waves: $c = f\lambda$ (speed = frequency × wavelength, $c = 3.00 \times 10^8$ m/s). Photon energy $E = hf = hc/\lambda$ where $h = 6.63 \times 10^{-34}$ J·s. In a medium, speed decreases to $v = c/n$, wavelength decreases, but frequency remains unchanged.
Pause, write the highlighted wave equation and photon energy formula into your book before the check below.
A radio station broadcasts at 100 MHz. The wavelength of these waves is _____ m. (Use $c = 3.0 \times 10^8$ m/s)
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Work photon calculations end to end
Follow three worked problems: red light, an X-ray in keV, and photons per second from a 100 W tower.
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Worked Example: Photon Calculations
+5 XP
From wavelength to energy across the spectrum
We just saw that $E = hf$ links frequency to photon energy. That raises a question: how do you apply this to concrete calculations involving wavelengths in different units or energies in electronvolts? This card answers it → a step-by-step approach to photon calculations across the spectrum.
Problem (a), Red light at 650 nm
Calculate the frequency and photon energy of red light with wavelength 650 nm. Express the energy in both joules and electronvolts.
Calculate the wavelength of a photon with energy 2.5 eV. In which region of the electromagnetic spectrum does this photon lie?
Photon calculation method: convert $\lambda$ to metres → $f = c/\lambda$ → $E = hf$ in joules → divide by $1.60 \times 10^{-19}$ for eV. Reverse: eV to J → $f = E/h$ → $\lambda = c/f$. Power emitted ÷ energy per photon = photons per second.
Add the highlighted calculation method to your notes before the check below.
A photon has energy $3.3\times10^{-19}$ J. What is its wavelength? (Use $hc = 1.99\times10^{-25}$ J·m)
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Explain why photon energy decides the effect
Separate ionising from non-ionising regions and match each one to its biological effect.
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Applications and Biological Effects
+5 XP
Why photon energy determines safety, not wave speed
We just saw how to calculate photon energy from wavelength or frequency. That raises a question: since all EM waves travel at $c$, why do some harm living tissue while others do not? This card answers it → it is photon energy $E = hf$, not speed, that determines ionisation and biological effects.
Because all EM waves travel at the same speed, it might seem they should all have the same effect on matter. The key difference is photon energy, $E = hf$. High-frequency photons carry enough energy to ionise atoms (strip electrons away), damaging molecules including DNA. Low-frequency photons deposit only tiny amounts of energy, too little to cause ionisation.
Figure 3, Approximate ionising and non-ionising regions of the EM spectrum. The boundary lies within the ultraviolet region and depends on the ionisation energy of the material; X-rays and gamma rays are ionising.Text description
The regions are listed in order of increasing photon energy: radio, microwave, infrared, visible, ultraviolet, X-ray, gamma.
Radio, microwave, infrared and visible are non-ionising.
An approximate ionisation threshold crosses the ultraviolet region; where exactly depends on the ionisation energy of the material.
X-rays and gamma rays are ionising.
Key biological effects to know:
UV: Damages DNA, causing sunburn and increasing skin cancer risk. Some UV is beneficial (vitamin D synthesis).
X-rays: Penetrate soft tissue; absorbed by dense bone. Ionising, dose must be carefully managed in medical imaging.
Gamma rays: Penetrate almost everything; produced in nuclear reactions. Used in cancer radiotherapy, targeted to tumour cells.
Infrared: Causes heating (molecular vibrations), not ionisation. Prolonged exposure can burn skin.
Radio/microwave: These photons are non-ionising. At sufficient intensity the fields can deposit energy and heat tissue; microwave ovens heat food through dielectric absorption.
HSC Tip: Unit Prefixes
The EM spectrum spans many orders of magnitude. Common traps: confusing nanometres ($10^{-9}$ m) with micrometres ($10^{-6}$ m), or forgetting to convert MHz ($10^6$ Hz) to Hz before using $E = hf$. Always write your answer with the correct unit. Use $hc = 1.99\times10^{-25}$ J·m to speed up calculations when given wavelength.
Ionisation depends on photon energy $E = hf$, not wave speed. X-rays and gamma rays are ionising; sufficiently short-wavelength ultraviolet can also ionise, while the UV boundary is not rigid. Radio, microwave, infrared and visible photons are non-ionising but can still cause harm through heating or photochemical effects depending on dose. Medical uses include X-ray imaging, gamma radiotherapy and UV sterilisation.
Add the highlighted principle to your notes before the check below.
All electromagnetic waves travel at different speeds in vacuum, which explains why gamma rays are more dangerous than radio waves.
When light enters glass, its frequency remains unchanged but its wavelength decreases.
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Apply and synthesise
Run your own spectrum comparisons, then decide what changes when light enters glass.
Activity 1, EM Spectrum Explorer
ApplyBand 3
Verify $c = f\lambda$ and compare photon energies across regions
For radio waves ($\lambda = 1.0$ m): calculate the frequency and photon energy. Verify $c = f\lambda$ by hand.
For visible light ($\lambda = 500$ nm): calculate how many times more energetic a visible photon is than a radio photon at 1.0 m.
For gamma rays ($\lambda = 1.0\times10^{-12}$ m): calculate how many visible photons (500 nm) would equal the energy of one gamma photon.
Explain why gamma photons can ionise tissue while radio photons at the same total power cannot, and state why total dose and heating still matter.
Activity 2, Concept Challenge
UnderstandBand 4
Explain what changes and what stays constant when light enters a medium
A beam of red light ($\lambda = 700$ nm in air) enters a glass block ($n = 1.5$).
Calculate the speed of light in the glass.
Calculate the wavelength of the light inside the glass.
Does the frequency change? Explain why or why not.
A student argues: "Since the light slows down, it must have less energy per photon inside the glass." Is this correct? Explain.
Synthesis, connect the ideas
In this lesson, three ideas lock together:
All EM waves obey $c = f\lambda$, one spectrum, seven regions differing only in wavelength/frequency.
Photon energy $E = hf$ scales with frequency: gamma photons carry $\sim10^{15}\times$ more energy than radio photons.
It is photon energy, not wave speed, that determines biological and physical effects of EM radiation.
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See how Maxwell predicted electromagnetic waves
Trace the theory that gave $c = 1/\sqrt{\varepsilon_0\mu_0}$ and the experiment that confirmed it.
Maxwell's Contribution to Electromagnetism
+5 XP
How theory predicted that light is an electromagnetic wave
James Clerk Maxwell unified electricity and magnetism into one theory and showed, from that theory alone, that light is an electromagnetic wave.
In the 1860s, James Clerk Maxwell unified electricity and magnetism into a single theory of electromagnetism (Maxwell's equations). His equations predicted that a changing electric field produces a changing magnetic field, and vice versa, so the two propagate together as a self-sustaining electromagnetic wave through a vacuum, with no medium required.
Maxwell calculated the speed of these waves as $c = 1/\sqrt{\varepsilon_0\mu_0} \approx 3\times10^8$ m/s, determined only by the electric permittivity ($\varepsilon_0$) and magnetic permeability ($\mu_0$) of the vacuum. This matched the measured speed of light, leading to the conclusion that light is an electromagnetic wave.
Maxwell's work therefore unified optics with electricity and magnetism. His prediction was confirmed experimentally about 20 years later by Heinrich Hertz, who generated and detected radio waves travelling at the speed of light.
Copy the highlighted points (Maxwell's prediction of EM waves and $c = 1/\sqrt{\varepsilon_0\mu_0}$) into your book.
Maxwell calculated the speed of electromagnetic waves and found it equalled the measured speed of _____.
Analyse6 marks
Analyse how Maxwell’s theory of electromagnetism led to the conclusion that light is an electromagnetic wave.
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Unpack the verb
Analyse means take the theory apart and show how each piece produces the next. A list of Maxwell’s achievements is an Outline answer; the marks here are for the chain from the equations to the conclusion.
6 marks. Plan four links in the chain, each worth about a mark and a half: mutual induction → a self-sustaining wave → a predicted speed → that speed matches the measured speed of light.
The question says “led to”. That phrase is asking for a sequence, and a marker awards the joins between the steps as much as the steps themselves.
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Name the governing principle
State the one physical idea the rest of the answer hangs off.
Maxwell’s equations show that a changing electric field produces a magnetic field and a changing magnetic field produces an electric field. Each field therefore regenerates the other, so a disturbance can propagate without any medium to carry it.
This sentence is what makes the answer an analysis rather than a history. Everything that follows is a consequence of it.
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Link 1 — from mutual induction to a travelling wave
Because the two fields regenerate each other, an oscillating charge produces perpendicular electric and magnetic fields that are in phase and mutually perpendicular, and this pattern travels outwards as a transverse wave. No mechanical medium is required, since the fields themselves are the disturbance.
Naming the geometry (perpendicular, in phase, transverse) is cheap and it is on the criteria sheet. Students who only say “the fields make a wave” lose the mark here.
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Link 2 — the theory predicts a speed, not just a wave
Maxwell’s equations give the speed of that wave as $c = 1/\sqrt{\varepsilon_0\mu_0}$, where $\varepsilon_0$ is the permittivity and $\mu_0$ the permeability of free space. Both are electrical and magnetic constants measured in laboratory experiments that had nothing to do with light.
This is the decisive point and it is the one most often left out. The constants came from capacitors and coils, so the speed was an independent prediction rather than a number fitted to optical data.
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Link 3 — the numerical coincidence that closes the argument
Substituting the measured constants gives approximately $3\times10^{8}$ m s$^{-1}$, which agreed with the speed of light already measured by Fizeau and Foucault. Since electromagnetic waves and light travel at the same speed in a vacuum, Maxwell concluded that light is an electromagnetic wave.
Quote the agreement as agreement between two separately obtained numbers. That is what makes it evidence and not a definition.
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Link back to the question
One closing sentence, in the question’s own words.
The theory therefore led to the conclusion by prediction rather than by observation: electromagnetism required a wave of a particular speed to exist, and light was found to be a wave of exactly that speed.
Markers work from criteria that repeat the question’s wording. Echoing “led to the conclusion” makes the final mark easy to award.
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What a full-mark answer looks like
Maxwell’s equations show that a changing electric field produces a magnetic field and a changing magnetic field produces an electric field, so each field regenerates the other and a disturbance can propagate with no medium to carry it. An oscillating charge therefore produces perpendicular electric and magnetic fields, in phase with each other, that travel outwards as a transverse wave. The equations also fix the speed of that wave as $c = 1/\sqrt{\varepsilon_0\mu_0}$, and both constants had been measured in purely electrical and magnetic experiments, so this was an independent prediction. Substituting them gives approximately $3\times10^{8}$ m s$^{-1}$, which matched the speed of light measured optically by Fizeau and Foucault. Since electromagnetic waves and light share the same vacuum speed, Maxwell concluded that light is an electromagnetic wave. The theory therefore led to the conclusion by prediction rather than observation: electromagnetism demanded a wave of a particular speed, and light turned out to be a wave of exactly that speed.
One principle, three links, one close. Every clause carries either a mechanism or a join, which is what an Analyse criteria sheet rewards.
Assess5 marks
Assess the significance of Hertz’s experiments for the acceptance of Maxwell’s electromagnetic theory.
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Unpack the verb
Assess means make a judgement about value or importance and support it. Describing what Hertz did earns almost nothing on its own; the marks are for weighing what his results settled and what they did not.
5 marks. Plan a judgement, two pieces of evidence that support it, and one honest limitation, then restate the judgement.
An Assess answer that never states a verdict cannot reach the top band no matter how much correct physics it contains.
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State the judgement first
Hertz’s experiments were decisive: they converted Maxwell’s theory from a mathematical prediction into an experimentally confirmed description of nature.
Putting the verdict in sentence one means every later paragraph reads as support for it. Markers can then find the judgement without hunting.
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Evidence 1 — he produced and detected the predicted waves
Using a spark gap driven by an induction coil, Hertz generated oscillating currents and detected a spark induced across a separate loop several metres away. This demonstrated that an accelerating charge radiates energy through space with no connecting medium, exactly as Maxwell had predicted.
Existence alone is the first half of the case. Say what the apparatus actually did — a named mechanism beats “he made radio waves”.
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Evidence 2 — he showed the waves behaved like light
Hertz then measured the wavelength from a standing wave pattern and, with the known frequency of his oscillator, calculated a speed close to $3\times10^{8}$ m s$^{-1}$. He also showed the waves could be reflected, refracted and polarised. The predicted speed and the optical behaviour together confirmed that these waves and light are the same phenomenon.
This is the part that does the real work. Producing an invisible wave proves the wave exists; measuring its speed and its polarisation proves it is light.
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The limitation that keeps the judgement honest
The confirmation was of the classical theory only, and it was not complete. Hertz himself noticed that ultraviolet light striking the spark gap made sparking easier — the photoelectric effect, which the wave theory he had just confirmed could not explain.
The best Assess answers name the boundary of what was established. This detail is also true and memorable: the same experiment contained the first hint of the theory’s eventual limits.
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Restate the judgement against the evidence
On balance the significance is high: Hertz supplied the only kind of evidence that could settle the question, an independent measurement of the predicted speed together with light-like behaviour, and Maxwell’s theory was accepted as a result — while the same apparatus quietly exposed the anomaly that would later limit it.
The closing sentence must weigh, not summarise. “Significance is high, because the evidence was of the decisive kind” is a judgement; “Hertz was very important” is not.
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What a full-mark answer looks like
Hertz’s experiments were decisive, converting Maxwell’s theory from a mathematical prediction into an experimentally confirmed description of nature. Using a spark gap driven by an induction coil, he generated oscillating currents and detected a spark induced across a loop metres away, showing that an accelerating charge radiates energy through space with no connecting medium, as Maxwell had predicted. He then measured the wavelength from a standing wave pattern and, with the known oscillator frequency, obtained a speed close to $3\times10^{8}$ m s$^{-1}$, and showed the waves could be reflected, refracted and polarised — establishing not merely that the waves existed but that they were light. The confirmation was of the classical theory only: Hertz also observed that ultraviolet light made the spark gap discharge more readily, the photoelectric effect, which that theory could not explain. On balance the significance is high, because Hertz supplied the one kind of evidence that could settle the question, while the same apparatus exposed the anomaly that would later limit the theory he had confirmed.
Judgement, two supports, one limitation, judgement restated. Five marks, five moves.
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Independent practice
Show what you can do without prompts
Answer a shuffled question-bank set, then write two HSC-style responses on the spectrum and photon energy.
Quick recall, the electromagnetic spectrum
+5 XP
A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
Short Answer
+5 XP
ApplyBand 4(3 marks) 1. A laser pointer emits red light at 650 nm with a power of 3.0 mW. (a) Calculate the energy of each photon in joules. (b) Calculate the number of photons emitted per second. (c) If a blue laser emitted the same power at 450 nm, would it emit more or fewer photons per second? Explain.
1 mark: correct photon energy · 1 mark: correct photons/s · 1 mark: correct comparison with explanation
AnalyseBand 5(5 marks) 2. (a) Explain why all electromagnetic waves travel at the same speed in a vacuum. (b) When visible light enters glass ($n = 1.5$), describe what happens to its speed, wavelength, and frequency. (c) A student argues that since frequency is unchanged in glass, the photon energy is unchanged too. Evaluate this claim. (d) Explain using $E = hf$ why gamma rays are ionising but radio waves are not.
1 mark each: (a) Maxwell's equations prediction, (b) all three quantities correct, (c) evaluation, (d) energy comparison
Show all answers
Multiple choice
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.
Short Answer, Model Answers
Q1 (3 marks): (a) $f = c/\lambda = 3.00\times10^8 / 650\times10^{-9} = 4.62\times10^{14}$ Hz. $E = hf = 6.63\times10^{-34} \times 4.62\times10^{14} = 3.06\times10^{-19}$ J (1 mark). (b) $N/s = P/E = 3.0\times10^{-3} / 3.06\times10^{-19} = 9.8\times10^{15}$ photons/s (1 mark). (c) Blue photons at 450 nm have higher frequency and therefore more energy per photon ($E = hf$). The same power divided by a larger energy per photon gives fewer photons per second for blue light (1 mark).
Q2 (5 marks): (a) Maxwell's equations predicted self-propagating EM waves with speed $c = 1/\sqrt{\varepsilon_0\mu_0}$, a constant determined by the properties of the vacuum, independent of frequency (1 mark). (b) Speed decreases: $v = c/n = 3.00\times10^8/1.5 = 2.0\times10^8$ m/s. Wavelength decreases: $\lambda' = \lambda/n = 700/1.5 = 467$ nm. Frequency is unchanged (1 mark). (c) The claim is correct: $E = hf$, frequency is unchanged in the medium, so photon energy is unchanged. What changes is the spatial extent of the wave (wavelength and speed), not the energy of individual photons (1 mark). (d) A gamma photon ($f \sim 3\times10^{20}$ Hz) has energy $E = hf \approx 2\times10^{-13}$ J, enough to ionise atoms (break chemical bonds). A radio photon ($f \sim 10^6$ Hz) has energy $\approx 7\times10^{-28}$ J, far too small to ionise any atom (1 mark).
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Final step
Retrieve, reflect and finish
Check what stuck, revisit your opening predictions and record the rule you will reuse.
Check what actually stuck
How did your thinking change?
At the start you compared a representative 66 cm radio wave ($f\approx454$ MHz) with a gamma ray ($\lambda \sim 10^{-12}$ m).
The answers: (1) The gamma ray has far higher frequency, since $c = f\lambda$ and both travel at $c$, the shorter-wavelength gamma ray must have higher frequency. (2) For the stated 66 cm radio wavelength, $E = hc/\lambda \approx 3.0\times10^{-25}$ J; for $\lambda = 10^{-12}$ m, the gamma photon has $E \approx 2.0\times10^{-13}$ J, about $6.6\times10^{11}$ times greater. (3) Same speed does not mean same interaction: photon energy determines whether a single photon can ionise, while absorbed dose and intensity determine the scale of heating and other biological effects. Hertz's work experimentally confirmed Maxwell's electromagnetic-wave prediction.