Electromagnetic Spectrum: How to Match Waves to Applications

A microwave oven heats yesterday’s rice, a visible-light camera cannot see through your hand, and an X-ray can reveal a bone. All three use electromagnetic waves. Why do they behave so differently?

Before reading on, predict this: does a microwave photon carry more energy than a visible-light photon because a microwave oven can heat food? No. A visible-light photon carries far more energy than a microwave photon. The oven heats food because electromagnetic energy is absorbed by the food and transferred into thermal motion, with enormous numbers of low-energy photons involved.

The useful wave is not automatically the one with the highest energy. You need to match the wave’s properties to the job, whether that job is transmitting information, detecting heat, producing an image, or changing a material.

01The quick decision rule

When a question asks which electromagnetic wave is suitable for an application, use this sequence.

1. Identify the job. Is the aim to communicate, heat something, detect radiation, see inside an object, or cause a chemical change?
2. Identify the useful interaction. Does the wave need to be absorbed, transmitted, reflected, or detected?
3. Match the properties. Consider wavelength, frequency, photon energy, penetration, and the material involved.
4. Check the limitations. Think about signal loss, image contrast, exposure time, and radiation safety.

As a starting point:

• Long-distance broadcasting and communication around obstacles often use radio waves.
• Wi-Fi, radar, satellite links, and microwave heating use microwaves.
• Thermal imaging and many remote controls use infrared.
• Cameras, human vision, and optical microscopes use visible light.
• Fluorescence, some industrial curing processes, and surface disinfection use suitable ultraviolet wavelengths.
• Medical radiography and CT use X-rays.
• Some cancer treatments and sterilisation processes use gamma rays.

These are starting choices, not universal rules. A wave only does its job if it interacts appropriately with the target and the equipment can detect the result.

01Read the electromagnetic spectrum quantitatively

The relationship between wavelength and frequency

Imagine watching waves pass a fixed point. If the crests are close together, more crests pass each second. If they are far apart, fewer pass each second.

Predict which has the higher frequency: a wave with a wavelength of 1 metre or one with a wavelength of 1 centimetre, assuming both travel through a vacuum.

The 1-centimetre wave has the higher frequency. Its wavelength is 100 times smaller, so 100 times as many cycles pass a fixed point each second.

Electromagnetic waves all travel at the same speed in a vacuum, regardless of their frequency. That speed is approximately \(3.00\times10^8\ \mathrm{m\,s^{-1}}\).

The relationship is

\[ c=f\lambda \]

where:

• \(c\) is the speed of light in a vacuum, measured in metres per second (\(\mathrm{m\,s^{-1}}\)).
• \(f\) is frequency, measured in hertz (\(\mathrm{Hz}\)).
• \(\lambda\) is wavelength, measured in metres (\(\mathrm{m}\)).

Rearranging gives

\[ f=\frac{c}{\lambda} \qquad \lambda=\frac{c}{f} \]

For electromagnetic waves in a vacuum, wavelength and frequency are inversely proportional. A shorter wavelength means a higher frequency.

The relationship between frequency and photon energy

Electromagnetic radiation transfers energy in discrete packets called photons. The energy of an individual photon depends on its frequency.

\[ E=hf \]

Combining this with \(c=f\lambda\) gives

\[ E=\frac{hc}{\lambda} \]

where:

• \(E\) is the energy of one photon, measured in joules (\(\mathrm{J}\)).
• \(h\) is Planck’s constant, \(6.626\times10^{-34}\ \mathrm{J\,s}\).
• \(f\) is the photon’s frequency in hertz.
• \(c\) and \(\lambda\) have the meanings defined above.

This gives you a useful chain of reasoning:

• Shorter wavelength means higher frequency.
• Higher frequency means greater energy per photon.
• Longer wavelength means lower frequency and less energy per photon.

For example, a blue-light photon has more energy than a red-light photon. A visible-light photon also has much more energy than a typical microwave photon.

Photon energies are often expressed in electronvolts (\(\mathrm{eV}\)), particularly when comparing different parts of the spectrum.

\[ 1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J} \]

Photon energy is not the same as beam power

This distinction explains the microwave-oven puzzle.

Photon energy tells you how much energy an individual photon carries. Power tells you how much energy a source transfers each second.

A beam containing many low-energy photons can deliver substantial power. A beam containing fewer high-energy photons can deliver the same power.

For a monochromatic beam, meaning one with a single photon energy,

\[ P=\dot{N}E \]

where \(P\) is the beam power in watts (\(\mathrm{W}\)), and \(\dot{N}\) is the number of photons arriving each second, measured in photons per second.

Therefore,

\[ \dot{N}=\frac{P}{E} \]

At equal power, the beam with lower-energy photons contains more photons per second.

Heating depends on how much energy the material absorbs and how that energy is transferred. It cannot be predicted from photon energy alone.

What happens when a wave enters a material?

There is one important qualification to the wavelength equation.

The equation \(c=f\lambda\) applies in a vacuum. In a material, the wave travels at a speed \(v\), so

\[ v=f\lambda \]

For a material with refractive index \(n\),

\[ v=\frac{c}{n} \]

If the wave enters the material without changing frequency, its wavelength decreases because its speed decreases. Its photon energy does not change, because \(E=hf\) and the frequency remains constant.

Electromagnetic spectrum bands are normally classified using their vacuum wavelengths or frequencies, not the wavelength measured inside glass or another material.

02Compare the seven regions of the spectrum

The boundaries below are approximate. The electromagnetic spectrum is continuous, and different references may use slightly different boundaries.

The photon energies are also approximate. They describe the energy of a single photon, not the total energy delivered by a beam.

The values follow from \(c=f\lambda\) and \(E=hf\), using \(1\ \mathrm{nm}=10^{-9}\ \mathrm{m}\).

Notice the overall trend: wavelength decreases as you move down the table, while frequency and photon energy increase.

Radio waves and microwaves are sometimes grouped together, since microwaves are commonly treated as a subregion of the radio spectrum. X-rays and gamma rays can also overlap in energy. In many contexts, their distinction depends partly on how the radiation is produced, not just on its wavelength.

03Choose the wave that matches the job

Radio waves: broadcasting and communication

Radio waves are useful for transmitting information over large areas. They are used in radio broadcasting and a range of communication systems.

Their relatively long wavelengths allow them to diffract around obstacles more effectively than shorter-wavelength visible light. Some lower-frequency radio waves can also be returned by the ionosphere under suitable conditions, extending communication range beyond the direct line of sight.

The trade-off is that longer wavelengths generally require larger antennas for comparable directional performance. The available bandwidth, antenna design, transmission power, and surrounding environment also affect communication quality.

Choose radio waves when: the communication system benefits from broad coverage, diffraction around obstacles, or the propagation properties of a particular radio-frequency band.

Microwaves: wireless links, radar, and heating

Microwaves are used in Wi-Fi, mobile communication, radar, satellite links, and microwave ovens.

For communication, their relatively short wavelengths allow compact antennas and directional beams. The available frequency range can also support substantial data bandwidth. However, walls, rain, and other materials can attenuate microwave signals, so a higher frequency does not automatically mean a longer or more reliable link.

Microwave ovens use an oscillating electromagnetic field to transfer energy to food. Polar molecules, such as water molecules, respond to the field, and dissolved ions also contribute to energy absorption. The absorbed energy is converted into thermal motion.

The heating effect depends on the food’s composition, the field distribution, the absorbed power, and the depth to which the radiation penetrates. It is not because individual microwave photons have unusually high energy.

Choose microwaves when: you need a suitable frequency for a wireless link, radar system, or material-heating process. Check the target material and operating conditions rather than assuming all microwaves penetrate or heat equally.

Infrared: thermal imaging, remote controls, and fibre optics

Infrared radiation sits between microwaves and visible light in the spectrum.

Warm objects emit electromagnetic radiation, often with substantial emission in the infrared region at ordinary temperatures. A thermal camera detects infrared radiation and uses its distribution to construct an image.

This is useful for locating hot components, identifying temperature patterns, and monitoring equipment without touching it. However, a thermal image is not a perfect temperature map: emissivity, reflected infrared radiation, and the camera’s calibration can affect the reading.

Infrared also appears in two other important applications:

• Remote controls transmit coded pulses of near-infrared radiation. The receiver detects the pattern and converts it into a command.
• Optical communication systems often use near-infrared light. Certain wavelengths, including those around 1.3 and 1.55 micrometres, experience relatively low attenuation in silica optical fibres.

Infrared heaters work when the radiation is absorbed and converted into internal energy. How effectively a surface heats depends on its absorptivity, the wavelength, and the incident power.

Visible light: vision, cameras, and microscopes

Visible light is the narrow region of the spectrum detected by the human eye. Ordinary cameras also use visible light to record images.

The key is that light reflected or transmitted by an object reaches a detector in different amounts and colours. Those differences allow the detector to reconstruct the object’s appearance.

Visible light has shorter wavelengths than infrared, which can help optical systems resolve finer details when other conditions, such as the aperture and optical quality, are comparable. Shorter wavelength alone does not guarantee a sharper image, however. Resolution also depends on the lens, detector, contrast, and geometry.

Visible light is useful for examining surfaces, viewing cells with optical microscopes, and recording scenes under suitable illumination. It cannot normally form a direct image of a bone through an intact hand because skin and other tissues strongly scatter and absorb visible light.

Choose visible light when: you need an image from reflected or transmitted light and the object is accessible to that light. Consider infrared when detecting thermal emission or using a suitable optical communication system.

Ultraviolet: fluorescence, curing, and disinfection

Ultraviolet (UV) radiation has shorter wavelengths and higher photon energies than visible light.

Its photons can excite electrons in atoms and molecules, allowing UV radiation to trigger fluorescence or photochemical reactions. Some UV wavelengths are therefore useful for curing specialised materials, revealing fluorescent markings, and disinfecting exposed surfaces.

UV-C is widely used in germicidal applications because it can damage the genetic material of microorganisms. But UV lamps must be selected and operated appropriately. Germicidal effectiveness depends on wavelength, exposure, and the amount of radiation that reaches the target.

A surface that looks exposed may still contain shaded regions where the radiation cannot reach. Dirt and other materials can also block or absorb UV. A UV treatment is therefore not automatically effective just because a lamp is switched on.

UV radiation can damage skin and eyes. The shortest-wavelength UV can also ionise matter, so do not treat the entire ultraviolet band as having one universal interaction or safety threshold.

Choose UV when: a suitable wavelength can trigger the required fluorescence, curing reaction, or biological effect. Check exposure, material compatibility, shadowing, and safety.

X-rays: medical imaging and internal inspection

X-rays have high-energy photons and can pass through many materials that strongly absorb visible light.

In medical radiography, an X-ray beam passes through the body and reaches a detector. Different tissues attenuate the beam by different amounts. Attenuation means a reduction in the intensity of a beam as it passes through matter, due to absorption and scattering.

Bone generally attenuates more X-rays than surrounding soft tissue. Consequently, fewer X-ray photons reach the detector through bone than through soft tissue. On a conventional radiograph, bone therefore appears lighter.

This difference in attenuation creates image contrast. CT scanners use X-ray measurements from multiple directions to reconstruct cross-sectional images of the body.

X-rays are also useful for inspecting luggage and materials, and for studying crystal structures. In all these cases, the application relies on how the material interacts with the radiation, not simply on the fact that X-rays have high photon energy.

X-rays are ionising radiation. Their photons can remove electrons from atoms and molecules, potentially damaging cells and DNA. Medical imaging therefore requires appropriate exposure control.

Choose X-rays when: you need to inspect internal structure and the differences in attenuation between materials can produce useful contrast. Greater penetration is helpful, but maximum penetration is not automatically the best choice for an image.

Gamma rays: radiotherapy and sterilisation

Gamma rays occupy the high-energy end of the electromagnetic spectrum in the conventional school classification. They are associated particularly with nuclear transitions and other nuclear or particle processes.

Their high photon energies make gamma rays strongly ionising, and many gamma-ray applications exploit their ability to penetrate matter.

Examples include:

• Certain forms of cancer radiotherapy, in which radiation is used to damage cancer cells.
• Sterilisation of suitable medical equipment, where radiation can damage microorganisms even in locations that surface-only treatments cannot reach.
• Nuclear medicine applications that detect gamma photons emitted by radioactive tracers or produced by particle interactions.

The same penetrating and ionising properties that make gamma rays useful also create risks. Exposure must be controlled, and appropriate shielding and handling procedures are essential.

The distinction between X-rays and gamma rays is not always a simple energy boundary. Their production mechanism often helps distinguish them.

Choose gamma rays when: a controlled, high-energy ionising-radiation process is appropriate. Their high energy does not make them a universal replacement for X-rays, UV, or visible light.

05Worked examples

06Common mistakes

• Using nanometres directly in \(f=c/\lambda\). The equation requires consistent units. Convert nanometres to metres before substituting.
• Assuming the shortest wavelength is always the best choice. High photon energy can improve penetration in some situations, but an image needs contrast, a communication link needs a detectable signal, and a heating process needs effective energy absorption.
• Confusing photon energy with beam power. A low-energy photon does not prevent a beam from delivering substantial power. Power depends on the energy transferred each second.
• Assuming a wave gains photon energy when it enters glass. Its wavelength decreases because its speed decreases. In ordinary refraction, its frequency and photon energy remain unchanged.
• Treating the boundaries between regions as exact. The ranges are approximate, and some classifications, especially the distinction between X-rays and gamma rays, depend on how the radiation is produced.
• Assuming every UV wavelength has the same effect. Fluorescence, curing, disinfection, and ionisation depend on the wavelength and the target material, not merely on the label ultraviolet.

07Questions and solutions

Question 1

A television remote control emits infrared radiation with a wavelength of \(940\ \mathrm{nm}\).

Calculate the frequency of the radiation and identify its region of the electromagnetic spectrum.

Solution 1

The frequency is \(3.19\times10^{14}\ \mathrm{Hz}\), which places the radiation in the infrared region.

Convert the wavelength to metres:

\[ \lambda=940\times10^{-9}\ \mathrm{m} =9.40\times10^{-7}\ \mathrm{m} \]

Use \(f=c/\lambda\):

\[ \begin{aligned} f&=\frac{3.00\times10^8} {9.40\times10^{-7}}\\ &=3.19\times10^{14}\ \mathrm{Hz} \end{aligned} \]

The calculated frequency falls within the approximate infrared range of \(3\times10^{11}\) to \(4.3\times10^{14}\ \mathrm{Hz}\).

This is why the remote can use infrared pulses to transmit information to a compatible receiver.

Question 2

A Wi-Fi link operates at \(2.40\ \mathrm{GHz}\). Treat the air as a vacuum for this calculation.

(a) Calculate the wavelength of the signal.

(b) Calculate the energy of one microwave photon in joules and electronvolts.

(c) A green-light photon has a wavelength of \(500\ \mathrm{nm}\). Calculate how many times greater its energy is than the energy of one Wi-Fi photon.

(d) Explain why the low energy of a microwave photon does not prevent a microwave oven from heating food.

Solution 2

The Wi-Fi wavelength is \(0.125\ \mathrm{m}\), and a green-light photon carries approximately \(2.50\times10^5\) times as much energy as a Wi-Fi photon.

(a) Calculate the wavelength.

Convert the frequency into hertz:

\[ f=2.40\times10^9\ \mathrm{Hz} \]

Use \(\lambda=c/f\):

\[ \begin{aligned} \lambda&=\frac{3.00\times10^8} {2.40\times10^9}\\ &=0.125\ \mathrm{m} \end{aligned} \]

The wavelength is \(0.125\ \mathrm{m}\), or \(12.5\ \mathrm{cm}\). This is in the microwave region.

(b) Calculate the microwave photon energy.

Use \(E=hf\):

\[ \begin{aligned} E&=(6.626\times10^{-34}) (2.40\times10^9)\\ &=1.59\times10^{-24}\ \mathrm{J} \end{aligned} \]

Convert to electronvolts:

\[ \begin{aligned} E&=\frac{1.59\times10^{-24}} {1.602\times10^{-19}}\\ &=9.93\times10^{-6}\ \mathrm{eV} \end{aligned} \]

The microwave photon energy is approximately \(9.93\times10^{-6}\ \mathrm{eV}\).

(c) Compare the photon energies.

First, convert the green-light wavelength:

\[ \lambda_{\text{green}}=5.00\times10^{-7}\ \mathrm{m} \]

Calculate its photon energy:

\[ \begin{aligned} E_{\text{green}}&=\frac{hc}{\lambda}\\ &=\frac{(6.626\times10^{-34}) (3.00\times10^8)} {5.00\times10^{-7}}\\ &=3.98\times10^{-19}\ \mathrm{J} \end{aligned} \]

In electronvolts, this is approximately \(2.48\ \mathrm{eV}\).

Now compare the energies:

\[ \begin{aligned} \frac{E_{\text{green}}}{E_{\text{microwave}}} &=\frac{3.98\times10^{-19}} {1.59\times10^{-24}}\\ &\approx2.50\times10^5 \end{aligned} \]

One green-light photon carries about 250,000 times the energy of one Wi-Fi photon.

(d) Explain the heating effect.

A microwave oven can still heat food because the energy delivered depends on the total energy absorbed, not the energy of one photon in isolation.

The oscillating electromagnetic field transfers energy to polar molecules and ions in the food. The absorbed energy becomes thermal energy. A large number of low-energy photons can therefore deliver substantial energy over time.

The trap is to confuse energy per photon with the total power absorbed by a material.

Question 3

Two monochromatic ultraviolet sources illuminate the same area. Each delivers an optical power of \(2.0\ \mathrm{mW}\) at the target.

• Source A has a wavelength of \(254\ \mathrm{nm}\).
• Source B has a wavelength of \(365\ \mathrm{nm}\).

Assume both sources deliver their stated power continuously, with no additional wavelengths present.

(a) Calculate the energy of a photon from each source in joules and electronvolts.

(b) Determine which source delivers more photons per second and calculate the ratio of their photon rates.

(c) Can you conclude that the source delivering more photons per second will be more effective at sterilising a surface? Explain.

Solution 3

Source B delivers approximately 1.44 times as many photons per second as Source A, but the photon count alone does not establish which source will be more effective at sterilisation.

(a) Calculate the photon energies.

For Source A, convert the wavelength:

\[ \lambda_A=254\times10^{-9}\ \mathrm{m} \]

Use \(E=hc/\lambda\):

\[ \begin{aligned} E_A&=\frac{(6.626\times10^{-34}) (3.00\times10^8)} {254\times10^{-9}}\\ &=7.83\times10^{-19}\ \mathrm{J} \end{aligned} \]

Converting to electronvolts gives

\[ E_A=\frac{7.83\times10^{-19}} {1.602\times10^{-19}} =4.89\ \mathrm{eV} \]

For Source B:

\[ \lambda_B=365\times10^{-9}\ \mathrm{m} \]

\[ \begin{aligned} E_B&=\frac{(6.626\times10^{-34}) (3.00\times10^8)} {365\times10^{-9}}\\ &=5.45\times10^{-19}\ \mathrm{J} \end{aligned} \]

Therefore,

\[ E_B=\frac{5.45\times10^{-19}} {1.602\times10^{-19}} =3.40\ \mathrm{eV} \]

Source A has the greater photon energy because it has the shorter wavelength.

(b) Compare the photon rates.

For a monochromatic beam,

\[ \dot{N}=\frac{P}{E} \]

Both sources deliver the same power, so the ratio of their photon rates is inversely proportional to their photon energies:

\[ \begin{aligned} \frac{\dot{N}_B}{\dot{N}_A} &=\frac{E_A}{E_B}\\ &=\frac{365}{254}\\ &\approx1.44 \end{aligned} \]

Source B delivers about 1.44 times as many photons per second.

The result makes sense: each Source B photon carries less energy, so more photons are needed to deliver the same total energy each second.

(c) Evaluate the sterilisation claim.

No. A greater photon rate does not automatically mean a greater germicidal effect.

The biological response depends on the wavelength and how the radiation interacts with the microorganism. A wavelength near \(254\ \mathrm{nm}\), in the UV-C region, is commonly used for germicidal applications because it can damage microbial genetic material. Radiation at \(365\ \mathrm{nm}\), in the UVA region, does not have the same germicidal effectiveness under otherwise comparable conditions.

Exposure time, distance, shadowing, and the cleanliness of the surface also matter.

The important distinction is between the number of photons delivered and the effect each photon can produce in the target.

Question 4

A conventional X-ray radiograph of a wrist shows the bones as pale regions against darker surrounding tissue.

A student argues that gamma rays should replace X-rays because gamma-ray photons are more energetic, penetrate further, and must therefore produce a sharper image.

Evaluate this argument. Explain how the radiograph forms, why bone appears pale, why greater photon energy does not guarantee better contrast, and one important safety consideration.

Solution 4

The argument is incorrect: greater penetration does not automatically produce a sharper or more useful image.

In a conventional radiograph, an X-ray beam passes through the wrist and reaches a detector. Bone attenuates more of the beam than the surrounding soft tissue, so fewer photons reach the detector behind the bone.

The detector therefore records a lower exposure in that region. Under the conventional image display, the bone appears pale.

The image depends on differential attenuation: the difference in how strongly the different tissues reduce the beam.

Increasing photon energy can improve penetration. However, over a relevant diagnostic energy range, the difference in attenuation between bone and soft tissue can become smaller as photon energy increases. If the two tissues transmit more similar fractions of the beam, their contrast can decrease.

Image sharpness also depends on the detector, the geometry of the imaging system, and other factors. Photon energy alone cannot determine the final image quality.

Gamma rays are not automatically a better substitute. They are also ionising radiation, and their use must be matched to the application.

An important safety consideration is that X-rays and gamma rays can damage cells and DNA. The exposure must be controlled to obtain useful information while limiting unnecessary radiation dose.

Question 5

An optical-fibre communication system uses a monochromatic laser with a wavelength of \(600\ \mathrm{nm}\) in air. Its input power is \(1.00\ \mathrm{mW}\).

The fibre has a refractive index of \(1.50\), and the output power measured in air is \(0.600\ \mathrm{mW}\). Assume that the fibre causes absorption and scattering losses but does not change the frequency of the transmitted photons.

A student makes two claims:

• Claim A: The wavelength becomes \(400\ \mathrm{nm}\) inside the fibre, so the photons gain energy and the light changes from orange to violet.
• Claim B: The output power is 40% lower, so each photon that emerges has 40% less energy.

Calculate the wavelength inside the fibre, the frequency and photon energy, and the ratio of the output photon rate to the input photon rate. Then evaluate both claims.

Solution 5

Neither claim is correct. The wavelength decreases to \(400\ \mathrm{nm}\) inside the glass, but the frequency and energy of each photon remain unchanged. The output contains 60% of the input photon rate.

Step 1: Calculate the frequency in air.

The wavelength in air is approximately the vacuum wavelength:

\[ \lambda_0=600\times10^{-9}\ \mathrm{m} \]

Using \(f=c/\lambda_0\):

\[ \begin{aligned} f&=\frac{3.00\times10^8} {600\times10^{-9}}\\ &=5.00\times10^{14}\ \mathrm{Hz} \end{aligned} \]

The frequency is \(5.00\times10^{14}\ \mathrm{Hz}\).

Step 2: Calculate the wavelength inside the fibre.

For refractive index \(n=1.50\),

\[ \lambda_{\text{glass}}=\frac{\lambda_0}{n} \]

Therefore,

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{600\ \mathrm{nm}}{1.50}\\ &=400\ \mathrm{nm} \end{aligned} \]

The wavelength inside the glass is \(400\ \mathrm{nm}\).

This does not mean the photon has gained energy. The wave travels more slowly in the glass, so the wavelength becomes shorter while the frequency remains the same.

Step 3: Calculate the photon energy.

Use the unchanged frequency:

\[ \begin{aligned} E&=hf\\ &=(6.626\times10^{-34}) (5.00\times10^{14})\\ &=3.31\times10^{-19}\ \mathrm{J} \end{aligned} \]

Converting to electronvolts:

\[ E=\frac{3.31\times10^{-19}} {1.602\times10^{-19}} =2.07\ \mathrm{eV} \]

The photon energy remains approximately \(2.07\ \mathrm{eV}\).

Claim A fails because it assumes that a shorter wavelength must mean a higher frequency even when the wave has entered a material. That inference would only follow if the wave speed remained constant.

The light has not been converted into violet or ultraviolet radiation. Spectrum classification uses the vacuum wavelength or frequency. When the light exits the fibre into air, its wavelength returns to approximately \(600\ \mathrm{nm}\), corresponding to orange visible light.

Step 4: Calculate the input and output photon rates.

The photon rate is the power divided by the energy per photon:

\[ \dot{N}=\frac{P}{E} \]

For the input:

\[ \begin{aligned} \dot{N}_{\text{in}} &=\frac{1.00\times10^{-3}} {3.31\times10^{-19}}\\ &=3.02\times10^{15}\ \mathrm{s^{-1}} \end{aligned} \]

For the output:

\[ \begin{aligned} \dot{N}_{\text{out}} &=\frac{0.600\times10^{-3}} {3.31\times10^{-19}}\\ &=1.81\times10^{15}\ \mathrm{s^{-1}} \end{aligned} \]

Now compare the rates:

\[ \begin{aligned} \frac{\dot{N}_{\text{out}}} {\dot{N}_{\text{in}}} &=\frac{1.81\times10^{15}} {3.02\times10^{15}}\\ &=0.600 \end{aligned} \]

The output photon rate is 60% of the input rate, meaning 40% fewer photons emerge each second.

Claim B fails because a reduction in power does not necessarily mean a reduction in the energy of each photon. In this situation, the photons that emerge retain their original energy. The reduction in power is explained by the loss of photons through absorption and scattering.

This example brings together three separate ideas: wavelength depends on wave speed and frequency, photon energy depends on frequency, and beam power depends on the energy transferred each second.

08Next step: Predict what happens when waves meet matter

Knowing the spectrum helps you choose a starting point, but the next decision is what happens when the wave reaches its target.

Practise distinguishing reflection, transmission, absorption, and scattering. These interactions explain why a microwave signal weakens through a wall, why infrared reveals a hot component, why bone appears pale on a radiograph, and why two UV sources with equal power can produce different biological effects.

That is the step from memorising the spectrum to predicting what an actual device will do.