Why Visible Light Is Only a Small Part of the Spectrum

A prism spreads sunlight into a rainbow. Now place a thermometer just beyond the red end, where no coloured light appears. What should happen? If the rainbow contains all the light there is, the thermometer should stay close to its starting temperature.

In 1800, William Herschel discovered that the thermometer beyond red actually warmed more than the thermometers placed in the visible colours. There was radiation reaching the thermometer that human eyes could not see.

This observation reveals the answer to our title: visible light is only a small part of the electromagnetic spectrum because human vision detects only a narrow range of electromagnetic wavelengths. Radiation exists at much longer wavelengths, such as radio waves and infrared, and much shorter wavelengths, such as ultraviolet, X-rays, and gamma rays.

The electromagnetic spectrum does not stop at the edges of a rainbow. Those edges mark the limits of ordinary human vision, not the limits of electromagnetic radiation.

01The experiment that discovered light beyond red

Herschel’s unexpected result

William Herschel was interested in how different colours of sunlight transferred heat. He wanted to investigate whether the red, green, and violet parts of a rainbow produced the same heating effect.

To investigate this, he arranged a simple experiment.

Sunlight entered through a narrow opening and passed through a glass prism. The prism separated the sunlight into its different colours, producing a visible spectrum on a nearby surface.

A card with a narrow slit allowed Herschel to select a particular colour. He positioned a thermometer bulb in the selected band and compared its temperature with another thermometer kept nearby in the shade.

He repeated the measurements for different colours, including violet, green, and red. He then investigated the region just beyond the red end of the spectrum.

Before revealing the result, consider what the experiment should show if visible light is the only radiation present.

The thermometer beyond red should behave like the shaded control. There should be no additional heating because no visible colour reaches it.

But that prediction failed.

Herschel found that the thermometer placed just beyond red experienced a particularly large temperature increase. The heating effect was greater than in the visible colour bands.

He had detected radiation outside the visible spectrum.

We now call this radiation infrared. The word literally means below red, referring to its position beyond the red end of the visible spectrum in wavelength.

The American Physical Society provides a useful historical account of Herschel’s investigation of invisible light.

What did Herschel actually prove?

It is important to separate the observation from its modern interpretation.

Herschel directly observed a temperature increase. His thermometer was a detector of heating, not a device that could identify the exact nature or wavelength of the radiation.

The result supported the existence of an invisible form of radiation associated with sunlight. The fact that the radiation was dispersed by a prism provided further evidence that it behaved in a similar way to visible light.

However, the experiment did not establish the entire electromagnetic spectrum.

There were also experimental limitations. For example, stray visible light, heat conducted through the apparatus, or differences between thermometers could affect the readings.

A more convincing investigation would use matched thermometers, identical starting temperatures, equal exposure times, and a carefully shielded control. An opaque screen could also help prevent stray visible light from reaching the beyond-red detector.

The key achievement was identifying a measurable effect outside the range of human vision.

Ritter discovers radiation beyond violet

Herschel’s discovery raised another possibility.

If radiation existed beyond the red end of the spectrum, could something also exist beyond violet?

In 1801, Johann Wilhelm Ritter investigated this question using a different detector.

He passed sunlight through a prism to produce a visible spectrum. He then exposed silver chloride, a light-sensitive chemical, to different regions of the spectrum.

Silver chloride darkens when exposed to light because it undergoes a chemical change. Ritter noticed that the chemical reaction was particularly strong just beyond the violet end, where no visible violet light was present.

This revealed another invisible region of electromagnetic radiation, now called ultraviolet.

Notice the difference between the two experiments:

Neither investigator needed to see the radiation directly. They detected its effects on other materials.

This is an important experimental idea: a detector can reveal something that our senses cannot.

Later developments extended the picture further. James Clerk Maxwell’s electromagnetic theory predicted the existence of electromagnetic waves beyond the optical region, and Heinrich Hertz demonstrated radio waves experimentally in 1887.

The historical progression established that the visible rainbow was just one small part of a much larger family of radiation. NASA provides a useful historical summary of the discovery of the electromagnetic spectrum.

02Why human vision covers such a narrow range

The electromagnetic spectrum is continuous. It contains radiation with an enormous range of wavelengths.

So why do our eyes detect only a small section?

The answer lies in the structure of the human eye.

The retina contains specialised light-sensitive cells called photoreceptors. These include rods and cones, which respond to photons absorbed by light-sensitive pigments.

However, these pigments do not respond equally to every electromagnetic wavelength.

Under ordinary viewing conditions, the human eye detects wavelengths of approximately:

\[ 400\text{ nm to }700\text{ nm} \]

Some references use approximately 380 nm as the lower boundary. The precise limits depend on factors such as light intensity and individual visual sensitivity.

The important point is that the boundaries are not perfectly sharp.

Photoreceptors have limited wavelength sensitivity. Radiation outside their effective response range does not produce ordinary human vision.

Consider two examples.

Infrared radiation: A warm person emits infrared radiation. A normal human eye cannot detect this radiation, even though it is reaching the eye. A thermal camera can detect it using an infrared-sensitive detector and convert the signal into a visible image.

Ultraviolet radiation: UV radiation can produce fluorescence in certain materials. You may have seen fluorescent markings or objects glow under a UV lamp. Your eye sees the visible light emitted by the material, not the ultraviolet radiation itself.

The eye is therefore a specialised detector, not a complete detector of electromagnetic radiation.

Other instruments extend our ability to observe the spectrum. Radio telescopes detect radio signals, infrared cameras detect thermal radiation, and X-ray detectors record radiation that cannot be seen directly.

03How small is the visible region?

Let’s put the visible region into perspective.

The conventional wavelength ranges below are approximate. Different references sometimes use slightly different boundaries.

Look at the visible range: it covers only about 300 nm of wavelength.

Now compare this with a radio wave of wavelength 1 m.

First, convert the radio wavelength into nanometres:

\[ 1\text{ m}=1\times10^9\text{ nm} \]

Compare this with visible red light at 700 nm:

\[ \frac{1\times10^9}{700} =1.43\times10^6 \]

The radio wavelength is approximately 1.43 million times longer than the wavelength of red light.

Now consider an X-ray with a wavelength of 0.01 nm.

Compared with violet light at 400 nm:

\[ \frac{400}{0.01}=40\,000 \]

The X-ray wavelength is 40,000 times shorter.

These comparisons show why the visible region is considered small. Electromagnetic radiation extends across many orders of magnitude in wavelength.

There is an important mathematical qualification here: we cannot calculate a meaningful percentage of the entire electromagnetic spectrum occupied by visible light. The full spectrum has no fixed longest or shortest wavelength, and its region boundaries are approximate.

The word small describes the narrow range of wavelengths involved compared with the enormous range of electromagnetic radiation.

Why different wavelengths have different applications

The wavelength of electromagnetic radiation affects how it interacts with matter and which detectors can measure it.

For example:

• Radio waves have long wavelengths and are useful for broadcasting and communication.
• Microwaves are useful in radar and wireless communication, and can transfer energy to suitable materials.
• Infrared radiation can be detected by thermal cameras to map temperature patterns.
• Visible light allows our eyes and optical cameras to form images.
• Ultraviolet radiation can trigger fluorescence and photochemical reactions.
• X-rays can pass through soft tissue more readily than bone, making them useful for medical imaging.
• Gamma rays are used in applications involving high-energy nuclear radiation.

These are representative applications, not rules that apply to every wavelength within a region. The material being exposed and the detector being used also matter.

For a more detailed comparison of why particular wavelengths suit particular jobs, see Electromagnetic Spectrum: How to Match Waves to Applications.

04Why shorter wavelengths have more energetic photons

There is another reason the electromagnetic spectrum is interesting.

Visible light sits between lower-energy infrared radiation and higher-energy ultraviolet radiation. This relationship can be explained using photon energy.

Electromagnetic radiation transfers energy in discrete packets called photons.

The energy of an individual photon depends on its frequency.

\[ E=hf \]

where:

• \(E\) is the energy of one photon, measured in joules (J).
• \(h\) is Planck’s constant, approximately \(6.63\times10^{-34}\ \text{J s}\).
• \(f\) is frequency, measured in hertz (Hz).

For electromagnetic radiation travelling through a vacuum:

\[ c=f\lambda \]

where:

• \(c\) is the speed of light in a vacuum, \(3.00\times10^8\ \text{m s}^{-1}\).
• \(f\) is frequency in hertz.
• \(\lambda\) is wavelength in metres.

Rearranging:

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

Substitute this into the photon energy equation:

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

This equation connects wavelength directly to photon energy.

A shorter wavelength means a higher frequency, which means greater energy per photon.

A longer wavelength means a lower frequency and less energy per photon.

05Photon energy and photon count are different quantities

This is where a common misconception appears.

Imagine two light sources:

• Source A emits a relatively small number of high-energy photons.
• Source B emits a much larger number of low-energy photons.

Which source transfers more energy?

You cannot answer without knowing both the energy per photon and the number of photons emitted.

Think of photons as coins passing through a collection point. The energy of each photon is like the value of each coin, while the photon emission rate is the number of coins passing each second.

The total value collected each second depends on both quantities.

The analogy is only a bookkeeping tool. Photons are quantum packets of electromagnetic energy, not physical coins.

Connecting photon count to power

Power measures the rate of energy transfer.

\[ P=\frac{E_{\text{total}}}{t} \]

For a monochromatic beam, every photon has the same energy. Therefore:

\[ \boxed{P=RE} \]

where:

• \(P\) is optical power in watts (W).
• \(R\) is the photon emission rate in photons per second.
• \(E\) is the energy of one photon in joules.

Rearranging:

\[ \boxed{R=\frac{P}{E}} \]

This gives us two important situations.

Notice the difference.

If two sources emit the same number of photons per second, the shorter-wavelength source transfers more energy each second.

However, if two sources have equal optical power, the longer-wavelength source must emit more photons each second.

A useful HSC decision rule

When a question involves electromagnetic radiation, identify which quantity is being held constant.

1. If the question asks about one photon, calculate \(E=hf\) or \(E=hc/\lambda\).
2. If it asks about the number of photons, use the photon emission rate.
3. If it gives optical power, use \(P=RE\) to connect energy and photon count.
4. If it involves a photoelectric effect, compare the energy of each photon with the metal’s work function before calculating electron emission.

For radiation containing multiple wavelengths, treat each wavelength component separately when the interaction depends on individual photon energies.

This last point becomes especially important when radiation interacts with matter.

06What happens when electromagnetic radiation enters a material?

There is one more distinction worth understanding before attempting the questions.

The wavelength of electromagnetic radiation can change when it enters a material, but its frequency does not change at the boundary.

For a material with refractive index \(n\), the wave speed is:

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

The wave relationship inside the material becomes:

\[ v=f\lambda \]

Therefore:

\[ \lambda=\frac{v}{f} \]

Since the frequency remains constant, reducing the wave speed also reduces its wavelength.

The energy of an individual photon remains unchanged because:

\[ E=hf \]

The frequency has not changed, so the photon energy has not changed either.

This matters when classifying electromagnetic radiation.

Spectrum regions are conventionally classified using their vacuum wavelengths or frequencies. A visible photon does not become an ultraviolet photon simply because its wavelength decreases inside glass.

Keep this distinction in mind:

• Wavelength can change when radiation enters a different medium.
• Frequency remains constant at the boundary.
• Photon energy remains constant.
• Electromagnetic spectrum classifications use vacuum wavelength or frequency.

07Questions and solutions

Question 1

A laser produces monochromatic light with a wavelength of \(480\text{ nm}\).

(a) Calculate the frequency of the light.

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

(c) Identify the region of the electromagnetic spectrum in which the radiation belongs.

Use \(c=3.00\times10^8\text{ m s}^{-1}\) and \(h=6.63\times10^{-34}\text{ J s}\).

Solution 1

The frequency is \(6.25\times10^{14}\text{ Hz}\), and each photon carries approximately \(4.14\times10^{-19}\text{ J}\). The radiation belongs to the visible region.

(a) Calculate the frequency.

First convert the wavelength into metres:

\[ \lambda=480\times10^{-9}\text{ m} \]

Use:

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

Substitute:

\[ \begin{aligned} f&=\frac{3.00\times10^8} {480\times10^{-9}}\\ &=6.25\times10^{14}\text{ Hz} \end{aligned} \]

(b) Calculate the photon energy.

Use:

\[ E=hf \]

Substitute:

\[ \begin{aligned} E&=(6.63\times10^{-34}) (6.25\times10^{14})\\ &=4.14\times10^{-19}\text{ J} \end{aligned} \]

(c) Identify the region.

The wavelength is \(480\text{ nm}\).

This lies between the approximate visible boundaries of 400 nm and 700 nm.

Therefore, the radiation is visible light.

The important relationship is that shorter wavelengths correspond to higher photon energies. However, the energy of one photon does not tell us the total power of the laser. We would also need its photon emission rate.

Question 2

Two monochromatic electromagnetic sources deliver the same optical power of \(0.80\text{ mW}\) to a detector.

• Source A emits radiation at \(700\text{ nm}\).
• Source B emits radiation at \(1400\text{ nm}\).

(a) Calculate the energy of one photon from each source.

(b) Calculate the photon emission rate of each source.

(c) Determine the ratio of their photon emission rates.

(d) A student claims that Source A must transfer more energy to the detector each second because its photons have greater energy. Explain why this conclusion is incorrect.

Assume the detector receives the full stated optical power from each source.

Solution 2

Source B emits twice as many photons per second as Source A, but both sources deliver the same energy each second because their optical powers are equal.

(a) Calculate the photon energies.

For Source A:

\[ \lambda_A=700\times10^{-9}\text{ m} \]

Use:

\[ E_A=\frac{hc}{\lambda_A} \]

\[ \begin{aligned} E_A&=\frac{(6.63\times10^{-34}) (3.00\times10^8)} {700\times10^{-9}}\\ &=2.84\times10^{-19}\text{ J} \end{aligned} \]

For Source B:

\[ \lambda_B=1400\times10^{-9}\text{ m} \]

\[ \begin{aligned} E_B&=\frac{(6.63\times10^{-34}) (3.00\times10^8)} {1400\times10^{-9}}\\ &=1.42\times10^{-19}\text{ J} \end{aligned} \]

Source A photons carry twice the energy of Source B photons.

(b) Calculate the photon emission rates.

Convert the power:

\[ P=0.80\times10^{-3}\text{ W} \]

For Source A:

\[ R_A=\frac{P}{E_A} \]

\[ \begin{aligned} R_A&=\frac{0.80\times10^{-3}} {2.84\times10^{-19}}\\ &=2.82\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

For Source B:

\[ R_B=\frac{P}{E_B} \]

\[ \begin{aligned} R_B&=\frac{0.80\times10^{-3}} {1.42\times10^{-19}}\\ &=5.63\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

(c) Compare the rates.

\[ \frac{R_B}{R_A} = \frac{5.63\times10^{15}} {2.82\times10^{15}} \]

\[ \frac{R_B}{R_A}=2.00 \]

Source B emits twice as many photons per second.

(d) Evaluate the student’s claim.

The student’s conclusion is incorrect because they have confused energy per photon with total energy transferred per second.

Both sources deliver \(0.80\text{ mJ}\) of energy every second.

Source A transfers this energy using fewer, higher-energy photons. Source B transfers the same energy using more, lower-energy photons.

The total power is therefore equal, even though the individual photon energies differ.

A detector that counts individual photons could record a different count from a power meter, provided its detection efficiency is suitable for both wavelengths.

Question 3

A student investigates Herschel’s experiment using a prism and five identical thermometers.

Each thermometer starts at the same temperature. After five minutes of exposure, the student records the following temperature increases.

Illustrative experimental data only; these are not Herschel’s historical measurements.

(a) Calculate the corrected temperature increase beyond red, accounting for the shaded control.

(b) Explain what the result suggests about the radiation beyond the visible spectrum.

(c) Explain why the shaded control is important.

(d) Identify two improvements that would make the experimental conclusion more reliable.

Solution 3

The corrected temperature increase beyond red is \(2.2^\circ\text{C}\), providing evidence of a heating effect outside the visible spectrum.

(a) Calculate the corrected temperature increase.

The shaded thermometer measures the background temperature increase that occurs without direct exposure to the projected spectrum.

Therefore:

\[ \Delta T_{\text{corrected}} = \Delta T_{\text{measured}} - \Delta T_{\text{control}} \]

Substitute:

\[ \Delta T_{\text{corrected}} =2.3-0.1 \]

\[ \boxed{\Delta T_{\text{corrected}}=2.2^\circ\text{C}} \]

The thermometer beyond red has experienced a temperature increase of \(2.2^\circ\text{C}\) above the background change.

(b) Interpret the result.

The result suggests that radiation beyond the red end of the visible spectrum is transferring energy to the thermometer.

Because no visible red light reaches this position, the heating effect is evidence of invisible radiation.

This is consistent with infrared radiation.

The result does not mean the thermometer has detected a visible colour beyond red. It has detected an energy transfer caused by radiation that the eye cannot see.

(c) Explain the purpose of the control.

The shaded thermometer provides a reference for background heating.

For example, the thermometers might warm slightly because of changes in room temperature or heating from nearby equipment.

Subtracting the control reading helps distinguish the effect of the radiation from these background changes.

Without a control, the student could incorrectly attribute all the temperature increase to the radiation being investigated.

(d) Suggest improvements.

Two suitable improvements are:

• Use an opaque screen to prevent stray visible light from reaching the thermometer positioned beyond red.
• Ensure all thermometers have identical initial temperatures, are calibrated, and receive equal exposure times.

The student could also thermally isolate the thermometer supports to reduce heat conduction through the apparatus.

The main experimental limitation is that temperature measurements alone do not identify the precise nature of the radiation. Additional evidence, such as its behaviour under refraction or measurements using an infrared-sensitive detector, would strengthen the conclusion.

Question 4

A beam of monochromatic light has a vacuum wavelength of \(540\text{ nm}\).

It enters a glass block with a refractive index of \(1.50\). Assume the refractive index is constant at this wavelength.

(a) Calculate the frequency of the light.

(b) Calculate its speed inside the glass.

(c) Calculate its wavelength inside the glass.

(d) Calculate the energy of one photon.

(e) A student claims that the light has become ultraviolet because its wavelength inside the glass is less than \(400\text{ nm}\). Explain why this is incorrect.

Solution 4

The light has a frequency of \(5.56\times10^{14}\text{ Hz}\), a speed of \(2.00\times10^8\text{ m s}^{-1}\), and a wavelength of \(360\text{ nm}\) inside the glass. Its photon energy remains unchanged, and it does not become ultraviolet.

(a) Calculate the frequency.

The frequency is determined using the vacuum wavelength:

\[ \lambda_0=540\times10^{-9}\text{ m} \]

Use:

\[ f=\frac{c}{\lambda_0} \]

\[ \begin{aligned} f&=\frac{3.00\times10^8} {540\times10^{-9}}\\ &=5.56\times10^{14}\text{ Hz} \end{aligned} \]

The frequency remains constant as the light enters the glass.

(b) Calculate the speed inside the glass.

Use:

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

Substitute:

\[ \begin{aligned} v&=\frac{3.00\times10^8}{1.50}\\ &=2.00\times10^8\text{ m s}^{-1} \end{aligned} \]

The light travels more slowly inside the glass.

(c) Calculate the wavelength inside the glass.

Use:

\[ v=f\lambda_{\text{glass}} \]

Rearrange:

\[ \lambda_{\text{glass}}=\frac{v}{f} \]

Substitute:

\[ \begin{aligned} \lambda_{\text{glass}} &=\frac{2.00\times10^8} {5.56\times10^{14}}\\ &=3.60\times10^{-7}\text{ m} \end{aligned} \]

Therefore:

\[ \boxed{\lambda_{\text{glass}}=360\text{ nm}} \]

The wavelength decreases because the speed decreases while frequency remains constant.

(d) Calculate the photon energy.

Use:

\[ E=hf \]

\[ \begin{aligned} E&=(6.63\times10^{-34}) (5.56\times10^{14})\\ &=3.69\times10^{-19}\text{ J} \end{aligned} \]

The photon energy is approximately \(3.69\times10^{-19}\text{ J}\).

(e) Evaluate the student’s claim.

The student is incorrect because electromagnetic spectrum classifications are based on vacuum wavelength or frequency.

Although the wavelength inside the glass is \(360\text{ nm}\), the frequency has not changed.

Therefore, the photon energy has not changed either.

The light remains visible radiation. A change in wavelength caused by entering a material does not mean the radiation has moved into a different part of the electromagnetic spectrum.

This is an important distinction between wave behaviour in a medium and the classification of electromagnetic radiation.

Question 5

Two light sources illuminate identical metal photocathodes in separate photoelectric cells.

Both cells use the same metal, which has a work function of \(2.00\text{ eV}\).

The sources deliver the following radiation to their respective photocathodes.

Source A:

• Monochromatic wavelength: \(600\text{ nm}\)
• Optical power: \(1.00\text{ mW}\)

Source B:

• First component: \(400\text{ nm}\), optical power \(0.500\text{ mW}\)
• Second component: \(800\text{ nm}\), optical power \(0.500\text{ mW}\)

Both sources therefore deliver the same total optical power.

Assume:

• Every incident photon reaches and is absorbed by the photocathode.
• Each photon with sufficient energy releases exactly one electron.
• All released electrons are collected.
• The ordinary one-photon photoelectric model applies, with no multiphoton effects.
• The metal work function is identical in both cells.

The collector is made negative relative to the photocathode when measuring the stopping potential. Report the magnitude of the stopping potential as a positive value.

(a) Calculate the energy of a photon at each of the three wavelengths in electronvolts.

(b) Determine which wavelengths can release photoelectrons from the metal.

(c) Calculate the total photon emission rate of Source A and the total photon emission rate of Source B.

(d) Calculate the photocurrent produced by each source.

(e) Calculate the stopping potential magnitude for each source.

(f) A student argues that both sources should produce identical photoelectric results because they have the same total power and the same average photon energy. Explain precisely why this reasoning fails.

(g) Suppose the \(800\text{ nm}\) component of Source B is doubled in power while the \(400\text{ nm}\) component remains unchanged. Predict what happens to the photocurrent and stopping potential. State the assumptions behind your prediction.

Solution 5

Source A produces a photocurrent of approximately \(0.483\text{ mA}\) and a stopping potential of \(0.072\text{ V}\). Source B produces a photocurrent of approximately \(0.161\text{ mA}\) and a stopping potential of \(1.11\text{ V}\).

The surprising result is that the sources have the same total photon emission rate and the same average photon energy, but their photoelectric responses are different.

The reason is that photoelectric emission depends on the energy of each individual photon, not the average energy across the entire beam.

(a) Calculate the photon energies.

Use:

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

For calculations in electronvolts, we can use the approximate conversion:

\[ E(\text{eV})=\frac{1242}{\lambda(\text{nm})} \]

For the \(400\text{ nm}\) component:

\[ E_{400}=\frac{1242}{400} \]

\[ \boxed{E_{400}=3.11\text{ eV}} \]

For the \(600\text{ nm}\) component:

\[ E_{600}=\frac{1242}{600} \]

\[ \boxed{E_{600}=2.07\text{ eV}} \]

For the \(800\text{ nm}\) component:

\[ E_{800}=\frac{1242}{800} \]

\[ \boxed{E_{800}=1.55\text{ eV}} \]

The three photon energies are therefore:

(b) Determine which photons cause photoelectric emission.

The work function is:

\[ \phi=2.00\text{ eV} \]

A photon must have energy equal to or greater than the work function to release an electron.

Compare each photon energy with the work function:

• \(400\text{ nm}\): \(3.11\text{ eV}>2.00\text{ eV}\), so photoelectric emission occurs.
• \(600\text{ nm}\): \(2.07\text{ eV}>2.00\text{ eV}\), so photoelectric emission occurs.
• \(800\text{ nm}\): \(1.55\text{ eV}<2.00\text{ eV}\), so photoelectric emission does not occur.

Therefore, Source A can release electrons using all its photons, while only the \(400\text{ nm}\) component of Source B produces photoelectrons.

The \(800\text{ nm}\) photons cannot release electrons under the stated one-photon model.

This is the first important difference between the sources.

(c) Calculate the total photon emission rate of each source.

First convert the optical powers into watts:

\[ P_A=1.00\times10^{-3}\text{ W} \]

For Source B, each component has:

\[ P_{400}=P_{800}=0.500\times10^{-3}\text{ W} \]

Use:

\[ R=\frac{P}{E} \]

where the photon energy must be in joules.

For Source A, the \(600\text{ nm}\) photon energy is:

\[ E_{600}=3.315\times10^{-19}\text{ J} \]

Therefore:

\[ \begin{aligned} R_A&=\frac{1.00\times10^{-3}} {3.315\times10^{-19}}\\ &=3.02\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

Now calculate the two components of Source B.

For \(400\text{ nm}\):

\[ E_{400}=4.97\times10^{-19}\text{ J} \]

\[ \begin{aligned} R_{400}&=\frac{0.500\times10^{-3}} {4.97\times10^{-19}}\\ &=1.01\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

For \(800\text{ nm}\):

\[ E_{800}=2.49\times10^{-19}\text{ J} \]

\[ \begin{aligned} R_{800}&=\frac{0.500\times10^{-3}} {2.49\times10^{-19}}\\ &=2.01\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

Add the two photon rates:

\[ R_B=R_{400}+R_{800} \]

\[ \begin{aligned} R_B&=1.01\times10^{15} +2.01\times10^{15}\\ &=3.02\times10^{15} \text{ photons s}^{-1} \end{aligned} \]

Therefore:

\[ \boxed{R_A=R_B} \]

Both sources emit approximately \(3.02\times10^{15}\) photons per second.

This is an important part of the problem. The sources have equal total power and equal total photon emission rates.

However, their photons are distributed differently across the spectrum.

(d) Calculate the photocurrent produced by each source.

Under the stated assumptions, every photon above the work function produces one collected electron.

The electric current is:

\[ I=\frac{Q}{t} \]

Since each electron carries a charge of magnitude:

\[ e=1.60\times10^{-19}\text{ C} \]

The photocurrent is:

\[ I=eR_{\text{electron}} \]

For Source A, every photon has sufficient energy.

Therefore:

\[ R_{\text{electron,A}}=3.02\times10^{15} \text{ electrons s}^{-1} \]

\[ \begin{aligned} I_A&=(1.60\times10^{-19}) (3.02\times10^{15})\\ &=4.83\times10^{-4}\text{ A} \end{aligned} \]

\[ \boxed{I_A=0.483\text{ mA}} \]

For Source B, only the \(400\text{ nm}\) photons release electrons.

Therefore:

\[ R_{\text{electron,B}} =1.01\times10^{15} \text{ electrons s}^{-1} \]

\[ \begin{aligned} I_B&=(1.60\times10^{-19}) (1.01\times10^{15})\\ &=1.61\times10^{-4}\text{ A} \end{aligned} \]

\[ \boxed{I_B=0.161\text{ mA}} \]

Source B produces only one-third of the photocurrent produced by Source A.

Even though both sources emit the same total number of photons per second, Source B has a larger proportion of photons that cannot release electrons.

(e) Calculate the stopping potential.

The stopping potential measures the maximum kinetic energy of the emitted electrons.

Einstein’s photoelectric equation is:

\[ E_{\text{photon}}=\phi+K_{\text{max}} \]

Rearranging:

\[ K_{\text{max}}=E_{\text{photon}}-\phi \]

The stopping potential satisfies:

\[ K_{\text{max}}=eV_s \]

where \(V_s\) is the magnitude of the stopping potential.

When kinetic energy is expressed in electronvolts, its numerical value equals the stopping potential in volts.

For Source A, the photons have an energy of \(2.07\text{ eV}\).

\[ K_{\text{max,A}}=2.07-2.00 \]

\[ K_{\text{max,A}}=0.07\text{ eV} \]

Therefore:

\[ \boxed{V_{s,A}=0.072\text{ V}} \]

For Source B, the most energetic photons have a wavelength of \(400\text{ nm}\).

These photons have an energy of \(3.11\text{ eV}\).

\[ K_{\text{max,B}}=3.11-2.00 \]

\[ K_{\text{max,B}}=1.11\text{ eV} \]

Therefore:

\[ \boxed{V_{s,B}=1.11\text{ V}} \]

The \(800\text{ nm}\) photons do not contribute to photoelectric emission, so they cannot determine the maximum kinetic energy.

The \(400\text{ nm}\) photons determine the maximum kinetic energy because they have the greatest energy among the photons that cause emission.

(f) Explain why equal power and equal average photon energy do not produce identical results.

The student’s argument fails because averaging photon energies hides the distribution of individual photon energies.

Both sources have the same total power and photon emission rate. Consequently, their average photon energy is also the same.

However, the photoelectric effect does not respond to the average photon energy.

Each photon interacts individually with the metal.

For Source A, every photon has enough energy to release an electron.

For Source B, the \(400\text{ nm}\) photons have sufficient energy, but the \(800\text{ nm}\) photons do not.

The energy of a low-energy photon cannot simply be combined with the energy of another photon to release an electron under the ordinary one-photon model.

This produces two different outcomes:

The most important insight is that the average photon energy cannot tell us whether individual photons cross the photoelectric threshold.

We must examine the energy of each wavelength component separately.

(g) Predict what happens if the \(800\text{ nm}\) component is doubled.

The photocurrent remains approximately \(0.161\text{ mA}\), and the stopping potential remains approximately \(1.11\text{ V}\).

Doubling the \(800\text{ nm}\) component increases the total optical power and the total number of photons arriving each second.

However, its photons still have an energy of only \(1.55\text{ eV}\), which is below the \(2.00\text{ eV}\) work function.

They therefore cannot release additional electrons under the stated assumptions.

The \(400\text{ nm}\) component has not changed, so:

• The rate of electron emission remains constant.
• The photocurrent remains constant.
• The maximum kinetic energy remains constant.
• The stopping potential remains constant.

This prediction assumes ordinary single-photon photoelectric emission. At sufficiently high intensities, multiphoton effects can occur, but those are outside the model used here.

This final result is worth remembering: increasing the total power of a light beam does not necessarily increase the photoelectric current. It depends on which photons carry that additional energy.

The next useful step is to understand how electromagnetic radiation interacts with different materials. Absorption, transmission, reflection, and the photoelectric effect all depend on the radiation’s wavelength and the properties of the material.

Once you can distinguish wavelength, energy per photon, and photon emission rate, you have the foundation for analysing those interactions without confusing the behaviour of individual photons with the behaviour of an entire beam.