What Happens When Two Colours Illuminate the Same Metal?

Shine red and blue light onto the same metal surface at the same time. The blue photons carry more energy individually, but the red beam might produce more photons per second. Does the metal combine the energy from both colours to give each electron a bigger kick?

No. In the standard single-photon model, each emitted electron absorbs energy from one photon. If both colours have enough energy to eject electrons, they produce separate populations of photoelectrons. Their contributions to the emission current add, but the higher-frequency colour determines the maximum kinetic energy and stopping potential.

This gives us three important results:

• Emission current: Both colours can contribute, so the total current can increase.
• Maximum kinetic energy: The highest-frequency photons determine the fastest emitted electrons.
• Stopping potential: It is determined by those fastest electrons, not by the sum or average of the two photon energies.

If one colour is below the metal’s threshold frequency, it contributes no photoelectrons at all.

01What happens to each photon?

Start by thinking about what happens when one photon strikes a metal.

An electron inside the metal needs a minimum amount of energy to escape. This minimum energy is called the work function, represented by \(\phi\).

Think of the work function as an entry fee. A photon arrives with a certain energy budget. The electron needs enough energy to pay the entry fee, and any energy left over becomes kinetic energy.

The analogy is useful, but there is an important limitation: in the single-photon model, two separate photons cannot pool their energy to pay the entry fee for one electron.

The energy of one photon is:

\[ E_{\text{photon}}=hf=\frac{hc}{\lambda} \]

where:

• \(E_{\text{photon}}\) is the energy of one photon, in joules (J).
• \(h\) is Planck’s constant, \(6.63\times10^{-34}\text{ J s}\).
• \(f\) is the frequency of the light, in hertz (Hz).
• \(c\) is the speed of light, \(3.00\times10^8\text{ m s}^{-1}\).
• \(\lambda\) is the wavelength, in metres (m).

An electron can be emitted when the photon supplies at least the work function:

\[ hf\geq\phi \]

Any energy remaining becomes the electron’s kinetic energy.

Einstein’s photoelectric equation is:

\[ \boxed{K_{\max}=hf-\phi} \]

Here, \(K_{\max}\) is the maximum kinetic energy of the emitted photoelectrons.

Now introduce two colours

Suppose red and blue light illuminate the same metal.

Blue light has a higher frequency and shorter wavelength than red light. Therefore:

\[ E_{\text{blue}}>E_{\text{red}} \]

Each colour can produce its own population of photoelectrons, provided its photons have enough energy to overcome the work function.

For a metal with work function \(\phi\):

\[ K_{\max,\text{red}}=hf_{\text{red}}-\phi \]

\[ K_{\max,\text{blue}}=hf_{\text{blue}}-\phi \]

If both colours produce photoelectrons, the blue-generated electrons can reach a greater maximum kinetic energy.

The overall maximum kinetic energy is therefore:

\[ \boxed{K_{\max}=hf_{\text{blue}}-\phi} \]

The important point is that the two photon energies are not added together.

Three possible situations

This is the first decision to make in any two-colour photoelectric question: compare the energy of each photon with the work function separately.

02How would we test this experimentally?

The question becomes more interesting when we consider how an experiment could distinguish between two competing explanations.

The historical problem began with observations that did not fit a simple classical wave description of light.

In 1887, Heinrich Hertz observed that ultraviolet light affected the sparks produced across a spark gap.

Subsequent experiments by Philipp Lenard, including his investigations in 1902, showed that increasing light intensity increased the rate of photoelectric emission but did not increase the maximum speed of the emitted electrons.

This was an important distinction.

If light transferred energy continuously to individual electrons, we might expect a sufficiently intense beam to produce increasingly energetic electrons.

Instead, the maximum electron energy depended on light frequency.

In 1905, Albert Einstein proposed an explanation based on light quanta, now called photons. Each photon carries energy \(hf\), and an individual photon transfers its energy to an electron.

The following two-colour arrangement provides a useful modern test of that explanation. It is a proposed controlled experiment, rather than a claim that this exact arrangement was used in the original historical investigations.

Setting up the two-colour experiment

Imagine an evacuated photoelectric cell.

Inside the cell are two electrodes:

• A photosensitive metal cathode, which releases electrons when illuminated.
• A collector anode, which attracts the emitted electrons.

Outside the cell, an ammeter measures the photocurrent. A variable power supply changes the potential difference between the electrodes, and a voltmeter measures that potential difference.

To investigate two colours, we introduce two independently controlled light sources.

1. A red light source passes through a wavelength filter and an adjustable intensity control.
2. A blue light source passes through its own filter and intensity control.
3. Mirrors or a beam splitter direct both beams onto the same region of the metal cathode.
4. The emitted electrons travel towards the collector.
5. The ammeter measures the resulting current.

Each beam can be switched on independently, allowing us to compare red-only, blue-only, and combined illumination.

What should the experiment show?

First, illuminate the cathode with red light alone.

Record the photocurrent as the potential difference between the electrodes changes.

Next, repeat the measurement using blue light alone.

Finally, illuminate the cathode with both colours simultaneously.

If both colours are above the threshold frequency, the single-photon model predicts:

• The red photons produce photoelectrons.
• The blue photons produce photoelectrons.
• The total emission rate increases because both populations contribute.
• The maximum kinetic energy remains determined by the blue photons.

The combined experiment should therefore produce a greater saturation current than either beam alone, while retaining the stopping potential associated with blue light.

This is the crucial experimental distinction: adding another colour can increase the number of emitted electrons without increasing the energy of the fastest electrons.

An important experimental flaw

Suppose we initially adjust the red and blue sources until they have equal optical power.

Is that a fair comparison?

Not necessarily.

Equal optical power does not mean equal numbers of photons arriving each second.

A red photon carries less energy than a blue photon. Therefore, at the same optical power, the red source delivers more photons per second.

The photon arrival rate is:

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

where:

• \(\dot N\) is the number of photons arriving each second.
• \(P\) is the optical power, in watts (W).
• \(hf\) is the energy of each photon, in joules.

A better experimental design measures or calibrates the photon flux of each beam. We should also keep the illuminated area, metal surface, and collection conditions constant.

There is another complication: different wavelengths can have different photoelectric emission efficiencies. Equal photon arrival rates do not necessarily produce equal emission currents.

The revised experiment therefore needs to control photon flux and measure the current produced by each colour independently before combining them.

03Emission current, stopping potential, and kinetic energy

These three quantities are related, but they describe different properties of the photoelectric effect.

Confusing them is one of the easiest ways to lose marks in a question involving multiple colours.

1. Emission current: how many electrons?

The emission current measures the rate at which charge is emitted.

Since each electron carries a charge with magnitude \(e\):

\[ I=e\frac{\Delta N_e}{\Delta t} \]

where:

• \(I\) is the current, in amperes (A).
• \(e=1.60\times10^{-19}\text{ C}\) is the magnitude of the electron’s charge.
• \(\Delta N_e\) is the number of emitted electrons.
• \(\Delta t\) is the time interval, in seconds.

If both colours produce photoelectrons, their emission rates contribute to the total current.

At a sufficiently positive anode potential, all emitted electrons can be collected. The current reaches its maximum value, called the saturation current.

Under normal HSC single-photon assumptions, with independent emission and effective collection:

\[ I_{\text{sat,total}} = I_{\text{sat,red}}+I_{\text{sat,blue}} \]

The saturation current therefore tells us about the total rate of collected electrons, not their maximum kinetic energy.

Remember that the current at an arbitrary applied voltage might be lower because some electrons fail to reach the collector. Saturation current is the cleanest measure of the total emission rate.

2. Maximum kinetic energy: how fast are the fastest electrons?

The maximum kinetic energy is determined by the energy of the most energetic photons that successfully produce photoelectrons.

For two colours above threshold:

\[ K_{\max}=hf_{\text{high}}-\phi \]

Adding a lower-frequency beam does not increase this maximum.

It may produce additional electrons, but those electrons cannot exceed the maximum kinetic energy supplied by the higher-frequency photons.

3. Stopping potential: what voltage stops the fastest electrons?

The stopping potential is the reverse potential difference needed to prevent even the fastest photoelectrons from reaching the collector.

The anode is made negative relative to the cathode. It repels the negatively charged electrons.

As the reverse voltage increases, progressively more electrons are prevented from reaching the anode.

Eventually, even the fastest electrons are stopped, and the photocurrent falls to zero.

The relationship is:

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

where \(V_s\) is the positive magnitude of the stopping potential, measured in volts.

For sign conventions, define the applied voltage as:

\[ V=V_A-V_C \]

where \(V_A\) is the anode potential and \(V_C\) is the cathode potential.

At the stopping point:

\[ V=-V_s \]

The negative sign describes the actual anode-to-cathode potential difference. We use the positive magnitude \(V_s\) in the kinetic energy equation.

If both colours produce electrons, the stopping potential is determined by the higher-frequency colour.

An interesting feature of the combined graph is that the red-generated electrons can be stopped before the blue-generated electrons.

As the reverse voltage increases, the slower electrons are turned back first. The remaining current comes from electrons with enough kinetic energy to overcome the reverse potential.

The combined curve may show a shoulder or change in gradient, although a distinct step is not guaranteed because the electrons have distributions of kinetic energies.

The final zero-current point is still determined by the fastest electrons.

For more practice interpreting these curves, see How to Sketch Photocurrent Against Applied Voltage.

04Worked examples

05The contrasting case: What if neither colour is energetic enough?

Consider a metal with a work function of \(2.10\text{ eV}\).

It is illuminated by:

• Red light with a wavelength of \(700\text{ nm}\).
• Green light with a wavelength of \(600\text{ nm}\).

The photon energies are:

\[ E_R=\frac{1240}{700} =1.77\text{ eV} \]

\[ E_G=\frac{1240}{600} =2.07\text{ eV} \]

Neither photon has enough energy to overcome the work function.

Notice something tempting:

\[ E_R+E_G=3.84\text{ eV} \]

This is greater than the work function.

Should photoelectrons be emitted?

No. In the standard single-photon model, each electron absorbs one photon. The red and green photons do not combine their energies to release a single electron.

Therefore, neither beam produces photoelectric emission, even when they illuminate the metal simultaneously.

There is an important limitation to this explanation.

At sufficiently high laser intensities, multiphoton photoemission can occur. In that different process, an electron can absorb energy from more than one photon during a single emission event.

That is beyond the standard HSC single-photon model. For ordinary photoelectric questions, do not assume multiphoton absorption.

A reliable method for any two-colour question

Use this sequence:

1. Calculate the energy of each photon separately.
2. Compare each photon energy with the metal’s work function.
3. Identify which colours can produce photoelectrons.
4. Calculate the maximum kinetic energy for each contributing colour.
5. Select the greatest maximum kinetic energy to determine the stopping potential.
6. Add the emission rates to determine the combined saturation current, if sufficient information is provided.

Never add the energies of two different photons when applying Einstein’s ordinary photoelectric equation.

06Questions and solutions

Question 1

A metal has a work function of \(2.20\text{ eV}\). It is illuminated simultaneously by red light of wavelength \(600\text{ nm}\) and violet light of wavelength \(400\text{ nm}\).

Calculate:

a. The energy of each photon.

b. Which colour produces photoelectrons.

c. The maximum kinetic energy of the emitted electrons.

d. The stopping potential.

Use \(hc=1240\text{ eV nm}\).

Solution 1

Only violet light produces photoelectrons. The maximum kinetic energy is \(0.90\text{ eV}\), giving a stopping potential of \(0.90\text{ V}\).

Step 1: Calculate the photon energies.

For red light:

\[ E_R=\frac{1240}{600} =2.07\text{ eV} \]

For violet light:

\[ E_V=\frac{1240}{400} =3.10\text{ eV} \]

Step 2: Compare each photon energy with the work function.

The work function is \(2.20\text{ eV}\).

Red:

\[ 2.07<2.20 \]

Red photons cannot eject electrons.

Violet:

\[ 3.10>2.20 \]

Violet photons can eject electrons.

Step 3: Calculate the maximum kinetic energy.

Only violet light contributes:

\[ K_{\max}=3.10-2.20 \]

\[ \boxed{K_{\max}=0.90\text{ eV}} \]

Step 4: Calculate the stopping potential.

\[ K_{\max}=eV_s \]

Therefore:

\[ \boxed{V_s=0.90\text{ V}} \]

The red light does not increase the emission current because it cannot eject electrons from this metal. Increasing its intensity does not change that conclusion within the single-photon model.

Question 2

A metal has a work function of \(1.90\text{ eV}\). It is illuminated by two beams, each with an optical power of \(3.0\text{ mW}\).

• Beam A has a wavelength of \(500\text{ nm}\).
• Beam B has a wavelength of \(700\text{ nm}\).

a. Calculate the energy of each photon in electronvolts.

b. Calculate the number of photons arriving each second from each beam.

c. Determine which beam produces photoelectrons.

d. Calculate the maximum kinetic energy and stopping potential.

e. Explain why the beam with the greater photon arrival rate does not necessarily produce the greater emission current.

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

Solution 2

Only the \(500\text{ nm}\) beam produces photoelectrons. The stopping potential is \(0.58\text{ V}\), even though the \(700\text{ nm}\) beam delivers more photons each second.

Step 1: Calculate the photon energies.

Using:

\[ E_{\text{photon}}=\frac{hc}{\lambda} \]

For beam A:

\[ E_A= \frac{6.63\times10^{-34} \times3.00\times10^8} {500\times10^{-9}} \]

\[ E_A=3.98\times10^{-19}\text{ J} \]

Converting into electronvolts:

\[ E_A=\frac{3.98\times10^{-19}} {1.60\times10^{-19}} \]

\[ E_A=2.49\text{ eV} \]

For beam B:

\[ E_B= \frac{6.63\times10^{-34} \times3.00\times10^8} {700\times10^{-9}} \]

\[ E_B=2.84\times10^{-19}\text{ J} \]

\[ E_B=1.78\text{ eV} \]

Step 2: Calculate the photon arrival rates.

The relationship between power and photon arrival rate is:

\[ P=\dot N E_{\text{photon}} \]

Rearranging:

\[ \dot N=\frac{P}{E_{\text{photon}}} \]

For beam A:

\[ \dot N_A= \frac{3.0\times10^{-3}} {3.98\times10^{-19}} \]

\[ \dot N_A=7.54\times10^{15} \text{ photons s}^{-1} \]

For beam B:

\[ \dot N_B= \frac{3.0\times10^{-3}} {2.84\times10^{-19}} \]

\[ \dot N_B=1.06\times10^{16} \text{ photons s}^{-1} \]

Beam B delivers approximately 1.4 times as many photons each second.

Step 3: Determine which beam produces photoelectrons.

The work function is \(1.90\text{ eV}\).

For beam A:

\[ 2.49>1.90 \]

Photoelectrons are produced.

For beam B:

\[ 1.78<1.90 \]

No photoelectrons are produced.

Step 4: Calculate the maximum kinetic energy.

Only beam A contributes:

\[ K_{\max}=2.49-1.90 \]

\[ \boxed{K_{\max}=0.59\text{ eV}} \]

Therefore:

\[ \boxed{V_s=0.59\text{ V}} \]

Step 5: Explain the difference between photon arrival rate and emission current.

Although beam B delivers more photons each second, its photons do not have enough energy to eject electrons.

Beam A produces photoelectrons despite having a lower photon arrival rate.

We also cannot calculate the exact emission current from the supplied optical powers alone. We would need the effective emission efficiencies and collection conditions.

The important distinction is that photon arrival rate and photon energy describe different properties of the light.

Question 3

A photocell is tested using red and blue light separately and then simultaneously.

The following illustrative results are recorded:

Both colours are above the threshold frequency.

a. Calculate the maximum kinetic energy of the electrons produced by each colour.

b. Predict the combined saturation current.

c. Determine the stopping potential when both colours illuminate the cathode.

d. Describe what happens to the photocurrent as the reverse potential increases from zero to the stopping potential.

Solution 3

The combined saturation current is \(19\ \mu\text{A}\), and the stopping potential magnitude is \(0.90\text{ V}\).

Step 1: Calculate the maximum kinetic energies.

Use:

\[ K_{\max}=eV_s \]

For red light:

\[ K_{\max,R}=1.60\times10^{-19}\times0.20 \]

\[ K_{\max,R}=3.20\times10^{-20}\text{ J} \]

This is equivalent to:

\[ \boxed{K_{\max,R}=0.20\text{ eV}} \]

For blue light:

\[ K_{\max,B}=1.60\times10^{-19}\times0.90 \]

\[ K_{\max,B}=1.44\times10^{-19}\text{ J} \]

Therefore:

\[ \boxed{K_{\max,B}=0.90\text{ eV}} \]

Step 2: Calculate the combined saturation current.

Both colours produce photoelectrons, so their currents add:

\[ I_{\text{total}}=I_R+I_B \]

\[ I_{\text{total}}=12+7 \]

\[ \boxed{I_{\text{total}}=19\ \mu\text{A}} \]

Step 3: Determine the stopping potential.

Blue produces the fastest electrons.

Therefore, the stopping potential is the same as for blue illumination alone:

\[ \boxed{V_s=0.90\text{ V}} \]

It is not \(0.20+0.90=1.10\text{ V}\).

Step 4: Describe the photocurrent as the reverse voltage increases.

Initially, both populations contribute to the photocurrent.

As the anode becomes increasingly negative:

• Lower-energy electrons are stopped first.
• The red-generated population is completely stopped when the reverse potential reaches \(0.20\text{ V}\).
• Some blue-generated electrons can still reach the collector.
• The photocurrent eventually reaches zero at a reverse potential magnitude of \(0.90\text{ V}\).

The graph therefore has a greater saturation current than either individual beam, but its final zero-current intercept is determined by blue light.

The important point is that the stopping potential is determined by the fastest electrons, not by the average kinetic energy of the entire population.

Question 4

A student wants to determine whether adding red light to an existing blue beam produces additional photoelectrons.

They perform two experiments.

The student initially measures only the stopping potential and concludes that red light has no effect because the stopping potential does not change.

a. Explain why this conclusion is incorrect.

b. Determine the additional current produced when red light is added.

c. Explain why the stopping potential remains unchanged.

d. Identify the most useful measurement for distinguishing between an increase in emission rate and an increase in maximum kinetic energy.

Solution 4

Red light does produce additional photoelectrons. It increases the saturation current by \(6.0\ \mu\text{A}\), while blue light continues to determine the stopping potential.

Step 1: Compare the saturation currents.

The additional current is:

\[ \Delta I=I_{\text{combined}}-I_{\text{blue}} \]

\[ \Delta I=15.0-9.0 \]

\[ \boxed{\Delta I=6.0\ \mu\text{A}} \]

This demonstrates that adding red light increases the rate at which electrons are collected.

Step 2: Explain why the stopping potential does not change.

Blue light already produces the fastest electrons.

The red-generated electrons have lower maximum kinetic energy, so they cannot increase the overall maximum kinetic energy of the population.

The stopping potential therefore remains:

\[ \boxed{V_s=0.85\text{ V}} \]

Step 3: Identify the appropriate experimental measurement.

The saturation current is the useful measurement for determining the total rate of electron emission.

The stopping potential is useful for determining the maximum kinetic energy.

Both measurements are needed to distinguish these effects.

The student’s original conclusion was based on measuring only one property of the emitted electrons. An unchanged stopping potential does not mean that the number of emitted electrons has remained unchanged.

Question 5

A researcher investigates a metal with a work function of \(2.10\text{ eV}\).

Three narrow-band light sources are available:

All three beams can be directed onto the same metal surface.

The experiment is conducted at low light intensities, with no multiphoton absorption or other emission mechanisms.

The following observations are recorded:

• Red alone produces no measurable photocurrent.
• Green alone produces no measurable photocurrent.
• Red and green together produce no measurable photocurrent.
• Blue alone produces a saturation current of \(8.0\ \mu\text{A}\) and a stopping potential magnitude of \(0.66\text{ V}\).
• All three beams together produce a saturation current of \(8.0\ \mu\text{A}\) and a stopping potential magnitude of \(0.66\text{ V}\).

A student makes two claims:

Claim 1: Red and green photons should be able to eject electrons together because their combined photon energy exceeds the work function.

Claim 2: When all three colours are present, the maximum kinetic energy should be determined by the sum of all three photon energies.

Evaluate both claims using the single-photon model.

A second researcher then proposes a much more intense pulsed-light experiment. They suggest that a current appearing only when the red and green pulses overlap would demonstrate that the ordinary single-photon model is correct.

Is that conclusion justified? Explain what additional evidence would be needed.

Solution 5

Both claims are incorrect under the standard single-photon model. Red and green cannot pool their photon energies, and blue alone determines the maximum kinetic energy when all three beams are present.

The proposed high-intensity experiment would require a different explanation if a genuine current appeared only when the red and green pulses overlapped.

Step 1: Calculate the energy of each photon.

For red light:

\[ E_R=\frac{1240}{700} \]

\[ E_R=1.77\text{ eV} \]

For green light:

\[ E_G=\frac{1240}{600} \]

\[ E_G=2.07\text{ eV} \]

For blue light:

\[ E_B=\frac{1240}{450} \]

\[ E_B=2.76\text{ eV} \]

Step 2: Compare each photon with the work function.

The work function is \(2.10\text{ eV}\).

For red:

\[ 1.77<2.10 \]

For green:

\[ 2.07<2.10 \]

Neither colour can independently eject electrons.

Blue, however, has sufficient energy:

\[ 2.76>2.10 \]

Step 3: Evaluate Claim 1.

The student calculates:

\[ E_R+E_G=1.77+2.07 \]

\[ E_R+E_G=3.84\text{ eV} \]

This is greater than the work function.

However, the calculation combines the energies of two separate photons.

In the standard single-photon model, one electron absorbs one photon. It does not collect unused energy from another independent photon.

Therefore, red and green together still produce no photoelectric emission.

Claim 1 is incorrect.

Step 4: Evaluate Claim 2.

Blue photons individually have enough energy to eject electrons.

Their maximum kinetic energy is:

\[ K_{\max}=E_B-\phi \]

\[ K_{\max}=2.76-2.10 \]

\[ \boxed{K_{\max}=0.66\text{ eV}} \]

This agrees with the measured stopping potential:

\[ K_{\max}=eV_s \]

\[ \boxed{V_s=0.66\text{ V}} \]

Red and green do not contribute photoelectrons, so they cannot increase the maximum kinetic energy.

Claim 2 is incorrect.

Step 5: Explain the combined saturation current.

The blue beam produces a saturation current of \(8.0\ \mu\text{A}\).

When red and green are added, the current remains \(8.0\ \mu\text{A}\).

This is consistent with the single-photon model because neither additional beam can eject electrons.

The blue beam is the only source of photoelectrons.

Step 6: Evaluate the proposed high-intensity experiment.

A genuine current appearing only when red and green pulses overlap would not be explained by the ordinary single-photon model.

It could indicate a multiphoton process in which an electron absorbs energy from more than one photon during a single emission event.

In a genuine two-photon process, energy from the two photons can contribute to the same electron’s emission. This is a different physical mechanism, not the ordinary one-photon photoelectric effect.

However, the observation alone would not establish that explanation.

The researcher would need to investigate possible alternative causes, including:

• Unwanted ultraviolet light in either beam.
• Stray light reaching the cathode.
• Heating or thermionic emission.
• Electronic noise or detector artefacts.
• A genuine nonlinear multiphoton process.

Useful additional tests would include varying the intensity of each beam independently, changing the time delay between the pulses, and checking whether the signal depends on their temporal overlap.

A genuine multiphoton contribution should show behaviour consistent with a nonlinear interaction, rather than ordinary independent single-photon emission.

The key conclusion is that a model must be applied within its assumptions. If an experiment produces a result the model cannot explain, we investigate whether the assumptions have failed rather than quietly changing the meaning of its equations.

The next useful skill is interpreting a photocurrent-versus-voltage graph when several electron populations are present. Always read its two important features separately: the saturation current tells you about the number of collected electrons, while the stopping potential tells you about the energy of the fastest ones.

For the related calculation of electron speed from maximum kinetic energy, see How to Calculate Maximum Photoelectron Speed from Wavelength.