How to Compare Work Functions from Photoelectric Graphs

Two photocurrent-voltage graphs are recorded for two different metal cathodes using the same monochromatic light. Metal A produces a much larger current, but metal B requires a larger reverse voltage to bring its current to zero. Which metal has the larger work function?

Predict before reading on: Metal A has the larger work function. Its stopping-potential magnitude is smaller, meaning its fastest electrons have less kinetic energy. Since both metals receive photons with the same energy, more energy must have been used to release electrons from metal A.

This is the key to comparing work functions from photoelectric graphs:

• With the same incident light frequency, the metal with the smaller stopping potential has the larger work function.
• The height of the photocurrent does not tell you which metal has the larger work function.
• If the metals are illuminated with different frequencies, you must account for the different photon energies before comparing them.

Let’s work through why this is true, how to read the graphs correctly, and what to do when an exam question makes the comparison less straightforward.

01Understand what the photoelectric graph measures

Before comparing two graphs, you need to know what the experiment actually measures.

A photoelectric experiment uses a photocell containing two electrodes:

• The cathode is the metal surface being investigated. Light shines onto this surface and can eject electrons.
• The collector (anode) attracts the emitted electrons across the gap.

An ammeter measures the resulting photocurrent, while a variable power supply changes the potential difference between the electrodes. A voltmeter measures this potential difference.

How the experiment produces a graph

The experiment follows a simple sequence.

Step 1: Illuminate the cathode.

Light strikes the metal surface. If the photon frequency is high enough, electrons are emitted.

Step 2: Collect the electrons.

The collector attracts the emitted electrons. Their movement through the external circuit produces a measurable current.

Step 3: Change the applied voltage.

Initially, make the collector positive relative to the cathode. This attracts electrons towards it.

As the positive potential increases, more emitted electrons reach the collector. Eventually, the current reaches a maximum value called the saturation current.

Step 4: Apply a reverse potential.

Now make the collector negative relative to the cathode.

The negative collector repels the emitted electrons. Some electrons can still reach it because they have sufficient kinetic energy, but others are turned back.

As the reverse potential increases, the photocurrent decreases.

Eventually, even the fastest emitted electrons cannot reach the collector. The photocurrent falls to zero.

The magnitude of the reverse potential at this point is called the stopping potential, \(V_s\).

Reading a photocurrent-voltage graph

There are two important features to identify.

Notice that these quantities describe different things.

The current tells you how much charge reaches the collector each second. The stopping potential tells you about the maximum kinetic energy of the emitted electrons.

This distinction is essential when comparing two metals.

Why does the current decrease gradually?

Photoelectrons do not all leave the metal with the same kinetic energy.

Some have relatively low kinetic energies, while others leave with much greater kinetic energies.

As the reverse potential increases, the slower electrons are turned back first. The current therefore decreases progressively.

Only when the reverse potential is large enough to stop the fastest electrons does the photocurrent reach zero.

This is why the stopping potential gives information about the maximum kinetic energy, rather than the average kinetic energy of the electrons.

02Connect stopping potential to work function

Now we can connect the graph to the energy transferred by light.

In the standard single-photon model, one photon transfers its energy to one electron.

The photon energy is

\[ E_{\text{photon}}=hf \]

where:

• \(h\) is Planck’s constant, \(6.63\times10^{-34}\ \mathrm{J\,s}\);
• \(f\) is the frequency of the incident light, measured in hertz.

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

Any photon energy remaining after the electron escapes becomes the electron’s kinetic energy.

The photoelectric equation is therefore

\[ hf=\phi+K_{\max} \]

Rearranging:

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

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

Where does stopping potential fit in?

The stopping potential measures the energy needed to prevent the fastest electrons from reaching the collector.

The relationship is

\[ K_{\max}=eV_s \]

where:

• \(e=1.60\times10^{-19}\ \mathrm{C}\) is the positive magnitude of the electron’s charge;
• \(V_s\) is the positive magnitude of the stopping potential, measured in volts.

Although an electron has charge \(-e\), we use the positive magnitude \(e\) in this equation.

Combining the two equations gives

\[ eV_s=hf-\phi \]

Rearranging:

\[ \boxed{\phi=hf-eV_s} \]

This is the equation you need when calculating a work function from a photocurrent-voltage graph.

The crucial comparison rule

Suppose two different metals, A and B, are illuminated with light of the same frequency.

Both metals receive photons with the same energy, \(hf\).

For metal A:

\[ \phi_A=hf-eV_{s,A} \]

For metal B:

\[ \phi_B=hf-eV_{s,B} \]

Subtracting the equations gives

\[ \phi_A-\phi_B =e(V_{s,B}-V_{s,A}) \]

This tells us exactly how to compare their work functions.

If both metals receive light of the same frequency:

• Larger stopping potential means larger maximum kinetic energy.
• Larger maximum kinetic energy means less photon energy was needed to release the electron.
• Therefore, the metal with the larger stopping potential has the smaller work function.

In other words:

\[ V_{s,A}<V_{s,B} \quad\Rightarrow\quad \phi_A>\phi_B \]

A useful shortcut using electronvolts

You will often see work functions expressed in electronvolts (eV), rather than joules.

Remember:

\[ 1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm{J} \]

An electron gains 1 eV of energy when it moves through a potential difference of 1 V.

Consequently, the numerical value of the maximum kinetic energy in eV equals the numerical value of the stopping potential in V:

\[ K_{\max}(\mathrm{eV})=V_s(\mathrm{V}) \]

For example, a stopping potential of \(0.85\ \mathrm{V}\) corresponds to a maximum kinetic energy of \(0.85\ \mathrm{eV}\).

This makes calculations much quicker when the work function is required in electronvolts.

03Do not confuse photocurrent with work function

This is one of the most tempting mistakes when comparing two photoelectric graphs.

Imagine two metals illuminated by the same light.

Metal A produces a saturation current of \(6.0\ \mathrm{\mu A}\), while metal B produces only \(2.0\ \mathrm{\mu A}\).

Does metal A have a smaller work function because it produces more current?

No.

Current measures the rate at which charge reaches the collector:

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

If \(N\) electrons reach the collector in a time interval \(\Delta t\), then

\[ I=\frac{Ne}{\Delta t} \]

A larger current means more electrons are reaching the collector per second.

It does not mean that each electron has more kinetic energy.

The current can depend on:

• The intensity of the incident light.
• The number of photons arriving each second.
• The fraction of photons that produce emitted electrons.
• The illuminated surface area.
• The efficiency of electron collection.

The work function, however, determines the minimum energy needed to release an electron.

Think of intensity as sending more identical energy packets towards the metal each second. More packets can produce more emitted electrons, but they do not increase the energy carried by each individual photon.

The analogy has a limit: photons are quantised packets of energy, not ordinary physical objects.

What happens if you increase light intensity?

At a fixed frequency, increasing light intensity increases the number of incident photons per second.

This can increase the photocurrent.

However, the energy of each photon remains unchanged because its frequency has not changed.

Therefore, in the standard single-photon model:

• Photocurrent can increase.
• Saturation current can increase.
• Maximum kinetic energy remains unchanged.
• Stopping potential remains unchanged.

This is another reason why current height cannot be used to compare work functions.

05Important exception: What if the light frequencies are different?

So far, both metals have received light of the same frequency.

This condition matters.

Suppose metal A receives higher-frequency light than metal B.

The photons striking metal A carry more energy. Consequently, metal A could produce electrons with greater kinetic energy even if its work function is also greater.

You can no longer rank the work functions by comparing stopping potentials alone.

Instead, calculate the work function for each metal separately:

\[ \phi_A=hf_A-eV_{s,A} \]

\[ \phi_B=hf_B-eV_{s,B} \]

You must account for both the photon energy and the maximum kinetic energy.

06What if the question gives a different type of photoelectric graph?

Not every photoelectric graph has photocurrent on the vertical axis.

Before calculating anything, check what the graph actually plots.

The threshold frequency, \(f_0\), is the minimum frequency required to produce photoelectric emission.

At this frequency, the photon has exactly enough energy to release an electron, leaving it with zero maximum kinetic energy.

Therefore:

\[ hf_0=\phi \]

Rearranging:

\[ \boxed{\phi=hf_0} \]

This gives another way to compare work functions.

If two metals have different threshold frequencies, the metal with the higher threshold frequency has the larger work function.

For a more detailed explanation of gradients, vertical intercepts, and threshold-frequency intercepts, see Photoelectric Effect Graphs: Gradients and Intercepts Explained.

07Experimental checks before comparing the graphs

A good comparison depends on more than just reading two numbers.

Consider these checks when interpreting experimental data.

1. Were both metals illuminated at the same frequency?

If yes, you can compare stopping potentials directly.

If not, calculate the photon energy for each metal separately.

2. Are the stopping potentials expressed using the same sign convention?

If the horizontal axis shows signed collector potential, the stopping point may be negative.

Use its positive magnitude when calculating maximum kinetic energy.

For example:

\[ V_{\text{stop}}=-0.85\ \mathrm{V} \]

means

\[ V_s=0.85\ \mathrm{V} \]

3. Are you reading the stopping potential or saturation current?

Do not use the maximum current as a substitute for stopping potential.

They represent different physical quantities.

4. Has the dark current been accounted for?

Real measuring equipment may register a small current even when the light is switched off.

In this situation, zero photocurrent means the measured current has returned to its dark-current baseline, not necessarily that the ammeter displays exactly zero.

5. Was the experimental setup controlled?

When comparing two surfaces, keep the light frequency constant. If you also want to compare photocurrents, control the light intensity, illuminated area, and electrode geometry.

The surface condition matters too. Contamination or changes to the metal surface can affect its measured work function.

08Questions and solutions

The following questions progress from straightforward graph interpretation to more demanding comparisons involving different frequencies, experimental limitations, and measurement uncertainty.

Question 1

Two metal cathodes, X and Y, are illuminated using the same monochromatic light.

Their photocurrent-voltage graphs show the following results:

Determine:

a) Which metal has the larger work function?

b) Calculate the difference between their work functions in electronvolts and joules.

c) Explain why the saturation currents do not determine your answer.

Solution 1

Metal X has the larger work function, by \(0.44\ \mathrm{eV}\), or \(7.04\times10^{-20}\ \mathrm{J}\).

Step 1: Identify the stopping-potential magnitudes.

\[ V_{s,X}=0.48\ \mathrm{V} \]

\[ V_{s,Y}=0.92\ \mathrm{V} \]

Because both metals receive light at the same frequency, their photon energies are equal.

Step 2: Calculate the difference in work functions.

Using the difference equation:

\[ \phi_X-\phi_Y=e(V_{s,Y}-V_{s,X}) \]

Substitute:

\[ \begin{aligned} \phi_X-\phi_Y &=e(0.92-0.48)\\ &=e(0.44\ \mathrm{V})\\ &=0.44\ \mathrm{eV} \end{aligned} \]

Convert to joules:

\[ \begin{aligned} \Delta\phi &=0.44(1.60\times10^{-19})\\ &=7.04\times10^{-20}\ \mathrm{J} \end{aligned} \]

Step 3: Explain the role of current.

Metal X has the higher saturation current, but this only tells us that more charge reaches its collector each second.

Metal Y has the larger stopping potential, meaning its fastest electrons have greater kinetic energy.

Since both metals receive photons with the same energy, metal X must require more energy to release its electrons.

Therefore, metal X has the larger work function.

Question 2

Two different metal cathodes, P and Q, are illuminated using monochromatic light of frequency \(6.8\times10^{14}\ \mathrm{Hz}\).

Their photocurrent-voltage graphs have the following features:

A student argues that metal P must have the smaller work function because it produces a greater photocurrent.

Calculate the work function of each metal in:

a) Electronvolts.

b) Joules.

c) Assess the student’s claim using the photoelectric model.

Use:

\[ h=4.14\times10^{-15}\ \mathrm{eV\,s} \]

\[ 1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm{J} \]

Solution 2

Metal P has a work function of approximately \(2.20\ \mathrm{eV}\), while metal Q has a work function of approximately \(1.80\ \mathrm{eV}\). The student’s claim is incorrect.

Step 1: Calculate the photon energy.

Both metals receive light of the same frequency.

\[ \begin{aligned} E_{\text{photon}}&=hf\\ &=(4.14\times10^{-15})(6.8\times10^{14})\\ &=2.8152\ \mathrm{eV} \end{aligned} \]

Step 2: Calculate the maximum kinetic energy of each metal’s electrons.

Use the positive magnitudes of the stopping potentials.

\[ K_{\max,P}=0.62\ \mathrm{eV} \]

\[ K_{\max,Q}=1.02\ \mathrm{eV} \]

Step 3: Calculate the work functions.

For metal P:

\[ \begin{aligned} \phi_P&=hf-K_{\max,P}\\ &=2.8152-0.62\\ &=2.1952\ \mathrm{eV}\\ &\approx2.20\ \mathrm{eV} \end{aligned} \]

For metal Q:

\[ \begin{aligned} \phi_Q&=hf-K_{\max,Q}\\ &=2.8152-1.02\\ &=1.7952\ \mathrm{eV}\\ &\approx1.80\ \mathrm{eV} \end{aligned} \]

Step 4: Convert the results into joules.

For metal P:

\[ \begin{aligned} \phi_P&=2.20(1.60\times10^{-19})\\ &=3.52\times10^{-19}\ \mathrm{J} \end{aligned} \]

For metal Q:

\[ \begin{aligned} \phi_Q&=1.80(1.60\times10^{-19})\\ &=2.88\times10^{-19}\ \mathrm{J} \end{aligned} \]

Step 5: Assess the student’s claim.

The student has confused the number of emitted electrons with their kinetic energy.

Metal P produces more current, but metal Q has the larger stopping potential.

Because the photon energy is identical for both metals, Q’s electrons retain more kinetic energy after escaping.

Consequently, P has the larger work function.

Question 3

Two metal surfaces, M and N, are tested using different monochromatic light sources.

The measured stopping potentials are:

A student claims that metal N must have the larger work function because its stopping potential is smaller.

a) Calculate the work function of each metal in electronvolts.

b) Determine which metal actually has the larger work function.

c) Explain why comparing stopping potentials alone produces an incorrect conclusion.

Use:

\[ h=4.14\times10^{-15}\ \mathrm{eV\,s} \]

Solution 3

Metal M has the larger work function, approximately \(2.13\ \mathrm{eV}\), compared with \(1.98\ \mathrm{eV}\) for metal N. The student’s conclusion is incorrect.

Step 1: Calculate the photon energy for metal M.

\[ \begin{aligned} E_M&=hf_M\\ &=(4.14\times10^{-15})(7.2\times10^{14})\\ &=2.9808\ \mathrm{eV} \end{aligned} \]

Step 2: Calculate the photon energy for metal N.

\[ \begin{aligned} E_N&=hf_N\\ &=(4.14\times10^{-15})(6.0\times10^{14})\\ &=2.484\ \mathrm{eV} \end{aligned} \]

Step 3: Calculate both work functions.

For metal M:

\[ \begin{aligned} \phi_M&=hf_M-K_{\max,M}\\ &=2.9808-0.85\\ &=2.1308\ \mathrm{eV} \end{aligned} \]

For metal N:

\[ \begin{aligned} \phi_N&=hf_N-K_{\max,N}\\ &=2.484-0.50\\ &=1.984\ \mathrm{eV} \end{aligned} \]

Step 4: Compare the results.

\[ \phi_M>\phi_N \]

The difference is approximately

\[ \Delta\phi=2.1308-1.984 =0.147\ \mathrm{eV} \]

Metal M has the larger work function.

The tempting approach is to assume that a smaller stopping potential always means a larger work function.

That only works when the photon energies are equal.

Metal M receives higher-energy photons, so its electrons can have greater kinetic energy despite the metal requiring more energy to release them.

Question 4

An experiment compares two metal cathodes using the same monochromatic light.

Before switching on the light, the ammeter records a background current of \(0.12\ \mathrm{\mu A}\). This is the dark-current baseline.

With the light switched on, the following readings are obtained:

A student argues that the stopping potential cannot be determined because neither graph reaches an actual ammeter reading of zero.

a) Explain how the dark-current baseline should be interpreted.

b) Determine which metal has the larger work function.

c) Calculate the difference between their work functions in electronvolts and joules.

d) Explain why the maximum measured currents do not affect your comparison.

Assume the dark-current baseline is constant throughout the voltage sweep.

Solution 4

Metal P has the larger work function, by \(0.30\ \mathrm{eV}\), or \(4.80\times10^{-20}\ \mathrm{J}\). The stopping potential is measured where the current returns to the dark-current baseline, not necessarily where the ammeter reads zero.

Step 1: Interpret the dark-current baseline.

The measured current contains two contributions:

• The background current present even without illumination.
• The additional photocurrent caused by emitted electrons.

Therefore:

\[ I_{\text{photocurrent}} =I_{\text{measured}}-I_{\text{dark}} \]

When the measured current returns to \(0.12\ \mathrm{\mu A}\), the photocurrent is zero.

The student is incorrectly treating the background current as evidence that photoelectrons are still reaching the collector.

Step 2: Identify the stopping-potential magnitudes.

For metal P:

\[ V_{s,P}=0.72\ \mathrm{V} \]

For metal Q:

\[ V_{s,Q}=1.02\ \mathrm{V} \]

Step 3: Calculate the difference in work functions.

Since both metals receive the same photon energy:

\[ \phi_P-\phi_Q=e(V_{s,Q}-V_{s,P}) \]

Substitute:

\[ \begin{aligned} \phi_P-\phi_Q &=e(1.02-0.72)\\ &=e(0.30\ \mathrm{V})\\ &=0.30\ \mathrm{eV} \end{aligned} \]

Convert to joules:

\[ \begin{aligned} \Delta\phi &=0.30(1.60\times10^{-19})\\ &=4.80\times10^{-20}\ \mathrm{J} \end{aligned} \]

Step 4: Explain the role of the maximum currents.

The maximum photocurrent indicates the rate of charge reaching the collector.

It does not determine the energy of the fastest emitted electrons.

The stopping potential provides that information, so the different current plateaus do not change the work-function comparison.

Question 5

Two experimental groups investigate different metal cathodes, X and Y.

The groups use different light frequencies and different conventions for plotting the collector voltage.

Their measurements are:

Both stopping-point readings have already been corrected for dark current.

Use:

\[ h=4.14\times10^{-15}\ \mathrm{eV\,s} \]

A third measurement is performed on cathode Y. The light intensity is doubled while its frequency remains unchanged. The saturation current doubles, but the stopping potential remains \(0.39\ \mathrm{V}\).

Answer the following:

a) Convert both stopping-point readings into positive stopping-potential magnitudes.

b) Calculate the best estimate of the work function for each cathode.

c) Considering the stated measurement uncertainties, can you confidently identify which cathode has the larger work function? Justify your conclusion quantitatively.

d) Explain what the additional intensity experiment establishes, and why the different saturation currents cannot establish which cathode has the smaller work function.

e) Do the results prove that X and Y are the same metal? Propose a more reliable experimental comparison.

Solution 5

The best estimates of the work functions are \(1.684\ \mathrm{eV}\) for X and \(1.680\ \mathrm{eV}\) for Y. The difference is too small to distinguish the work functions confidently within the stated measurement uncertainties.

The important challenge is to account for the different photon energies, convert the voltage conventions correctly, and avoid claiming more precision than the experiment supports.

Step 1: Convert both stopping-potential readings into positive magnitudes.

Cathode X uses a signed collector-potential axis:

\[ V_{s,X}=|-0.80|=0.80\ \mathrm{V} \]

Cathode Y uses a positive reverse-potential magnitude:

\[ V_{s,Y}=0.39\ \mathrm{V} \]

Both are now expressed using the same convention.

Step 2: Calculate the work function of cathode X.

Use:

\[ \phi_X=hf_X-eV_{s,X} \]

Since we are working in electronvolts, the stopping-potential magnitude has the same numerical value as the maximum kinetic energy in eV.

Therefore:

\[ \begin{aligned} \phi_X &=(4.14\times10^{-15})(6.00\times10^{14})-0.80\\ &=2.484-0.80\\ &=1.684\ \mathrm{eV} \end{aligned} \]

Step 3: Calculate the work function of cathode Y.

\[ \begin{aligned} \phi_Y &=(4.14\times10^{-15})(5.00\times10^{14})-0.39\\ &=2.070-0.39\\ &=1.680\ \mathrm{eV} \end{aligned} \]

The best estimates are therefore:

\[ \phi_X=1.684\ \mathrm{eV} \]

\[ \phi_Y=1.680\ \mathrm{eV} \]

The difference is only

\[ \Delta\phi=0.004\ \mathrm{eV} \]

Step 4: Account for measurement uncertainty.

A calculated work function depends on both the measured frequency and the measured stopping potential.

A small uncertainty in either measurement changes the calculated work function.

Using a conservative worst-case addition of the uncertainties:

\[ \delta\phi\approx h\delta f+\delta V_s \]

Here, the frequency uncertainty contributes:

\[ \begin{aligned} h\delta f &=(4.14\times10^{-15})(0.05\times10^{14})\\ &=0.0207\ \mathrm{eV} \end{aligned} \]

The stopping-potential uncertainty contributes another \(0.03\ \mathrm{eV}\).

Therefore:

\[ \delta\phi\approx0.0207+0.03 \]

\[ \delta\phi\approx0.051\ \mathrm{eV} \]

The results can be expressed approximately as:

\[ \phi_X=1.684\pm0.051\ \mathrm{eV} \]

\[ \phi_Y=1.680\pm0.051\ \mathrm{eV} \]

These uncertainty ranges overlap substantially.

The calculated difference of \(0.004\ \mathrm{eV}\) is much smaller than the uncertainty in either result.

We therefore cannot confidently state that X has a larger work function than Y, or vice versa.

The measurements are consistent with the two surfaces having the same work function.

Step 5: Interpret the additional intensity experiment.

Doubling the light intensity doubles the saturation current, but leaves the stopping potential unchanged.

This supports the single-photon model.

Increasing intensity increases the number of photons arriving each second. It can therefore increase the number of emitted electrons and the resulting photocurrent.

However, the energy carried by each photon remains unchanged because the frequency is unchanged.

Consequently, the maximum kinetic energy and stopping potential remain unchanged.

The fourfold difference between the original saturation currents of X and Y cannot be used to rank their work functions.

Their current difference could reflect different emission efficiencies, incident intensities, or other experimental factors.

Step 6: Decide what the experiment actually establishes.

The measurements suggest that the two surfaces have approximately equal work functions.

However, this does not prove that X and Y are the same metal. Different materials can have similar work functions.

A better comparison would use:

• The same calibrated monochromatic light source for both cathodes.
• A frequency above both estimated threshold frequencies.
• The same voltage-axis convention.
• The same procedure for correcting dark current.
• Repeated stopping-potential measurements to reduce uncertainty.

For example, a frequency of \(6.0\times10^{14}\ \mathrm{Hz}\) would be above the estimated thresholds of both surfaces.

Using the same frequency would allow the difference in work functions to be determined directly from the difference in stopping potentials.

The key lesson is that a graph provides evidence about a physical quantity, but the precision of that evidence depends on the measurements and the assumptions used to interpret it.

Once you can distinguish the rate of electron emission from the energy of the fastest electrons, the next useful step is studying photocurrent-versus-intensity graphs. These reveal how changing the number of incident photons affects electron emission, while stopping-potential graphs reveal how photon frequency affects electron energy.