How to Sketch Photocurrent Against Applied Voltage

A photocell is illuminated, and its ammeter reads a current even when the applied voltage is zero. Make the anode positive, and the current rises before flattening out. Reverse the supply, and the current falls to zero at a negative voltage.

How should you sketch this relationship, and what do its two important features actually tell you?

For a standard photocurrent-versus-applied-voltage graph, draw a rising curve that approaches a horizontal plateau at positive voltage. On the negative side, the curve reaches zero current at the stopping potential.

Using the sign convention \(V=V_A-V_C\), where \(V_A\) is the anode potential and \(V_C\) is the cathode potential:

• The zero-current intercept is \(V=-V_s\), where \(V_s\) is the positive magnitude of the stopping potential.
• The horizontal plateau is the saturation current, \(I_s\).
• The stopping potential determines the maximum kinetic energy of the emitted electrons: \(K_{\max}=eV_s\).
• The saturation current tells you the rate at which electrons are being emitted.

These last two points are easy to confuse. One describes the energy of the fastest electrons; the other describes how many electrons are produced each second.

01Understand what the experiment measures

Before drawing the graph, picture the apparatus that produces it.

A photoelectric experiment uses an evacuated photocell containing two electrodes:

• Cathode (C): The metal surface being investigated. Light strikes this surface and can release electrons.
• Anode (A): The collector that attracts the emitted electrons.

An ammeter measures the photocurrent in the external circuit. A variable power supply changes the potential difference between the electrodes, and a voltmeter measures that potential difference.

The power supply must be able to reverse its polarity so that the anode can be made either positive or negative relative to the cathode.

What happens when the voltage changes?

Keep the light frequency and intensity constant. Now vary the applied voltage and record the current.

There are two different situations to investigate.

Situation 1: The anode is positive

The positive anode attracts the negatively charged photoelectrons.

As the positive voltage increases, more of the emitted electrons reach the anode. The photocurrent increases.

Eventually, every emitted electron that can be collected reaches the anode. Increasing the voltage further cannot increase the number of electrons being emitted by the cathode.

The current reaches a maximum value called the saturation current.

Situation 2: The anode is negative

Now the anode repels the emitted electrons.

Some electrons still have enough kinetic energy to reach the anode, but others are turned back towards the cathode.

As the negative voltage becomes increasingly large in magnitude, fewer electrons reach the collector. The photocurrent decreases.

Eventually, even the fastest emitted electrons cannot reach the anode. The current becomes zero.

The magnitude of the reverse voltage at this point is called the stopping potential.

A useful experimental improvement is therefore to use a reversible power supply. A circuit that only allows a positive anode voltage can measure the saturation current, but it cannot determine the stopping potential or the maximum kinetic energy.

02Establish the correct graph shape

The horizontal axis represents the applied voltage:

\[
V=V_A-V_C
\]

The vertical axis represents the magnitude of the photocurrent, measured in amperes (A), milliamperes (mA), or microamperes (\(\mu\text{A}\)).

For the standard graph, plot the photocurrent as a positive quantity.

With this sign convention:

Notice something important: the current does not necessarily equal zero when the applied voltage is zero.

Why not?

Photoelectrons already have kinetic energy when they leave the metal. Some can travel across the gap without needing an additional accelerating voltage.

This means your graph should not automatically pass through the origin.

An illustrative set of readings

Suppose a photocell produces the following illustrative data. These values are invented to demonstrate the shape of the graph, rather than representing measurements from a particular experiment.

Read the table from left to right:

1. At \(-0.80\text{ V}\), the current is zero. This is the stopping point.
2. As the voltage becomes less negative, the current increases because more electrons reach the anode.
3. At zero applied voltage, the current is already \(1.42\ \mu\text{A}\).
4. At positive voltage, the current increases towards \(2.00\ \mu\text{A}\).
5. Beyond approximately \(+1.00\text{ V}\), the current remains constant.

The resulting graph has two features you must always identify:

• The negative-voltage intercept, which gives the stopping potential.
• The positive-voltage plateau, which gives the saturation current.

The exact curvature between these features depends on the distribution of photoelectron energies and directions. A qualitative sketch should show the correct overall behaviour rather than attempt to reproduce an exact mathematical curve.

03Stopping potential: finding maximum kinetic energy

Why does the stopping potential tell us about the fastest photoelectrons?

Consider what happens when the anode becomes negative.

The negative anode creates an electrical potential-energy barrier. Electrons need sufficient kinetic energy to overcome this barrier and reach the collector.

You can picture this as an uphill climb. Increasing the reverse voltage makes the hill harder to climb. Slower electrons turn back first, while faster electrons can continue towards the anode.

The analogy is about energy, not an actual gravitational force: the electron is being affected by an electric field.

At the stopping potential, even the fastest electrons can no longer reach the anode.

The electrical potential-energy increase is equal to the maximum initial kinetic energy.

Therefore:

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

Where:

• \(K_{\max}\) is the maximum kinetic energy of the emitted electrons, in joules (J).
• \(e\) is the magnitude of the electron’s charge, \(1.60\times10^{-19}\text{ C}\).
• \(V_s\) is the positive magnitude of the stopping potential, in volts (V).

Remember that the electron has a negative charge, but we use the positive magnitude \(e\) in this equation.

If kinetic energy is expressed in electronvolts, the numerical value of the stopping potential in volts is the same.

For example:

\[
V_s=0.80\text{ V}
\]

means:

\[
K_{\max}=0.80\text{ eV}
\]

or, in joules:

\[
K_{\max}=0.80(1.60\times10^{-19})
\]

\[
K_{\max}=1.28\times10^{-19}\text{ J}
\]

Connecting stopping potential to the single-photon model

The single-photon model explains why changing the light frequency changes the stopping potential.

One photon transfers its energy to one electron. Some energy is needed to release the electron from the metal, and the remaining energy becomes kinetic energy.

The energy of a photon is:

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

where \(h\) is Planck’s constant and \(f\) is the light frequency.

The minimum energy required to release an electron is called the work function, \(\phi\).

Conservation of energy gives:

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

Therefore:

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

Combining this with the stopping-potential relationship:

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

This equation connects the light source, the metal surface, and the graph’s negative-voltage intercept.

A higher light frequency produces more energetic photons. For the same metal, this increases the maximum kinetic energy and increases the magnitude of the stopping potential.

Importantly, the stopping potential measures the maximum kinetic energy, not the average kinetic energy of all emitted electrons.

04Saturation current: finding the emission rate

Now consider the other important feature of the graph.

The saturation current tells you how much charge reaches the anode each second when all emitted photoelectrons are collected.

This is different from the stopping potential, which tells you about the energy of the fastest electrons.

Current is the rate of charge flow:

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

If \(N\) electrons are emitted each second, the total charge emitted per second is:

\[
Q=Ne
\]

Therefore, the emission current is:

\[
\boxed{I_{\text{em}}=Ne}
\]

Here, \(N\) is the number of electrons emitted per second, and \(e\) is the magnitude of the electron charge.

At saturation, all emitted electrons are collected, assuming negligible losses. Therefore:

\[
\boxed{I_s=I_{\text{em}}}
\]

This is why the plateau gives the maximum photocurrent.

At lower applied voltages, some emitted electrons fail to reach the anode, so the measured photocurrent is smaller than the emission current.

What happens if the light intensity increases?

Suppose you illuminate the same metal with light of the same frequency but increase its intensity.

What changes?

In the single-photon model, each photon has the same energy as before. However, a greater photon flux means more photons arrive each second.

Assuming the probability of emission per incident photon remains constant, more electrons are emitted each second.

The result is:

• The saturation current increases.
• The stopping potential remains unchanged.
• The maximum kinetic energy remains unchanged.

The graph’s plateau becomes higher, but its negative-voltage intercept stays in the same position.

Intensity versus frequency: the important distinction

There is an important qualification when comparing frequencies.

Light intensity is often measured as energy arriving per second per unit area. If the optical power is held constant while frequency increases, the number of photons arriving each second decreases because each photon carries more energy.

Consequently, the saturation current can decrease even while the stopping potential increases.

We’ll use this less obvious situation in the final practice question.

What if the frequency is below the threshold?

If the photon energy is less than the metal’s work function:

\[
hf<\phi
\]

No electrons are emitted in the single-photon model.

Increasing the light intensity does not solve the problem. You are simply supplying more photons, each of which still has insufficient energy to release an electron.

This is a useful distinction:

• Intensity controls the number of photons arriving.
• Frequency controls the energy carried by each photon.

05Worked examples

06A reliable method for sketching the graph

When an exam question asks you to sketch photocurrent against applied voltage, follow this sequence.

1. Label both axes.

• Horizontal: applied voltage, \(V_A-V_C\), in volts.
• Vertical: photocurrent magnitude, in amperes or microamperes.

2. Identify the stopping potential.

If the stopping-potential magnitude is \(V_s\), mark the zero-current intercept at \(-V_s\).

3. Identify the saturation current.

Mark the maximum current, \(I_s\), as a horizontal plateau at positive applied voltage.

4. Draw the curve.

Start at zero current at the negative stopping-potential intercept. Increase the current smoothly as the applied voltage increases, then approach the saturation plateau.

The current will generally be non-zero at zero applied voltage.

5. Label the physical meaning.

Write:

\[
K_{\max}=eV_s
\]

beside the stopping potential, and identify the plateau as the emission current under complete collection.

One final sign-convention warning: some questions plot the magnitude of the retarding potential rather than \(V_A-V_C\). In that case, the stopping potential may appear as a positive intercept. Always check what the horizontal axis represents before assigning a sign.

07Questions and solutions

Question 1

A photocell is illuminated with monochromatic light. Its photocurrent reaches zero when the applied voltage is \(-0.72\text{ V}\), and its saturation current is \(1.8\ \mu\text{A}\).

The horizontal axis is defined as \(V_A-V_C\).

Sketch the photocurrent-versus-applied-voltage graph, label the stopping potential and saturation current, and calculate the maximum kinetic energy of the emitted electrons in both eV and joules.

Solution 1

The stopping-potential magnitude is \(0.72\text{ V}\), and the maximum kinetic energy is \(0.72\text{ eV}\), or approximately \(1.15\times10^{-19}\text{ J}\).

Step 1: Identify the stopping potential.

The graph reaches zero current at:

\[
V=-0.72\text{ V}
\]

Therefore:

\[
V_s=0.72\text{ V}
\]

Step 2: Calculate maximum kinetic energy.

Use:

\[
K_{\max}=eV_s
\]

In electronvolts:

\[
\boxed{K_{\max}=0.72\text{ eV}}
\]

In joules:

\[
K_{\max}=0.72(1.60\times10^{-19})
\]

\[
\boxed{K_{\max}=1.15\times10^{-19}\text{ J}}
\]

Step 3: Sketch the graph.

Your graph should:

• Reach zero current at \(-0.72\text{ V}\).
• Rise as the applied voltage increases.
• Approach a horizontal plateau at \(1.8\ \mu\text{A}\).
• Generally show a non-zero current at zero applied voltage.

The negative intercept measures electron energy, while the plateau measures the rate of electron collection. They describe different properties of the experiment.

Question 2

A metal cathode is illuminated using monochromatic light of fixed frequency.

Two photocurrent-voltage curves are recorded:

The quantum efficiency and the metal are unchanged.

A student claims that Curve B must represent electrons with greater maximum kinetic energy because its saturation current is larger.

Calculate the ratio of the electron emission rates for the two curves, determine the maximum kinetic energy in each case, and explain why the student’s conclusion is incorrect.

Solution 2

Curve B has an electron emission rate 2.5 times that of Curve A, but both curves have the same maximum kinetic energy of \(0.60\text{ eV}\).

Step 1: Relate saturation current to electron emission rate.

At saturation:

\[
I_s=Ne
\]

Therefore:

\[
\frac{N_B}{N_A}=\frac{I_{s,B}}{I_{s,A}}
\]

Substitute:

\[
\frac{N_B}{N_A}=\frac{2.00}{0.80}
\]

\[
\boxed{\frac{N_B}{N_A}=2.5}
\]

Curve B produces 2.5 times as many electrons per second.

Step 2: Calculate the maximum kinetic energy.

Both curves have:

\[
V_s=0.60\text{ V}
\]

Using:

\[
K_{\max}=eV_s
\]

Therefore:

\[
\boxed{K_{\max}=0.60\text{ eV}}
\]

for both curves.

In joules:

\[
K_{\max}=0.60(1.60\times10^{-19})
\]

\[
K_{\max}=9.6\times10^{-20}\text{ J}
\]

Step 3: Explain the misconception.

The student has confused the number of emitted electrons with the energy carried by each electron.

Increasing light intensity at fixed frequency increases the photon flux. More photons arrive each second, so more electrons are emitted.

However, the energy of each photon remains unchanged.

Consequently, the saturation current increases while the stopping potential and maximum kinetic energy remain constant.

Question 3

A metal has a work function of \(2.00\text{ eV}\).

It is illuminated separately with two monochromatic light sources:

• Source A: wavelength \(500\text{ nm}\).
• Source B: wavelength \(400\text{ nm}\).

Both sources have the same photon flux, and the probability of electron emission per incident photon is the same for both.

Calculate the stopping-potential magnitude for each source. Then describe how the two photocurrent-voltage curves compare, including their saturation currents.

Solution 3

Source B produces the greater stopping-potential magnitude, \(1.10\text{ V}\), while Source A produces \(0.48\text{ V}\). Their saturation currents are equal under the stated assumptions.

Step 1: Calculate the photon energy for Source A.

Using:

\[
E_{\text{photon}}=\frac{1240}{\lambda}
\]

\[
E_A=\frac{1240}{500}
\]

\[
E_A=2.48\text{ eV}
\]

Step 2: Calculate its maximum kinetic energy.

\[
K_{\max,A}=E_A-\phi
\]

\[
K_{\max,A}=2.48-2.00
\]

\[
K_{\max,A}=0.48\text{ eV}
\]

Therefore:

\[
\boxed{V_{s,A}=0.48\text{ V}}
\]

Its graph reaches zero current at \(-0.48\text{ V}\).

Step 3: Repeat for Source B.

\[
E_B=\frac{1240}{400}
\]

\[
E_B=3.10\text{ eV}
\]

Therefore:

\[
K_{\max,B}=3.10-2.00
\]

\[
K_{\max,B}=1.10\text{ eV}
\]

So:

\[
\boxed{V_{s,B}=1.10\text{ V}}
\]

Its graph reaches zero current at \(-1.10\text{ V}\).

Step 4: Compare the saturation currents.

Both sources have the same photon flux and the same probability of producing an emitted electron.

Therefore, they produce the same number of emitted electrons per second.

Their saturation currents are equal.

Step 5: Compare the graph shapes.

Both curves approach the same saturation-current plateau.

However, Source B has a more negative stopping-potential intercept because its photons have greater energy.

This is a useful way to separate the two effects:

• Photon frequency determines the maximum kinetic energy.
• Photon arrival rate, together with emission efficiency, determines the saturation current.

Question 4

A metal has a work function of \(2.35\text{ eV}\).

It is illuminated separately by two monochromatic light sources:

• Source A has a wavelength of \(560\text{ nm}\) and a photon flux 100 times greater than Source B.
• Source B has a wavelength of \(500\text{ nm}\).

For each source, determine whether photoelectrons are emitted.

For the source that produces emission, calculate the maximum kinetic energy and stopping potential.

Describe the qualitative photocurrent-voltage graph for each source. Explain why the much greater intensity of Source A does not guarantee a photocurrent.

Solution 4

Source A produces no photoelectrons, despite its much greater photon flux. Source B produces photoelectrons with a maximum kinetic energy of \(0.13\text{ eV}\) and a stopping-potential magnitude of \(0.13\text{ V}\).

Step 1: Calculate the photon energy from Source A.

\[
E_A=\frac{1240}{560}
\]

\[
E_A=2.21\text{ eV}
\]

Compare this with the work function:

\[
2.21<2.35
\]

Each photon has insufficient energy to release an electron.

Therefore:

\[
\boxed{\text{No photoelectrons are emitted by Source A.}}
\]

Even increasing its photon flux by a factor of 100 does not change this conclusion.

Step 2: Calculate the photon energy from Source B.

\[
E_B=\frac{1240}{500}
\]

\[
E_B=2.48\text{ eV}
\]

Since:

\[
2.48>2.35
\]

photoelectric emission occurs.

Step 3: Calculate maximum kinetic energy.

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

\[
K_{\max}=2.48-2.35
\]

\[
\boxed{K_{\max}=0.13\text{ eV}}
\]

Step 4: Calculate stopping potential.

Using the numerical equivalence between eV and volts:

\[
\boxed{V_s=0.13\text{ V}}
\]

With the chosen voltage convention, the photocurrent reaches zero at:

\[
V=-0.13\text{ V}
\]

Step 5: Describe the two graphs.

For Source A, there is no photocurrent at any applied voltage because no photoelectrons are produced.

For Source B, the current is zero at \(-0.13\text{ V}\), increases as the applied voltage becomes more positive, and eventually reaches a saturation plateau.

The actual plateau height cannot be calculated without the absolute photon flux and emission probability for Source B.

The important principle is that photon energy and photon number are not interchangeable. A large number of photons cannot compensate for insufficient energy per photon in the single-photon model.

Question 5

A photocell has a metal cathode with a work function of \(1.80\text{ eV}\).

Two monochromatic light sources are available:

• Source A: wavelength \(400\text{ nm}\).
• Source B: wavelength \(600\text{ nm}\).

Each source delivers an optical power of \(1.20\text{ mW}\) to the cathode.

For both wavelengths, assume that 25% of incident photons produce an emitted photoelectron. This emission probability is constant, and every emitted electron can be collected when the anode is sufficiently positive.

Assume there are no contact-potential effects or significant collection losses.

(a) Calculate the maximum kinetic energy and stopping-potential magnitude for each source.

(b) Calculate the saturation current produced by each source.

(c) Both light sources are now switched on simultaneously. Calculate the combined saturation current and determine the stopping-potential intercept of the combined graph.

(d) A student argues that Source A must produce the greater saturation current because its photons have greater energy. Explain why this prediction is wrong.

(e) When both sources are operating, the anode is set to \(-0.50\text{ V}\). Which light source can still contribute to the photocurrent? Can the exact current at this voltage be calculated from the information provided? Explain.

Solution 5

Source A produces more energetic electrons and has a stopping-potential magnitude of \(1.30\text{ V}\). However, Source B produces the greater saturation current because its lower-energy photons arrive in greater numbers per second at the same optical power.

When both sources operate simultaneously, the saturation current is approximately \(242\ \mu\text{A}\), and the stopping-potential intercept remains \(-1.30\text{ V}\).

Step 1: Calculate the photon energy for each source.

For Source A:

\[
E_A=\frac{1240}{400}
\]

\[
E_A=3.10\text{ eV}
\]

For Source B:

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

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

Both photon energies exceed the work function of \(1.80\text{ eV}\), so both sources produce photoelectrons.

Step 2: Calculate the maximum kinetic energies.

For Source A:

\[
K_{\max,A}=3.10-1.80
\]

\[
\boxed{K_{\max,A}=1.30\text{ eV}}
\]

For Source B:

\[
K_{\max,B}=2.07-1.80
\]

\[
\boxed{K_{\max,B}=0.27\text{ eV}}
\]

Step 3: Determine the stopping potentials.

The numerical value of maximum kinetic energy in eV equals the stopping-potential magnitude in volts.

Therefore:

\[
V_{s,A}=1.30\text{ V}
\]

\[
V_{s,B}=0.27\text{ V}
\]

The separate graphs reach zero current at:

\[
\boxed{V_A=-1.30\text{ V}}
\]

and:

\[
\boxed{V_B=-0.27\text{ V}}
\]

Source A produces the faster photoelectrons.

Step 4: Calculate the saturation current for Source A.

We need to calculate how many photons arrive each second.

First, convert the photon energy into joules:

\[
E_A=3.10(1.602\times10^{-19})
\]

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

The photon arrival rate is:

\[
N_{\gamma,A}=\frac{P}{E_A}
\]

Substitute the optical power:

\[
N_{\gamma,A}
=\frac{1.20\times10^{-3}}{4.97\times10^{-19}}
\]

\[
N_{\gamma,A}=2.41\times10^{15}\text{ photons/s}
\]

Only 25% of these photons produce an emitted electron:

\[
N_{e,A}=0.25(2.41\times10^{15})
\]

\[
N_{e,A}=6.03\times10^{14}\text{ electrons/s}
\]

The saturation current is:

\[
I_{s,A}=N_{e,A}e
\]

\[
I_{s,A}
=(6.03\times10^{14})(1.602\times10^{-19})
\]

\[
\boxed{I_{s,A}\approx96.6\ \mu\text{A}}
\]

Step 5: Calculate the saturation current for Source B.

First, convert its photon energy into joules:

\[
E_B=2.07(1.602\times10^{-19})
\]

\[
E_B\approx3.31\times10^{-19}\text{ J}
\]

Calculate its photon arrival rate:

\[
N_{\gamma,B}
=\frac{1.20\times10^{-3}}{3.31\times10^{-19}}
\]

\[
N_{\gamma,B}=3.63\times10^{15}\text{ photons/s}
\]

The electron emission rate is:

\[
N_{e,B}=0.25(3.63\times10^{15})
\]

\[
N_{e,B}=9.08\times10^{14}\text{ electrons/s}
\]

Therefore:

\[
I_{s,B}=N_{e,B}e
\]

\[
I_{s,B}
=(9.08\times10^{14})(1.602\times10^{-19})
\]

\[
\boxed{I_{s,B}\approx145\ \mu\text{A}}
\]

Source B produces more electrons each second, despite having less energetic photons.

Step 6: Calculate the combined saturation current.

When both sources operate simultaneously, their electron emission rates add.

Therefore:

\[
I_{s,\text{total}}=I_{s,A}+I_{s,B}
\]

\[
I_{s,\text{total}}=96.6+145
\]

\[
\boxed{I_{s,\text{total}}\approx242\ \mu\text{A}}
\]

This is the combined plateau, assuming all emitted electrons are collected.

Step 7: Determine the combined stopping potential.

The stopping potential is determined by the fastest emitted electrons.

Source A produces electrons with a maximum kinetic energy of \(1.30\text{ eV}\), while Source B produces electrons with a maximum kinetic energy of \(0.27\text{ eV}\).

Adding Source B does not remove the more energetic electrons produced by Source A.

Therefore, the combined photocurrent reaches zero at:

\[
\boxed{V=-1.30\text{ V}}
\]

The combined graph has a higher saturation plateau than either individual graph, but its stopping-potential intercept is determined by Source A.

Step 8: Analyse the current at \(-0.50\text{ V}\).

Source B has a maximum kinetic energy of only \(0.27\text{ eV}\).

A reverse potential of \(0.50\text{ V}\) creates an energy barrier of \(0.50\text{ eV}\) for an electron.

Since:

\[
0.27<0.50
\]

no electrons from Source B can reach the anode.

Source A, however, produces electrons with maximum kinetic energy of \(1.30\text{ eV}\). Some of these electrons can overcome the \(0.50\text{ eV}\) barrier.

Therefore, only Source A can contribute to the photocurrent at this voltage.

We cannot calculate the exact current from the information supplied. Although we know the maximum kinetic energy and total emission rate, we do not know the distribution of electron energies. We would need that additional information to determine how many Source A electrons have sufficient energy to reach the anode.

This is the important final distinction: knowing the stopping potential tells you the energy of the fastest electrons, but it does not tell you the full distribution of electron energies.

Step 9: Explain why the student’s original prediction was wrong.

The student assumed that greater photon energy must produce greater photocurrent.

But the two sources have the same optical power.

Source A delivers more energy per photon, so fewer photons arrive each second.

Source B delivers less energy per photon, so more photons arrive each second.

Because the emission probability is the same, Source B produces more electrons per second and therefore has the greater saturation current.

The two graph features respond to different properties of the light:

• Photon energy determines the maximum kinetic energy and stopping potential.
• Photon arrival rate determines the electron emission rate and saturation current, given a fixed emission probability.

That distinction remains important even when several light sources operate simultaneously.

08Where to go next

Once you can interpret the two features of a photocurrent-voltage graph, you can use the stopping potential to investigate electron motion and the properties of the metal.

Two useful next steps are:

• How to Calculate Maximum Photoelectron Speed from Wavelength – convert maximum kinetic energy into the speed of the fastest emitted electron.
• How to Compare Work Functions from Photoelectric Graphs – use stopping potentials and photon energies to compare different metal surfaces.

The key decision when interpreting any photoelectric graph is simple: read the plateau to understand the emission rate, and read the stopping-potential intercept to understand the maximum kinetic energy.