Why Stopping Potential Is Independent of Light Intensity

Turn up the brightness of a light shining on a metal surface. More electrons are emitted each second, so the photocurrent increases. But there is something unexpected: the reverse voltage needed to stop the fastest electrons does not change.

Why not? Surely brighter light should give electrons more energy?

In the simple single-photon model, increasing light intensity increases the number of photoelectrons emitted, not the maximum kinetic energy of each electron. If the light frequency and metal are unchanged, the stopping potential stays constant, even though the emission current increases.

The key is to separate three quantities that are easy to confuse: the number of electrons emitted, the maximum kinetic energy of an emitted electron, and the stopping potential.

01What is actually happening in the experiment?

Picture a photoelectric tube. It contains two metal electrodes inside an evacuated glass enclosure:

• A photosensitive cathode is illuminated by monochromatic light.
• An anode collects electrons emitted from the cathode.
• An adjustable power supply controls the potential difference between the electrodes.
• An ammeter measures the current in the external circuit.
• A voltmeter measures the potential difference between the electrodes.

The light travels towards the cathode. If its photons have enough energy, electrons escape from the metal and travel towards the anode.

To investigate the effect of intensity, keep the metal and light frequency constant, then increase the intensity of the incoming light. Measure the current and the reverse voltage required to reduce that current to zero.

There are two different measurements here, and they tell us different things.

1. Emission current: how many electrons are emitted?

Emission current describes the rate at which electric charge leaves the photosensitive surface.

If more electrons are emitted each second, the emission current increases. Under suitable collection conditions, the measured photocurrent approaches this emission current because the anode collects essentially all the emitted electrons.

A brighter light of the same frequency supplies more photons each second. This allows more electrons to be emitted each second, so the photocurrent increases.

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

Not all emitted photoelectrons have the same kinetic energy. Some lose energy inside the metal before escaping.

The maximum kinetic energy, \(K_{\max}\), belongs to the fastest emitted photoelectrons. It depends on the energy of the incoming photons and the work function of the metal.

For the same metal and the same light frequency, this maximum kinetic energy does not change when intensity increases.

3. Stopping potential: what reverse voltage prevents even the fastest electrons from arriving?

Now make the anode negative relative to the cathode. The electric field opposes the motion of the emitted electrons.

As the magnitude of this reverse voltage increases, fewer electrons have enough kinetic energy to reach the anode. Eventually, even the fastest photoelectrons are turned back, and the measured photocurrent falls to zero.

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

An important detail: a stopping potential of this size does not prevent electrons from being emitted. Electrons still leave the cathode. They simply cannot reach the anode.

The controlled comparison: what should change?

The experiment has a specific question:

If the light becomes brighter but its frequency stays constant, does the maximum energy of the emitted electrons increase?

There are two competing predictions.

• If brighter light gives each electron more energy, the stopping potential should increase.
• If brighter light simply produces more photoelectrons, the stopping potential should remain unchanged while the current increases.

To distinguish these possibilities, the experiment must change intensity without changing frequency.

A clean way to do this is to use a monochromatic light source and place neutral-density filters in its path. Different filters reduce the number of photons reaching the cathode while keeping their frequency unchanged.

Keep the same cathode, illuminated area, and electrode arrangement. For each intensity, gradually increase the reverse voltage until the photocurrent reaches zero.

A potential problem with a less controlled experiment is changing the power of a broadband lamp. Its spectral output may also change, meaning frequency and intensity are no longer independent variables. Using a fixed-frequency source avoids this confusion.

The important observation is that brighter light produces a greater current, but the zero-current endpoint occurs at the same reverse voltage.

This supports the single-photon explanation: the number of emitted electrons changes, but the maximum kinetic energy available to each electron does not.

02Why photon energy, rather than intensity, determines stopping potential

The single-photon model provides the explanation.

Each photon carries a specific amount of energy:

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

Here:

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

When a photon transfers its energy to an electron in the metal, some energy is needed to free the electron.

This minimum escape energy is the metal’s work function, \(\phi\).

Any remaining energy becomes the kinetic energy of the emitted electron. For the fastest emitted electrons:

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

Notice what is missing from this equation: intensity.

For the same metal, \(\phi\) is constant. If frequency \(f\) is also constant, the maximum kinetic energy must remain constant.

Connecting maximum kinetic energy to stopping potential

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

An electron moving against a reverse potential difference loses kinetic energy as it moves towards the anode. At the stopping potential, the fastest electrons have just enough kinetic energy to be stopped.

The electrical energy required to stop an electron is:

\[ E_{\text{electrical}}=eV_s \]

where:

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

At the stopping potential:

\[ K_{\max}=eV_s \]

Combining this with the photoelectric equation gives:

\[ eV_s=hf-\phi \]

Rearranging:

\[ \boxed{V_s=\frac{hf-\phi}{e}} \]

This is the central result.

For a fixed metal and frequency, the stopping potential is fixed. Increasing intensity cannot change it because intensity does not appear in the equation.

There is also a useful shortcut when working in electron volts:

• If \(K_{\max}=1.5\ \text{eV}\), then \(V_s=1.5\ \text{V}\).
• If \(K_{\max}=2.0\ \text{eV}\), then \(V_s=2.0\ \text{V}\).

The numerical values are equal because one electron gains or loses 1 eV of energy when moving through a potential difference of 1 V.

For a broader explanation of the photon model, see The Photoelectric Effect: Understanding the Photon Model. For more practice using the energy equation, see The Photoelectric Equation: HSC Physics Explained Clearly.

03Why the current-voltage graph changes shape without moving its endpoint

This is where the distinction between photocurrent and stopping potential becomes particularly useful.

Consider an experiment in which the frequency is kept constant while the light intensity is increased.

The experimenter measures photocurrent at different applied voltages.

Use the sign convention:

\[ V=V_A-V_C \]

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

A negative value means the anode is negative relative to the cathode, creating a retarding potential.

At sufficiently positive voltages, the anode collects essentially all emitted electrons. The current reaches a maximum called the saturation current.

As the anode becomes more negative, fewer electrons can reach it, so the measured current decreases.

Eventually, the applied voltage reaches \(-V_s\), and the current becomes zero.

The graph reveals two separate effects:

1. Increasing intensity raises the saturation current because more electrons are emitted each second.
2. Increasing intensity does not move the zero-current intercept because the maximum kinetic energy is unchanged.

Illustrative experimental data

Suppose the following measurements are obtained using the same metal and monochromatic light source.

Illustrative ideal data, not measurements from a real experiment.

Notice that doubling the intensity doubles the saturation current, while quadrupling the intensity quadruples it.

The stopping potential remains \(1.50\ \text{V}\) in every case.

Why can the current be non-zero at a negative voltage?

A negative anode potential does not automatically stop every electron.

Photoelectrons leave the cathode with a range of kinetic energies. Some are relatively slow, while others are much faster.

At a moderate reverse voltage, the slower electrons turn back towards the cathode. The faster electrons can still reach the anode.

As the reverse voltage increases, an increasing proportion of the emitted electrons are prevented from reaching the anode.

At the stopping potential, even the fastest electrons cannot reach it.

This explains an easily missed point:

Zero photocurrent does not mean zero electron emission.

The cathode can continue emitting electrons even when none reach the anode.

That is why emission current and measured photocurrent should not automatically be treated as the same thing.

04When does the stopping potential actually change?

The statement that stopping potential is independent of intensity has an important condition: the frequency must remain constant.

If frequency changes, photon energy changes.

Recall:

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

Increasing frequency increases the energy of each photon. For the same metal, this increases the maximum kinetic energy of the emitted electrons and therefore increases the stopping potential.

Consider three different situations.

The last row is particularly important.

Suppose a metal has a work function of \(2.2\ \text{eV}\), but the incoming photons each have only \(2.0\ \text{eV}\).

Even if the light intensity increases by a factor of 100, the photons still do not individually have enough energy to release electrons.

In the simple single-photon model, no photoelectrons are emitted.

Now increase the photon energy to \(2.8\ \text{eV}\).

The maximum kinetic energy becomes:

\[ K_{\max}=2.8-2.2 \]

\[ K_{\max}=0.6\ \text{eV} \]

The stopping potential is now \(0.6\ \text{V}\).

It is the change in photon energy, not the change in brightness, that produces the change in stopping potential.

A useful decision process for HSC questions

When a question describes a change to a photoelectric experiment, identify exactly what has changed.

1. Has the frequency changed? If yes, calculate the new photon energy.
2. Has the metal changed? If yes, check the new work function.
3. Has only the intensity changed? If yes, the photon energy per photon is unchanged.
4. Is the photon energy above the work function? If not, there is no ordinary single-photon emission.
5. Is the question asking about current or stopping potential? Current concerns the rate of charge collection; stopping potential concerns the maximum kinetic energy.

There is one further complication worth understanding: if light contains several frequencies, total intensity alone is not enough to predict the stopping potential. The highest-frequency component that produces photoelectrons determines their maximum possible kinetic energy, provided those electrons can be detected.

The ideal model also assumes ordinary single-photon emission. Extremely high intensities can introduce multiphoton effects, while real apparatus can have collection limitations and finite measurement sensitivity. Those effects are outside the simple HSC model.

05Questions and solutions

Question 1

A metal has a work function of \(1.8\ \text{eV}\). It is illuminated with monochromatic light whose photons each have an energy of \(3.0\ \text{eV}\).

Calculate:

a) The maximum kinetic energy of the emitted photoelectrons.

b) The stopping potential.

c) The effect of doubling the light intensity on the maximum kinetic energy and saturation current.

Solution 1

The maximum kinetic energy is \(1.2\ \text{eV}\), the stopping potential is \(1.2\ \text{V}\), and doubling the intensity doubles the saturation current without changing the stopping potential.

Step 1: Calculate maximum kinetic energy.

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

\[ K_{\max}=3.0-1.8 \]

\[ K_{\max}=1.2\ \text{eV} \]

Step 2: Calculate stopping potential.

\[ K_{\max}=eV_s \]

Therefore:

\[ V_s=1.2\ \text{V} \]

Step 3: Consider doubling the intensity.

Doubling the intensity doubles the number of photons arriving per second because the frequency is unchanged.

Assuming constant quantum efficiency, twice as many electrons are emitted each second. The saturation current therefore doubles.

However, each photon still carries \(3.0\ \text{eV}\). The maximum kinetic energy remains \(1.2\ \text{eV}\), so the stopping potential remains \(1.2\ \text{V}\).

Question 2

Monochromatic light with a frequency of \(7.50\times10^{14}\ \text{Hz}\) shines on a metal surface.

The measured stopping potential is \(0.90\ \text{V}\).

Given:

• Planck’s constant, \(h=6.63\times10^{-34}\ \text{J s}\).
• Electron charge magnitude, \(e=1.60\times10^{-19}\ \text{C}\).

Calculate:

a) The maximum kinetic energy in electron volts.

b) The work function of the metal in electron volts.

c) The new saturation current if the original saturation current is \(0.25\ \mu\text{A}\) and the light intensity increases by a factor of four.

d) The new stopping potential.

Solution 2

The maximum kinetic energy is \(0.90\ \text{eV}\), the work function is approximately \(2.21\ \text{eV}\), the new saturation current is \(1.00\ \mu\text{A}\), and the stopping potential remains \(0.90\ \text{V}\).

Step 1: Calculate the maximum kinetic energy.

Use the stopping potential relationship:

\[ K_{\max}=eV_s \]

\[ K_{\max}=(1.60\times10^{-19})(0.90) \]

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

Convert into electron volts:

\[ K_{\max}= \frac{1.44\times10^{-19}} {1.60\times10^{-19}} \]

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

Step 2: Calculate photon energy.

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

\[ E_{\text{photon}}= (6.63\times10^{-34})(7.50\times10^{14}) \]

\[ E_{\text{photon}}=4.9725\times10^{-19}\ \text{J} \]

Convert to electron volts:

\[ E_{\text{photon}}= \frac{4.9725\times10^{-19}} {1.60\times10^{-19}} \]

\[ E_{\text{photon}}=3.11\ \text{eV} \]

Step 3: Calculate the work function.

Rearrange the photoelectric equation:

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

Substitute:

\[ \phi=3.11-0.90 \]

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

Step 4: Calculate the new saturation current.

The intensity increases by a factor of four.

Therefore:

\[ I_{\text{new}}=4I_{\text{original}} \]

\[ I_{\text{new}}=4(0.25) \]

\[ I_{\text{new}}=1.00\ \mu\text{A} \]

Step 5: Determine the new stopping potential.

The frequency and metal have not changed.

Therefore, photon energy and work function remain constant. Maximum kinetic energy is unchanged, so:

\[ V_{s,\text{new}}=0.90\ \text{V} \]

The key is not to confuse the increase in the number of emitted electrons with an increase in the energy of each electron.

Question 3

A metal has a work function of \(2.20\ \text{eV}\).

It is first illuminated with monochromatic light of wavelength \(600\ \text{nm}\). A student increases the intensity by a factor of 50 but observes no photoelectric emission.

The student then replaces the light source with one producing a wavelength of \(500\ \text{nm}\).

Use:

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

where photon energy is in electron volts and wavelength is in nanometres.

a) Explain why increasing the intensity of the original light does not produce photoelectrons.

b) Calculate the maximum kinetic energy produced by the \(500\ \text{nm}\) light.

c) Calculate the new stopping potential.

Solution 3

The \(600\ \text{nm}\) light cannot eject electrons, even at 50 times its original intensity. The \(500\ \text{nm}\) light produces photoelectrons with a maximum kinetic energy of \(0.28\ \text{eV}\) and a stopping potential of \(0.28\ \text{V}\).

Step 1: Calculate the original photon energy.

\[ E_{\text{photon}}=\frac{1240}{600} \]

\[ E_{\text{photon}}=2.07\ \text{eV} \]

Compare this with the work function:

\[ 2.07<2.20 \]

Each photon has insufficient energy to release an electron.

Step 2: Explain why increasing intensity does not help.

Increasing intensity increases the number of photons arriving per second.

However, each photon still carries only \(2.07\ \text{eV}\). It cannot overcome the \(2.20\ \text{eV}\) work function.

In the simple single-photon model, increasing the number of insufficient-energy photons does not produce photoelectric emission.

Step 3: Calculate the energy of the replacement light.

\[ E_{\text{photon}}=\frac{1240}{500} \]

\[ E_{\text{photon}}=2.48\ \text{eV} \]

This is greater than the work function, so electrons can now escape.

Step 4: Calculate maximum kinetic energy.

\[ K_{\max}=2.48-2.20 \]

\[ K_{\max}=0.28\ \text{eV} \]

Step 5: Calculate stopping potential.

Using the numerical relationship between electron energy in eV and stopping potential in V:

\[ V_s=0.28\ \text{V} \]

Changing the wavelength changed the energy per photon. That is why photoelectric emission became possible and a stopping potential could now be measured.

Question 4

A photoelectric tube is illuminated by monochromatic light. The experimenter performs two tests using the same frequency but different intensities.

The following illustrative measurements are obtained:

Answer the following:

a) Determine the stopping potential magnitude.

b) Calculate the maximum kinetic energy of the photoelectrons in joules and electron volts.

c) Calculate the ratio of the electron emission rates between the two intensity settings. Determine both rates in electrons per second.

d) Explain why the current is non-zero at \(-0.80\ \text{V}\), but zero at \(-1.30\ \text{V}\).

e) Does the zero current at \(-1.30\ \text{V}\) mean that no electrons are being emitted? Explain.

Solution 4

The stopping potential is \(1.30\ \text{V}\), and the maximum kinetic energy is \(2.08\times10^{-19}\ \text{J}\), or \(1.30\ \text{eV}\). The electron emission rate triples when the intensity increases from \(I\) to \(3I\). Emission continues at the stopping potential, even though the measured photocurrent is zero.

Step 1: Determine the stopping potential.

The current becomes zero at an applied potential of:

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

The stopping potential is the magnitude of this reverse voltage:

\[ V_s=1.30\ \text{V} \]

Step 2: Calculate maximum kinetic energy.

\[ K_{\max}=eV_s \]

\[ K_{\max}=(1.60\times10^{-19})(1.30) \]

\[ K_{\max}=2.08\times10^{-19}\ \text{J} \]

In electron volts:

\[ K_{\max}=1.30\ \text{eV} \]

Step 3: Calculate the electron emission rates.

At saturation, assume all emitted electrons are collected.

The emission rate is:

\[ \dot N_e=\frac{I_{\text{emit}}}{e} \]

For intensity \(I\):

\[ \dot N_{e,1}= \frac{0.24\times10^{-6}} {1.60\times10^{-19}} \]

\[ \dot N_{e,1}=1.5\times10^{12}\ \text{s}^{-1} \]

For intensity \(3I\):

\[ \dot N_{e,2}= \frac{0.72\times10^{-6}} {1.60\times10^{-19}} \]

\[ \dot N_{e,2}=4.5\times10^{12}\ \text{s}^{-1} \]

The ratio is:

\[ \frac{\dot N_{e,2}}{\dot N_{e,1}} = \frac{4.5\times10^{12}} {1.5\times10^{12}} \]

\[ \text{Ratio}=3:1 \]

The brighter light produces three times as many emitted electrons per second.

Step 4: Explain the current at the two reverse voltages.

At \(-0.80\ \text{V}\), some photoelectrons still have enough kinetic energy to reach the anode.

The reverse electric field turns back the slower electrons, but the faster ones can still reach the collector. This produces a non-zero current.

At \(-1.30\ \text{V}\), even the fastest electrons cannot reach the anode. The measured photocurrent becomes zero.

Step 5: Explain whether electron emission stops.

No. The cathode can continue emitting photoelectrons at \(-1.30\ \text{V}\).

The reverse electric field prevents those electrons from reaching the anode. It does not prevent them from initially escaping the metal.

This question tests an important distinction: emission is an event at the cathode, whereas photocurrent is measured through the external circuit.

Question 5

A metal has a work function of \(2.10\ \text{eV}\).

Initially, it is illuminated by monochromatic light with:

• Wavelength: \(500\ \text{nm}\).
• Irradiance: \(20.0\ \text{W m}^{-2}\).

An experimenter then makes two changes.

Stage 1: A second beam of wavelength \(650\ \text{nm}\) and irradiance \(80.0\ \text{W m}^{-2}\) is added. The original \(500\ \text{nm}\) beam remains unchanged.

Stage 2: The \(500\ \text{nm}\) beam is removed and replaced with a \(400\ \text{nm}\) beam of irradiance \(0.040\ \text{W m}^{-2}\). The \(650\ \text{nm}\) beam remains unchanged.

Both beams illuminate the same area of the metal.

Assume:

• The single-photon model applies.
• There are no thermal or multiphoton effects.
• The quantum efficiency is the same for the \(400\ \text{nm}\) and \(500\ \text{nm}\) photons.
• The \(400\ \text{nm}\) beam produces enough photoelectrons for its maximum-energy electrons to be detected.

Use:

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

where energy is in electron volts and wavelength is in nanometres.

Answer the following:

a) Calculate the initial stopping potential.

b) Determine the stopping potential and relative emission current during Stage 1.

c) Determine the stopping potential during Stage 2.

d) Calculate the ratio of the Stage 2 emission current to the initial emission current.

e) Explain why the total light irradiance is a poor predictor of both the emission current and stopping potential in this experiment.

Solution 5

The initial stopping potential is \(0.38\ \text{V}\). It remains \(0.38\ \text{V}\) during Stage 1, and the emission current is unchanged. During Stage 2, the stopping potential increases to \(1.00\ \text{V}\), but the emission current falls to approximately \(1/625\) of its initial value.

This is a particularly useful example because the total irradiance increases even while the emission current decreases.

Step 1: Calculate the photon energies of all three wavelengths.

For the original \(500\ \text{nm}\) beam:

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

\[ E_{500}=2.48\ \text{eV} \]

For the \(650\ \text{nm}\) beam:

\[ E_{650}=\frac{1240}{650} \]

\[ E_{650}=1.91\ \text{eV} \]

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

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

\[ E_{400}=3.10\ \text{eV} \]

Compare these energies with the work function:

\[ \phi=2.10\ \text{eV} \]

The \(500\ \text{nm}\) and \(400\ \text{nm}\) photons have enough energy to release electrons.

The \(650\ \text{nm}\) photons do not.

Step 2: Calculate the initial stopping potential.

Initially, only the \(500\ \text{nm}\) beam is present.

The maximum kinetic energy is:

\[ K_{\max}=2.48-2.10 \]

\[ K_{\max}=0.38\ \text{eV} \]

Therefore:

\[ V_s=0.38\ \text{V} \]

Step 3: Analyse Stage 1.

The \(650\ \text{nm}\) beam has a photon energy of only \(1.91\ \text{eV}\).

Since:

\[ 1.91<2.10 \]

these photons cannot eject electrons in the simple single-photon model.

Adding this beam increases the total irradiance from:

\[ 20.0\ \text{W m}^{-2} \]

to:

\[ 20.0+80.0=100.0\ \text{W m}^{-2} \]

However, the additional light produces no additional photoelectrons.

The original \(500\ \text{nm}\) beam is unchanged, so:

• The emission current remains unchanged.
• The maximum kinetic energy remains \(0.38\ \text{eV}\).
• The stopping potential remains \(0.38\ \text{V}\).

The total irradiance has increased fivefold, but the photoelectric emission has not changed.

Step 4: Analyse Stage 2.

The \(500\ \text{nm}\) beam is removed and replaced with \(400\ \text{nm}\) light.

The maximum kinetic energy is now:

\[ K_{\max}=3.10-2.10 \]

\[ K_{\max}=1.00\ \text{eV} \]

Therefore:

\[ V_s=1.00\ \text{V} \]

The stopping potential increases because the \(400\ \text{nm}\) photons have more energy than the original \(500\ \text{nm}\) photons.

The \(650\ \text{nm}\) beam still produces no photoelectrons.

Step 5: Compare the photon arrival rates.

This is the crucial part of the problem.

The irradiance tells us how much energy arrives per second per unit area. It does not directly tell us how many photons arrive.

For monochromatic light:

\[ \text{Photon flux}= \frac{\text{Irradiance}}{\text{Energy per photon}} \]

The ratio of the new photon flux to the original photon flux is:

\[ \frac{\dot N_{400}}{\dot N_{500}} = \frac{0.040/E_{400}}{20.0/E_{500}} \]

Substitute the photon energies:

\[ = \frac{0.040/3.10}{20.0/2.48} \]

Rearranging:

\[ = \frac{0.040(2.48)} {20.0(3.10)} \]

\[ =0.00160 \]

Therefore:

\[ \frac{\dot N_{400}}{\dot N_{500}} = \frac{1}{625} \]

Because the quantum efficiency is assumed to be the same at both wavelengths, the electron emission rate also falls to \(1/625\) of its initial value.

The emission current therefore becomes approximately 625 times smaller.

Step 6: Explain the overall result.

The total irradiance during Stage 2 is:

\[ 80.0+0.040=80.04\ \text{W m}^{-2} \]

This is four times the original irradiance, yet the emission current is approximately 625 times smaller.

The reason is that most of the incident energy now comes from \(650\ \text{nm}\) photons, which cannot overcome the work function.

The much weaker \(400\ \text{nm}\) beam produces the photoelectrons. Its higher photon energy determines their maximum kinetic energy and therefore the stopping potential.

The tempting assumption is that greater total irradiance must mean more emitted electrons and a greater stopping potential.

But that ignores the energy carried by each photon.

The experiment depends on two separate things:

• How many photons arrive above the threshold frequency determines how many electrons can be emitted.
• The highest photon energy that produces photoelectrons determines the maximum kinetic energy and stopping potential.

That distinction is the deeper reason stopping potential is independent of intensity in the simple photoelectric model.

06Where this leads next

Once you can separate photocurrent from stopping potential, interpreting photoelectric current-voltage graphs becomes much easier.

You can identify the stopping potential from the zero-current intercept, compare saturation currents to investigate emission rates, and use the stopping potential to calculate maximum kinetic energy.

The next useful connection is How to Sketch Photocurrent Against Applied Voltage, where these same relationships are used to construct and interpret photoelectric current-voltage curves.