Why Photoelectrons Have a Range of Kinetic Energies

Shine monochromatic light on a metal surface. Every photon has the same energy, so you might expect every emitted electron to leave with exactly the same kinetic energy.

That isn’t what happens. Photoelectrons emerge with a range of kinetic energies, even when the incident light has a single frequency.

The reason is that electrons inside a metal do not all start with the same energy. Some need more energy to escape than others, and some lose energy through collisions before reaching the surface. Consequently, only a small group of electrons can emerge with the maximum possible kinetic energy.

This is an important distinction in the photoelectric effect: the energy of each photon is fixed, but the kinetic energy of each emitted electron is not.

01The single-photon model: where does the energy go?

Start with the single-photon model.

When a photon is absorbed by an electron in a metal, it transfers its energy to that electron. The energy of the photon is:

\[ E_{\gamma}=hf \]

where:

• \(E_{\gamma}\) is photon energy, measured in joules (J).
• \(h\) is Planck’s constant, \(6.626\times10^{-34}\text{ J s}\).
• \(f\) is the frequency of the incident light, measured in hertz (Hz).

The important point is that, for monochromatic light, every photon has the same energy.

However, the electrons absorbing those photons are not all in identical energy states.

Electrons need different amounts of energy to escape

A metal contains electrons occupying a range of energy states.

Some electrons are relatively close to the surface and have higher initial energies. Others occupy lower energy states and are more strongly bound within the metal.

To escape, an electron must receive enough energy to overcome the metal’s work function.

The work function, represented by \(\phi\), is the minimum energy required to remove an electron from the metal.

Think of it as the minimum exit cost. An electron starting in a lower energy state effectively has an additional energy requirement before it can escape.

There is another complication. An electron travelling towards the surface may collide with other particles inside the metal and lose some of its energy.

These two effects explain why photoelectrons do not all emerge with identical kinetic energies.

Why some electrons are faster than others

Consider two electrons absorbing identical photons.

• Electron A starts at a relatively high energy state and escapes without losing energy inside the metal.
• Electron B starts at a lower energy state and loses additional energy through collisions before escaping.

Both receive the same photon energy, but Electron A has more energy available for kinetic energy after escaping.

This can be represented using an energy balance:

\[ K=hf-\phi-E_b-\Delta E_{\text{loss}} \]

where:

• \(K\) is the kinetic energy of the emitted electron.
• \(E_b\) is the additional energy required because the electron started below the highest occupied energy level.
• \(\Delta E_{\text{loss}}\) is energy lost through collisions inside the metal.

Both additional terms are non-negative.

For the electron with the greatest possible kinetic energy, neither additional energy requirement reduces the available energy:

\[ E_b=0 \]

\[ \Delta E_{\text{loss}}=0 \]

Therefore:

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

This is Einstein’s photoelectric equation.

The maximum kinetic energy belongs to electrons that start with the highest available initial energy and escape without losing energy along the way.

Other emitted electrons have less kinetic energy.

Importantly, the electrons do not necessarily have a uniform distribution of kinetic energies. The equation gives the upper limit, not the number of electrons at each energy.

02How do we know photoelectrons have a range of energies?

The photoelectric effect can be investigated using an evacuated photocell.

The apparatus is designed to measure the current produced by emitted electrons while the voltage between two electrodes is changed.

Understanding how the apparatus works is important because it reveals the difference between the number of electrons being collected and the energy of the fastest electrons.

The photoelectric tube experiment

The apparatus contains two electrodes inside an evacuated glass tube:

1. A metal cathode is illuminated by monochromatic light.
2. Photoelectrons are emitted from the cathode.
3. A collector electrode (anode) is positioned opposite the cathode.
4. A microammeter measures the current produced as electrons reach the collector.
5. A variable voltage supply changes the potential difference between the electrodes.

The vacuum allows electrons to travel between the electrodes without colliding with air molecules.

The experimenter can now investigate what happens as the collector becomes increasingly negative.

This creates a retarding potential: the electric field opposes the motion of the emitted electrons towards the collector.

What would happen if every electron had the same kinetic energy?

Imagine that every emitted electron had exactly the same kinetic energy and travelled directly towards the collector.

In an idealised one-dimensional arrangement, increasing the reverse voltage would eventually stop all the electrons at approximately the same voltage.

The current would then drop sharply to zero.

But the actual photocurrent generally decreases progressively as the retarding voltage increases.

Why?

Because electrons with lower kinetic energies are stopped first. Electrons with greater kinetic energies can overcome larger opposing potentials and continue reaching the collector.

As the reverse voltage increases, fewer electrons make it across the gap.

Eventually, even the fastest electrons can no longer reach the collector.

The photocurrent becomes zero.

Understanding the stopping potential

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

Let:

• \(V_{\text{stop}}\) be the signed collector potential at which current becomes zero.
• \(V_s=|V_{\text{stop}}|\) be the positive magnitude of the stopping potential.
• \(e=1.602\times10^{-19}\text{ C}\) be the magnitude of the electron charge.

The relationship is:

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

Why does this work?

An electron has a negative charge. When the collector is made negative relative to the cathode, the electron must gain electrical potential energy to move towards it.

That energy comes from the electron’s kinetic energy.

At the stopping potential, the reverse electric field removes exactly enough kinetic energy to stop the fastest electrons.

For example, if the stopping potential is \(1.2\text{ V}\), then:

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

The signed stopping voltage would be \(-1.2\text{ V}\) under our chosen convention, but its magnitude is \(1.2\text{ V}\).

Be careful with this distinction in HSC calculations.

What the experiment actually tells us

Consider the following illustrative photocurrent measurements.

All measurements in this table are illustrative, not historical experimental data.

The current reaches a saturation value of approximately \(1.00\text{ μA}\) when the collector is sufficiently positive.

As the collector becomes negative, the current decreases until it reaches zero at approximately \(-1.10\text{ V}\).

Therefore:

\[ V_s=1.10\text{ V} \]

And:

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

Notice what we have established.

The stopping potential tells us the maximum kinetic energy of the emitted electrons. It does not tell us that every electron has that energy.

The gradual current decrease is consistent with a spread of electron energies. In a real photocell, emission angles and collection geometry also influence the shape of the current-voltage curve, so it should not be treated as a direct histogram of electron kinetic energies.

The distinction between the shape of the photocurrent curve and its stopping-voltage intercept is explored further in How to Sketch Photocurrent Against Applied Voltage.

03Emission current, stopping potential and maximum kinetic energy

These three quantities are related, but they describe different things.

Mixing them up is one of the easiest ways to lose marks in photoelectric-effect questions.

Emission current: how many electrons are arriving?

Electric current measures the rate at which electric charge flows.

For a photocell, the current depends on how many emitted electrons reach the collector each second.

If more electrons arrive per second, the measured current increases.

The relationship is:

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

where:

• \(I\) is current in amperes (A).
• \(Q\) is charge in coulombs (C).
• \(t\) is time in seconds (s).

Since each electron carries a charge of magnitude \(e\), the number of electrons collected per second is:

\[ \boxed{\frac{\Delta N}{\Delta t}=\frac{I}{e}} \]

This is why increasing light intensity can increase the photocurrent.

At a fixed frequency, greater intensity means more photons arrive each second. If the frequency is above the threshold frequency, more photons can produce more photoelectrons.

However, increasing intensity does not increase the energy of an individual photon.

Therefore, it does not increase the maximum kinetic energy.

Stopping potential: how much voltage is needed to stop the fastest electrons?

Stopping potential is about energy, not the number of electrons.

A larger stopping-potential magnitude means the fastest electrons have greater kinetic energy.

For example:

• A stopping potential of \(0.5\text{ V}\) corresponds to \(0.5\text{ eV}\) maximum kinetic energy.
• A stopping potential of \(1.5\text{ V}\) corresponds to \(1.5\text{ eV}\) maximum kinetic energy.

The current might be large or small in either case.

Maximum kinetic energy: what determines it?

The maximum kinetic energy depends on the energy of each photon and the work function of the metal.

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

Increasing photon frequency increases photon energy.

Consequently, the maximum kinetic energy increases.

Increasing intensity at the same frequency does not change the photon energy, so the maximum kinetic energy remains unchanged.

The important comparison

*Assuming the probability of emission and collection efficiency remain unchanged. In real materials, emission efficiency can depend on frequency.

This gives us an important contrasting example.

A bright light source can produce a larger photocurrent than a dim light source, while the dim source produces photoelectrons with greater maximum kinetic energy.

The number of electrons and the energy of the fastest electron are separate properties.

04The historical significance: why the single-photon model mattered

The experimental observations created a problem for the classical wave model of light.

If light transferred energy continuously to electrons, increasing its intensity should increase the energy transferred to individual electrons.

That would suggest that brighter light should produce faster photoelectrons.

However, photoelectric experiments showed a different pattern:

• Increasing intensity increased the photocurrent.
• Increasing frequency increased the maximum kinetic energy.
• Below the threshold frequency, ordinary one-photon photoemission did not occur, regardless of intensity.

In 1905, Albert Einstein proposed that light transfers energy in individual packets, now called photons.

Each photon carries energy \(hf\), and its energy is transferred to an electron during absorption.

This explained why frequency determines the maximum kinetic energy, while intensity primarily determines the number of photons arriving per second.

Subsequent measurements of stopping potential against frequency, including Robert Millikan’s experimental investigations, supported the predicted linear relationship.

Starting with:

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

And using:

\[ K_{\max}=eV_s \]

We obtain:

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

This equation predicts a straight-line relationship between stopping potential and frequency.

Its gradient is \(h/e\), while its vertical intercept is related to the work function.

The important experimental distinction is that the stopping potential identifies the upper limit of the electron energy distribution. It does not mean every electron emerges with the same kinetic energy.

05Common misconceptions to avoid

Before moving into practice, check these four statements.

One more detail is worth remembering.

The equation \(K_{\max}=hf-\phi\) assumes the ordinary single-photon photoelectric effect. At sufficiently high light intensities, specialised multiphoton processes can occur, but these are outside the simplified HSC model.

For standard HSC calculations, use the single-photon model unless the question explicitly introduces another process.

06Questions and solutions

Question 1

A metal surface has a work function of \(2.35\text{ eV}\). It is illuminated with monochromatic light of wavelength \(400\text{ nm}\).

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

Calculate:

a) The energy of each incident photon.

b) The maximum kinetic energy of the emitted photoelectrons.

c) The magnitude of the stopping potential.

Solution 1

The photon energy is \(3.10\text{ eV}\), the maximum kinetic energy is \(0.75\text{ eV}\), and the stopping potential magnitude is \(0.75\text{ V}\).

Part a: Photon energy

Use:

\[ E_{\gamma}=\frac{hc}{\lambda} \]

Substitute:

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

\[ \boxed{E_{\gamma}=3.10\text{ eV}} \]

Part b: Maximum kinetic energy

Apply Einstein’s photoelectric equation:

\[ K_{\max}=E_{\gamma}-\phi \]

\[ K_{\max}=3.10-2.35 \]

\[ \boxed{K_{\max}=0.75\text{ eV}} \]

Part c: Stopping potential

Use:

\[ K_{\max}=eV_s \]

When energy is expressed in electronvolts:

\[ \boxed{V_s=0.75\text{ V}} \]

The fastest photoelectrons have \(0.75\text{ eV}\) of kinetic energy. A reverse potential of \(0.75\text{ V}\) is therefore sufficient to stop even these electrons.

Question 2

An evacuated photocell is illuminated with monochromatic light.

An experimenter records the following photocurrent as the collector potential is changed.

The values are illustrative.

a) Determine the stopping potential magnitude.

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

c) Explain why the photocurrent decreases progressively rather than remaining constant until it suddenly reaches zero.

d) Explain why these measurements do not establish that the kinetic energy distribution is uniform.

Solution 2

The stopping potential magnitude is \(0.90\text{ V}\), corresponding to a maximum kinetic energy of \(0.90\text{ eV}\), or approximately \(1.44\times10^{-19}\text{ J}\).

Part a: Stopping potential

The photocurrent reaches zero when the collector potential is:

\[ V_{\text{stop}}=-0.90\text{ V} \]

Therefore, the positive magnitude is:

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

Part b: Maximum kinetic energy

Use:

\[ K_{\max}=eV_s \]

In electronvolts:

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

Convert to joules:

\[ K_{\max}=0.90(1.602\times10^{-19}) \]

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

Part c: Explain the progressive decrease in current.

The photoelectrons have a range of kinetic energies.

As the collector becomes increasingly negative, electrons with lower kinetic energies are stopped first because they cannot overcome the increasing electrical potential-energy barrier.

Electrons with greater kinetic energies continue reaching the collector.

Consequently, the measured current decreases progressively as fewer electrons are collected.

Part d: Explain the limitation of the measurements.

The current-voltage curve does not directly show the number of electrons at each kinetic energy.

Its shape is influenced by the energy distribution, emission directions, and collection geometry.

Therefore, a progressive or approximately linear decrease in current does not establish a uniform distribution of electron kinetic energies.

The stopping potential identifies the maximum kinetic energy, but further information would be needed to determine the complete distribution.

Question 3

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

Two monochromatic light sources illuminate identical samples of this metal:

• Source A has a wavelength of \(500\text{ nm}\).
• Source B has a wavelength of \(400\text{ nm}\).

The intensities are adjusted so that both sources produce the same emission rate of \(3.0\times10^{12}\) electrons per second.

Assume all emitted electrons are collected at saturation.

Calculate:

a) The saturation current produced by each source.

b) The maximum kinetic energy produced by each source.

c) The stopping potential magnitude for each source.

d) Explain why both sources produce the same current despite producing photoelectrons with different maximum kinetic energies.

Solution 3

Both sources produce a saturation current of approximately \(0.48\text{ μA}\). However, Source A produces a maximum kinetic energy of \(0.28\text{ eV}\), while Source B produces \(0.90\text{ eV}\).

Part a: Saturation current

Use:

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

Substitute:

\[ I=(1.602\times10^{-19})(3.0\times10^{12}) \]

\[ I=4.806\times10^{-7}\text{ A} \]

Therefore:

\[ \boxed{I_A=I_B=0.48\text{ μA}} \]

Both sources produce the same number of emitted electrons per second, so they produce the same saturation current.

Part b: Maximum kinetic energy

For Source A:

\[ E_{\gamma,A}=\frac{1240}{500} \]

\[ E_{\gamma,A}=2.48\text{ eV} \]

Therefore:

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

\[ \boxed{K_{\max,A}=0.28\text{ eV}} \]

For Source B:

\[ E_{\gamma,B}=\frac{1240}{400} \]

\[ E_{\gamma,B}=3.10\text{ eV} \]

Therefore:

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

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

Part c: Stopping potentials

Using \(K_{\max}=eV_s\):

\[ \boxed{V_{s,A}=0.28\text{ V}} \]

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

Part d: Explain the difference.

Photocurrent depends on the number of electrons collected per second.

Both sources produce the same emission rate, so their saturation currents are equal.

However, Source B has a shorter wavelength and therefore a higher frequency. Its photons carry more energy.

Consequently, Source B produces photoelectrons with greater maximum kinetic energy.

This demonstrates that equal photocurrents do not imply equal photoelectron kinetic energies.

Question 4

An experimenter investigates the photoelectric effect using a metal cathode and a variable reverse potential.

At a fixed light frequency, the photocurrent decreases approximately linearly as the collector becomes more negative, eventually reaching zero.

The experimenter concludes:

“The photoelectrons must have a uniform distribution of kinetic energies because the current decreases linearly with reverse voltage.”

Evaluate this conclusion.

Then describe an experiment that could distinguish the effects of light frequency and intensity on the maximum kinetic energy and photocurrent.

Your answer should identify:

• The variables that should be controlled.
• The measurements that should be recorded.
• How the maximum kinetic energy would be determined.
• The results expected from the single-photon model.

Solution 4

The experimenter’s conclusion is not justified. A linear decrease in photocurrent does not prove that photoelectrons have a uniform kinetic energy distribution.

Why the conclusion is incorrect

Photocurrent measures the rate at which electrons reach the collector.

As the reverse potential increases, progressively fewer electrons have sufficient kinetic energy in the direction of the collector to overcome the potential barrier.

However, the relationship between current and voltage is also affected by electron emission angles and the geometry of the apparatus.

Therefore, the current-voltage curve is not automatically a direct representation of the kinetic energy distribution.

A linear graph does not establish a uniform distribution.

Designing a suitable experiment

Use an evacuated photocell with a fixed metal cathode, a collector electrode, a variable voltage supply, and a microammeter.

First, illuminate the cathode with monochromatic light of a known frequency.

Record the photocurrent at different collector potentials, including sufficiently positive and negative potentials.

Determine the saturation current and the stopping potential.

Repeat the experiment using different light frequencies.

Keep the cathode material, surface condition, electrode arrangement, and temperature constant. Control the photon flux when comparing frequencies so that changes in the number of incident photons do not confuse the interpretation of current.

Interpreting the results

For each frequency, calculate maximum kinetic energy using:

\[ K_{\max}=eV_s \]

The single-photon model predicts:

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

Therefore, increasing frequency should increase maximum kinetic energy and stopping potential.

Next, keep the frequency fixed and increase the light intensity.

The expected result is an increase in saturation current, while maximum kinetic energy and stopping potential remain unchanged.

This separates the two effects:

• Frequency determines the energy available from each photon.
• Intensity determines the number of photons arriving per second.

The experiment supports the single-photon model without requiring every emitted electron to have identical kinetic energy.

Question 5

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

Three monochromatic light beams illuminate the same metal cathode simultaneously.

Their wavelengths and photon fluxes are:

Here, \(\Phi\) is a reference photon flux in photons per second.

Assume:

• The photocell operates in the ordinary single-photon regime.
• Multiphoton absorption does not occur.
• Each above-threshold photon has the same probability of producing an electron that reaches the collector at saturation.
• The green and blue electron populations each contain electrons with kinetic energies extending up to their respective maximum values, including some blue electrons with kinetic energies above \(0.50\text{ eV}\).
• The retarding field is idealised so that an electron must have sufficient kinetic energy in the direction of the collector to overcome the potential barrier.

Answer the following.

a) Calculate the energy of photons from each beam.

b) Determine which beams produce photoelectrons.

c) At saturation, determine the relative photocurrent contributions from the three beams.

d) Calculate the maximum kinetic energy of each emitted electron population and determine the overall stopping potential magnitude when all three beams are present.

e) The collector is now made \(0.50\text{ V}\) negative relative to the cathode. Which electron population can still contribute to the photocurrent? Explain why the magnitude of this current cannot be calculated from the information provided.

f) A student argues that because the green beam produces eight times as many electrons as the blue beam, the stopping potential should be determined by the green electrons. Evaluate this argument.

g) Predict the effect on the saturation current and stopping potential if the blue beam is blocked. Separately, predict what happens if the red beam’s intensity is increased by a factor of ten while the other beams remain unchanged.

Solution 5

The blue beam determines the overall stopping potential because it produces the highest-energy photoelectrons. The much larger red photon flux produces no photoelectrons, while the green beam produces the majority of emitted electrons but does not determine the maximum kinetic energy.

Part a: Calculate photon energies

Use:

\[ E_{\gamma}=\frac{1240}{\lambda} \]

For the red beam:

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

\[ \boxed{E_R=1.91\text{ eV}} \]

For the green beam:

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

\[ \boxed{E_G=2.48\text{ eV}} \]

For the blue beam:

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

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

Part b: Determine which beams produce photoelectrons

The work function is:

\[ \phi=2.20\text{ eV} \]

A photon must have at least this much energy to cause emission in the single-photon model.

Compare each photon energy with the work function:

• Red: \(1.91<2.20\text{ eV}\). No photoemission.
• Green: \(2.48>2.20\text{ eV}\). Photoemission occurs.
• Blue: \(3.10>2.20\text{ eV}\). Photoemission occurs.

Therefore, only the green and blue beams produce photoelectrons.

The red beam cannot contribute to photoemission, even though its photon flux is much greater.

Its photons do not have sufficient energy to overcome the work function.

Importantly, a red photon and a blue photon cannot simply combine their energies to eject a single electron in the ordinary single-photon model.

Part c: Determine relative photocurrent contributions

The green beam has a photon flux of \(8\Phi\), while the blue beam has a photon flux of \(\Phi\).

Since both have the same probability of producing collected electrons, their emission rates are in the ratio:

\[ 8:1 \]

The red beam produces no electrons.

Therefore, the relative photocurrent contributions are:

\[ \boxed{I_R:I_G:I_B=0:8:1} \]

At saturation, the green beam produces eight times as much current as the blue beam.

The green beam therefore accounts for most of the emitted electrons.

Part d: Calculate maximum kinetic energies and stopping potential

Use:

\[ K_{\max}=E_{\gamma}-\phi \]

For green light:

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

\[ \boxed{K_{\max,G}=0.28\text{ eV}} \]

For blue light:

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

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

The red beam produces no photoelectrons, so it has no emitted-electron kinetic energy distribution to consider.

When both green and blue beams are present, the overall maximum kinetic energy is the greater of their two maximum values:

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

Therefore:

\[ K_{\max}=eV_s \]

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

Notice that the overall stopping potential is determined by the blue beam, even though it produces fewer electrons.

Part e: What happens at a reverse potential of 0.50 V?

The retarding potential creates an energy barrier of:

\[ E_{\text{barrier}}=0.50\text{ eV} \]

The maximum green photoelectron kinetic energy is only:

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

Therefore, every green photoelectron is stopped.

The red beam produces no photoelectrons in the first place.

However, some blue photoelectrons have kinetic energies greater than \(0.50\text{ eV}\).

These electrons can still overcome the retarding potential and reach the collector.

Therefore, only the blue population contributes to the remaining photocurrent.

We cannot calculate the magnitude of this current because we have not been given the distribution of blue photoelectron kinetic energies.

Knowing the maximum kinetic energy does not tell us what fraction of electrons have energies above \(0.50\text{ eV}\).

Part f: Evaluate the student’s argument

The student’s argument is incorrect.

The stopping potential is determined by the maximum kinetic energy of the emitted electrons, not by the most numerous population.

Although the green beam produces eight times as many electrons as the blue beam, the blue beam produces some electrons with greater kinetic energies.

The fastest blue electrons therefore determine the stopping potential.

This is an important distinction between an average or majority property and a maximum value.

A small population of high-energy electrons can determine the stopping potential even when most electrons have considerably less kinetic energy.

Part g: Predict the effects of blocking the blue beam and increasing red intensity

If the blue beam is blocked, only the green beam produces photoelectrons.

The maximum kinetic energy is now:

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

Therefore:

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

The saturation current decreases because the blue contribution is removed.

Previously, the relative contributions were:

\[ I_G:I_B=8:1 \]

After removing blue light, only the green contribution remains.

Now consider increasing the red beam intensity by a factor of ten.

Its photon flux increases from \(100\Phi\) to \(1000\Phi\).

However, its photon energy remains \(1.91\text{ eV}\), which is still below the work function.

Therefore, it produces no additional photoelectrons.

The saturation current and stopping potential remain unchanged.

This final result brings together the central ideas of the photoelectric effect:

• Photon energy determines whether emission occurs and the maximum kinetic energy available.
• Photon flux influences how many electrons are emitted.
• Different electron energy states and energy losses produce a range of kinetic energies.
• Stopping potential measures the maximum kinetic energy, not the average or most common kinetic energy.

Once you can distinguish these quantities, the next useful step is converting maximum kinetic energy into maximum photoelectron speed using \(K_{\max}=\frac{1}{2}mv_{\max}^2\). See How to Calculate Maximum Photoelectron Speed from Wavelength.