Why No Photoelectrons Appear Below the Threshold Frequency

Leave a bright lamp shining on a clean metal surface for an hour. If each photon has less energy than the metal’s work function, no photoelectrons will appear. Leave it shining for another hour, increase the brightness, or use a thousand times more photons. The result is still the same.

Why? Because, in the single-photon model, an electron must receive enough energy from an individual photon to escape the metal. It cannot gradually collect small amounts of energy from thousands of separate photons.

This is the reason for the threshold frequency in the photoelectric effect. The important quantity is not how much total light energy reaches the metal. It is how much energy each photon carries.

01The experiment that exposed the problem

Physicists did not initially know that light behaved as a stream of individual energy packets.

In 1887, Heinrich Hertz observed that ultraviolet light affected the behaviour of sparks in an electrical experiment. A year later, Wilhelm Hallwachs showed that ultraviolet light could cause a negatively charged metal plate to lose its charge.

These observations suggested that light could cause negative charge to leave a metal surface. However, they did not explain how much energy the escaping electrons carried. 【turn891100search10】

How the photoelectric experiment works

A more detailed investigation required an apparatus that could measure both the number of escaping electrons and their kinetic energy.

Consider an evacuated glass tube containing two metal electrodes.

• The metal cathode is positioned on one side of the tube.
• Light enters through a quartz window and strikes the cathode.
• If photoemission occurs, electrons leave the cathode and travel across the vacuum towards the collector electrode, called the anode.
• An external power supply controls the potential difference between the electrodes.
• An ammeter measures the photocurrent, while a voltmeter measures the potential difference.

The vacuum is important because it prevents electrons from losing kinetic energy through collisions with gas molecules before reaching the collector. 【turn891100search0】

Initially, the anode is made positive relative to the cathode. This attracts emitted electrons, allowing them to reach the collector and produce a current.

Now reverse the power supply so that the anode becomes negative relative to the cathode.

The electric field opposes the electrons’ motion. Only electrons with sufficient kinetic energy can reach the anode.

As the reverse voltage increases, fewer electrons reach the collector. Eventually, even the fastest photoelectrons cannot reach it, and the measured current becomes zero.

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

This modification was important because it allowed physicists to investigate the energy of the fastest photoelectrons, rather than simply observing whether electrons were emitted.

What did the experiments reveal?

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

If light transferred energy continuously, you might expect a sufficiently intense light source, or a sufficiently long exposure, to eventually give electrons enough energy to escape.

However, experiments revealed a different pattern:

• Below a certain frequency, no photoelectrons were emitted, regardless of light intensity.
• Above that frequency, increasing light intensity increased the number of emitted electrons.
• Increasing light frequency increased the maximum kinetic energy of the emitted electrons.
• Photoemission occurred without the significant time delay expected from gradual energy accumulation.

Philipp Lenard’s investigations in 1902 were particularly important. He found that increasing light intensity affected the rate of electron emission but did not increase the electrons’ velocities. Their velocities depended on the frequency of the incident light. 【turn891100search10】

The observations suggested that two different properties of light controlled two different outcomes:

• Frequency controlled the energy available to each emitted electron.
• Intensity controlled how many photons arrived per second.

But why should these properties behave independently?

02Einstein’s explanation: one photon, one electron

In 1905, Albert Einstein proposed an explanation based on light quanta, now called photons.

Instead of treating light energy as continuously distributed, Einstein treated it as arriving in individual packets.

Each photon carries an energy determined by its frequency:

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

Where:

• \(E_{\text{photon}}\) 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 frequency of the light, measured in hertz (Hz).

In the basic photoelectric model, one photon transfers its energy to one electron.

However, the electron is initially bound to the metal. It needs a minimum amount of energy to escape.

This minimum energy is called the work function, represented by \(\phi\).

The work function depends on the metal and its surface condition.

Where does the photon’s energy go?

When a photon interacts with an electron, its energy has two possible destinations:

1. The energy required to release the electron from the metal.
2. The kinetic energy remaining after the electron escapes.

Conservation of energy gives:

\[
E_{\text{photon}}=\phi+E_k
\]

Substituting the photon energy equation:

\[
hf=\phi+E_k
\]

The fastest emitted electrons have the greatest possible kinetic energy. They are associated with the minimum energy required to escape, so their maximum kinetic energy is:

\[
\boxed{E_{k,\max}=hf-\phi}
\]

This is Einstein’s photoelectric equation.

Notice something important: the equation only describes physically emitted electrons when the photon supplies at least the work function.

Why waiting longer cannot help

Suppose an electron needs 3.0 eV to escape a particular metal.

You shine light on it with a photon energy of only 2.0 eV.

Each photon is short by 1.0 eV.

You might predict that sending more photons towards the metal will eventually provide the missing energy.

But that prediction assumes the electron can accumulate energy from separate photon interactions.

That is not how the standard single-photon photoelectric process works.

Each interaction involves an individual photon transferring its energy to an individual electron. If the photon does not have enough energy to release the electron, that interaction cannot produce a photoelectron.

Sending another photon does not automatically add its energy to the previous interaction.

Think of it like trying to pay a $3 entry fee using separate $2 tickets when the system only accepts one ticket per entry. Having more tickets does not make an individual ticket worth more.

The analogy is not exact, because photons are physical energy packets rather than tickets. However, it captures the important idea: the required energy must be supplied in a single photon-electron interaction in the basic model.

Increasing intensity increases the number of photons, not the energy of each photon.

There is an important qualification. Extremely intense laser light can produce multiphoton photoemission, in which an electron absorbs multiple photons during an interaction. Thermal emission can also occur if the metal becomes sufficiently hot. These are different physical processes and are outside the ordinary single-photon HSC model.

03What happens at the threshold frequency?

The threshold frequency is the minimum frequency of light required to cause photoelectric emission from a particular metal.

It is represented by \(f_0\).

At the threshold, the photon supplies exactly the energy needed to release an electron, leaving no kinetic energy.

Therefore:

\[
hf_0=\phi
\]

Rearranging:

\[
\boxed{f_0=\frac{\phi}{h}}
\]

This relationship explains why different metals have different threshold frequencies: they have different work functions.

There are three possible situations.

At exactly the threshold, emission is possible in the ideal model, although the resulting current can be difficult to detect in a real experiment.

Below the threshold, however, the prediction is clear: no photoelectrons are produced by the single-photon process.

04Three quantities you must keep separate

A common source of mistakes in HSC questions is treating photocurrent, maximum kinetic energy, and stopping potential as though they measure the same thing.

They do not.

1. Emission current: how many electrons are produced?

The photocurrent measures the rate at which charge carried by photoelectrons reaches the collector.

Current is defined as:

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

Where:

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

At a fixed frequency above the threshold, increasing light intensity means more photons arrive per second.

More photons can produce more photoelectrons, so the photocurrent increases, provided the collection efficiency remains unchanged.

But below the threshold, increasing intensity does not produce photoelectrons in the single-photon model.

The photocurrent therefore remains zero.

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

The maximum kinetic energy is determined by the photon frequency and the work function:

\[
E_{k,\max}=hf-\phi
\]

At a fixed frequency, increasing intensity does not change this value.

The reason is simple: each photon still carries exactly the same energy.

You have increased the number of photons, not the energy of an individual photon.

Also, remember that \(E_{k,\max}\) describes the fastest emitted electrons. Not every photoelectron necessarily leaves the surface with this energy.

3. Stopping potential: how much reverse voltage stops the fastest electrons?

The stopping potential measures the maximum kinetic energy indirectly.

When the collector is made negative relative to the cathode, photoelectrons must work against the electric field to reach it.

Eventually, the reverse voltage becomes large enough to stop even the fastest electrons.

At that point, the photocurrent falls to zero.

Using conservation of energy:

\[
\boxed{E_{k,\max}=eV_s}
\]

Where:

• \(e\) is the magnitude of the electron charge, \(1.60\times10^{-19}\,\text{C}\).
• \(V_s\) is the stopping potential in volts (V).
• \(E_{k,\max}\) is the maximum kinetic energy in joules.

Here, \(V_s\) is defined as a positive magnitude. If the cathode is at 0 V, the collector is at \(-V_s\) at the stopping point.

Combining the equations:

\[
eV_s=hf-\phi
\]

This gives us a useful distinction.

At a fixed frequency above threshold:

• Increasing intensity increases photocurrent.
• Increasing intensity does not change maximum kinetic energy.
• Increasing intensity does not change stopping potential.

One subtle point is worth remembering: when the photocurrent reaches zero at the stopping potential, electrons may still be leaving the cathode. They simply cannot reach the collector.

This is different from below-threshold illumination, where no photoelectrons are emitted in the first place.

For a more detailed treatment of these current-voltage curves, see How to Sketch Photocurrent Against Applied Voltage.

05How to decide whether photoelectrons will be emitted

When you encounter a photoelectric question, follow this sequence.

Step 1: Find the energy of an individual photon.

If frequency is given:

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

If wavelength is given:

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

A useful shortcut when using electronvolts and nanometres is:

\[
E_{\text{photon}}(\text{eV})
=\frac{1240}{\lambda(\text{nm})}
\]

Step 2: Compare photon energy with the work function.

• If \(E_{\text{photon}}<\phi\), no photoelectric emission occurs.
• If \(E_{\text{photon}}=\phi\), the photon is at the threshold.
• If \(E_{\text{photon}}>\phi\), photoelectric emission is possible.

Step 3: If emission occurs, calculate maximum kinetic energy.

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

Step 4: If required, calculate stopping potential.

\[
V_s=\frac{E_{k,\max}}{e}
\]

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

For example, a maximum kinetic energy of 0.75 eV corresponds to a stopping potential of 0.75 V.

Never calculate a negative maximum kinetic energy and report it as a physical result. If the photon energy is below the work function, the correct conclusion is that no photoelectric emission occurs.

06Worked examples

07Questions and solutions

Question 1

A metal has a work function of 2.5 eV. It is illuminated with light containing photons of energy 2.0 eV.

A student suggests leaving the light on for 30 minutes and then increasing its intensity by a factor of 100 to produce photoelectrons.

Will this work? Explain using the single-photon model.

Solution 1

No photoelectrons will be produced by the single-photon process, regardless of the exposure time or increase in intensity.

The work function is:

\[
\phi=2.5\,\text{eV}
\]

The photon energy is:

\[
E_{\text{photon}}=2.0\,\text{eV}
\]

Since:

\[
E_{\text{photon}}<\phi
\]

Each photon has insufficient energy to release an electron.

Increasing intensity increases the number of photons arriving per second, but it does not increase the energy carried by an individual photon.

Therefore, the photocurrent remains zero in the stated model.

The student’s mistake is assuming that electrons accumulate energy from separate photon interactions until they have enough to escape. That is not the single-photon photoelectric process.

Question 2

Two different metals, A and B, are illuminated with the same monochromatic light of wavelength 430 nm.

Their work functions are:

• Metal A: 2.0 eV
• Metal B: 3.0 eV

Calculate the maximum kinetic energy and stopping potential for metal A.

Explain what happens to metal B if the light intensity is increased by a factor of 20.

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

Solution 2

Metal A produces photoelectrons with a maximum kinetic energy of approximately 0.88 eV and a stopping potential of approximately 0.88 V. Metal B produces no photoelectrons, even when the intensity is increased.

Step 1: Calculate photon energy.

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

\[
E_{\text{photon}}=\frac{1240}{430}
\]

\[
E_{\text{photon}}=2.88\,\text{eV}
\]

Step 2: Consider metal A.

Its work function is 2.0 eV.

Since:

\[
2.88>2.0
\]

Photoelectric emission occurs.

Calculate maximum kinetic energy:

\[
E_{k,\max}=2.88-2.0
\]

\[
\boxed{E_{k,\max}=0.88\,\text{eV}}
\]

The stopping potential is therefore:

\[
\boxed{V_s=0.88\,\text{V}}
\]

Step 3: Consider metal B.

Its work function is 3.0 eV.

However:

\[
2.88<3.0
\]

Each photon has insufficient energy to release an electron.

Therefore, no photoelectric emission occurs.

Increasing the intensity by a factor of 20 does not change the energy of each photon. Metal B still produces no photoelectrons.

This comparison shows why the work function matters. The same light can cause photoelectric emission from one metal but not another.

Question 3

An evacuated photoelectric cell is used to investigate an unknown metal.

The following illustrative data are collected. The photocurrent is measured with a collecting voltage sufficient to collect all emitted electrons. The stopping potential is recorded as a positive magnitude.

Assume the metal surface remains unchanged and contact-potential effects are negligible.

(a) Use trial B to calculate the work function of the metal in joules and electronvolts.

(b) Calculate the threshold frequency.

(c) Explain whether trial A is consistent with your calculated threshold frequency.

(d) Explain what trials B and C demonstrate about the effects of light intensity.

Use \(h=6.63\times10^{-34}\,\text{J s}\) and \(e=1.60\times10^{-19}\,\text{C}\).

Solution 3

The work function is approximately \(3.45\times10^{-19}\,\text{J}\), or 2.16 eV. The threshold frequency is approximately \(5.21\times10^{14}\,\text{Hz}\).

(a) Calculate the work function.

First, use the stopping potential from trial B to calculate maximum kinetic energy.

\[
E_{k,\max}=eV_s
\]

\[
E_{k,\max}
=(1.60\times10^{-19})(0.12)
\]

\[
E_{k,\max}=1.92\times10^{-20}\,\text{J}
\]

Next, calculate the energy of an incident photon.

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

\[
E_{\text{photon}}
=(6.63\times10^{-34})(5.5\times10^{14})
\]

\[
E_{\text{photon}}=3.6465\times10^{-19}\,\text{J}
\]

Using Einstein’s equation:

\[
\phi=hf-E_{k,\max}
\]

\[
\phi=3.6465\times10^{-19}
-1.92\times10^{-20}
\]

\[
\boxed{\phi=3.45\times10^{-19}\,\text{J}}
\]

Convert to electronvolts:

\[
\phi=\frac{3.4545\times10^{-19}}
{1.60\times10^{-19}}
\]

\[
\boxed{\phi=2.16\,\text{eV}}
\]

(b) Calculate threshold frequency.

Using:

\[
f_0=\frac{\phi}{h}
\]

\[
f_0=
\frac{3.4545\times10^{-19}}
{6.63\times10^{-34}}
\]

\[
\boxed{f_0=5.21\times10^{14}\,\text{Hz}}
\]

(c) Check trial A.

Trial A uses a frequency of:

\[
4.5\times10^{14}\,\text{Hz}
\]

This is below the calculated threshold frequency:

\[
4.5\times10^{14}<5.21\times10^{14}
\]

Therefore, no photoelectric emission is expected.

The zero photocurrent in trial A is consistent with the calculated work function and threshold frequency.

(d) Compare trials B and C.

Both trials use exactly the same frequency.

However, trial C has twice the light intensity and twice the photocurrent.

The stopping potential remains 0.12 V.

This demonstrates that increasing intensity increases the rate of photoelectron emission without changing the maximum kinetic energy of the electrons.

Since stopping potential measures maximum kinetic energy, it also remains unchanged.

Question 4

A photoelectric cell is tested under two different illumination conditions.

The collector voltage is measured relative to the illuminated cathode, which is maintained at 0 V.

Trial 1: The metal is illuminated with light above its threshold frequency.

Trial 2: The metal is illuminated with light below its threshold frequency. The photocurrent remains zero at every tested collector voltage, including +2.0 V and -1.0 V.

Answer the following:

(a) Calculate the maximum kinetic energy of the photoelectrons in trial 1.

(b) Explain why the current becomes zero at -0.60 V in trial 1.

(c) Explain why the current is zero in trial 2.

(d) Can a stopping potential be determined from trial 2? Explain.

Solution 4

The maximum kinetic energy in trial 1 is 0.60 eV. The zero current in trial 1 results from the reverse electric field preventing emitted electrons from reaching the collector, whereas trial 2 has no photoelectric emission at all.

(a) Calculate maximum kinetic energy.

The stopping potential is the magnitude of the reverse voltage at which the photocurrent reaches zero.

Therefore:

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

Using:

\[
E_{k,\max}=eV_s
\]

\[
E_{k,\max}
=(1.60\times10^{-19})(0.60)
\]

\[
\boxed{E_{k,\max}=9.60\times10^{-20}\,\text{J}}
\]

In electronvolts:

\[
\boxed{E_{k,\max}=0.60\,\text{eV}}
\]

(b) Explain the zero current in trial 1.

The cathode is still producing photoelectrons because the incident light has a frequency above the threshold.

However, the negative collector repels the emitted electrons.

As the reverse voltage increases, progressively fewer electrons have sufficient kinetic energy to reach the collector.

At -0.60 V, even the fastest electrons are stopped.

The photocurrent therefore becomes zero, even though photoemission from the cathode continues.

(c) Explain the zero current in trial 2.

In trial 2, the light frequency is below the threshold frequency.

Each photon has insufficient energy to overcome the work function.

Consequently, no photoelectrons are produced.

This remains true even when the collector is made positive and would otherwise attract electrons.

The zero current has a completely different physical cause from the zero current in trial 1.

(d) Can a stopping potential be determined?

No. A stopping potential cannot be experimentally determined from trial 2.

A stopping potential measures the reverse voltage needed to prevent already-emitted photoelectrons from reaching the collector.

In trial 2, no photoelectrons are emitted in the first place.

The current is already zero before any retarding voltage is applied. There is therefore no photocurrent to gradually reduce and no meaningful stopping-potential measurement.

This is an important distinction: zero photocurrent does not always mean the same physical thing.

It can mean either that no photoelectrons were emitted or that emitted photoelectrons could not reach the collector.

Question 5

A clean metal surface has a work function of 2.0 eV.

It is illuminated simultaneously by two monochromatic light sources:

• Red source: photon energy of 2.4 eV, with \(1.0\times10^8\) photons reaching the surface per second.
• Violet source: photon energy of 3.0 eV, with \(1.0\times10^5\) photons reaching the surface per second.

Assume both sources have the same quantum efficiency, the same collection geometry, and operate in the single-photon regime. Neglect contact potentials and assume the collector initially collects all emitted electrons.

The collector is then made progressively more negative relative to the cathode.

Answer the following:

(a) Calculate the maximum kinetic energy of electrons produced by each light source.

(b) Which source produces the greater contribution to the initial photocurrent? Explain.

(c) Determine the stopping potential for the combined illumination.

(d) Describe what happens to the photocurrent as the reverse voltage increases from 0 V to 1.0 V. In particular, explain what happens at 0.4 V and 0.7 V.

(e) A student argues that the stopping potential must be 0.4 V because the red source produces most of the photoelectrons. Explain precisely why this conclusion is incorrect.

(f) Predict what happens to the stopping potential if the red source is made ten times more intense while the violet source remains unchanged.

Solution 5

The combined illumination has a stopping potential of 1.0 V, even though the red source produces approximately 1000 times as many photoelectrons as the violet source.

The reason is that stopping potential is determined by the fastest emitted electrons, not the number of electrons produced by each source.

(a) Calculate maximum kinetic energy for each source.

For the red source:

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

\[
E_{k,\max}=2.4-2.0
\]

\[
\boxed{E_{k,\max,\text{red}}=0.4\,\text{eV}}
\]

For the violet source:

\[
E_{k,\max}=3.0-2.0
\]

\[
\boxed{E_{k,\max,\text{violet}}=1.0\,\text{eV}}
\]

The violet photons produce photoelectrons with a greater maximum kinetic energy.

(b) Determine which source dominates the initial photocurrent.

The red source supplies:

\[
1.0\times10^8
\]

photons per second.

The violet source supplies:

\[
1.0\times10^5
\]

photons per second.

The ratio of photon arrival rates is:

\[
\frac{1.0\times10^8}{1.0\times10^5}
=1000
\]

Because the quantum efficiencies are assumed equal, the red source produces approximately 1000 times as many photoelectrons per second.

Therefore, the red source dominates the initial photocurrent.

However, producing more electrons does not mean producing faster electrons.

(c) Determine the stopping potential.

The stopping potential is determined by the maximum kinetic energy of all emitted photoelectrons.

The violet source produces electrons with a maximum kinetic energy of 1.0 eV.

Therefore:

\[
eV_s=E_{k,\max}
\]

\[
\boxed{V_s=1.0\,\text{V}}
\]

The combined photocurrent reaches zero when the reverse voltage is large enough to stop the fastest violet-generated electrons.

(d) Describe the changes in photocurrent.

Initially, at zero reverse voltage, both groups of photoelectrons can reach the collector.

As the reverse voltage increases, electrons with lower kinetic energies are stopped first.

At 0.4 V, the red-generated electrons can no longer reach the collector because their maximum kinetic energy is only 0.4 eV.

The violet-generated electrons can still contribute to the current because some have kinetic energies greater than 0.4 eV.

At 0.7 V, the red-generated electrons remain completely stopped. However, the fastest violet-generated electrons can still reach the collector, provided their kinetic energies exceed 0.7 eV.

At 1.0 V, even the fastest violet-generated electrons are stopped.

The total photocurrent therefore becomes zero.

The exact current remaining between 0.4 V and 1.0 V cannot be calculated from the information provided. We would need the energy distribution of the emitted electrons.

(e) Explain the student’s mistake.

The student’s conclusion of 0.4 V confuses the most numerous photoelectrons with the most energetic photoelectrons.

The red source produces most of the electrons, but the violet source produces the fastest electrons.

The stopping potential must be large enough to stop even the fastest electrons, regardless of how few of them are produced.

Therefore, the correct stopping potential is 1.0 V.

This is a particularly useful distinction in experiments involving multiple light frequencies. A small number of high-energy photoelectrons can determine the stopping potential even when the majority of the photocurrent comes from lower-energy electrons.

(f) Predict the effect of increasing red intensity.

The stopping potential remains 1.0 V.

Increasing red intensity increases the number of red photons arriving per second. This increases the red contribution to the photocurrent.

However, the energy of each red photon remains 2.4 eV.

Its maximum photoelectron kinetic energy therefore remains 0.4 eV.

The fastest electrons are still produced by the violet source, so the combined stopping potential remains unchanged.

08What to remember for your HSC exam

The entire threshold-frequency problem comes down to separating the energy of an individual photon from the number of photons arriving per second.

Keep these relationships clear:

The most important equation is:

\[
\boxed{E_{k,\max}=hf-\phi}
\]

And the most important threshold condition is:

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

If the photon energy is below the work function, no photoelectrons are produced in the single-photon model, regardless of how long the light shines.

Once you understand this distinction, the next useful step is interpreting a graph of stopping potential against light frequency. Its gradient is \(h/e\), and its frequency-axis intercept is the threshold frequency. This is how experimental photoelectric data can be used to determine Planck’s constant.

See How to Find Planck’s Constant from Stopping-Voltage Data for the next stage.