Understand the photoelectric effect the easy way, with plain English intuition, an interactive simulation, photons and the work function, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.
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Shine the right colour of light on a clean piece of metal and electrons jump straight off the surface. That is the photoelectric effect, and the strange part is the colour matters far more than the brightness. A dim blue light can free electrons while a blindingly bright red light frees none at all. The only way to make sense of this is to picture light not as a smooth wave but as a stream of tiny energy packets.
Light comes in packets
Forget the idea of light as a continuous wave for a moment. Light arrives in little bundles of energy called photons, and each photon carries a fixed amount of energy set entirely by its frequency.
A high frequency photon (towards the blue and ultraviolet end) is a big packet of energy.
A low frequency photon (towards the red end) is a small packet of energy.
The energy of one photon is
E=hf=λhc
where h=6.63×10−34 J s is Planck’s constant. Brightness does not change the size of each packet. A brighter light simply sends more photons, not bigger ones.
Freeing an electron costs energy
An electron sitting in a metal is held in place, a bit like a coin stuck at the bottom of a well. To get it out you have to pay an energy toll. That toll is the work function, written W, and it is the minimum energy needed to free one electron from that particular metal.
Here is the key rule. When a photon hits the surface, it hands its whole packet of energy to a single electron. The electron spends W just to escape, and whatever is left over becomes its kinetic energy:
KEmax=hf−W
A photon hands its energy to an electron, which escapes carrying the leftover as kinetic energy. The graph on the right shows the rule KEmax=hf−W as a straight line. It only lifts off the axis once the frequency passes the threshold frequencyf0, and from there it climbs with a gradient equal to Planck’s constanth.
Below the threshold, nothing happens
There is a frequency below which no electrons come off at all, however bright the light. This is the threshold frequencyf0, and it is the frequency at which a single photon carries exactly the work function and no more.
The reason is simple once you think in packets. If each photon is too small to pay the full work function, the electron cannot escape, and it cannot save up energy from two photons at once. So a dim light of high enough frequency works, while a brilliant light of too low a frequency does nothing.
Intensity versus frequency
These two are the heart of every photoelectric question, so keep them apart:
Frequency controls the energy of each photon, and so the maximum kinetic energy of the electrons. Raise the frequency and the electrons come off faster.
Intensity (brightness) controls the number of photons, and so the number of electrons, which is the photocurrent. Raise the intensity and more electrons come off, but at the same speed.
A smooth wave could never behave like this. A wave would let any colour of light, however dim, gradually pour in energy until an electron broke free. The fact that frequency draws a hard line, with nothing happening below it, is exactly the particle behaviour the wave model cannot explain.
See it for yourself
Change the colour (frequency) of the light, turn the brightness up and down, and adjust the voltage. Below the threshold frequency nothing happens no matter how bright the light gets. Above it, electrons fly off, and only raising the frequency gives them more energy. The reverse voltage that just stops even the fastest electrons, the stopping voltageV0, is how their maximum kinetic energy is measured, since KEmax=qV0.
Most photoelectric questions are one of a few short calculations.
Find the energy of one photon with E=hf, or E=λhc if you are given the wavelength.
Subtract the work function to get the maximum kinetic energy: KEmax=hf−W.
If the answer is needed in electron-volts, divide the joules by 1.6×10−19.
To test whether any electrons come off at all, check that hf>W, which is the same as the frequency being above the threshold.
Watch the powers of ten. These energies are tiny, so keep your scientific notation tidy and add the exponents carefully.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
Write the energy of a single photon two ways.
E=hf=λhc, where h=6.63×10−34 J s. It depends only on frequency, not brightness.
What is the work functionW?
The minimum energy needed to free one electron from the surface of a particular metal. A photon must supply at least W before any electron can leave.
Write the maximum kinetic energy of a photoelectron.
KEmax=hf−W — the photon energy minus the work function (the toll paid to escape).
Below the threshold frequency, what happens as you turn up the brightness?
Still no electrons are emitted. Each photon is too weak to pay the work function, and an electron cannot save up energy from two photons — brightness only adds more weak photons.
Above threshold, raising the intensity changes what?
The number of electrons (the photocurrent) rises, but the maximum kinetic energy is unchanged. Only raising the frequency raises KEmax.
Why can the wave model not explain the photoelectric effect?
A wave would let dim light slowly pour energy into an electron until it escaped, so any frequency should eventually work. The hard threshold only makes sense if light arrives as photons.
Recall · Atomic Spectra and Energy Levels
How is the energy of a photon related to its frequency in an atomic transition?
ΔE=hf — the same Planck relation. The photon energy equals the gap between two energy levels, Ehigh−Elow.
Recall · The Wave Model of Light
What is the speed of every electromagnetic wave in a vacuum?
c=3.0×108 m/s, for radio, light, X-rays and all the rest, regardless of frequency.
See the recipe in action in the Worked Examples tab, then test yourself in Try It.
Worked examples
Worked Example 1Maximum kinetic energy of a photoelectron
Light of frequency 1.0×1015 Hz strikes a metal with a work function of 3.0×10−19 J. Taking h=6.63×10−34 J s, find the maximum kinetic energy of the ejected photoelectrons.
1
Each photon delivers a fixed packet of energy that depends only on the frequency. Work out that packet first.
E=hf=(6.63×10−34)(1.0×1015)=6.63×10−19 J
2
The electron must first spend the work function just to escape the metal. Whatever energy is left over becomes kinetic energy.
KEmax=hf−W
3
Substitute the photon energy and the work function.
KEmax=6.63×10−19−3.0×10−19=3.63×10−19 J
Answer
KEmax≈3.6×10−19 J
Worked Example 2Energy of a photon in joules and electron-volts
Find the energy of a photon of wavelength 500 nm (5.0×10−7 m), in joules and in electron-volts. Take h=6.63×10−34 J s, c=3.0×108 m/s and 1 eV =1.6×10−19 J.
1
When you are given wavelength instead of frequency, use the form of the photon energy that uses the speed of light.
E=λhc
2
Substitute the numbers.
E=5.0×10−7(6.63×10−34)(3.0×108)=3.98×10−19 J
3
To convert to electron-volts, divide by the size of one electron-volt in joules.
E=1.6×10−193.98×10−19=2.5 eV
Answer
E=3.98×10−19 J=2.5 eV
Practice questions
Practice test
Try it yourself
6 questions, 9 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.A metal is lit with light below its threshold frequency. The brightness of the light is then increased greatly. The number of electrons ejected from the metal is:
1mark
Need a hint?
Below the threshold frequency, no single photon carries enough energy to free an electron. Brightness only adds more photons of the same energy.
Show worked solution
Below the threshold frequency, every photon is too weak to free an electron on its own, and an electron cannot absorb two photons at once. Adding more photons by turning up the brightness just means more photons that are each still too weak, so the number of ejected electrons stays zero. This result is exactly what the wave model could not explain.
Q2.The intensity of light above the threshold frequency is increased. What happens to the photoelectrons?
1mark
Need a hint?
Intensity sets the number of photons. The energy of each photon, and so the kinetic energy of each electron, depends only on frequency.
Show worked solution
Brighter light means more photons per second, so more electrons are knocked out and the photocurrent rises. But each photon still carries the same energy hf, so the maximum kinetic energy of each electron does not change. To raise the kinetic energy you must raise the frequency, not the intensity.
Q3.A photon has frequency 2.0×1015 Hz. Taking h=6.63×10−34 J s, its energy is closest to:
1mark
Need a hint?
Use E=hf and multiply carefully.
Show worked solution
E=hf=(6.63×10−34)(2.0×1015)=1.326×10−18 J, which rounds to 1.3×10−18 J. Forgetting to add the powers of ten correctly, or dividing instead of multiplying, gives the other options.
Q4.A photon of energy 8.0×10−19 J strikes a metal with a work function of 5.0×10−19 J. The maximum kinetic energy of the ejected electron is:
1mark
Need a hint?
Use KEmax=hf−W, where hf is the photon energy.
Show worked solution
KEmax=hf−W=8.0×10−19−5.0×10−19=3.0×10−19 J. Adding the two energies instead of subtracting gives 1.3×10−18 J, a common slip.
Q5.Light of frequency 1.2×1015 Hz falls on a metal with a work function of 4.0×10−19 J. Taking h=6.63×10−34 J s, find the energy of each photon and the maximum kinetic energy of the ejected photoelectrons. Show your working.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
The energy of each photon is
E=hf=(6.63×10−34)(1.2×1015)=7.96×10−19 J.
The maximum kinetic energy is the photon energy minus the work function,
KEmax=hf−W=7.96×10−19−4.0×10−19=3.96×10−19 J≈4.0×10−19 J.
Q6.In a photoelectric experiment, the maximum kinetic energy of the emitted electrons is plotted against the frequency of the light, giving a straight line. The line crosses the energy axis at −2.2 eV. Determine the work function of the metal, and state what the gradient of the line represents.
2marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
The graph follows KEmax=hf−W. The vertical intercept is −W, so the work function is W=2.2 eV. The gradient of the line is Planck's constant h.
Adapted from VCAA 2025 Physics Exam, Section B Q16
Frequently asked questions
Why does turning up the brightness not give faster electrons?
Brightness only adds more photons, and each photon still carries the same energy because that energy depends on frequency alone. More photons means more electrons knocked out, but each electron still gets the same leftover energy, so the maximum kinetic energy is unchanged.
What is the work function?
The work function is the minimum energy an electron needs to escape the surface of a particular metal. A photon must supply at least this much energy before any electron can leave, and whatever is left over after paying the work function becomes the kinetic energy of the electron.
Why can the wave model not explain the photoelectric effect?
A wave would let a dim light slowly pour energy into an electron until it built up enough to escape, so even low frequency light should eventually work. In reality nothing happens below the threshold frequency no matter how long you wait, which only makes sense if light arrives as discrete packets of energy called photons.
