What is the kinetic-energy-against-frequency graph?

The single most examined graph in this topic plots maximum kinetic energy E_k, against photon frequency f. From E_k,=hf-W this is a straight line of the form y=mx+c, so the gradient equals Planck's constant h and is the same for every metal. The horizontal intercept (where E_k,=0) is the threshold frequency f_0=W/h, and the vertical intercept is -W. Different metals therefore give parallel lines (same gradient h) shifted horizontally according to their work function: a metal with a larger work function has a higher threshold frequency and its line crosses the axis further to the right.

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How does the photoelectric effect reveal the particle nature of light?

Explain the photoelectric effect using photons, work function, threshold frequency and Einstein's equation

A focused answer to the WACE Year 12 Physics Unit 4 dot point on the photoelectric effect. Photon energy, work function, threshold frequency, Einstein's photoelectric equation, stopping voltage and why the wave model fails.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

WACE wants you to describe the photoelectric effect, explain why it forced a particle (photon) model of light, and apply Einstein's photoelectric equation including the work function, threshold frequency and stopping voltage. The central argument is that the experimental results contradict the wave model and are explained naturally if light is quantised.

Photons

Light is emitted and absorbed in discrete packets called photons, each carrying energy

E=hf=\frac{hc}{\lambda},

where $h=6.63\times10^{-34}\ \text{J s}$ is Planck's constant. A brighter beam contains more photons but each photon still has the same energy fixed by the frequency. This is the key break from classical physics, where energy was thought to be spread continuously across a wave.

The photoelectric effect

When light of high enough frequency strikes a metal surface, electrons are ejected. The experimental facts are:

Emission happens only above a threshold frequency $f_0$ , no matter how intense the light.
Above threshold, increasing intensity increases the number of electrons but not their maximum kinetic energy.
The maximum kinetic energy rises linearly with frequency.
Emission is essentially instantaneous.

A wave model cannot explain these: it predicts that any frequency should work given enough time and intensity, and that brighter light should give faster electrons. None of that is observed.

The work function and threshold

The work function $W$ is the minimum energy needed to free an electron from the surface. A single photon must supply at least this much in one interaction. The threshold frequency is the frequency at which a photon just provides $W$ ,

f_0=\frac{W}{h}.

Below $f_0$ no electrons escape however bright the light, because no single photon has enough energy.

Einstein's photoelectric equation

For photons above threshold, the photon energy splits into the work function plus the kinetic energy of the freed electron. The most energetic electrons (those at the surface) have

E_{k,max}=hf-W.

Plotting $E_{k,max}$ against $f$ gives a straight line of gradient $h$ and frequency-intercept $f_0$ , a classic graph question. The vertical intercept is $-W$ .

Stopping voltage

The maximum kinetic energy can be measured with a stopping voltage $V_0$ , the reverse potential difference that just halts the fastest electrons,

eV_0=E_{k,max}=hf-W.

Because $V_0$ depends on frequency but not intensity, it provides direct evidence for the photon model.

Maximum kinetic energy and stopping voltage

Photoelectrons from a metal with W = 2.3 eV

Ultraviolet light of wavelength $250\ \text{nm}$ strikes a metal whose work function is $W=2.3\ \text{eV}=3.68\times10^{-19}\ \text{J}$ .

Photon energy:

E=\frac{hc}{\lambda}=\frac{(6.63\times10^{-34})(3.0\times10^{8})}{2.50\times10^{-7}}=7.96\times10^{-19}\ \text{J}.

Maximum kinetic energy of the electrons:

E_{k,max}=hf-W=7.96\times10^{-19}-3.68\times10^{-19}=4.3\times10^{-19}\ \text{J}.

Stopping voltage:

V_0=\frac{E_{k,max}}{e}=\frac{4.3\times10^{-19}}{1.60\times10^{-19}}=2.7\ \text{V}.

The kinetic-energy-against-frequency graph

The single most examined graph in this topic plots maximum kinetic energy $E_{k,\max}$ against photon frequency $f$ . From $E_{k,\max}=hf-W$ this is a straight line of the form $y=mx+c$ , so the gradient equals Planck's constant $h$ and is the same for every metal. The horizontal intercept (where $E_{k,\max}=0$ ) is the threshold frequency $f_0=W/h$ , and the vertical intercept is $-W$ . Different metals therefore give parallel lines (same gradient $h$ ) shifted horizontally according to their work function: a metal with a larger work function has a higher threshold frequency and its line crosses the axis further to the right. Being asked to extract $h$ from the gradient, or $W$ from an intercept, is a standard data-analysis question.

Working in joules and electronvolts

Work functions are usually quoted in electronvolts, where $1\ \text{eV}=1.60\times10^{-19}\ \text{J}$ . Decide on one unit and convert everything to it before substituting into $E_{k,max}=hf-W$ . Keep frequencies in hertz and wavelengths in metres.

Exam-style practice questions

Practice questions written in the style of SCSA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

WACE 20227 marksIn a photoelectric experiment, light of wavelength

4.0\times10^{-7}\ \text{m}

is shone on a metal of work function

1.9\ \text{eV}

. (a) Calculate the energy of each photon in joules. (b) Calculate the maximum kinetic energy of the emitted photoelectrons. (c) Calculate the stopping voltage required to halt the fastest electrons.

Show worked answer →

A 7 mark calculation rewards the photon energy, a consistent unit conversion and Einstein's equation.

(a) Photon energy: $E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3.0\times10^{8})}{4.0\times10^{-7}}=4.97\times10^{-19}\ \text{J}$ .
(b) Maximum kinetic energy: Convert the work function: $W=1.9\times1.6\times10^{-19}=3.04\times10^{-19}\ \text{J}$ . Then
$E_{k,\max}=hf-W=4.97\times10^{-19}-3.04\times10^{-19}=1.93\times10^{-19}\ \text{J}.$
(c) Stopping voltage: $V_0=\dfrac{E_{k,\max}}{e}=\dfrac{1.93\times10^{-19}}{1.6\times10^{-19}}=1.2\ \text{V}$ .

Markers reward $E=hc/\lambda$ , converting the work function to joules, $E_{k,\max}=hf-W$ and $V_0=E_{k,\max}/e$ giving about $1.2\ \text{V}$ .

WACE 20216 marksExplain how the experimental results of the photoelectric effect cannot be explained by the wave model of light but are explained by the photon model. Refer to threshold frequency and the effect of intensity.

Show worked answer →

A 6 mark explanation needs the failing wave predictions and the matching photon explanation.

Threshold frequency. Experiments show no electrons are emitted below a threshold frequency, however intense or prolonged the light. The wave model predicts that any frequency should eventually free electrons given enough intensity or time, which contradicts this. The photon model explains it: one electron absorbs one photon of energy $E=hf$ , and if $hf<W$ no single photon can free an electron regardless of how many arrive.

Effect of intensity. Experiments show that increasing intensity above threshold raises the current (number of electrons) but not their maximum kinetic energy. The wave model predicts brighter light should give faster electrons, which is wrong. The photon model explains it: intensity is the number of photons per second, so more photons free more electrons, but each electron's energy is still set by the single photon's frequency through $E_{k,\max}=hf-W$ .

Markers reward the threshold and intensity facts, the failed wave predictions and the one-photon-one-electron photon explanation.