NSWPhysicsSyllabus dot point

Inquiry Question 2: What is observed when light interacts with matter?

Analyse the photoelectric effect, including Einstein's photon equation hf = phi + KE_max, the role of Planck's constant, and the inability of the wave model to explain the threshold frequency and the kinetic-energy results

Q: 2022 HSC HSC: Light of wavelength 300 nm is incident on a metal surface with work function 3.5 eV. Calculate the energy of an incident photon in eV, the maximum kinetic energy of the ejected photoelectrons, and the stopping voltage. State what would happen if the intensity of the light were doubled and what would happen if the wavelength were increased to 400 nm.

Photon energy: $E = h c / \lambda = (6.626 \times 10^{-34})(3.00 \times 10^8) / (3.00 \times 10^{-7}) = 6.63 \times 10^{-19}$ J $= 4.14$ eV. Maximum kinetic energy of photoelectrons (Einstein's equation): $KE_{\max} = h f - \phi = 4.14 - 3.5 = 0.64$ eV. Stopping voltage: $V_s = KE_{\max} / e = 0.64$ V. Doubling the intensity: doubles the photon flux, so doubles the photocurrent (more electrons ejected per second), but does not change $KE_{\max}$ or $V_s$ (each electron still absorbs only one photon of the same energy). At $\lambda = 400$ nm: photon energy $E = 1240 / 400$ nm-eV $= 3.10$ eV, which is below the $3.5$ eV work function. No photoelectrons are emitted regardless of how bright the light is. This is the threshold-frequency phenomenon. Markers reward correct photon energy in eV, the Einstein equation, the stopping voltage, and qualitative answers on intensity and wavelength changes consistent with the photon model.

Q: 2017 HSC HSC: Outline two experimental observations of the photoelectric effect that the classical wave model of light could not explain, and describe how Einstein's photon hypothesis accounts for them.

Observation 1: threshold frequency. Below a certain frequency $f_0$, no photoelectrons are emitted no matter how bright the light. A classical wave would deliver more energy when more intense, so it should always eject electrons given enough intensity. Einstein's explanation: light is a stream of photons, each with energy $h f$. A single photon delivers its energy to one electron all at once. If $h f < \phi$ (the work function), no electron can escape; if $h f \geq \phi$, electrons are ejected with $KE_{\max} = h f - \phi$. Intensity controls photon number, not energy per photon. Observation 2: maximum kinetic energy depends on frequency, not intensity. Brighter light of the same colour gives more photoelectrons per second but no extra energy per electron. Brighter blue light ejects faster electrons than dim blue light only in number, not in speed. Einstein's explanation: each electron absorbs one photon and carries away energy $h f - \phi$. Doubling intensity doubles the number of absorbing electrons but each still gets exactly $h f$ of input. Observation 3 (bonus): essentially zero delay between switching on the light and ejection of photoelectrons, even at very low intensity. A wave model would predict a build-up time; the photon model predicts instantaneous absorption. Markers reward two observations correctly described and a clear photon-by-photon explanation linked to $h f = \phi + KE_{\max}$.

A focused answer to the HSC Physics Module 7 dot point on the quantum model of light. Photon energy E = hf, Einstein's photoelectric equation hf = phi + KE_max, Planck's constant, threshold frequency and stopping voltage, and why the wave model cannot explain the observations.

Generated by Claude OpusReviewed by Better Tuition Academy10 min answerUpdated 2026-05-18

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What this dot point is asking

NESA wants you to use Einstein's photon model to explain the photoelectric effect, write and apply $h f = \phi + KE_{\max}$ , calculate the threshold frequency and stopping voltage, and clearly state which observations the classical wave model cannot account for.

The answer

Setting the scene

By the late 1800s the wave model of light was the standard. But in 1887 Heinrich Hertz noticed UV light striking metal electrodes increased the spark distance in his radio-wave apparatus. Lenard's careful experiments (1902) revealed three features the wave model could not explain:

Threshold frequency. Below a metal-specific frequency $f_0$ , no electrons are ejected no matter how intense the light or how long it shines.
Frequency, not intensity, sets electron energy. $KE_{\max}$ depends linearly on frequency. Intensity sets the number of photoelectrons per second, not their energies.
No measurable time delay. Electrons are ejected effectively instantaneously when the light starts, even at very low intensity. A classical wave would need time to accumulate enough energy in one electron.

Einstein's photon hypothesis (1905)

Building on Planck's 1900 idea that energy is exchanged in discrete amounts $h f$ , Einstein proposed that light itself is made of discrete energy packets:

E_{\text{photon}} = h f

A single photon is absorbed by a single electron all at once. If the electron is bound to the metal by an energy $\phi$ (the work function, the minimum energy to remove the least tightly bound electron), then:

\boxed{h f = \phi + KE_{\max}}

This is Einstein's photoelectric equation. $KE_{\max}$ is the maximum kinetic energy of the ejected photoelectron; electrons deeper in the metal lose more energy before escape and emerge with less.

Threshold frequency

The minimum frequency that can eject any electron is the one where $KE_{\max} = 0$ :

f_0 = \frac{\phi}{h}

Below $f_0$ , photons simply do not carry enough energy to free an electron, no matter how many arrive. This is the killer observation for the wave model: a classical wave of any frequency should eventually deliver enough energy if intense or sustained enough.

Stopping voltage

In the standard experiment, a positive collector electrode is gradually reverse-biased until the most energetic photoelectrons are turned back. The reversing voltage at which photocurrent vanishes is the stopping voltage $V_s$ :

e V_s = KE_{\max} = h f - \phi

A plot of $V_s$ vs $f$ is a straight line with gradient $h / e$ (giving Planck's constant) and $x$ -intercept $f_0$ (giving the threshold) and $y$ -intercept $-\phi / e$ (giving the work function). This is Millikan's 1916 experiment, which confirmed Einstein's equation to high precision and helped earn both their Nobel Prizes.

Worked example: caesium photocell

Caesium has work function $\phi = 2.10$ eV. Calculate the maximum kinetic energy and stopping voltage when light of wavelength $400$ nm illuminates the cathode.

Photon energy: $E = h c / \lambda$ . Using the shortcut $h c = 1240$ eV nm:

$E = 1240 / 400 = 3.10$ eV.

$KE_{\max} = 3.10 - 2.10 = 1.00$ eV.

Stopping voltage: $V_s = KE_{\max} / e = 1.00$ V.

If the wavelength is increased to $700$ nm: $E = 1240 / 700 = 1.77$ eV $< \phi$ , so no photoelectrons are emitted regardless of intensity.

Why the wave model fails

Three predictions of the classical wave model are flatly contradicted:

Wave model prediction	Observation
Any frequency works given enough intensity	A sharp threshold frequency $f_0$ exists
IMATH_31 increases with intensity	IMATH_32 depends only on $f$ , not intensity
Time lag at low intensity (energy builds up)	No measurable delay

The wave model is rescued for interference and diffraction, but for absorption and emission by atoms, the quantum (photon) model is needed. This duality is the heart of "wave-particle duality": light behaves as a wave in propagation and as a particle in interaction.

Planck's constant

$h = 6.626 \times 10^{-34}$ J s is the universal constant relating frequency to energy quantum. It also appears in:

The energy of a photon: $E = h f$ .
The momentum of a photon: $p = h / \lambda$ .
De Broglie's matter-wave relation: $\lambda = h / p$ .
The Heisenberg uncertainty principle: $\Delta x \Delta p \geq \hbar / 2$ .

Millikan's measurement of the slope $h / e$ in the photoelectric stopping-voltage plot gave $h = 6.57 \times 10^{-34}$ J s, agreeing with Planck's blackbody value.

Common traps

Writing $h f = \phi - KE_{\max}$ . The signs are: photon energy in, work function out, kinetic energy out. So $h f = \phi + KE_{\max}$ .

Saying intensity increases the electron energy. Intensity sets the photon flux (number per second), so it sets the current, not the energy per electron.

Mixing up frequency and wavelength. Higher frequency = shorter wavelength = more energetic photon. Threshold frequency $f_0$ corresponds to maximum (cut-off) wavelength $\lambda_0 = c / f_0$ .

Quoting $\phi$ as the ionisation energy of the metal atom. It is the work function of the bulk metal: the energy needed to remove an electron from the metal surface to a stationary state just outside.

Saying photoelectrons are "freed" when photons "shake" the electrons loose. Each photon is absorbed entire by one electron in a single quantum event; there is no shaking.

In one sentence

Einstein's photon model treats light as discrete packets of energy $E = h f$ that are absorbed one-for-one by electrons, giving $h f = \phi + KE_{\max}$ , which explains the photoelectric effect's threshold frequency, intensity-independent electron energies and zero time delay - none of which the wave model can produce.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC5 marksLight of wavelength 300 nm is incident on a metal surface with work function 3.5 eV. Calculate the energy of an incident photon in eV, the maximum kinetic energy of the ejected photoelectrons, and the stopping voltage. State what would happen if the intensity of the light were doubled and what would happen if the wavelength were increased to 400 nm.

Show worked answer →

Photon energy:

$E = h c / \lambda = (6.626 \times 10^{-34})(3.00 \times 10^8) / (3.00 \times 10^{-7}) = 6.63 \times 10^{-19}$ J $= 4.14$ eV.

Maximum kinetic energy of photoelectrons (Einstein's equation):

$KE_{\max} = h f - \phi = 4.14 - 3.5 = 0.64$ eV.

Stopping voltage:

$V_s = KE_{\max} / e = 0.64$ V.

Doubling the intensity: doubles the photon flux, so doubles the photocurrent (more electrons ejected per second), but does not change $KE_{\max}$ or $V_s$ (each electron still absorbs only one photon of the same energy).

At $\lambda = 400$ nm: photon energy $E = 1240 / 400$ nm-eV $= 3.10$ eV, which is below the $3.5$ eV work function. No photoelectrons are emitted regardless of how bright the light is. This is the threshold-frequency phenomenon.

Markers reward correct photon energy in eV, the Einstein equation, the stopping voltage, and qualitative answers on intensity and wavelength changes consistent with the photon model.

2017 HSC4 marksOutline two experimental observations of the photoelectric effect that the classical wave model of light could not explain, and describe how Einstein's photon hypothesis accounts for them.

Show worked answer →

Observation 1: threshold frequency. Below a certain frequency $f_0$ , no photoelectrons are emitted no matter how bright the light. A classical wave would deliver more energy when more intense, so it should always eject electrons given enough intensity.

Einstein's explanation: light is a stream of photons, each with energy $h f$ . A single photon delivers its energy to one electron all at once. If $h f < \phi$ (the work function), no electron can escape; if $h f \geq \phi$ , electrons are ejected with $KE_{\max} = h f - \phi$ . Intensity controls photon number, not energy per photon.

Observation 2: maximum kinetic energy depends on frequency, not intensity. Brighter light of the same colour gives more photoelectrons per second but no extra energy per electron. Brighter blue light ejects faster electrons than dim blue light only in number, not in speed.

Einstein's explanation: each electron absorbs one photon and carries away energy $h f - \phi$ . Doubling intensity doubles the number of absorbing electrons but each still gets exactly $h f$ of input.

Observation 3 (bonus): essentially zero delay between switching on the light and ejection of photoelectrons, even at very low intensity. A wave model would predict a build-up time; the photon model predicts instantaneous absorption.

Markers reward two observations correctly described and a clear photon-by-photon explanation linked to $h f = \phi + KE_{\max}$ .