Year 12 Physics Module 7 · The Nature of Light ⏱ ~45 min 5 MC · 2 Short Answer Lesson 13 of 14

The Photoelectric Effect

On 17 March 1905, Albert Einstein, a patent examiner at the Swiss Patent Office, Bern, published "On a Heuristic Point of View Concerning the Production and Transformation of Light." He proposed that light arrives in discrete quanta, each carrying energy $E = hf$. His equation $E_{k,\text{max}} = hf - \phi$ precisely predicted the linear relationship between photon frequency and maximum electron kinetic energy, confirmed quantitatively by Millikan in 1916. For this paper, not for relativity, Einstein received the Nobel Prize in 1921.

Today's hook: On 17 March 1905, Albert Einstein at the Swiss Patent Office Bern published the paper that would win him the 1921 Nobel Prize. He predicted that the maximum kinetic energy of ejected electrons should be $E_{k,\text{max}} = hf - \phi$, a linear function of frequency, independent of intensity. A dim UV lamp ejects electrons; a very bright red lamp ejects none. How does treating light as discrete quanta explain this, when the wave model cannot?

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Orient and predict

Predict which lamp ejects electrons from zinc, then set your goals and vocabulary for the photoelectric effect.

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.

Before you read, predict

A dim UV lamp and a bright red lamp both shine on a zinc plate. The UV lamp causes electrons to be emitted; the red lamp does not.

According to the wave model of light, which lamp should cause electron emission, the brighter one or the higher-frequency one?
What does the actual result tell you about how light energy is delivered to the metal?
If you double the intensity of the UV lamp, what happens to the number of emitted electrons? What happens to their maximum kinetic energy?

Write your predictions before reading on, you will revisit them at the end.

Warm-up, which observation of the photoelectric effect CANNOT be explained by the classical wave model?

Learning Intentions

goals

Know, The Photoelectric Effect

Light ejects electrons from metal surfaces
Only frequencies above threshold $f_0$ work
Instantaneous emission contradicts wave model

Understand, Photon Model

$E = hf$, energy quantised in photons
$E_{k,max} = hf - \phi$
Work function $\phi$ is material-dependent

Can Do, Analyse Photoelectric Data

Calculate $E_{k,max}$ from frequency
Determine work function from graphs
Explain why wave model fails

Scan these before reading

vocab

PhotonA quantum of electromagnetic energy with $E = hf$. Light comes in discrete packets, not a continuous wave.

Work function ($\phi$)The minimum energy needed to eject an electron from a metal surface; different for each metal.

Threshold frequency ($f_0$)The minimum frequency of light that can cause photoemission: $hf_0 = \phi$. Below this, no emission occurs regardless of intensity.

Stopping voltage ($V_s$)The reverse voltage needed to stop the most energetic photoelectrons: $eV_s = E_{k,max}$.

PhotocurrentThe electric current produced by the flow of photoelectrons. Proportional to intensity (number of photons per second) when $f > f_0$.

Cross-lesson links: L10 showed the experimental failures of the wave model, Hertz 1887 Karlsruhe, Hallwachs 1888 Dresden. L16 provides Einstein's 1905 Swiss Patent Office Bern resolution: $E_{k,\text{max}} = hf - \phi$, confirmed quantitatively by Millikan 1916. This is the most-tested equation in Module 7. L20 completes wave-particle duality by showing that not just light, but electrons too exhibit wave behaviour, the Davisson–Germer 1927 Bell Labs diffraction experiment.

Core Content

Observe what the wave model cannot explain

Meet the three experimental results, threshold frequency, instantaneous emission and frequency-dependent $E_{k,max}$, that break classical wave theory.

The Experimental Observations

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What the wave model cannot explain

When light shines on a clean metal surface in a vacuum, electrons may be ejected. This is the photoelectric effect, first observed by Hertz in 1887. Lenard's systematic experiments (1902) revealed three features that the wave theory of light could not explain.

1. Threshold frequency: For any given metal, there is a minimum frequency $f_0$ below which no electrons are emitted, regardless of how intense the light is. A dim UV lamp causes emission; a bright infrared lamp does not.

2. Instantaneous emission: If $f > f_0$, electrons are emitted immediately, even at very low light intensity. The wave model predicts that dim light should require time to build up enough energy to eject an electron, but no delay is observed.

3. Maximum kinetic energy depends on frequency, not intensity: $E_{k,max}$ increases linearly with frequency. Increasing the intensity only increases the number of electrons emitted, not their energy. The wave model cannot explain this, it predicts that brighter light should give electrons more energy.

Figure 1, Photoelectric effect apparatus: light strikes the cathode, ejecting electrons that are collected by the anode, creating a measurable photocurrent.

Stop & Check

A metal has work function 2.3 eV. Light of frequency $8.0\times10^{14}$ Hz shines on it. Using $h = 6.63\times10^{-34}$ J·s, calculate the maximum kinetic energy of emitted electrons. What is the threshold frequency for this metal?

Photoelectric effect, three observations the wave model cannot explain: (1) threshold frequency $f_0$ (below which no emission at any intensity); (2) instantaneous emission even at low intensity; (3) $E_{k,max}$ depends on frequency, not intensity. Wave model fails because it predicts energy builds up continuously, but one photon interacts with one electron instantly.

Write the three observations and one sentence each on why the wave model fails each one.

The photoelectric effect shows that doubling the intensity of light above the threshold frequency:

Apply Einstein's photon model

Use $E = hf$ and $E_{k,max} = hf - \phi$, and read the $E_{k,max}$ against $f$ graph the model predicts.

Einstein's Photon Explanation

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Light as quantised energy packets

We just saw three experimental observations that the wave model cannot explain. That raises a question: what single physical idea did Einstein propose in 1905 that explains all three observations at once? This card answers it → the photon model: one photon, one electron, energy $E = hf$.

In 1905, Einstein proposed that light consists of discrete packets of energy called photons, each with energy $E = hf$. One photon interacts with one electron, either it has enough energy to eject the electron, or it doesn't. No accumulation is possible. This single insight explains all three observations that stumped the wave model.

When a photon strikes the metal, it transfers all its energy to a single electron. If this energy exceeds the metal's work function $\phi$, the electron escapes with kinetic energy:

Einstein's Photoelectric Equations

$$E = hf = \dfrac{hc}{\lambda}$$

$$E_{k,max} = hf - \phi$$

$$hf_0 = \phi \quad \Rightarrow \quad f_0 = \dfrac{\phi}{h}$$

$$eV_s = E_{k,max}$$

Why the wave model fails, photon model succeeds:

Threshold frequency: Emission only occurs when $hf > \phi$, so $f > f_0 = \phi/h$. Below this, each photon is too weak, no accumulation possible.
Instantaneous emission: Energy is delivered in a single photon-electron interaction, not accumulated over time. At $f > f_0$, emission is immediate even for dim light.
Frequency dependence of $E_{k,max}$: $E_{k,max} = hf - \phi$ depends only on $f$. Higher intensity means more photons, so more electrons, but each photon still carries $hf$, so $E_{k,max}$ is unchanged.

Figure 2, $E_{k,max}$ vs frequency: linear above threshold $f_0$ with slope = $h$ (Planck's constant). Different metals have different work functions (different $f_0$) but the same slope $h$, confirming the universal photon model.

Stop & Check

Sodium has $\phi = 2.28$ eV. Light of wavelength 450 nm shines on it. (Use $hc = 1240$ eV·nm as a shortcut.) Calculate $E_{k,max}$ in eV. What stopping voltage is required? What is the threshold wavelength?

Einstein's four equations: $E = hf = hc/\lambda$ (shortcut: $E_{\text{eV}} = 1240/\lambda_{\text{nm}}$); $E_{k,max} = hf - \phi$; $f_0 = \phi/h$; $eV_s = E_{k,max}$. The $E_{k,max}$–$f$ graph is linear (slope $= h$, x-intercept $= f_0$) and confirms the photon model, different metals give different $f_0$ but the same slope $h$.

Write all four equations and note what slope and intercepts mean on the $E_{k,max}$ vs $f$ graph.

A metal has work function $\phi = 3.0$ eV. Light of frequency $9.0\times10^{14}$ Hz shines on it ($h = 4.14\times10^{-15}$ eV·s). The maximum kinetic energy of emitted electrons is approximately:

Derive Planck's constant from Millikan's data

Work the stopping-voltage graph step by step to get $h$ and the threshold frequency from real measurements.

Worked Example, Millikan's Confirmation

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From stopping voltage to Planck's constant

We just saw Einstein's four photoelectric equations and the $E_{k,max}$–$f$ graph. That raises a question: how did Millikan use stopping-voltage measurements to determine Planck's constant and confirm Einstein's equation experimentally? This card answers it → a four-step worked example deriving $h$ from real data.

Problem

A student measures the stopping voltage $V_s$ for light of different frequencies incident on a sodium cathode ($\phi = 2.28$ eV).

$f$ ($\times10^{14}$ Hz)	6.0	7.0	8.0	9.0	10.0
$V_s$ (V)	0.20	0.61	1.03	1.44	1.85

Show that $eV_s = hf - \phi$.
Plot $V_s$ vs $f$ and determine the gradient.
Use the gradient to find Planck's constant.
Find the threshold frequency from the graph.

Step 1, Derivation

The stopping voltage is the voltage needed to stop the most energetic electrons: $eV_s = E_{k,max}$. From Einstein's equation, $E_{k,max} = hf - \phi$. Therefore:

$$eV_s = hf - \phi \quad \Rightarrow \quad V_s = \dfrac{h}{e}f - \dfrac{\phi}{e}$$

This is a linear equation in $f$ with gradient $h/e$ and y-intercept $-\phi/e$.

Step 2, Gradient

$V_s$ vs $f$ is linear. Using the first and last data points:

$$\text{gradient} = \dfrac{\Delta V_s}{\Delta f} = \dfrac{1.85 - 0.20}{(10.0 - 6.0)\times10^{14}} = \dfrac{1.65}{4.0\times10^{14}} = 4.13\times10^{-15} \text{ V·s}$$

Step 3, Planck's constant

Since the gradient is $h/e$:

$$h = \text{gradient} \times e = 4.13\times10^{-15} \times 1.60\times10^{-19} = \mathbf{6.6\times10^{-34} \text{ J·s}}$$

(Accepted value: $6.63\times10^{-34}$ J·s, excellent agreement. This is how Millikan confirmed Einstein's theory in 1916.)

Step 4, Threshold frequency

When $V_s = 0$, all photon energy goes into the work function: $hf_0 = \phi$.

$$f_0 = \dfrac{\phi}{h} = \dfrac{2.28 \text{ eV}}{4.13\times10^{-15} \text{ eV·s}} = 5.52\times10^{14} \text{ Hz}$$

From the graph: extrapolate the line to $V_s = 0$, x-intercept $\approx 5.5\times10^{14}$ Hz.

Figure 3, Millikan's data: stopping voltage $V_s$ vs frequency $f$ for sodium. The gradient = $h/e = 4.13\times10^{-15}$ V·s, so $h = 6.6\times10^{-34}$ J·s, confirming Einstein's photon model.

Stop & Check

Potassium has $\phi = 2.30$ eV. Light of wavelength 400 nm shines on it. Calculate $E_{k,max}$ and the stopping voltage. What colour is this light?

Millikan's method: $V_s = (h/e)f - \phi/e$, linear graph. Gradient $= h/e = 4.13\times10^{-15}$ V·s $\Rightarrow h = 6.6\times10^{-34}$ J·s. x-intercept gives $f_0 \approx 5.5\times10^{14}$ Hz. Millikan intended to disprove Einstein but instead confirmed him perfectly. Graph features: slope $= h/e$; x-int $= f_0$; y-int $= -\phi/e$.

Write Millikan's linear equation $V_s = (h/e)f - \phi/e$ and label what each feature of the graph tells you.

A graph of $V_s$ vs $f$ for a metal has gradient $4.0\times10^{-15}$ V·s. The work function can be found from:

Avoid the common photoelectric mistakes

Separate intensity from frequency, keep your units straight, and work through the five-step exam checklist.

Common Mistakes and HSC Strategies

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Avoiding the traps that cost marks

We just saw how Millikan derived Planck's constant from a $V_s$–$f$ graph. That raises a question: what are the most common exam errors on photoelectric effect questions? This card answers it → three mistakes to avoid and a five-step strategy.

Mistake 1: Confusing intensity with frequency.
Intensity = number of photons per second. Frequency = energy per photon. Doubling intensity at fixed frequency above $f_0$ doubles the photocurrent but does NOT change $E_{k,max}$. Changing frequency changes $E_{k,max}$ (and may stop emission if $f$ drops below $f_0$).

Mistake 2: Forgetting to convert units.
Using $E = hc/\lambda$ but not converting nm to m. Quick check: visible light ($\lambda \approx 500$ nm) has $E \approx 1240/500 = 2.5$ eV. Use $hc = 1240$ eV·nm as a shortcut.

Mistake 3: Confusing $E_{k,max}$ and average kinetic energy.
$E_{k,max} = hf - \phi$ is the maximum KE, for electrons emitted from the surface with no energy lost. Electrons deeper in the metal lose energy escaping and have lower KE. Only surface electrons achieve $E_{k,max}$.

HSC Exam Strategy, Photoelectric Questions

Check if emission occurs first: Is $hf > \phi$? (Or is $\lambda < hc/\phi$?) If not, $E_{k,max} = 0$ and you stop there.
Apply Einstein's equation: $E_{k,max} = hf - \phi$ (convert eV to J or stay in eV consistently)
Stopping voltage: $V_s = E_{k,max}/e$ (joules) or numerically equal to $E_{k,max}$ in eV
Graph interpretation: slope of $V_s$–$f$ graph = $h/e$; x-intercept = $f_0$; y-intercept gives $\phi$
Photocurrent vs intensity: proportional (more photons = more electrons). $E_{k,max}$ is independent of intensity.

Three exam traps: (1) Intensity $\uparrow$ → photocurrent $\uparrow$, $E_{k,max}$ unchanged. (2) Convert nm to m before using $E = hc/\lambda$ in J, or use shortcut $E_{\text{eV}} = 1240/\lambda_{\text{nm}}$. (3) $E_{k,max}$ is the maximum (surface electrons), electrons from deeper in the metal have lower KE. HSC strategy: check $hf > \phi$ first; if not, $E_{k,max} = 0$.

Write the three mistakes and the five-step strategy checklist.

Apply and consolidate

Run your own photoelectric calculations, then argue in full why the wave model fails and the photon model succeeds.

Activities

Activity 1, Photoelectric Effect Calculations

ApplyBand 5

Apply $E_{k,max} = hf - \phi$ across a range of scenarios

Set $\phi = 2.3$ eV and $f = 5.0\times10^{14}$ Hz. Is there emission? Show using $E_{k,max} = hf - \phi$. What minimum frequency is needed?
For sodium ($\phi = 2.28$ eV), calculate $f_0$ and the threshold wavelength. What region of the spectrum is this?
For $\phi = 2.0$ eV and $f = 10.0\times10^{14}$ Hz, calculate $E_{k,max}$ and $V_s$. If the intensity is doubled, what changes and what doesn't?
A $V_s$–$f$ graph has gradient $4.14\times10^{-15}$ V·s and x-intercept $5.5\times10^{14}$ Hz. Determine $h$ and $\phi$ for this metal.

Activity 2, Wave Model Failure Analysis

EvaluateBand 6

Analyse why classical physics fails and how the photon model succeeds

The classical wave model predicts that dim light should eventually cause emission if left long enough, because the wave continuously delivers energy. Using classical wave theory, estimate how long it would take a 1 W/m² lamp to deliver 2.3 eV to a single electron in an area of $10^{-20}$ m². Compare this to observed instantaneous emission.
Explain, using the photon model, why zinc ($\phi = 4.3$ eV) requires UV light for emission, while caesium ($\phi = 2.0$ eV) can be triggered by visible blue light.
In Millikan's experiment, he measured the gradient of the $V_s$–$f$ graph to be $4.13\times10^{-15}$ V·s. (a) Calculate $h$. (b) Explain why this was important scientific evidence, what was Millikan trying to disprove, and what did he actually find?
Zinc is illuminated by UV light of wavelength 250 nm ($\phi_{Zn} = 4.3$ eV). (a) Calculate $E_{k,max}$. (b) The intensity is then doubled. Describe the effect on: photocurrent, $E_{k,max}$, $V_s$, $f_0$.

Independent practice

Show what you can do without prompts

Answer a shuffled question-bank set, then write two HSC-style responses on the photoelectric effect.

Quick recall, the photoelectric effect

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A fresh five-question set drawn from this lesson's bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

Short Answer, 8 marks

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ApplyBand 5(4 marks) 1. (a) Describe three experimental observations of the photoelectric effect that cannot be explained by the wave model of light. (b) Explain how Einstein's photon model accounts for each observation. (c) A metal has work function 2.5 eV. Calculate the threshold frequency and threshold wavelength. (d) Light of wavelength 350 nm shines on this metal, calculate $E_{k,max}$ and the stopping voltage.

1 mark: three observations · 1 mark: photon explanations · 1 mark: $f_0$ and $\lambda_0$ · 1 mark: $E_{k,max}$ and $V_s$

EvaluateBand 6(4 marks) 2. (a) A student plots $V_s$ vs $f$ for photoelectric experiments on two different metals and gets two parallel lines with the same gradient but different x-intercepts. Explain what information can be extracted from this graph and what the parallel lines tell you about Planck's constant and the work functions. (b) Metal A has $\phi = 2.1$ eV; Metal B has $\phi = 4.5$ eV. Light of wavelength 300 nm shines on both. For each metal, calculate $E_{k,max}$ (or state no emission), the stopping voltage, and explain what happens to the photocurrent if intensity is halved.

1 mark: gradient → $h$; x-intercept → $f_0$; parallel → same $h$ · 1 mark: $E_{k,max}$ Metal A · 1 mark: Metal B (no emission reasoning) · 1 mark: photocurrent argument

Show all answers

Multiple choice

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer, Model Answers

Q1 (a): (i) Threshold frequency: no emission below $f_0$ regardless of intensity. (ii) Instantaneous emission: no time delay even for very dim light above $f_0$. (iii) $E_{k,max}$ depends on frequency only, increasing intensity does not increase $E_{k,max}$. (1 mark)

Q1 (b): (i) Each photon has energy $hf$; below $f_0$, $hf < \phi$, so no photon can eject an electron. (ii) Energy transferred in a single photon-electron interaction, no accumulation needed. (iii) $E_{k,max} = hf - \phi$ depends only on $f$; intensity determines number of photons (electrons), not energy per photon. (1 mark)

Q1 (c): $f_0 = \phi/h = (2.5\times1.60\times10^{-19})/(6.63\times10^{-34}) = 6.03\times10^{14}$ Hz. $\lambda_0 = c/f_0 = 3.0\times10^8/6.03\times10^{14} = 498$ nm. (1 mark)

Q1 (d): $E = hc/\lambda = 1240/350 = 3.54$ eV. $E_{k,max} = 3.54 - 2.5 = 1.04$ eV. $V_s = 1.04$ V (numerically equal to $E_{k,max}$ in eV). (1 mark)

Q2 (a): Gradient = $h/e$, from gradient, determine Planck's constant $h$. x-intercept = $f_0$, from x-intercept, determine work function $\phi = hf_0$. y-intercept = $-\phi/e$, gives another route to $\phi$. Parallel lines mean identical gradient = same $h$ for both metals, confirming Planck's constant is a universal constant, not metal-dependent. Different x-intercepts mean different $f_0$, i.e., different work functions. (1 mark)

Q2 (b): $E_{photon} = 1240/300 = 4.13$ eV. Metal A ($\phi = 2.1$ eV): $E_{k,max} = 4.13 - 2.1 = 2.03$ eV; $V_s = 2.03$ V; intensity halved → photocurrent halved (fewer photons per second), $E_{k,max}$ unchanged. (1 mark) Metal B ($\phi = 4.5$ eV): $E_{photon} = 4.13$ eV $< \phi = 4.5$ eV, no emission occurs. The photon does not have enough energy to overcome the work function; no stopping voltage is meaningful; halving intensity makes no difference (still no emission). (2 marks)

Final step

Retrieve, reflect and finish

Check what stuck, revisit your opening predictions about the two lamps, and record the rule you will reuse.

Check what actually stuck

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quiz

A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.

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How did your thinking change?

Look back at your Think First answers:

Did you predict that the wave model would favour the brighter (red) lamp? The wave model predicts that brighter light delivers more energy per unit time, so it should eventually eject electrons. But experiment shows that even a faint UV lamp works while bright red light does not, contradicting the wave model.
Did you predict that light energy is delivered in discrete packets (photons)? This is Einstein's key insight: one photon interacts with one electron. If $hf < \phi$, no number of photons can eject an electron, no accumulation is possible.
Did you predict that doubling UV intensity doubles the number of electrons but keeps $E_{k,max}$ the same? Correct, more photons means more electrons, but each photon still carries energy $hf$, so each electron gets the same maximum energy.

The historical anchor: on 17 March 1905, Albert Einstein at the Swiss Patent Office, Bern, published "On a Heuristic Point of View Concerning the Production and Transformation of Light." His equation $E_{k,\text{max}} = hf - \phi$ predicted the linear relationship between photon frequency and maximum electron kinetic energy. Robert Millikan spent ten years trying to disprove it experimentally (1916), and instead confirmed it to high precision. Einstein received the 1921 Nobel Prize in Physics for this paper specifically, not for special relativity, because the photoelectric equation gave the first quantitative proof that light energy is quantised.

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