Physics • Year 12 • Module 7 • Lesson 10
Synthesis — The Wave Model of Light
Apply the photoelectric effect equations, interpret real data, and compare wave vs photon model predictions across experimental scenarios.
1. Interpret photoelectric data — sodium surface
A student illuminates a sodium surface (φ = 2.28 eV) with several different light sources and records the results below. 8 marks
| Light source | Wavelength (nm) | Intensity (W/m²) | Electrons ejected? | Kmax of electrons (eV) |
|---|---|---|---|---|
| Red LED | 650 | 500 | ||
| Green laser | 532 | 1 | ||
| Blue LED | 450 | 100 | ||
| UV lamp | 300 | 0.1 |
Data: sodium work function φ = 2.28 eV; h = 4.14 × 10−15 eV·s; c = 3.00 × 108 m/s.
1.1 For each light source, calculate the photon energy in eV (using E = hc/λ). Complete the “Electrons ejected?” and “Kmax” columns. Show a sample calculation for one source. 5 marks
1.2 Compare the red LED (500 W/m²) with the UV lamp (0.1 W/m²). The red LED is 5000 times more intense, yet no electrons are ejected from sodium. Explain this result using the photon model. 2 marks
1.3 What does the classical (wave) model predict would happen with the very intense red LED? Why is this prediction wrong? 1 mark
2. Interpret a Kmax vs frequency graph
A student measures the maximum kinetic energy of photoelectrons for several frequencies of light shining on an unknown metal. The graph below shows the results. 7 marks
Figure 2. Maximum kinetic energy of photoelectrons vs frequency of incident light for an unknown metal. Illustrative data consistent with h = 4.14 × 10−15 eV·s.
2.1 From the graph, read off the threshold frequency f0 and calculate the work function of the metal in eV. 2 marks
2.2 The gradient of the graph should equal Planck’s constant h. Using any two points, estimate the gradient (in eV/Hz) and compare it to the accepted value of h = 4.14 × 10−15 eV·s. 3 marks
2.3 Explain what it means physically that the line intercepts the x-axis (frequency axis) at a positive value rather than at zero. 2 marks
3. Compare wave and photon model predictions
Complete the two-column table below. For each observable, state what each model predicts and whether the prediction matches experiment. 10 marks (1 per cell)
| Observable | Wave model predicts… | Photon model predicts… |
|---|---|---|
| Effect of increasing frequency (intensity constant) | ||
| Effect of increasing intensity (frequency constant) | ||
| Threshold frequency | ||
| Time lag before electrons are ejected | ||
| What determines Kmax of electrons |
4. Predict and justify — Australian astronomical context
The Australian Synchrotron in Clayton, Victoria, produces intense X-rays used in materials science. Researchers study the Compton effect by firing high-energy X-rays at thin metal foils and measuring the scattered X-ray wavelength at various angles.
5 marks
4.1 An X-ray photon has wavelength 0.071 nm. (a) Calculate its momentum using p = h/λ (h = 6.63 × 10−34 J·s). (b) After scattering off an electron, the X-ray’s wavelength increases slightly. Explain, using the equation p = h/λ, whether the scattered photon has more or less momentum than the incident photon, and where the missing momentum goes. 3 marks
4.2 Explain why the Compton effect could not be explained by the classical wave model of light. What property of photons does it specifically demonstrate? 2 marks
5. Diagram critique — what is wrong with this student’s explanation?
A Year 12 student wrote the following explanation of the photoelectric effect: “When bright red light shines on the metal, the electrons absorb the large amount of energy in the intense wave. This makes the electrons gain enough kinetic energy to escape the surface. Brighter light means more kinetic energy, so the electrons escape with higher speed.”
Identify three errors in the student’s explanation and write the correction for each. 6 marks (2 per error: 1 identify, 1 correct)
5.1 Error 1: What is wrong?
Correction:
5.2 Error 2: What is wrong?
Correction:
5.3 Error 3: What is wrong?
Correction:
Q1.1 — Photon energies and ejection table
Using E = hc/λ with hc = 1242 eV·nm:
Red LED (650 nm): E = 1242/650 = 1.91 eV < 2.28 eV. No electrons ejected. Kmax = N/A.
Green laser (532 nm): E = 1242/532 = 2.34 eV > 2.28 eV. Electrons ejected. Kmax = 2.34 − 2.28 = 0.06 eV.
Blue LED (450 nm): E = 1242/450 = 2.76 eV. Electrons ejected. Kmax = 2.76 − 2.28 = 0.48 eV.
UV lamp (300 nm): E = 1242/300 = 4.14 eV. Electrons ejected. Kmax = 4.14 − 2.28 = 1.86 eV.
Marking notes. 1 mark per correct row (Yes/No + Kmax); 1 mark for sample calculation showing method. Accept minor rounding (±0.02 eV).
Q1.2 — Intense red vs dim UV (2 marks)
Each red photon has energy 1.91 eV, which is less than the work function of 2.28 eV. No matter how many red photons arrive (i.e. how bright the light), no individual photon can liberate an electron — the energy is not pooled [1]. The dim UV lamp delivers photons of 4.14 eV each, which exceeds the work function; even a single UV photon can eject one electron [1].
Q1.3 — Classical wave model prediction (1 mark)
The wave model predicts that with very intense red light, energy builds up continuously at the metal surface until eventually sufficient energy accumulates to eject an electron (possibly after a time lag). This is observed to be wrong: no electrons are ever ejected by red light on sodium, regardless of intensity or exposure time [1].
Q2.1 — Threshold frequency and work function (2 marks)
From the graph, the line crosses the x-axis (Kmax = 0) at f0 ≈ 5.5 × 1014 Hz [1]. Work function: φ = h·f0 = (4.14 × 10−15)(5.5 × 1014) = 2.28 eV [1].
Q2.2 — Gradient = Planck’s constant (3 marks)
Using data points at f = 6 × 1014 Hz (Kmax = 0.25 eV) and f = 10 × 1014 Hz (Kmax = 1.89 eV): gradient = (1.89 − 0.25) / [(10 − 6) × 1014] = 1.64 / (4 × 1014) = 4.10 × 10−15 eV·s [2 marks for correct calculation]. This is close to the accepted value of 4.14 × 10−15 eV·s; the small discrepancy is due to reading uncertainty from the graph [1 mark for comparison and comment].
Q2.3 — Meaning of positive x-intercept (2 marks)
The positive x-intercept is the threshold frequency f0. It means that photons must have a minimum frequency before any electrons can be ejected [1]. This minimum frequency corresponds to photon energy exactly equal to the work function (hf0 = φ); at this frequency the electron is liberated with zero kinetic energy. Below f0, each individual photon has insufficient energy to overcome the work function, regardless of how many photons arrive [1].
Q3 — Wave vs photon model table
Effect of increasing frequency: Wave: higher frequency = higher energy wave; should increase Kmax but not in a simple threshold way. Photon: Kmax = hf − φ; increasing frequency directly increases Kmax. (Observation matches photon model.)
Effect of increasing intensity: Wave: more energy arrives; should increase Kmax. Photon: more photons per second → more electrons ejected (bigger photocurrent); Kmax unchanged. (Observation matches photon model.)
Threshold frequency: Wave: no threshold; given enough time, any frequency should work. Photon: strict threshold at f0 = φ/h; below this no ejection ever. (Observation matches photon model.)
Time lag: Wave: predicts a time lag while energy builds up. Photon: instantaneous ejection (one photon interacts with one electron immediately). (Observation matches photon model.)
What determines Kmax: Wave: amplitude (intensity) of the wave. Photon: frequency of the photon only (Kmax = hf − φ). (Observation matches photon model.)
Marking notes. 1 mark per correct cell (2 cells per row × 5 rows = 10 marks).
Q4.1 — Photon momentum and Compton scattering (3 marks)
(a) p = h/λ = 6.63 × 10−34 / (0.071 × 10−9) = 9.34 × 10−24 kg m s−1 [1].
(b) Since p = h/λ, a larger wavelength means a smaller momentum. The scattered photon has less momentum than the incident photon [1]. The missing momentum has been transferred to the recoiling electron — the photon behaves like a particle colliding elastically with the electron, transferring momentum in a way that the wave model cannot predict [1].
Q4.2 — Why the Compton effect disproves the wave model (2 marks)
The wave model treats light as a continuous wave; when a wave interacts with an electron it should not change wavelength — the electron may oscillate but the scattered wave should have the same wavelength (this is classical Thomson scattering) [1]. The observed wavelength increase with scattering angle can only be explained if light behaves as a discrete particle (photon) that collides elastically with the electron, transferring momentum according to p = h/λ. This demonstrates that photons carry momentum — a fully particle-like property [1].
Q5 — Diagram critique (6 marks)
5.1 Error 1 — “bright red light” can eject electrons: Red light has a frequency below the threshold frequency of most metals; no matter how bright or intense, each red photon has insufficient energy to overcome the work function [1]. Correction: For electrons to be ejected, the frequency (not brightness) of light must exceed the threshold frequency f0 = φ/h. Red light cannot eject electrons from metals with work functions above 1.77 eV regardless of intensity [1].
5.2 Error 2 — “brighter light means higher kinetic energy”: The student confuses intensity with frequency. Intensity (brightness) controls how many electrons are ejected (photocurrent), not their kinetic energy [1]. Correction: The maximum kinetic energy is Kmax = hf − φ, which depends only on photon frequency. Brighter light ejects more electrons per second but each electron has the same maximum kinetic energy as with dim light of the same frequency [1].
5.3 Error 3 — “absorb the large amount of energy in the intense wave” (wave model used): This describes the classical wave model, where energy is distributed across the whole wave. In the photon model, one photon interacts with one electron; the electron does not “accumulate” wave energy [1]. Correction: Each photon interacts with a single electron. The photon’s entire quantum of energy hf is absorbed by that electron; if hf ≥ φ, the electron is ejected instantly with Kmax = hf − φ. Energy from many photons is never pooled to eject one electron [1].