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The photoelectric effect and wave-particle duality represent a profound revolution in physics — the realisation that light and matter have both wave-like and particle-like properties. Einstein's explanation of the photoelectric effect earned him the Nobel Prize in 1921 and provided key evidence for quantum theory. This lesson covers the experimental observations, Einstein's photon model, the photoelectric equation, and de Broglie's hypothesis. This is assessed in AQA section 3.2.2.
Specification mapping. This lesson develops AQA A-Level Physics (7408) Particles and Radiation sub-strands 3.2.2.1 (the photoelectric effect: photon model, work function, photoelectric equation, threshold frequency, stopping potential) and 3.2.2.4 (wave–particle duality and de Broglie wavelength, with electron diffraction as supporting evidence). Refer to the official AQA 7408 specification document for the authoritative wording. The lesson is sequenced as the capstone of the 3.2 particles strand because it requires students to have already absorbed the photon concept (introduced in particles-antiparticles-and-photons, order 2) and the framework of quantum-level interactions developed across orders 3–6.
Synoptic links. Three principal synoptic connections run through this content. First, photons as exchange particles (3.2.1.4, our order 4) — the same photon that mediates electromagnetic forces at the Feynman-diagram level is the real, on-shell photon liberated when light passes through space; the photoelectric effect gives the kinematic accounting (E = hf) that fixes its energy. Second, wave properties of light — diffraction, interference, polarisation (course 3, waves) — wave-particle duality forces a synthesis between the classical wave evidence of the wave course and the particle evidence of this lesson. Third, electric current and circuits (course 5) — the photocurrent in a photocell is measured electrically; the stopping-potential method links macroscopic potential differences directly to single-photon energies via eV_s = ½mv²_max.
When electromagnetic radiation (typically ultraviolet light) is shone onto a clean metal surface, electrons may be emitted from the surface. These emitted electrons are called photoelectrons. The key experimental observations are:
Classical wave theory predicts that:
Einstein proposed that light consists of discrete packets of energy called photons. Each photon carries energy:
E = hf
where h = 6.63 × 10⁻³⁴ J s is the Planck constant and f is the frequency of the light.
Einstein's key insight: the photoelectric effect is a one-to-one interaction — each photon interacts with exactly one electron. Either the photon has enough energy to liberate the electron, or it does not. There is no accumulation of energy from multiple photons.
hf = φ + E_k(max)
or equivalently:
E_k(max) = hf − φ
where:
At the threshold frequency f₀, the photon has just enough energy to release an electron with zero kinetic energy:
hf₀ = φ
Therefore: φ = hf₀ and f₀ = φ/h
The maximum kinetic energy of the photoelectrons can be measured using a stopping potential V_s:
E_k(max) = eV_s
where e = 1.60 × 10⁻¹⁹ C. Combining with the photoelectric equation:
eV_s = hf − φ
A graph of V_s against f gives a straight line with gradient h/e and x-intercept equal to the threshold frequency f₀.
The work function of sodium is 2.28 eV. Find the threshold frequency.
Convert to joules: φ = 2.28 × 1.60 × 10⁻¹⁹ = 3.648 × 10⁻¹⁹ J
f₀ = φ / h = 3.648 × 10⁻¹⁹ / 6.63 × 10⁻³⁴
f₀ = 5.50 × 10¹⁴ Hz
This is in the visible/UV range — sodium has a relatively low work function.
Ultraviolet light of frequency 8.0 × 10¹⁴ Hz is incident on a sodium surface (φ = 2.28 eV). Find the maximum kinetic energy of the emitted photoelectrons in eV and in joules.
Energy of photon: E = hf = 6.63 × 10⁻³⁴ × 8.0 × 10¹⁴ = 5.304 × 10⁻¹⁹ J
Convert to eV: 5.304 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 3.315 eV
E_k(max) = hf − φ = 3.315 − 2.28 = 1.04 eV = 1.04 × 1.60 × 10⁻¹⁹ = 1.66 × 10⁻¹⁹ J
Find the stopping potential for the electrons in Worked Example 2.
eV_s = E_k(max)
V_s = E_k(max) / e = 1.66 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹
V_s = 1.04 V
(When working in eV, the stopping potential in volts is numerically equal to the maximum kinetic energy in eV.)
A graph of stopping potential V_s against frequency f gives a straight line with gradient 4.14 × 10⁻¹⁵ V Hz⁻¹ and x-intercept 5.50 × 10¹⁴ Hz. Determine h and φ.
Gradient = h/e, so h = gradient × e = 4.14 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.62 × 10⁻³⁴ J s ✓
Threshold frequency f₀ = 5.50 × 10¹⁴ Hz
φ = hf₀ = 6.62 × 10⁻³⁴ × 5.50 × 10¹⁴ = 3.64 × 10⁻¹⁹ J = 2.28 eV ✓
Exam Tip: Photoelectric effect questions often ask you to explain why increasing the intensity does not increase the maximum kinetic energy. The key answer: each photon interacts with one electron. Intensity determines the number of photons per second (and hence the number of photoelectrons), but each photon still has the same energy (hf). Only increasing the frequency increases the photon energy and hence the maximum KE.
Louis de Broglie proposed that all matter has wave-like properties. A particle with momentum p has an associated wavelength:
λ = h / p = h / (mv)
where λ is the de Broglie wavelength, h is the Planck constant, m is the mass, and v is the velocity. This applies to all particles — electrons, protons, neutrons, atoms, and even macroscopic objects (though their wavelengths are immeasurably small).
When a beam of electrons is directed at a thin polycrystalline graphite target, a pattern of concentric rings is observed on a fluorescent screen behind the target. This is a diffraction pattern — exactly the kind of pattern produced by waves passing through a set of slits (the gaps between atoms in the crystal lattice act as the slits).
Key observations:
An electron is accelerated through a potential difference of 5.0 kV. Find its de Broglie wavelength.
Kinetic energy gained: E_k = eV = 1.60 × 10⁻¹⁹ × 5000 = 8.00 × 10⁻¹⁶ J
Speed: E_k = ½mv² → v = √(2E_k / m) = √(2 × 8.00 × 10⁻¹⁶ / 9.11 × 10⁻³¹)
v = √(1.757 × 10¹⁵) = 4.19 × 10⁷ m s⁻¹
(This is about 14% of the speed of light — relativistic effects are small but not negligible at higher voltages.)
Momentum: p = mv = 9.11 × 10⁻³¹ × 4.19 × 10⁷ = 3.82 × 10⁻²³ kg m s⁻¹
De Broglie wavelength: λ = h / p = 6.63 × 10⁻³⁴ / 3.82 × 10⁻²³
λ = 1.74 × 10⁻¹¹ m = 0.0174 nm
This is comparable to the spacing between atoms in a crystal (~0.1–0.3 nm), which is why electron diffraction by crystals is observable.
A tennis ball (mass 0.058 kg) is travelling at 50 m s⁻¹. Find its de Broglie wavelength.
λ = h / (mv) = 6.63 × 10⁻³⁴ / (0.058 × 50) = 6.63 × 10⁻³⁴ / 2.9
λ = 2.3 × 10⁻³⁴ m
This is far smaller than any atomic dimension — wave behaviour of macroscopic objects is completely undetectable. This is why classical physics works perfectly for everyday objects.
| Metal | Work function (eV) | Threshold frequency (×10¹⁴ Hz) | Threshold wavelength (nm) |
|---|---|---|---|
| Caesium | 1.95 | 4.71 | 636 (visible red) |
| Sodium | 2.28 | 5.51 | 544 (visible green) |
| Zinc | 4.30 | 10.4 | 289 (UV) |
| Platinum | 5.65 | 13.7 | 220 (UV) |
Common Misconception: Students sometimes think wave-particle duality means that a particle "switches" between being a wave and a particle. This is not correct. Quantum objects always have both wave-like and particle-like properties simultaneously. Which property is most apparent depends on the experiment: diffraction experiments reveal wave properties, while detection events (individual clicks on a detector) reveal particle properties. The de Broglie wavelength determines the scale at which wave effects become significant.
| Equation | Meaning |
|---|---|
| E = hf | Energy of a photon |
| hf = φ + ½mv²_max | Photoelectric equation |
| φ = hf₀ | Work function and threshold frequency |
| E_k(max) = eV_s | Stopping potential |
| λ = h / (mv) | De Broglie wavelength |
Exam Tip: In photoelectric effect calculations, be very careful with unit conversions. Work functions are often given in eV but equations require joules. Always convert: 1 eV = 1.60 × 10⁻¹⁹ J. Similarly, wavelengths may be in nm (10⁻⁹ m). A common error is to forget to convert eV to joules before using E = hf.
Specimen question modelled on the AQA paper format (9 marks).
A clean zinc surface has a work function of 4.30 eV. Ultraviolet light of wavelength 200 nm is incident on the surface.
(a) Calculate the energy of an incident photon in joules and in eV. [2 marks]
(b) Calculate the maximum kinetic energy of photoelectrons emitted from the surface, and determine the stopping potential needed to bring them to rest. [3 marks]
(c) Explain one observation of the photoelectric effect that cannot be accounted for by classical wave theory, and show how Einstein's photon model resolves the discrepancy. [2 marks]
(d) Calculate the de Broglie wavelength of the most energetic photoelectron emitted in (b), and explain why electron diffraction can be observed using crystal lattices but not using a standard double-slit apparatus. [2 marks]
For this 9-mark item, AO1 (knowledge) carries roughly 2 marks for recall of the photoelectric equation, de Broglie equation, and the qualitative observations of the photoelectric effect. AO2 (application) carries roughly 5 marks for the calculations in parts (a), (b) and (d). AO3 (analysis and evaluation) carries roughly 2 marks for explaining why classical theory fails and how the photon model resolves it. Calculation parts must show full working; "show your method" mark schemes typically award substitution marks even if arithmetic is slipped.
Grade C response (5 marks out of 9).
(a) E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (200 × 10⁻⁹) = 9.95 × 10⁻¹⁹ J = 6.22 eV.
(b) E_k(max) = hf − φ = 6.22 − 4.30 = 1.92 eV. Stopping potential V_s = 1.92 V.
(c) Classical wave theory says any frequency of light should eventually emit electrons if intense enough. Einstein said photons have energy hf so if it's below threshold no emission happens.
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