OCR A Level Physics A: Photons, photoelectric effect and wave-particle duality (H556)

This guide covers section 4.5 (Quantum physics: 4.5.1 Photons, 4.5.2 The photoelectric effect and 4.5.3 Wave-particle duality) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. H556 is the full A level.

See the OCR A Level Physics hub and the printable checklist. Pair this page with the revision notes and the practice questions. Try the free diagnostics.

Values used on this page: c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg and the Planck constant 6.63 × 10⁻³⁴ J s. In an exam, use the values the question or your own paper gives you.

What you must be able to do

Statement	Content
4.5.1(a)-(b)	Photon model of electromagnetic radiation; the photon as a quantum of energy
4.5.1(c)-(d)	E = hf and E = hc/λ; the electronvolt
4.5.1(e)	Estimate h with LEDs using eV = hc/λ (practical activity group 6)
4.5.2(a)-(b)	Photoelectric effect; gold-leaf electroscope and zinc demonstration; one-to-one photon-electron interaction
4.5.2(c)-(d)	hf = φ + KEmax; work function; threshold frequency
4.5.2(e)-(f)	KEmax independent of intensity; emission rate proportional to intensity above threshold
4.5.3(a)-(c)	Electron diffraction; polycrystalline graphite; the de Broglie equation λ = h/p

4.5.1(a)-(c) The photon model and E = hf

In the photon model, electromagnetic radiation travels as discrete packets, each carrying a fixed amount of energy. One packet is a photon, and a photon is a quantum (smallest unit) of electromagnetic energy. The energy of a single photon is

E = hf = hc/λ

where h is the Planck constant, f the frequency, c the speed of light and λ the wavelength. Higher frequency means a shorter wavelength and a more energetic photon, whatever the brightness of the beam.

Worked example 1. Find the energy of a photon of wavelength 485 nm.

E = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 485 × 10⁻⁹ = 4.10 × 10⁻¹⁹ J (3 s.f.).

Worked example 2. A laser emits 4.0 mW of light at 532 nm. How many photons leave it each second?

Energy per photon = hc/λ = 3.74 × 10⁻¹⁹ J. Power = energy per second = N × E, so N = 4.0 × 10⁻³ / 3.74 × 10⁻¹⁹ = 1.1 × 10¹⁶ s⁻¹.

Intensity is power per unit area: photons arriving per second per square metre, times the energy of each photon.

4.5.1(d) The electronvolt

One electronvolt (eV) is the energy an electron gains when it is accelerated across a potential difference of one volt. Using W = QV, 1 eV = 1.60 × 10⁻¹⁹ × 1 V = 1.60 × 10⁻¹⁹ J. To convert eV to J, multiply by 1.60 × 10⁻¹⁹. To convert J to eV, divide by it.

Worked example 3. (a) Convert 3.4 eV to joules. (b) An electron is accelerated from rest through 2.5 kV. Find its kinetic energy in eV and J, and its speed.

(a) 3.4 × 1.60 × 10⁻¹⁹ = 5.4 × 10⁻¹⁹ J. (b) Kinetic energy = 2.5 keV = 2500 eV = 2500 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻¹⁶ J. ½mv² = 4.0 × 10⁻¹⁶, so v = √(2 × 4.0 × 10⁻¹⁶ / 9.11 × 10⁻³¹) = 3.0 × 10⁷ m s⁻¹.

4.5.1(e) Estimating h with LEDs (PAG6)

Practical activity group 6 lists determining the Planck constant using different coloured LEDs. No knowledge of semiconductor theory is required.

The model: an LED begins to emit light when each electron passing through it has gained about eV of energy, and that energy appears as one photon. So

eV = hc/λ, which gives V = (hc/e) × (1/λ).

A graph of V against 1/λ is a straight line through the origin with gradient hc/e, so h = gradient × e/c.

Method.

1. Connect an LED, a protective resistor, a variable power supply (or potentiometer), a voltmeter across the LED and an ammeter in series. Check the LED polarity.
2. Shield the LED (a dark tube or darkened room helps) and raise the potential difference slowly until the LED just begins to glow. Record V.
3. Repeat for LEDs of different colours, taking each wavelength from the supplier’s data. Repeat each reading and average.
4. Plot V against 1/λ and draw a line of best fit.

Worked example 4. Five LEDs give these onset values.

λ / nm	640	590	525	470	405
V / V	1.95	2.09	2.36	2.65	3.05

The gradient of the best-fit line is 1.23 × 10⁻⁶ V m. Then h = gradient × e/c = 1.23 × 10⁻⁶ × 1.60 × 10⁻¹⁹ / 3.00 × 10⁸ = 6.56 × 10⁻³⁴ J s.

Judging when the LED “just” lights is subjective, which limits accuracy.

4.5.2(a)-(b) The photoelectric effect

The photoelectric effect is the emission of electrons from a metal surface when electromagnetic radiation of high enough frequency shines on it. The electrons emitted are called photoelectrons.

Demonstration with a gold-leaf electroscope and zinc plate.

1. Clean the zinc plate with emery paper to remove the oxide layer, and fix it to the cap of the electroscope.
2. Charge the plate negatively. The gold leaf rises because the leaf and stem repel.
3. Shine ultraviolet light on the plate. The leaf falls: negative charge is leaving as photoelectrons.
4. Repeat with a very bright lamp of visible light. The leaf does not fall.
5. Place glass between the UV lamp and the plate (it absorbs most of the ultraviolet). The leaf stops falling.
6. A positively charged plate does not lose charge, because emitted electrons are pulled back by the positive plate.

This provides evidence for the particulate nature of electromagnetic radiation.

The one-to-one interaction (4.5.2(b))

Each photoelectron is released by a single photon. One photon gives up all its energy to one surface electron, in a one-to-one interaction. An electron cannot save up energy from several photons. If one photon cannot supply enough energy, no electron escapes, however many photons arrive.

4.5.2(c)-(d) Einstein’s equation, work function and threshold frequency

The work function φ is the minimum energy needed to remove an electron from the surface of a metal. The threshold frequency f₀ is the minimum frequency of radiation that causes emission, found when the photon energy equals the work function: hf₀ = φ.

For a photon of frequency f, energy is shared out as

hf = φ + KEmax

Here KEmax is the maximum kinetic energy of the photoelectrons. It is a maximum because electrons from slightly deeper lose extra energy before escaping.

Worked example 5. Light of wavelength 400 nm falls on potassium, work function 2.3 eV. Find (a) the maximum kinetic energy in J, (b) the maximum speed, (c) the threshold frequency.

(a) Photon energy = hc/λ = 4.97 × 10⁻¹⁹ J (3.11 eV). φ = 2.3 × 1.60 × 10⁻¹⁹ = 3.68 × 10⁻¹⁹ J. KEmax = 4.97 × 10⁻¹⁹ − 3.68 × 10⁻¹⁹ = 1.29 × 10⁻¹⁹ J (0.81 eV). (b) v = √(2 × 1.29 × 10⁻¹⁹ / 9.11 × 10⁻³¹) = 5.3 × 10⁵ m s⁻¹. (c) f₀ = φ/h = 3.68 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.55 × 10¹⁴ Hz.

Graph of KEmax against frequency

Rearranging gives KEmax = hf − φ. A graph of KEmax against f is a straight line with gradient h, a frequency-axis intercept f₀ and a KEmax-axis intercept of −φ. Different metals give parallel lines (same gradient h) with different intercepts.

Worked example 6. Measurements for one metal:

f / 10¹⁴ Hz	6.5	7.5	8.5	9.5	10.5
KEmax / eV	0.17	0.58	1.00	1.41	1.83

Gradient (first and last points) = (1.83 − 0.17) / (4.0 × 10¹⁴) = 4.15 × 10⁻¹⁵ eV s. Convert: h = 4.15 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.64 × 10⁻³⁴ J s. φ = hf − KEmax at the first point = 4.15 × 10⁻¹⁵ × 6.5 × 10¹⁴ − 0.17 = 2.53 eV. f₀ = φ/gradient = 2.53 / 4.15 × 10⁻¹⁵ = 6.1 × 10¹⁴ Hz.

4.5.2(e)-(f) Intensity

• KEmax does not depend on intensity (4.5.2(e)). Each electron interacts with one photon, and photon energy depends only on frequency. More intense light of the same frequency means more photons per second, not more energy per photon.
• Emission rate is proportional to intensity above threshold (4.5.2(f)). More photons per second means more electrons released per second, so the photoelectric current is directly proportional to intensity, provided f > f₀.

Worked example 7. Light of 400 nm and power 1.0 mW strikes the potassium surface above, then the power is doubled. Photon rate: 1.0 × 10⁻³ / 4.97 × 10⁻¹⁹ = 2.0 × 10¹⁵ s⁻¹, rising to 4.0 × 10¹⁵ s⁻¹. The number of electrons released per second doubles, while KEmax stays at 0.81 eV.

Why the wave model fails

A wave model predicts that energy builds up over time and depends on intensity. Observations show:

• No emission below f₀, however intense the light. The wave model predicts enough intensity would free electrons.
• Emission is immediate, with no measurable delay at low intensity. The wave model predicts a delay while energy accumulates.
• KEmax rises with frequency but not with intensity. The wave model predicts it should rise with intensity.

The photon model explains all three.

4.5.3(a)-(c) Wave-particle duality

Electron diffraction provides evidence for wave-like behaviour of particles. In the experiment, a beam of electrons accelerated through a high potential difference passes through a thin slice of polycrystalline graphite and hits a fluorescent screen. A pattern of concentric rings appears.

Diffraction by the regularly spaced atoms of graphite spreads the beam out. The crystals are randomly oriented, so every diffraction angle is available and each gives a ring. Rings occur for different spacings between planes of atoms. A ring pattern is only possible if electrons diffract and interfere as waves.

The de Broglie equation

A particle of momentum p has an associated wavelength

λ = h/p = h/mv

Worked example 8. Electrons are accelerated through 3.5 kV. Find their de Broglie wavelength.

Kinetic energy = eV = 1.60 × 10⁻¹⁹ × 3.5 × 10³ = 5.6 × 10⁻¹⁶ J. Since KE = p²/2m, p = √(2m × KE) = √(2 × 9.11 × 10⁻³¹ × 5.6 × 10⁻¹⁶) = 3.19 × 10⁻²³ kg m s⁻¹. λ = h/p = 6.63 × 10⁻³⁴ / 3.19 × 10⁻²³ = 2.1 × 10⁻¹¹ m.

This is comparable to atomic spacings, so diffraction is visible.

Worked example 9. Find λ for a 0.057 kg ball moving at 40 m s⁻¹: λ = 6.63 × 10⁻³⁴ / (0.057 × 40) = 2.9 × 10⁻³⁴ m. This is far too small to show diffraction, so everyday objects show no wave behaviour.

Effect of the accelerating voltage. Raising V raises p, so λ falls (p ∝ √V). Doubling V reduces λ by a factor of √2 (to 0.71 of the original). The rings shrink as the diffraction angles get smaller.

Common errors

• Writing E = hf with the wavelength in nanometres. Convert to metres first.
• Mixing eV and J in hf = φ + KEmax. Use one unit throughout.
• Saying “more intense light gives faster electrons”. Intensity changes the number of electrons, not their maximum energy.
• Saying that low-frequency light “does not have enough energy” without referring to the photon: it is the energy of each photon that matters.
• Forgetting the unit of the LED gradient (V m) before multiplying by e/c.

Move on to the revision notes and then test yourself with the practice questions.

Official syllabus

OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Section 4.5 Quantum physics (4.5.1 Photons, 4.5.2 The photoelectric effect, 4.5.3 Wave-particle duality).