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

These revision notes condense section 4.5 (Quantum physics: 4.5.1, 4.5.2 and 4.5.3) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Read the study guide for full explanations and the practice questions to test yourself.

Course pages: OCR A Level Physics hub | checklist | free diagnostics. Values used here: 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.

Definitions to recall

Term	Statement
Photon (4.5.1(a)-(b))	A quantum (packet) of electromagnetic energy
Electronvolt (4.5.1(d))	The energy an electron gains when it is accelerated across 1 V; 1 eV = 1.60 × 10⁻¹⁹ J
Photoelectric effect (4.5.2(a))	Emission of electrons from a metal surface when radiation of high enough frequency falls on it
Photoelectron	An electron released by the photoelectric effect
Work function φ (4.5.2(d))	Minimum energy needed to remove an electron from the surface of the metal
Threshold frequency f₀ (4.5.2(d))	Minimum frequency of radiation that causes emission; hf₀ = φ
de Broglie wavelength (4.5.3(c))	The wavelength associated with a moving particle, λ = h/p

Equations and units

Quantity	Equation	Notes
Photon energy	E = hf	f in Hz, E in J
Photon energy	E = hc/λ	λ in m (convert from nm)
Einstein’s equation	hf = φ + KEmax	One unit throughout, J or eV
Threshold	hf₀ = φ	At threshold, KEmax = 0
LED estimate	eV = hc/λ	Gradient of V against 1/λ is hc/e
de Broglie	λ = h/p = h/mv	p = momentum in kg m s⁻¹
Electron through V	p = √(2m × eV)	Starting from rest
Photon rate	N = P/E	P in W, N in photons per second

Handy conversions: 1 eV = 1.60 × 10⁻¹⁹ J. With the values above, hc = 1.99 × 10⁻²⁵ J m, so E (eV) = 1243 / λ (nm).

Photon model (4.5.1(a)-(c))

• Electromagnetic radiation behaves as discrete photons, each of energy hf.
• Photon energy depends only on frequency. Brightness changes the number of photons per second.
• Intensity = power per unit area, which is photons per second per square metre multiplied by the energy of each photon.

Reminder. A 2.0 mW beam of 500 nm light: E = 3.98 × 10⁻¹⁹ J per photon, so N = 2.0 × 10⁻³ / 3.98 × 10⁻¹⁹ = 5.0 × 10¹⁵ s⁻¹.

Method: LED estimate of h (4.5.1(e), PAG6)

1. Build the circuit with the LED the right way round, a protective resistor and a voltmeter across the LED.
2. Shield the LED from stray light and raise the potential difference slowly until it just glows. Record V.
3. Repeat for several colours, using each colour’s wavelength from the supplier. Repeat readings and average.
4. Plot V against 1/λ, draw a best-fit line, and find the gradient.
5. h = gradient × e / c.

No semiconductor theory is required. The main uncertainty is judging when the LED “just” lights.

Method: photoelectric calculations (4.5.2(c))

1. Find the photon energy, hc/λ or hf.
2. Convert the work function to the same unit.
3. If the photon energy is smaller than φ, there is no emission.
4. Otherwise KEmax = photon energy − φ.
5. For speed, use KEmax = ½mv².

Reminder. Light of 300 nm on a metal of work function 3.5 eV: photon energy = 4.14 eV, so KEmax = 0.64 eV (1.0 × 10⁻¹⁹ J). The threshold frequency is 3.5 × 1.60 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 8.4 × 10¹⁴ Hz.

Gold-leaf electroscope with zinc (4.5.2(a))

Step	Observation	Meaning
Clean zinc, charge negatively	Leaf rises	Like charges repel
Ultraviolet on zinc	Leaf falls	Electrons leave the plate
Bright visible light only	Leaf stays up	Below threshold frequency, however intense
Glass between UV lamp and plate	Leaf stays up	Glass absorbs the UV
Plate charged positively, UV on	No fall	Emitted electrons are pulled back

Graph of KEmax against f (4.5.2(c)-(d))

• Straight line, KEmax = hf − φ.
• Gradient = h. Frequency-axis intercept = f₀. KEmax-axis intercept = −φ.
• A metal with a bigger φ gives a parallel line, shifted to the right.
• If the axis is in eV, multiply the gradient by 1.60 × 10⁻¹⁹ to get h in J s.

Intensity (4.5.2(e)-(f))

• KEmax is independent of intensity: photon energy, and so electron energy, depends only on frequency.
• Above f₀, emission rate (photoelectric current) is directly proportional to intensity.
• Below f₀ there is no emission at any intensity.

Wave-particle duality (4.5.3)

• Photoelectric effect: evidence for the particle nature of electromagnetic radiation.
• Electron diffraction: evidence for the wave nature of particles.
• Polycrystalline graphite: electrons pass through a thin slice onto a fluorescent screen, and rings appear. The atoms and their spacing act as the diffracting object.
• Faster electrons (higher accelerating voltage) have a smaller λ, so the rings have a smaller radius.
• Everyday objects have wavelengths far too small to see: a 0.057 kg ball moving at 40 m s⁻¹ has λ of order 10⁻³⁴ m.

Must-know distinctions

Idea	Contrast
Intensity vs frequency	Intensity sets how many photons per second; frequency sets the energy of each
Work function vs threshold frequency	φ is an energy; f₀ is the frequency with hf₀ = φ
KEmax vs mean kinetic energy	KEmax belongs to the electrons that lose least energy escaping
Wave model vs photon model	The wave model cannot explain the threshold, instant emission or the lack of dependence on intensity
Photons vs electrons	A photon has energy hf; an electron has momentum p and wavelength h/p

Quick self-test

1. Convert 5.0 eV to joules.
2. A photon has energy 2.4 × 10⁻¹⁹ J. Give its energy in eV.
3. State what is meant by the work function of a metal.
4. Light of wavelength 300 nm falls on a metal of work function 3.5 eV. Calculate the maximum kinetic energy of the photoelectrons in eV.
5. Calculate the threshold frequency for a metal of work function 3.5 eV.
6. An LED turns on at 1.8 V and its wavelength is 690 nm. Estimate h.
7. State the effect on the photoelectric current, and on KEmax, of doubling the intensity of light above the threshold frequency.
8. Calculate the number of photons per second in a 2.0 mW beam of wavelength 500 nm.
9. Calculate the de Broglie wavelength of an electron moving at 3.0 × 10⁶ m s⁻¹.
10. Calculate the de Broglie wavelength of electrons accelerated through 1.0 kV.
11. State what observation with polycrystalline graphite shows that electrons behave as waves.

Answers

1. 5.0 × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹⁹ J.
2. 2.4 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 1.5 eV.
3. The minimum energy needed to remove an electron from the surface of the metal.
4. Photon energy = 1243 / 300 = 4.14 eV. KEmax = 4.14 − 3.5 = 0.64 eV.
5. f₀ = φ/h = 3.5 × 1.60 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 8.4 × 10¹⁴ Hz.
6. h = eVλ/c = 1.60 × 10⁻¹⁹ × 1.8 × 690 × 10⁻⁹ / 3.00 × 10⁸ = 6.62 × 10⁻³⁴ J s.
7. Photoelectric current doubles, because the number of photons per second doubles. KEmax is unchanged.
8. N = P/E = 2.0 × 10⁻³ / 3.98 × 10⁻¹⁹ = 5.0 × 10¹⁵ s⁻¹.
9. p = 9.11 × 10⁻³¹ × 3.0 × 10⁶ = 2.73 × 10⁻²⁴ kg m s⁻¹, so λ = 6.63 × 10⁻³⁴ / 2.73 × 10⁻²⁴ = 2.4 × 10⁻¹⁰ m.
10. eV = 1.60 × 10⁻¹⁶ J, so p = √(2m × eV) = 1.71 × 10⁻²³ kg m s⁻¹, so λ = 3.9 × 10⁻¹¹ m.
11. A pattern of concentric rings on the screen, which is a diffraction pattern.

Where marks are usually lost

• Leaving the wavelength in nm inside E = hc/λ.
• Combining eV and J in one line of hf = φ + KEmax.
• Saying brighter light gives electrons more energy, when it gives more electrons.
• Stating that the photon “does not have enough energy” for low-frequency light but writing about the beam, not each photon.
• Explaining the one-to-one interaction without saying one photon releases one electron.
• Forgetting the cleaning, negative charging and ultraviolet in the electroscope description.
• Calling the intercept on the KEmax axis φ rather than −φ.
• Giving the de Broglie wavelength using mass instead of momentum, or forgetting that p = √(2m × eV) comes from eV = ½mv².

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).