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

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 4.5 (Quantum physics: photons, the photoelectric effect and wave-particle duality) of the OCR Advanced GCE in Physics A specification (H556), Version 3.0 (March 2026), for first assessment in 2017. Use c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg and h = 6.63 × 10⁻³⁴ J s. Revise first with the study guide or revision notes.

Questions

1. (a) Define the electronvolt. [1] (b) Convert a photon energy of 3.6 × 10⁻¹⁹ J to eV. [1] (c) Calculate the kinetic energy, in J, gained by an electron accelerated from rest through 450 V. [1]

2. A photon has energy 3.2 × 10⁻¹⁹ J. Calculate its frequency and its wavelength. [3]

3. Calculate the energy, in eV, of a photon of ultraviolet radiation of wavelength 280 nm. [3]

4. A laser pointer emits 5.0 mW of light of wavelength 650 nm. Calculate the number of photons emitted per second. [3]

5. A red LED just begins to emit light when the potential difference across it is 1.9 V. Its wavelength is 655 nm. Use eV = hc/λ to estimate the Planck constant. [3]

6. Describe how you would use a set of LEDs of different colours to determine the Planck constant, including how you would analyse the results. [5]

7. Sodium has a work function of 2.36 eV.

(a) Calculate its threshold frequency. [2] (b) Light of wavelength 450 nm falls on a clean sodium surface. Calculate the maximum kinetic energy of the photoelectrons, in eV. [3] (c) Explain why light of wavelength 650 nm releases no electrons, however bright it is. [1]

8. Explain three observations about the photoelectric effect that the wave model of light cannot explain, and how the photon model accounts for each. [6]

9. The maximum kinetic energy of photoelectrons from a metal was measured at four frequencies.

f / 10¹⁴ Hz	7.5	8.5	9.5	10.5
KEmax / eV	0.51	0.92	1.34	1.75

(a) Using hf = φ + KEmax, explain why a graph of KEmax against f is a straight line, and state what its gradient and its intercept on the KEmax axis represent. [2] (b) Use the data to determine the Planck constant in J s. [3] (c) Determine the work function of the metal in eV. [2]

10. Describe a demonstration of the photoelectric effect using a gold-leaf electroscope and a zinc plate. Include the preparation of the plate, what you would see, and what each observation shows. [5]

11. Light of wavelength 520 nm falls on a caesium surface of work function 2.10 eV.

(a) Calculate the maximum kinetic energy of the photoelectrons, in J. [3] (b) Calculate their maximum speed. [2] (c) The intensity of the light is tripled. State and explain the effect on the maximum kinetic energy and on the photoelectric current. [2]

12. In an electron diffraction tube, electrons are accelerated from rest through a potential difference of 150 V and pass through a thin slice of polycrystalline graphite onto a fluorescent screen.

(a) Calculate the de Broglie wavelength of the electrons. [3] (b) Describe the pattern seen on the screen and explain how it provides evidence for the wave nature of electrons. [3] (c) The potential difference is increased to 600 V. State the new wavelength and the effect on the pattern. [2]

Answers

1. (a) The energy an electron gains when accelerated across a potential difference of 1 V [1] [1] (b) 3.6 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 2.25 eV [1] [1] (c) E = eV = 1.60 × 10⁻¹⁹ × 450 = 7.2 × 10⁻¹⁷ J [1] [1] Examiner insight: Give the unit with each answer; a bare number in a mixed eV and J question is easy to misread.

2. f = E/h = 3.2 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ [1] = 4.8 × 10¹⁴ Hz [1]. λ = c/f = 3.00 × 10⁸ / 4.83 × 10¹⁴ = 6.2 × 10⁻⁷ m (620 nm) [1] [3] Examiner insight: Carry unrounded values into the second part; rounding f to 4.8 first changes the third significant figure of λ.

3. E = hc/λ [1] = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 280 × 10⁻⁹ = 7.10 × 10⁻¹⁹ J [1]. In eV: 7.10 × 10⁻¹⁹ / 1.60 × 10⁻¹⁹ = 4.44 eV [1] [3] Examiner insight: Convert nanometres to metres at the start; a 10⁹ slip is the most common cause of a wrong answer here.

4. Photon energy = hc/λ = 3.06 × 10⁻¹⁹ J [1]. N = P/E = 5.0 × 10⁻³ / 3.06 × 10⁻¹⁹ [1] = 1.6 × 10¹⁶ s⁻¹ [1] [3] Examiner insight: Show the energy of one photon as its own step. Power divided by a wavelength is a common wrong route.

5. h = eVλ/c [1] = 1.60 × 10⁻¹⁹ × 1.9 × 655 × 10⁻⁹ / 3.00 × 10⁸ [1] = 6.6 × 10⁻³⁴ J s [1] [3] Examiner insight: Rearrange before you substitute, and give the unit J s with the answer.

6. Connect an LED (correct polarity) in series with a protective resistor and a variable supply, with a voltmeter across the LED [1]. Shield from stray light and raise the potential difference slowly until the LED just glows; record V [1]. Repeat for LEDs of different colours, taking λ from the supplier’s data, and repeat each reading to average [1]. Plot V against 1/λ and draw a best-fit line [1]. The gradient is hc/e, so h = gradient × e/c [1] [5] Examiner insight: State what is plotted and what the gradient equals; “plot a graph” alone leaves out the link to the equation.

7. (a) f₀ = φ/h = 2.36 × 1.60 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ [1] = 5.7 × 10¹⁴ Hz [1] [2] (b) Photon energy = hc/λ = 4.42 × 10⁻¹⁹ J [1] = 2.76 eV [1]. KEmax = 2.76 − 2.36 = 0.40 eV [1] [3] (c) A 650 nm photon has energy 1.91 eV, which is less than the work function, and one photon interacts with one electron, so no electron can escape [1] [1] Examiner insight: In (c), talk about the energy of a single photon. Referring to “the light” or its brightness does not explain the absence of emission.

8. (i) No emission below the threshold frequency, however intense the light [1]. A wave model would predict that enough intensity delivers enough energy; the photon model says each photon must carry at least φ [1]. (ii) Emission is immediate even at low intensity [1]. The wave model predicts a delay while energy builds up; in the photon model one photon gives its energy to one electron at once [1]. (iii) KEmax is independent of intensity but increases with frequency [1]. The wave model predicts more intense light gives more energetic electrons; photon energy is hf and intensity only changes the number of photons [1] [6] Examiner insight: Each observation needs both halves: what is seen, and why the wave model fails or the photon model works.

9. (a) KEmax = hf − φ has the form y = mx + c [1]. The gradient is h; the intercept on the KEmax axis is −φ [1] [2] (b) Gradient = (1.75 − 0.51) / (3.0 × 10¹⁴) [1] = 4.13 × 10⁻¹⁵ eV s [1]. h = 4.13 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.6 × 10⁻³⁴ J s [1] [3] (c) φ = hf − KEmax = 4.13 × 10⁻¹⁵ × 7.5 × 10¹⁴ − 0.51 [1] = 2.6 eV [1] [2] Examiner insight: Take the gradient from points far apart on the line, and convert eV to J before quoting h.

10. Clean the zinc plate with emery paper and fix it to the electroscope; charge it negatively so the leaf rises [1]. Shine ultraviolet light on it and the leaf falls [1]: electrons are emitted from the zinc, so the charge is lost [1]. A bright visible lamp does not make the leaf fall, so there is a threshold frequency [1]. If the plate is charged positively the leaf does not fall, because emitted electrons are attracted back [1] [5] Examiner insight: Say that the plate is negative and what is leaving it; “the leaf falls” without a cause earns little.

11. (a) Photon energy = hc/λ = 3.83 × 10⁻¹⁹ J [1]. φ = 2.10 × 1.60 × 10⁻¹⁹ = 3.36 × 10⁻¹⁹ J [1]. KEmax = 4.7 × 10⁻²⁰ J [1] [3] (b) ½mv² = 4.65 × 10⁻²⁰ [1]. v = √(2 × 4.65 × 10⁻²⁰ / 9.11 × 10⁻³¹) = 3.2 × 10⁵ m s⁻¹ [1] [2] (c) KEmax is unchanged because photon energy depends only on frequency [1]. The photoelectric current triples because three times as many photons per second release three times as many electrons per second [1] [2] Examiner insight: Put both quantities in the answer, and give the reason in terms of photons.

12. (a) eV = 1.60 × 10⁻¹⁹ × 150 = 2.4 × 10⁻¹⁷ J, and p = √(2m × eV) [1] = 6.6 × 10⁻²⁴ kg m s⁻¹ [1]. λ = h/p = 6.63 × 10⁻³⁴ / 6.61 × 10⁻²⁴ = 1.0 × 10⁻¹⁰ m [1] [3] (b) A set of concentric rings appears [1]. The atoms in the graphite, and the spacing between them, diffract the electrons, and diffraction and interference are wave properties [1]. The pattern is only possible if the electrons behave as waves [1] [3] (c) p ∝ √V, so four times V doubles p and halves λ to 5.0 × 10⁻¹¹ m [1]. The rings become smaller in radius, closer to the centre [1] [2] Examiner insight: Link the observation (rings) to the property (diffraction) and then to the conclusion (wave behaviour).

Where marks are usually lost

• Using wavelength in nanometres without converting to metres.
• Mixing eV and J in one equation.
• Answering “why no emission” with the brightness of the beam rather than the energy of one photon.
• Saying more intense light gives faster electrons.
• Describing the electroscope without cleaning the plate or stating the charge on it.
• Quoting h in eV s or without a unit.
• Using p = mv for electrons that were accelerated, when the voltage is given and p = √(2meV) is the route.
• Stating that rings “prove” waves without naming diffraction by the graphite.

Next steps

• Read the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick items off the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

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).