IB DP Physics: Photoelectric effect, matter waves and Compton scattering (DP Physics) – 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 – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions cover sub-topic E.2, Quantum physics, from the IB Diploma Programme Physics guide, first assessment 2025. E.2 is HL only: every question below is for HL candidates. Learn the content first from the study guide and the revision notes. Course hub: /boards/ib/ib-dp/physics/. Checklist: /checklists/ib/ib-dp/physics/.

Data for every question: Planck constant 6.63 × 10⁻³⁴ J s; speed of light 3.00 × 10⁸ m s⁻¹; elementary charge 1.60 × 10⁻¹⁹ C; electron mass 9.11 × 10⁻³¹ kg. Answer to three significant figures unless told otherwise.

Questions

1. (HL only) A metal has work function 4.30 eV.

(a) Calculate its threshold frequency. [2] (b) Ultraviolet light of wavelength 220 nm falls on the metal. Find the greatest kinetic energy an emitted electron can have, in J and in eV. [3]

2. (HL only) Discuss three features of the photoelectric effect that classical wave theory cannot explain. State the wave prediction and the observation for each. [4]

3. (HL only) Light of frequency 9.00 × 10¹⁴ Hz is shone on a metal. The stopping potential is found to be 1.45 V.

(a) Determine the work function of the metal in eV. [3] (b) Calculate the threshold frequency. [1]

4. (HL only) A student measures the stopping potential Vs for a different metal at four frequencies f.

f / 10¹⁴ Hz	5.5	6.5	7.5	8.5
Vs / V	0.18	0.59	1.01	1.42

(a) Find the gradient of the line using the outermost data points. [2] (b) Hence determine a value for the Planck constant. [2] (c) Determine the threshold frequency and the work function in eV. [2]

5. (HL only) A metal has work function 1.90 eV. A beam of green light of wavelength 520 nm and power 1.2 mW falls on it.

(a) Calculate the number of photons arriving each second. [2] (b) On average 1 photon in 400 releases an electron. Calculate the photocurrent. [2] (c) Calculate the maximum kinetic energy of the electrons in J. [1] (d) The green light is replaced by red light of wavelength 700 nm and the same power. Deduce what happens to the current. [1]

6. (HL only) Calculate the de Broglie wavelength of

(a) an electron of kinetic energy 3.0 keV. [3] (b) an alpha particle of mass 6.64 × 10⁻²⁷ kg and kinetic energy 5.0 MeV. [2]

7. (HL only) A beam of electrons is directed at a narrow gap of width 1.5 nm. The first intensity minimum is observed at 9.5° from the central axis.

(a) Determine the de Broglie wavelength of the electrons. [2] (b) Calculate their speed. [2] (c) The electrons are now sent at twice the speed. Calculate the new angle of the first minimum. [2]

8. (HL only) Describe an experiment in which electron diffraction is observed, and explain what it shows about matter. [4]

9. (HL only) A photon of wavelength 0.0720 nm is Compton scattered by an electron through 150°. Use h/mec = 2.43 × 10⁻¹² m.

(a) Calculate the change in wavelength. [2] (b) State the wavelength of the scattered photon. [1] (c) Explain why the wavelength increases. [2] (d) Explain why Compton scattering supports the photon model of light more strongly than the photoelectric effect does. [2]

10. (HL only) A photon of wavelength 0.0975 nm scatters from a stationary electron and emerges with wavelength 0.1000 nm. Use h/mec = 2.43 × 10⁻¹² m.

(a) Calculate the angle through which the photon is scattered. [3] (b) Calculate the kinetic energy gained by the electron, in eV. [3]

11. (HL only) An electron and a photon each have wavelength 0.250 nm.

(a) Calculate the momentum and the kinetic energy of the electron, in eV. [3] (b) Calculate the energy of the photon. [2] (c) Compare the two energies. [1]

Answers

1. (a) Φ = 4.30 × 1.60 × 10⁻¹⁹ = 6.88 × 10⁻¹⁹ J [1]. f₀ = Φ/h = 6.88 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 1.04 × 10¹⁵ Hz [1]. (b) Photon energy = hc/λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 220 × 10⁻⁹ = 9.04 × 10⁻¹⁹ J [1]. Emax = 9.04 × 10⁻¹⁹ - 6.88 × 10⁻¹⁹ = 2.16 × 10⁻¹⁹ J [1] = 1.35 eV [1]. Examiner insight: Show the conversion from eV to J before using h; an answer with a unit mix-up usually cannot show the earlier steps were right.

2. Any three of the following features, each with its prediction and observation, earn one mark each. The fourth mark is for the photon view. Threshold: a wave of any frequency should eventually free electrons if intense enough; observed, no emission below f₀ whatever the intensity [1]. Emax: a more intense wave should give electrons more energy; observed, Emax depends on frequency and not on intensity [1]. Delay: a dim wave should need time to accumulate energy; observed, emission is immediate [1]. The photon model fits all three, since one photon of energy hf gives its energy to one electron [1]. Examiner insight: Pair each feature with both a prediction and an observation; naming a feature alone usually leaves the argument incomplete.

3. (a) Photon energy = hf = 6.63 × 10⁻³⁴ × 9.00 × 10¹⁴ = 5.97 × 10⁻¹⁹ J [1]. Emax = eVs = 1.45 × 1.60 × 10⁻¹⁹ = 2.32 × 10⁻¹⁹ J [1]. Φ = 5.97 × 10⁻¹⁹ - 2.32 × 10⁻¹⁹ = 3.65 × 10⁻¹⁹ J = 2.28 eV [1]. (b) f₀ = 3.65 × 10⁻¹⁹ / 6.63 × 10⁻³⁴ = 5.50 × 10¹⁴ Hz [1]. Examiner insight: A stopping potential in volts equals Emax in eV numerically, so you can skip the conversion, but state that you are doing it.

4. (a) Gradient = (1.42 - 0.18) / ((8.5 - 5.5) × 10¹⁴) [1] = 4.13 × 10⁻¹⁵ V Hz⁻¹ [1]. (b) Gradient = h/e, so h = 4.13 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ [1] = 6.61 × 10⁻³⁴ J s [1]. (c) Vs = 0 at f₀ = 5.5 × 10¹⁴ - 0.18 / 4.13 × 10⁻¹⁵ = 5.06 × 10¹⁴ Hz [1]. Φ/e = gradient × f₀ = 4.13 × 10⁻¹⁵ × 5.06 × 10¹⁴, so Φ ≈ 2.09 eV [1]. Examiner insight: Use two points well separated on the line and show the subtraction; a value near 2.1 eV from sensible working is a fair reading of a graph.

5. (a) Photon energy = hc/λ = 3.83 × 10⁻¹⁹ J [1]. N = 1.2 × 10⁻³ / 3.83 × 10⁻¹⁹ = 3.14 × 10¹⁵ s⁻¹ [1]. (b) Electrons per second = 3.14 × 10¹⁵ / 400 = 7.84 × 10¹² [1]. I = 7.84 × 10¹² × 1.60 × 10⁻¹⁹ = 1.25 × 10⁻⁶ A [1]. (c) Φ = 1.90 × 1.60 × 10⁻¹⁹ = 3.04 × 10⁻¹⁹ J. Emax = 3.83 × 10⁻¹⁹ - 3.04 × 10⁻¹⁹ = 7.85 × 10⁻²⁰ J [1]. (d) Photon energy at 700 nm = 2.84 × 10⁻¹⁹ J (1.78 eV), which is below Φ, so there is no emission and the current falls to zero [1]. Examiner insight: In (d), the argument is the comparison of photon energy with Φ; stating that “red light has less intensity” would not answer the question.

6. (a) KE = 3.0 × 10³ × 1.60 × 10⁻¹⁹ = 4.8 × 10⁻¹⁶ J [1]. p = √(2 × 9.11 × 10⁻³¹ × 4.8 × 10⁻¹⁶) = 2.96 × 10⁻²³ kg m s⁻¹ [1]. λ = h/p = 2.24 × 10⁻¹¹ m [1]. (b) KE = 5.0 × 10⁶ × 1.60 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J; p = √(2mKE) = 1.03 × 10⁻¹⁹ kg m s⁻¹ [1]; λ = 6.43 × 10⁻¹⁵ m [1]. Examiner insight: Convert keV or MeV to joules first, and write the momentum on its own line before dividing into h.

7. (a) sin θ = λ/b, so λ = 1.5 × 10⁻⁹ × sin 9.5° [1] = 2.48 × 10⁻¹⁰ m [1]. (b) p = h/λ = 2.68 × 10⁻²⁴ kg m s⁻¹ [1]; v = p/m = 2.94 × 10⁶ m s⁻¹ [1]. (c) Doubling the speed doubles p and halves λ, so sin θ = 0.165 / 2 = 0.0825 [1], giving θ = 4.73° [1]. Examiner insight: Halving the angle is the usual slip; halve the sine of the angle, not the angle itself.

8. A beam of electrons of known kinetic energy (set by an accelerating voltage) travels in a vacuum [1]. It passes through a narrow gap or a thin crystal whose spacing is comparable with the de Broglie wavelength [1]. A screen or detector records intensity against angle, and shows maxima and minima [1]. A diffraction pattern is a wave property, so this shows matter has a wave nature, and the minima fit λ = h/p [1]. Examiner insight: Mention the vacuum and the spacing compared with λ, and link the pattern to a wave property.

9. (a) 1 - cos 150° = 1.866 [1]; Δλ = 2.43 × 10⁻¹² × 1.866 = 4.53 × 10⁻¹² m [1]. (b) λf = 7.20 × 10⁻¹¹ + 0.453 × 10⁻¹¹ = 7.65 × 10⁻¹¹ m [1]. (c) The photon gives some of its energy to the electron [1]. Since E = hc/λ, less energy means a longer wavelength [1]. (d) The photon behaves as a particle in a collision, exchanging momentum as well as energy [1]. The measured shift fits a formula in the scattering angle, so it is a quantitative test [1]. Examiner insight: The negative cosine matters: 1 - cos 150° is greater than 1, not less.

10. (a) Δλ = 0.1000 - 0.0975 = 2.50 × 10⁻¹² m [1]. cos θ = 1 - 2.50 / 2.43 = -0.0288 [1]. θ = 91.7° [1]. (b) hc/λi = 2.04 × 10⁻¹⁵ J [1]; hc/λf = 1.99 × 10⁻¹⁵ J [1]; difference = 5.10 × 10⁻¹⁷ J = 319 eV [1]. Examiner insight: A small change in the constant used shifts the angle in the third figure, so use the value the question gives and keep extra digits until the end.

11. (a) p = h/λ = 6.63 × 10⁻³⁴ / 0.250 × 10⁻⁹ = 2.65 × 10⁻²⁴ kg m s⁻¹ [1]. KE = p²/2m = 3.86 × 10⁻¹⁸ J [1] = 24.1 eV [1]. (b) E = hc/λ = 7.96 × 10⁻¹⁶ J [1] = 4.97 keV [1]. (c) The photon’s energy is about 206 times larger than the electron’s, although the two have the same momentum [1]. Examiner insight: Use p²/2m for the electron and hc/λ for the photon; applying the same energy formula to both is the standard mistake.

Where marks are usually lost

• Not converting eV to J, or keV and MeV to J, before using the Planck constant.
• Using Emax = hf instead of Emax = hf - Φ.
• Linking intensity to the maximum energy rather than to the number of electrons.
• Stating a classical feature without naming both the prediction and the observation.
• Halving an angle rather than the sine of the angle when the wavelength changes.
• Using the photon energy formula for an electron, or p²/2m for a photon.
• Getting the sign wrong in the Compton factor 1 - cos θ for angles above 90°.
• Rounding the constant h/mec early and losing accuracy in an angle.
• Describing a diffraction experiment without the detector or the gap size.

Next steps

• Re-read the revision notes and the study guide.
• Go back to the DP Physics hub and tick the topic off on the checklist.
• Try all free 10-minute diagnostics at /diagnostics/.
• Book a free trial class: /trial/.

Official syllabus

IB Diploma Programme Physics guide, first assessment 2025, published by the International Baccalaureate Organization. This page covers sub-topic E.2 (HL only).