OxfordAQA International AS and A-level Physics: Electron collisions, photoelectric effect and duality (9630) – 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 sections 3.5.9 to 3.5.11 of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4: collisions of electrons with atoms, the photoelectric effect and wave-particle duality. This is shared AS and A-level content. Use h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, electron mass 9.11 × 10⁻³¹ kg, neutron mass 1.675 × 10⁻²⁷ kg.

Revise first with the study guide and revision notes. See also the OxfordAQA A-level Physics hub and the printable checklist.

Questions

1. (a) Convert 2.7 eV to joules. [1] (b) Convert 5.6 × 10⁻¹⁹ J to electronvolts. [1] (c) Calculate the kinetic energy in joules gained by an electron accelerated from rest through 2500 V. [1]

2. Explain how a fluorescent tube produces visible light. [4]

3. An atom Z has energy levels of -8.5 eV, -3.6 eV and -1.9 eV.

(a) State its ionisation energy in eV. [1] (b) Calculate the longest wavelength it can emit by transitions between these levels. [3] (c) An electron of kinetic energy 5.5 eV collides with an atom in its ground state. State and explain what can happen. [2]

4. An X-ray tube operates at 80 kV.

(a) Calculate the maximum photon energy in joules. [1] (b) Calculate the minimum wavelength produced. [2] (c) Explain why the spectrum contains sharp peaks whose positions depend on the target metal. [2]

5. A laser emits light of wavelength 633 nm with power 1.5 mW. Calculate the number of photons emitted per second. [3]

6. The threshold wavelength of a metal surface is 520 nm.

(a) Calculate its work function in joules and in eV. [2] (b) Light of wavelength 400 nm falls on the surface. Calculate the maximum kinetic energy of the photoelectrons in joules. [2] (c) State the stopping potential. [1] (d) Calculate the maximum speed of the photoelectrons. [1]

7. Violet light above the threshold frequency releases photoelectrons from a clean metal plate.

(a) The light intensity is increased to four times its value. State and explain the effect on the photocurrent and on the maximum kinetic energy. [2] (b) Red light of any intensity releases no electrons. Explain this using the photon model. [2]

8. A student measures the stopping potential V(s) for a metal at different frequencies f.

f / 10¹⁴ Hz	7.0	8.0	9.0	10.0	11.0
V(s) / V	0.24	0.65	1.06	1.47	1.88

A best-fit line passes close to (7.0, 0.24) and (11.0, 1.88).

(a) Use hf = ϕ + Ek(max) to show that the gradient of V(s) against f is h/e. [2] (b) Find the gradient. [2] (c) Hence find a value for h. [1] (d) Find the threshold frequency. [2] (e) Find the work function in eV. [1]

9. An electron is accelerated from rest through 400 V.

(a) Calculate its de Broglie wavelength. [3] (b) The p.d. is raised to 1600 V. State the new wavelength and explain. [1]

10. A beam of electrons passes through a thin layer of graphite and produces rings on a screen.

(a) State what the rings suggest about electrons. [1] (b) The accelerating p.d. is increased. Describe and explain the change in the rings. [3] (c) Explain why a moving tennis ball shows no such effect. [1]

11. An electron and a photon each have a wavelength of 0.20 nm.

(a) Calculate the energy of the photon in keV. [2] (b) Calculate the kinetic energy of the electron in eV. [3] (c) Comment on the difference. [1] (d) A neutron has kinetic energy 0.025 eV. Calculate its de Broglie wavelength. [2]

12. Describe how ideas about the nature of light and of electrons changed over time. [3]

Answers

1. (a) 2.7 × 1.60 × 10⁻¹⁹ = 4.3 × 10⁻¹⁹ J [1] (b) 5.6 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 3.5 eV [1] (c) Ek = eV = 1.60 × 10⁻¹⁹ × 2500 = 4.0 × 10⁻¹⁶ J [1] Examiner insight: conversions are quick marks that are easy to lose by multiplying when you should divide; check that the number is smaller in joules (a power of ten near -19) than in eV.

2. Electrons are accelerated by the p.d. across the tube [1]. They collide with mercury atoms and excite them (some collisions ionise, keeping electrons available) [1]. Excited atoms de-excite, emitting photons mainly in the ultraviolet [1]. The phosphor coating absorbs the ultraviolet and re-emits visible light [1]. [4] Examiner insight: answers that stop at “excited atoms emit light” usually miss the phosphor step, which carries its own point.

3. (a) 8.5 eV [1] (b) Longest wavelength is the smallest gap: 3.6 to 1.9 gives 1.7 eV = 2.72 × 10⁻¹⁹ J [1]. λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/(2.72 × 10⁻¹⁹) [1] = 7.31 × 10⁻⁷ m (731 nm) [1] (c) The 1 to 2 gap is 4.9 eV, so the atom can be excited to the second level [1]. The 1 to 3 gap is 6.6 eV, more than 5.5 eV, so that is impossible. The electron leaves with 0.6 eV [1]. [6] Examiner insight: in part (c), quote the gaps you compared and state the kinetic energy left over; a bare “it excites the atom” is incomplete.

4. (a) E = eV = 1.60 × 10⁻¹⁹ × 80 × 10³ = 1.28 × 10⁻¹⁴ J [1] (b) λ = hc/E = 1.989 × 10⁻²⁵/1.28 × 10⁻¹⁴ [1] = 1.55 × 10⁻¹¹ m [1] (c) Incoming electrons eject inner-shell electrons from target atoms [1]. Outer electrons drop into the vacancies, emitting photons of fixed energy equal to the level difference, which depends on the metal [1]. [5] Examiner insight: the minimum wavelength is linked to the full energy of one electron; show that link before substituting.

5. Photon energy = hc/λ = 1.989 × 10⁻²⁵/633 × 10⁻⁹ = 3.14 × 10⁻¹⁹ J [1]. Photons per second = P/E = 1.5 × 10⁻³/3.14 × 10⁻¹⁹ [1] = 4.8 × 10¹⁵ s⁻¹ [1] [3] Examiner insight: convert nanometres to metres and milliwatts to watts at the start; a slip of a factor of a thousand is easy to make.

6. (a) ϕ = hc/λ₀ = 1.989 × 10⁻²⁵/520 × 10⁻⁹ = 3.83 × 10⁻¹⁹ J [1] = 3.83 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 2.39 eV [1] (b) Photon energy at 400 nm = 4.97 × 10⁻¹⁹ J [1]. Ek(max) = 4.97 × 10⁻¹⁹ - 3.83 × 10⁻¹⁹ = 1.15 × 10⁻¹⁹ J [1] (c) V(s) = 1.15 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 0.72 V [1] (d) v = √(2Ek/m) = √(2 × 1.15 × 10⁻¹⁹/9.11 × 10⁻³¹) = 5.0 × 10⁵ m s⁻¹ [1] [6] Examiner insight: to avoid rounding drift in a chain of steps, carry an extra digit and round only the final answer.

7. (a) The photocurrent increases (four times) because four times as many photons arrive per second, so more electrons are released per second [1]. Ek(max) is unchanged because each photon has the same energy, depending on frequency only [1]. (b) Each electron absorbs one photon only [1]. A red photon has hf less than the work function, so no electron can escape, however many photons arrive [1]. [4] Examiner insight: “one photon to one electron” is the key idea, and answers about energy building up over time describe the wave model instead.

8. (a) hf = ϕ + eV(s), so V(s) = (h/e)f - ϕ/e [1]. This has the form y = mx + c, so the gradient is h/e [1]. (b) Gradient = (1.88 - 0.24)/((11.0 - 7.0) × 10¹⁴) [1] = 4.10 × 10⁻¹⁵ V s [1] (c) h = gradient × e = 4.10 × 10⁻¹⁵ × 1.60 × 10⁻¹⁹ = 6.56 × 10⁻³⁴ J s [1] (d) At V(s) = 0: f₀ = 7.0 × 10¹⁴ - 0.24/4.10 × 10⁻¹⁵ [1] = 6.4 × 10¹⁴ Hz [1] (e) ϕ = hf₀/e = 4.10 × 10⁻¹⁵ × 6.42 × 10¹⁴ = 2.6 eV [1] [8] Examiner insight: take the gradient from points far apart on the best-fit line, not from table values that sit off the line, and show the multiplication by e.

9. (a) Ek = eV = 1.60 × 10⁻¹⁹ × 400 = 6.4 × 10⁻¹⁷ J [1]. p = √(2mEk) = √(2 × 9.11 × 10⁻³¹ × 6.4 × 10⁻¹⁷) = 1.08 × 10⁻²³ kg m s⁻¹ [1]. λ = h/p = 6.1 × 10⁻¹¹ m [1] (b) Four times the p.d. gives twice the momentum, so the wavelength halves to 3.1 × 10⁻¹¹ m [1] [4] Examiner insight: the route from p.d. to wavelength needs an energy step and a momentum step; skipping either is the usual cause of a wrong answer.

10. (a) Electrons have wave properties, because diffraction is a wave effect [1]. (b) Higher p.d. gives greater speed and momentum [1]. So λ = h/p decreases [1]. A smaller wavelength diffracts through smaller angles, so the rings shrink towards the centre [1]. (c) Its wavelength is far too small compared with any gap, so no diffraction is observable [1] [5] Examiner insight: chain the reasoning in order: momentum, wavelength, angle. Saying the rings “change” without direction gains little.

11. (a) E = hc/λ = 1.989 × 10⁻²⁵/0.20 × 10⁻⁹ = 9.9 × 10⁻¹⁶ J [1] = 6.2 keV [1] (b) p = h/λ = 6.63 × 10⁻³⁴/0.20 × 10⁻⁹ = 3.3 × 10⁻²⁴ kg m s⁻¹ [1]. Ek = p²/2m = 6.0 × 10⁻¹⁸ J [1] = 38 eV [1] (c) For the same wavelength, the photon has much more energy than the electron, because photon energy is proportional to 1/λ while electron energy is proportional to 1/λ² [1]. (d) Ek = 0.025 × 1.60 × 10⁻¹⁹ = 4.0 × 10⁻²¹ J, p = √(2 × 1.675 × 10⁻²⁷ × 4.0 × 10⁻²¹) = 3.66 × 10⁻²⁴ [1], λ = 1.8 × 10⁻¹⁰ m [1] [8] Examiner insight: a comment usually earns credit when it uses numbers from parts (a) and (b), not a general statement about duality.

12. Light was explained as a wave (interference and diffraction) [1]. The photoelectric effect needed photons, so light has a particle nature too [1]. Electrons first seemed to be particles, but de Broglie’s wavelength and electron diffraction showed wave behaviour [1]. [3] Examiner insight: name the evidence that forced each change, not just the change.

Where marks are usually lost

• Converting between eV and J in the wrong direction.
• Leaving wavelengths in nm inside hc/λ.
• Writing that an electron “absorbs” a photon of any energy; excitation needs a matching gap.
• Saying more intense light gives faster electrons.
• Not turning a stopping potential in volts into energy with e.
• Forgetting to multiply the V(s) gradient by e.
• Using kinetic energy where momentum is required in λ = h/mv.
• Describing a diffraction change without linking it to momentum and wavelength.
• Omitting the phosphor in the fluorescent tube.

Next steps

• Revise with the revision notes and study guide.
• Tick off the OxfordAQA A-level Physics checklist and visit the course hub.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), version 4.4, for International AS and A-level exams from May/June 2020, published by OxfordAQA.