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AQA A-Level Physics: Particles and radiation (7408) – Practice Questions

Original practice questions with marked answers on AQA A-level Physics 7408 Particles and radiation, from specific charge to de Broglie wavelength.

Subject
Physics
Level
A LEVELS
Topic
Particles and radiation
Updated

Aligned to AQA A Level Physics (7408), For first teaching 2015. Official specification .

Syllabus page (what it covers and how it is assessed): AQA A Level Physics.

Syllabus points this page covers

7408

  • 2 Particles and radiation (whole topic)
  • 3.2.1 Particles
  • 3.2.2 Electromagnetic radiation and quantum phenomena

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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 3.2 Particles and radiation of the AQA AS and A-level Physics (7407/7408) specification, version 1.4 (AS and A-level exams June 2016 onwards): 3.2.1 Particles and 3.2.2 Electromagnetic radiation and quantum phenomena. It is AS and A-level content, examined in A-level Paper 1 and assumed in Paper 2. Sections 3.9-3.13 are options (you study one, assessed in Paper 3 Section B); this topic is compulsory.

Learn the content first with the study guide and the revision notes. Course pages: AQA A-level Physics hub and printable checklist.

Use: h = 6.63 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹, e = 1.60 × 10⁻¹⁹ C, mₑ = 9.11 × 10⁻³¹ kg, nucleon mass = 1.67 × 10⁻²⁷ kg.

Questions

1. An oxide ion is written ¹⁶₈O²⁻.

(a) State the number of protons, neutrons and electrons in the ion. [2] (b) Calculate the specific charge of the ion. Ignore the mass of the electrons. [2]

2. Describe how the strong nuclear force between two nucleons depends on their separation, and explain how this keeps a nucleus stable. [3]

3. Carbon-14, ¹⁴₆C, decays by β⁻ emission to an isotope of nitrogen.

(a) Write the complete decay equation. [2] (b) Explain why the existence of the neutrino was hypothesised. [2]

4. A photon produces a proton-antiproton pair. The rest energy of a proton is 938.3 MeV.

(a) Calculate the minimum photon energy in joules. [2] (b) Calculate the wavelength of a photon with this energy. [2]

5. In electron capture, p + e⁻ → n + νₑ.

(a) Name the interaction responsible and its exchange particle. [2] (b) Describe a diagram for this interaction in terms of incoming particles, outgoing particles and exchange particle. [2] (c) State the quark change that takes place. [1]

6. Use conservation laws to decide whether each interaction is possible. If not, name the law broken.

(a) μ⁻ → e⁻ + ν̄ₑ + ν_μ [2] (b) p + ν̄ₑ → n + e⁺ [2] (c) p → n + e⁺ + ν̄ₑ [2]

7. Kaons and pions are mesons.

(a) State the quark composition of a K⁺ and of a π⁻. [2] (b) Show that the charge of the K⁺ is +1. [1] (c) A K⁻ decays: K⁻ → π⁻ + π⁰. Identify the interaction responsible, giving a reason. [2]

8. The lambda particle Λ⁰ has quark composition uds, charge 0, baryon number +1 and strangeness −1. It is made in the reaction π⁻ + p → K⁰ + Λ⁰.

(a) Show that charge, baryon number and strangeness are conserved in this reaction. [3] (b) Explain how the reaction is consistent with the way strange particles are produced. [1] (c) The Λ⁰ later decays: Λ⁰ → p + π⁻. State and explain which interaction causes this decay. [2]

9. Ultraviolet radiation of wavelength 320 nm falls on a clean metal surface. The stopping potential is 1.05 V.

(a) Calculate the energy of one photon in joules. [2] (b) Calculate the work function of the metal in eV. [3] (c) Calculate the threshold frequency of the metal. [2] (d) The intensity of the radiation is doubled. Explain the effect on the stopping potential and on the photoelectric current. [3]

10. An atom has energy levels at −10.4 eV (ground state), −5.5 eV, −3.7 eV and −1.6 eV.

(a) State how many different photon energies can be emitted by atoms excited to the −1.6 eV level. [1] (b) Calculate the longest wavelength these atoms can emit. [3] (c) Calculate the ionisation energy of the atom in joules. [1] (d) An electron with kinetic energy 6.0 eV collides with an atom in its ground state. Explain what can happen to the atom, and state the electron’s kinetic energy after an inelastic collision. [3]

11. Electrons are accelerated from rest through a potential difference of 600 V and directed at a thin graphite film.

(a) Show that the speed of the electrons is about 1.45 × 10⁷ m s⁻¹. [2] (b) Calculate their de Broglie wavelength. [2] (c) The potential difference is increased to 2400 V. Explain the effect on the diffraction pattern. [3]

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

Answers

1. (a) 8 protons and 8 neutrons [1]; 10 electrons [1]. (b) Charge = 2 × 1.60 × 10⁻¹⁹ = 3.20 × 10⁻¹⁹ C and mass = 16 × 1.67 × 10⁻²⁷ = 2.672 × 10⁻²⁶ kg [1]; specific charge = 1.20 × 10⁷ C kg⁻¹ [1]. Examiner insight: The unit C kg⁻¹ is needed for the final mark, and using 8 (the proton number) instead of 2 for the ion’s charge gets no credit.

2. Attractive up to about 3 fm, which overcomes the electrostatic repulsion between protons [1]. Repulsive at separations below about 0.5 fm, so the nucleons are not pulled into each other [1]. Negligible beyond about 3 fm, so it only acts between nearby nucleons [1]. Examiner insight: Each range needs its distance and its effect; “strong force holds the nucleus together” alone scores nothing.

3. (a) ¹⁴₆C → ¹⁴₇N + ⁰₋₁β, with correct A and Z [1]; + ν̄ₑ [1]. (b) Beta particles are emitted with a range of kinetic energies up to a maximum [1]. Another particle was needed to carry away the remaining energy so that energy is conserved [1]. Examiner insight: A neutrino (rather than an antineutrino) in a β⁻ equation loses the second mark.

4. (a) Minimum energy = 2 × 938.3 = 1876.6 MeV [1]; × 10⁶ × 1.60 × 10⁻¹⁹ = 3.00 × 10⁻¹⁰ J [1]. (b) λ = hc/E = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/3.003 × 10⁻¹⁰ [1] = 6.62 × 10⁻¹⁶ m [1]. Examiner insight: Using one rest energy instead of two is a method error; part (b) then usually earns error carried forward.

5. (a) Weak interaction [1]; W⁺ boson (or W⁻ if drawn from the electron to the proton) [1]. (b) The proton enters at one vertex and leaves as a neutron, emitting a W⁺ [1]. The W⁺ meets the incoming electron at the second vertex, and an electron neutrino leaves [1]. (c) u → d [1]. Examiner insight: The W charge must balance at both vertices; a W⁻ leaving the proton contradicts charge conservation and loses the mark.

6. (a) Charge −1 → −1 + 0 + 0; Lₑ 0 → +1 − 1 = 0; L_μ +1 → +1 [1]. All conserved: possible [1]. (b) Charge +1 → 0 + 1; B 1 → 1 [1]; Lₑ −1 → −1: possible [1]. (c) Charge and baryon number are conserved [1]; Lₑ 0 → −1 + (−1) = −2, so not possible: lepton number not conserved [1]. Examiner insight: A bare “possible” scores nothing; show the totals for each quantity before stating a conclusion.

7. (a) K⁺ = us̄ [1]; π⁻ = dū [1]. (b) +⅔ + (+⅓) = +1 [1]. (c) Strangeness changes from −1 to 0 [1]; so it is the weak interaction, because strangeness is conserved in strong interactions but may change by 1 in weak ones [1]. Examiner insight: For (b), remember s̄ carries +⅓, not −⅓; a sign slip here loses the mark.

8. (a) Charge: −1 + 1 = 0 and 0 + 0 = 0 [1]. Baryon number: 0 + 1 = 1 and 0 + 1 = 1 [1]. Strangeness: 0 + 0 = 0 and +1 + (−1) = 0 [1]. (b) Two strange particles (K⁰ with S = +1, Λ⁰ with S = −1) are made together by the strong interaction, so total strangeness stays zero [1]. (c) Strangeness changes from −1 to 0 [1], so it must be the weak interaction [1]. Examiner insight: Each conservation law needs both sides written out to earn its mark.

9. (a) E = hc/λ = (6.63 × 10⁻³⁴ × 3.00 × 10⁸)/320 × 10⁻⁹ [1] = 6.22 × 10⁻¹⁹ J [1]. (b) Ek(max) = eVs = 1.05 eV [1]; photon energy = 6.22 × 10⁻¹⁹/1.60 × 10⁻¹⁹ = 3.88 eV [1]; ϕ = 3.88 − 1.05 = 2.83 eV [1]. (c) ϕ = 6.22 × 10⁻¹⁹ − 1.68 × 10⁻¹⁹ = 4.54 × 10⁻¹⁹ J, and f₀ = ϕ/h = 4.54 × 10⁻¹⁹/6.63 × 10⁻³⁴ [1] = 6.84 × 10¹⁴ Hz (6.83 × 10¹⁴ Hz from the rounded 2.83 eV is also fine) [1]. (d) Stopping potential unchanged [1], because each photon still has the same energy, so Ek(max) is the same [1]. Current increases: more photons per second release more electrons per second [1]. Examiner insight: Allow error carried forward from (a) into (b) and (c), but a stopping potential “increasing with intensity” in (d) gets no credit.

10. (a) 6 [1]. (b) Smallest gap is −3.7 to −5.5 eV = 1.8 eV [1]; = 1.8 × 1.60 × 10⁻¹⁹ = 2.88 × 10⁻¹⁹ J [1]; λ = hc/E = 6.91 × 10⁻⁷ m [1]. (c) 10.4 × 1.60 × 10⁻¹⁹ = 1.66 × 10⁻¹⁸ J [1]. (d) It can be excited to the −5.5 eV level, which needs 4.9 eV [1]. It cannot reach −3.7 eV, which needs 6.7 eV [1]. The electron leaves with 6.0 − 4.9 = 1.1 eV [1]. Examiner insight: The longest wavelength comes from the smallest energy gap; choosing the largest gap is a common way to lose all of (b).

11. (a) Ek = 600 × 1.60 × 10⁻¹⁹ = 9.60 × 10⁻¹⁷ J [1]; v = √(2 × 9.60 × 10⁻¹⁷/9.11 × 10⁻³¹) = 1.452 × 10⁷ m s⁻¹ [1]. (b) λ = h/mv = 6.63 × 10⁻³⁴/(9.11 × 10⁻³¹ × 1.452 × 10⁷) [1] = 5.01 × 10⁻¹¹ m [1]. (c) Kinetic energy is four times larger, so momentum doubles [1]; λ halves to about 2.5 × 10⁻¹¹ m [1]; there is less diffraction, so the rings have a smaller radius [1]. Examiner insight: On a “show that”, give the answer to more significant figures than quoted (here 1.452 × 10⁷) or show every substitution; just restating the given value earns nothing.

12. A high voltage accelerates free electrons through mercury vapour, and they collide with mercury atoms [1]. Electrons in the mercury atoms are excited to higher energy levels [1]. The atoms de-excite and emit ultraviolet photons [1]. The phosphor coating absorbs the UV and re-emits visible light as its atoms de-excite in smaller steps [1]. Examiner insight: Four separate, ordered points are needed for full marks; missing the phosphor step caps the answer at three marks.

Where marks are usually lost

  • Using the proton number instead of the ion’s net charge in specific charge.
  • Writing ν instead of ν̄ₑ in β⁻ decay equations.
  • Using one rest energy for pair production instead of two.
  • Drawing a W⁻ leaving a proton that turns into a neutron.
  • Applying strangeness conservation to a weak decay.
  • Mixing eV and J inside one equation.
  • Saying more intense light gives faster photoelectrons.
  • Taking the largest energy gap for the longest wavelength.
  • Missing that a larger accelerating voltage means shorter wavelength and less diffraction.

Next steps

Official syllabus

AQA AS and A-level Physics (7407/7408) specification, version 1.4, July 2026, AS and A-level exams June 2016 onwards, published by AQA. Section 3.2 Particles and radiation.

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