Edexcel IAL Physics: Nuclear and Particle 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 — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Nuclear and Particle Physics revision notes

Scope note: every question in this paper tests Unit 4 sub-topic 4.5, Nuclear and Particle Physics. Radioactive decay, half-life, nuclear decay equations, mass deficit, binding energy, fission and fusion belong to Unit 5 sub-topic 5.4 and are practised in the Nuclear Decay practice questions.

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Section A

1. State what is meant by the nucleon number and the proton number of a nucleus, and give both for the nucleus ⁵⁶₂₆Fe. [3]

2. State the quark composition of a proton and a neutron. [2]

Section B

3. Electrons are released from a heated filament by thermionic emission and are accelerated from rest through a potential difference of 5.0 kV. (Mass of an electron = 9.11 × 10⁻³¹ kg; e = 1.60 × 10⁻¹⁹ C.)

(a) State the kinetic energy of each electron in keV, and calculate it in joules. [2] (b) Calculate the speed of the electrons. [2] (c) The beam then enters a uniform magnetic field at right angles to its velocity. Explain why the field changes the direction of the electrons but not their speed. [2]

4. Muons are created by cosmic rays about 15 km above the Earth’s surface and travel towards the ground at a speed very close to the speed of light (take this speed as 3.0 × 10⁸ m s⁻¹). A muon at rest has a mean lifetime of 2.2 × 10⁻⁶ s.

(a) Calculate the time, measured on Earth, for a muon to travel 15 km. [1] (b) Compare this time with the muon’s lifetime at rest, and state what the comparison would suggest about the number of muons reaching the ground. [2] (c) Large numbers of muons are detected at ground level. Explain this observation. [3]

5. In a particle accelerator, protons are each given a kinetic energy of 7.0 GeV.

(a) Convert this energy to joules. [2] (b) Explain why such high energies are needed to investigate the structure of nucleons. [3]

6. Check whether this interaction is possible, testing charge, baryon number and lepton number: p + p → p + n + e⁺ + ν_e. [4]

7. State what the Geiger-Marsden alpha-particle scattering experiment provided evidence for, and what model of the atom it replaced. [2]

8. State the equation for the radius of a charged particle’s circular path in a magnetic field, and explain how this can be used to identify a particle’s momentum and the sign of its charge. [3]

9. An electron and a positron, each of rest mass 9.11 × 10⁻³¹ kg, annihilate at rest, producing two identical gamma-ray photons. Calculate the energy of each photon. [3]

10. Classify each of the following as a baryon, a meson or a lepton: proton, pion, electron. [3]

11. Describe how thermionic emission is used to produce a beam of electrons for injection into a linear accelerator (linac), and explain briefly how a cyclotron differs from a linac in the way it accelerates particles. [4]

12. A particle has a rest mass of 1.67 × 10⁻²⁷ kg. (a) Convert this mass to units of MeV/c². (b) A different particle has a rest energy of 105 MeV. State its mass in MeV/c², and explain why MeV/c² is a convenient unit for particle rest mass. [4]

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Answers

1. The nucleon number (mass number) is the total number of protons and neutrons in the nucleus [1]; the proton number (atomic number) is the number of protons [1]. For ⁵⁶₂₆Fe: nucleon number 56, proton number 26 (so 30 neutrons) [1].

2. Proton = uud [1]; neutron = udd [1].

3. (a) Each electron gains 5.0 keV [1]; E = eV = 1.60 × 10⁻¹⁹ × 5.0 × 10³ = 8.0 × 10⁻¹⁶ J [1]. (b) ½mv² = 8.0 × 10⁻¹⁶, so v = √(2 × 8.0 × 10⁻¹⁶ ÷ 9.11 × 10⁻³¹) [1] = 4.2 × 10⁷ m s⁻¹ [1]. (c) The magnetic force (F = Bev) is perpendicular to the velocity at every instant [1], so it does no work on the electrons: their kinetic energy, and so their speed, is unchanged and only their direction changes [1].

4. (a) t = 15 × 10³ ÷ 3.0 × 10⁸ = 5.0 × 10⁻⁵ s [1]. (b) 5.0 × 10⁻⁵ ÷ 2.2 × 10⁻⁶ ≈ 23, so the journey lasts about 23 lifetimes [1]; this suggests almost all the muons would decay before reaching the ground [1]. (c) The muons move at speeds close to c, where the relativistic increase in particle lifetime is significant [1]. Measured by an observer on Earth, the lifetime of a fast-moving muon is much longer than its lifetime at rest [1]. So the journey lasts far fewer of these extended lifetimes, and a significant fraction of the muons survive to reach the ground [1].

5. (a) 7.0 GeV = 7.0 × 10⁹ eV [1]; × 1.60 × 10⁻¹⁹ J eV⁻¹ = 1.1 × 10⁻⁹ J [1]. (b) To resolve structure inside a nucleon, the probing particle’s de Broglie wavelength must be comparable with or smaller than the nucleon (about 10⁻¹⁵ m) [1]. Since λ = h/p, a very short wavelength needs a very large momentum [1], and therefore a very high energy [1].

6. Charge: left 2, right 1 + 0 + 1 + 0 = 2 ✓ [1]. Baryon number: left 2, right 1 + 1 + 0 + 0 = 2 ✓ [1]. Lepton number: left 0, right 0 + 0 + (−1) + (+1) = 0 ✓ [1]. All three quantities are conserved, so the interaction is possible [1].

7. It provided evidence for a small, dense, positively charged nucleus [1] (from the observation of large-angle scattering), replacing the earlier “plum pudding” model of the atom [1].

8. r = p ÷ BQ [1]. The radius of the curved path reveals the particle’s momentum (from r, B and Q) [1], and the direction of curvature reveals the sign of its charge [1].

9. Total mass converted: Δm = 2 × 9.11 × 10⁻³¹ = 1.822 × 10⁻³⁰ kg [1]. Total energy released: ΔE = c²Δm = (3.00 × 10⁸)² × 1.822 × 10⁻³⁰ = 1.640 × 10⁻¹³ J [1]. Since two identical photons share this energy equally, each photon has energy = 1.640 × 10⁻¹³ ÷ 2 = 8.20 × 10⁻¹⁴ J (about 0.51 MeV) [1].

10. Proton — baryon (three quarks) [1]; pion — meson (a quark and an antiquark) [1]; electron — lepton (a fundamental particle) [1].

11. A heated filament (cathode) releases electrons from its surface by thermionic emission [1]; these electrons are then accelerated by a potential difference towards an anode to form the initial beam injected into the linac [1]. In a linac, the particle is accelerated in a straight line through a series of tubes with an alternating field switching sign as it crosses each gap [1]. In a cyclotron, the particle instead spirals outward inside a magnetic field, crossing the same alternating-field gap repeatedly as its radius grows [1].

12. (a) E = mc² = 1.67 × 10⁻²⁷ × (3.00 × 10⁸)² = 1.503 × 10⁻¹⁰ J [1]; ÷ 1.60 × 10⁻¹³ = 939 MeV, so mass = 939 MeV/c² [1]. (b) Mass = 105 MeV/c² [1]. MeV/c² is convenient because rest energy in MeV can be converted to mass by simply dividing by c², avoiding repeated use of very small SI masses and the large factor of c² in every calculation [1].

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Where marks are usually lost

• Giving the charge of a positron as −1.
• Forgetting that an antineutrino has lepton number −1.
• Using grams instead of kilograms in E = mc².
• Forgetting to convert GeV or MeV to eV (× 10⁹ or × 10⁶) before multiplying by 1.60 × 10⁻¹⁹ J eV⁻¹.
• Saying a muon’s lifetime at rest changes: it is the lifetime measured by an observer for whom the particle moves close to c that increases.
• Forgetting that the top quark’s existence was predicted by the symmetry of the standard model before it was observed.
• Confusing a baryon (three quarks) with a meson (a quark and an antiquark).