OCR A Physics · Module 6: Particles and medical physics
Nuclear & particle physics: practice questions with answers
6 exam-style questions on nuclear & particle physics. Try each one on paper first, then open the answer to see the full working and the common slips that cost marks.
- Using the right equation3 marks
Cobalt-60 has a half-life of 5.3 years. A source has an initial activity of 800 kBq. Calculate its activity after 12.0 years, in kBq.
A = kBq
Show the answer and working
Answer A = 167 kBq
Number of half-lives = 12.0 ÷ 5.3 = 2.26. A = 800 × 0.5^2.26 = 167 kBq.
- Got 200 kBq? You rounded 12.0 years to exactly 2 half-lives. The number of half-lives is 12.0 ÷ 5.3 = 2.26, so A = 800 × 0.5^2.26 = 167 kBq.
- Got 353 kBq? You scaled the activity linearly. Decay is exponential: A = A₀ × 0.5^(t/T½) = 167 kBq.
- eV and joules3 marks
The mass of a helium-4 nucleus is 0.03037 u less than the total mass of its separate nucleons. Calculate the binding energy of the nucleus, in MeV.
E = MeV
Show the answer and working
Answer E = 28.4 MeV
Δm = 0.03037 × 1.661 × 10⁻²⁷ = 5.04 × 10⁻²⁹ kg. E = Δmc² = 4.54 × 10⁻¹² J = 28.4 MeV.
- Got 4.54 × 10⁻¹² MeV? That is the energy in joules. Divide by 1.60 × 10⁻¹⁹ for eV, then by 10⁶ for MeV: 28.4 MeV.
- Got 1.71 × 10²⁸ MeV? You did not convert the mass defect from u to kg. Δm = 0.03037 × 1.661 × 10⁻²⁷ = 5.04 × 10⁻²⁹ kg, so E = Δmc² = 28.4 MeV.
- Unit prefixes2 marks
Technetium-99m has a half-life of 6.0 hours. Calculate its decay constant, in s⁻¹.
λ = s⁻¹
Show the answer and working
Answer λ = 3.21 × 10⁻⁵ s⁻¹
T½ = 6.0 × 3600 = 21 600 s. λ = ln 2 ÷ T½ = 3.21 × 10⁻⁵ s⁻¹.
- Got 0.116 s⁻¹? You left the half-life in hours, giving λ in h⁻¹. Convert: T½ = 6.0 × 3600 = 21 600 s, so λ = ln 2 ÷ T½ = 3.21 × 10⁻⁵ s⁻¹.
- Got 0.00193 s⁻¹? You converted hours to seconds using 60. 1 hour = 3600 s, so T½ = 21 600 s and λ = 3.21 × 10⁻⁵ s⁻¹.
- eV and joules2 marks
An electron and a positron annihilate at rest, producing two photons of equal energy. Calculate the energy of one photon, in MeV.
E = MeV
Show the answer and working
Answer E = 0.512 MeV
Each particle has rest energy mc² = 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J = 0.512 MeV. Two photons carry the total, so each carries mc² = 0.512 MeV.
- Got 1.02 MeV? That is the total energy of both photons. Each photon takes half: E = mc² = 0.512 MeV.
- Got 8.20 × 10⁻¹⁴ MeV? That is in joules (8.2 × 10⁻¹⁴ J). Divide by 1.60 × 10⁻¹³ J per MeV: E = 0.512 MeV.
- Using the right equation1 mark
What happens to the quarks in β⁻ decay?
- A A down quark changes into an up quark.
- B An up quark changes into a down quark.
- C A strange quark changes into an up quark.
Show the answer and working
Answer A: A down quark changes into an up quark.
A neutron (udd) becomes a proton (uud), so a down quark changes to an up quark via the weak interaction.
- Why not B? That is what happens in β⁺ decay, where a proton becomes a neutron.
- Why not C? Strange quarks are not involved in beta decay of ordinary nuclei.
- Reading graphs1 mark
Why does nuclear fusion of light nuclei release energy?
- A The products have a lower binding energy per nucleon.
- B The products have a higher binding energy per nucleon.
- C The products have fewer nucleons than the reactants.
Show the answer and working
Answer B: The products have a higher binding energy per nucleon.
Moving up the binding energy curve towards iron-56 means more energy per nucleon is bound, and the difference is released.
- Why not A? Lower binding energy per nucleon would mean energy must be put in.
- Why not C? The number of nucleons is conserved in fusion.