Flashcards · 9702 · A Level Topic 23

Nuclear physics (A Level) flashcards

12 cards on A Level Topic 23, subtopics 23.1 to 23.2. Pick a subtopic or study them all.

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Mass–energy equivalence: E = mc²: mass and energy are equivalent. In a nuclear reaction, the energy released is ΔE = Δmc², where Δm is the decrease in mass.
Mass defect: The total mass of the separate nucleons minus the mass of the nucleus. The separate nucleons are heavier than the bound nucleus.
Binding energy of a nucleus: The energy needed to separate a nucleus completely into its individual protons and neutrons. Binding energy = mass defect × c².
Helium-4 nucleus 4.00150 u; proton 1.00728 u; neutron 1.00867 u (1 u = 931.5 MeV). Find the binding energy per nucleon.: Nucleons: 2(1.00728) + 2(1.00867) = 4.03190 u. Δm = 0.03040 u. Binding energy = 0.03040 × 931.5 = 28.3 MeV. Per nucleon: 28.3 ÷ 4 = 7.08 MeV.
Shape of the graph of binding energy per nucleon against nucleon number: It rises steeply for light nuclei to a peak near iron-56, then falls slowly for heavier nuclei. A higher binding energy per nucleon means a more stable nucleus.
Why do both fusion and fission release energy?: Fusion of light nuclei and fission of heavy nuclei both give products with a higher binding energy per nucleon (closer to iron). The increase in binding energy is released.

What do fluctuations in count rate show about radioactive decay?: That decay is random (you cannot predict which nucleus decays next, or when) and spontaneous (not affected by external factors such as temperature or pressure).
Activity: The number of nuclei decaying per unit time. A = λN, where N is the number of undecayed nuclei and λ the decay constant (the probability that a nucleus decays per unit time). Unit becquerel (1 Bq = 1 decay per second).
Exponential decay equation: x = x₀e^(−λt), where x is the activity, the number of undecayed nuclei, or the received count rate.
Relationship between decay constant and half-life: t½ = ln 2 ÷ λ ≈ 0.693 ÷ λ. A long half-life means a small decay constant.
Half-life 8.0 days, initial activity 4.8 × 10⁵ Bq. Find the activity after 20 days.: λ = 0.693 ÷ 8.0 = 0.0866 day⁻¹. A = 4.8 × 10⁵ × e^(−0.0866 × 20) = 8.5 × 10⁴ Bq.
Changes in nucleon number A and proton number Z in α and β⁻ decay: α: A falls by 4, Z falls by 2. β⁻: A unchanged, Z rises by 1. (β⁺: A unchanged, Z falls by 1.)