Nuclear physics
Mass–energy equivalence, mass defect and binding energy, the binding energy per nucleon curve and why fission and fusion release energy, and the random, spontaneous nature of radioactive decay with activity, decay constant and half-life.
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In 30 seconds
- Mass and energy are equivalent: E = mc². A nucleus has less mass than its separate nucleons; the difference is the mass defect.
- Binding energy is the energy needed to separate a nucleus into its nucleons. Higher binding energy per nucleon means a more stable nucleus.
- Fission and fusion both release energy because the products have a higher binding energy per nucleon than the starting nuclei (iron-56 is near the peak).
- Decay is random and spontaneous. Activity A = λN, and N = N₀e−λt, with half-life t½ = ln 2/λ.
By the end you can
- Understand mass defect and binding energy, and use E = mc².
- Sketch and interpret the binding energy per nucleon curve, explaining fission and fusion.
- Explain that radioactive decay is random and spontaneous.
- Define and use activity, decay constant and half-life; use A = λN and x = x₀e−λt.
1Mass and energy
Binding energy
Mass defect — the difference between the total mass of the separate nucleons and the mass of the nucleus.
Binding energy — the minimum energy required to separate a nucleus into its constituent protons and neutrons.
Find the binding energy per nucleon of helium-4. Masses: proton 1.007276 u, neutron 1.008665 u, He-4 nucleus 4.001506 u.
- Δm = 2(1.007276) + 2(1.008665) − 4.001506 = 0.030376 u C1
- E = 0.030376 × 1.66 × 10⁻²⁷ × (3.00 × 10⁸)² = 4.5 × 10⁻¹² J (28.3 MeV) C1 A1
- per nucleon: 1.1 × 10⁻¹² J ≈ 7.1 MeV A1
- Fission: a heavy nucleus (e.g. uranium-235) splits into two medium nuclei, which have greater binding energy per nucleon. Fusion: light nuclei (hydrogen isotopes) join, also moving up the curve.
2Radioactive decay
Random and spontaneous
- Random: you can't predict which nucleus will decay next or when; each nucleus has a constant probability of decay per unit time. Evidence: fluctuations in count rate.
- Spontaneous: decay isn't affected by external conditions (temperature, pressure, chemical state).
Activity — the number of nuclei decaying per unit time: A = λN (unit: becquerel, Bq = s⁻¹).
Decay constant λ — the probability that an individual nucleus will decay per unit time.
Half-life — the time taken for the number of undecayed nuclei (or the activity) to halve.
Iodine-131 has a half-life of 8.0 days. A sample has activity 4.0 × 10⁶ Bq. Find (a) λ, (b) the number of nuclei, (c) the activity after 20 days, (d) the time for the activity to fall to 10%.
- (a) λ = 0.693/(8.0 × 86 400) = 1.0 × 10⁻⁶ s⁻¹ A1
- (b) N = A/λ = 4.0 × 10¹² nuclei A1
- (c) A = 4.0 × 10⁶ × 0.52.5 = 7.1 × 10⁵ Bq C1 A1
- (d) t = ln 10/λ = 27 days A1
Test yourself — cover the answers first
Quick check
- Define binding energy.
- Which nucleus is near the peak of the binding energy per nucleon curve?
- What does "spontaneous" mean for radioactive decay?
- Find λ for a half-life of 10 s.
- What fraction remains after 3 half-lives?
Exam-style
- Find the energy equivalent of 1.0 u in MeV.
- A source has 3.0 × 10¹⁵ nuclei and λ = 2.0 × 10⁻⁹ s⁻¹. Find its activity.
- The count rate (corrected) falls from 800 to 200 in 6.0 h. Find the half-life.
Show answers
- the minimum energy needed to separate a nucleus into its nucleons
- iron-56
- not affected by external factors
- 0.0693 s⁻¹
- 1/8
- 1.66 × 10⁻²⁷ × 9.0 × 10¹⁶ = 1.49 × 10⁻¹⁰ J = 934 MeV (931.5 with precise constants)
- A = 6.0 × 10⁶ Bq
- two half-lives in 6.0 h ⇒ t½ = 3.0 h
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