What this chapter is about
This chapter explores the structure of the atomic nucleus, the tiny core at the centre of every atom. After studying atomic models and electron behaviour in earlier chapters, you now turn inward to understand what the nucleus itself is made of, how it holds together, and why some nuclei are stable while others spontaneously break apart.
You will learn about protons and neutrons (collectively called nucleons), how nuclear size and density are measured, and the concept of nuclear binding energy that explains why nuclei exist at all. The chapter introduces radioactivity — the spontaneous emission of particles or radiation from unstable nuclei — and the laws that govern radioactive decay. You will also meet nuclear fission and fusion, the two processes that release enormous energy and power nuclear reactors and stars respectively.
By the end, you should be able to calculate binding energies, predict decay products, work with half-life problems, and explain why mass converts to energy in nuclear reactions using Einstein's mass-energy relation.
Key ideas
- Nuclear composition: A nucleus contains Z protons and N neutrons, where A = Z + N is the mass number. Nuclei with the same Z but different N are called isotopes.
- Nuclear size: The radius of a nucleus follows R = R₀ × A^(1/3), where R₀ ≈ 1.2 × 10⁻¹⁵ m. This means nuclear density is nearly constant for all nuclei, about 2.3 × 10¹⁷ kg/m³.
- Mass defect and binding energy: The mass of a nucleus is always less than the sum of masses of its free nucleons. This difference, Δm, when multiplied by c², gives the binding energy that holds the nucleus together.
- Binding energy per nucleon: This quantity peaks around A ≈ 56 (iron region). Nuclei with higher binding energy per nucleon are more stable. This curve explains why fusion of light nuclei and fission of heavy nuclei both release energy.
- Radioactive decay: Unstable nuclei emit alpha particles (helium nuclei), beta particles (electrons or positrons), or gamma rays (high-energy photons) to reach more stable configurations.
- Decay law: The number of undecayed nuclei decreases exponentially: N(t) = N₀ × e^(−λt), where λ is the decay constant. Half-life T₁/₂ = 0.693/λ is the time for half the nuclei to decay.
- Nuclear reactions: In fission, a heavy nucleus splits into lighter fragments; in fusion, light nuclei combine. Both processes convert mass to energy according to E = Δm × c².
Formulas and facts to remember
- Mass-energy relation: E = m × c², where c = 3 × 10⁸ m/s. One atomic mass unit (1 u) = 931.5 MeV/c².
- Nuclear radius: R = R₀ × A^(1/3), with R₀ ≈ 1.2 fm (1 fm = 10⁻¹⁵ m).
- Mass defect: Δm = [Z × mₚ + N × mₙ] − M_nucleus, where mₚ and mₙ are proton and neutron masses.
- Binding energy: B.E. = Δm × c², often expressed in MeV.
- Radioactive decay law: N(t) = N₀ × e^(−λt); Activity A = λN, measured in becquerel (1 Bq = 1 decay/s).
- Half-life relation: T₁/₂ = 0.693/λ = ln 2/λ.
- Mean life: τ = 1/λ.
- Alpha decay: Parent nucleus emits ⁴He; atomic number decreases by 2, mass number by 4.
- Beta-minus decay: A neutron converts to a proton, emitting an electron and an antineutrino; Z increases by 1.
Worked examples
Example 1: Calculating binding energy
Find the binding energy of the helium-4 nucleus (⁴He). Given: mass of proton = 1.00728 u, mass of neutron = 1.00867 u, mass of ⁴He nucleus = 4.00150 u.
Solution: Helium-4 has Z = 2 protons and N = 2 neutrons. Sum of free nucleon masses = 2 × 1.00728 + 2 × 1.00867 = 2.01456 + 2.01734 = 4.03190 u. Mass defect Δm = 4.03190 − 4.00150 = 0.03040 u. Binding energy = 0.03040 × 931.5 MeV = 28.32 MeV. Binding energy per nucleon = 28.32/4 = 7.08 MeV.
Example 2: Half-life problem
A radioactive sample has a half-life of 20 minutes. If it initially contains 6.4 × 10⁸ atoms, how many remain after 1 hour?
Solution: Time elapsed = 60 minutes = 3 half-lives. After each half-life, the number halves. After 1 half-life: 6.4 × 10⁸ / 2 = 3.2 × 10⁸. After 2 half-lives: 3.2 × 10⁸ / 2 = 1.6 × 10⁸. After 3 half-lives: 1.6 × 10⁸ / 2 = 0.8 × 10⁸ = 8.0 × 10⁷ atoms.
Alternatively: N = N₀ × (1/2)^n = 6.4 × 10⁸ × (1/2)³ = 8.0 × 10⁷ atoms.
Example 3: Alpha decay equation
Write the nuclear reaction when radium-226 undergoes alpha decay.
Solution: Radium has Z = 88. In alpha decay, an alpha particle (⁴₂He) is emitted. The daughter nucleus has Z = 88 − 2 = 86 (radon) and A = 226 − 4 = 222. Reaction: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He.
Conservation check: Mass number 226 = 222 + 4 ✓; Atomic number 88 = 86 + 2 ✓.
Common mistakes
- Confusing atomic mass number A with atomic number Z → A is total nucleons, Z is only protons.
- Using total binding energy instead of binding energy per nucleon to compare stability → Always compare binding energy per nucleon; higher value means more stable.
- Forgetting that mass defect must be converted to energy units using 931.5 MeV per u → Without this conversion, the answer has wrong units.
- Assuming half-life means all nuclei decay in 2 × T₁/₂ → Decay is exponential; some nuclei always remain.
- Writing wrong daughter nucleus in decay equations by not conserving both Z and A → Check that both mass number and atomic number balance on each side.
Quick revision
- Nucleus = Z protons + N neutrons; A = Z + N.
- Nuclear radius ∝ A^(1/3); nuclear density is roughly constant for all nuclei.
- Binding energy = mass defect × c²; higher B.E. per nucleon means greater stability.
- Radioactive decay is random but follows N = N₀ × e^(−λt); half-life T₁/₂ = 0.693/λ.
- Fission splits heavy nuclei; fusion joins light nuclei — both release energy because products have higher B.E. per nucleon.
- Always conserve mass number and charge in nuclear reactions.