Understand why nuclei decay, the N-Z stability curve, and the different types of radioactive decay including the role of neutrinos.
AQA A-Level Physics · Section 8 · Nuclear Physics
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Describe and use the N-Z stability curve
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Write equations for alpha, beta-minus and beta-plus decay
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Explain electron capture and gamma emission
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State the role of neutrinos and antineutrinos in beta decay
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Predict which decay mode a nucleus will undergo from its position on the N-Z curve
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Apply conservation laws to decay equations
The N-Z Stability Curve
A plot of neutron number N against proton number Z for stable nuclei forms a "valley of stability" (also called the belt of stability).
For light nuclei (Z < 20): N ≈ Z (approximately equal numbers of protons and neutrons)
For heavier nuclei: N > Z — more neutrons are needed to overcome the greater proton-proton repulsion
No stable nuclei exist with Z > 82 (lead)
A nucleus is unstable if it lies outside the band of stability. The type of decay it undergoes depends on which side of the stable band it falls on.
Position relative to stability
Decay mode
Too many neutrons (above the curve)
Beta-minus (β⁻)
Too many protons (below the curve)
Beta-plus (β⁺) or electron capture
Very heavy (Z > 82)
Alpha (α)
Excited nuclear state
Gamma (γ)
Alpha Decay (α)
An alpha particle consists of 2 protons and 2 neutrons (a helium-4 nucleus: ⁴₂He). Alpha decay occurs in very heavy nuclei (Z > 82) where the nucleus is too large.
ᴬ_Z X → ᴬ⁻⁴_(Z−2) Y + ⁴₂He
Properties of alpha decay:
Z decreases by 2, A decreases by 4
Alpha particles are monoenergetic (single discrete energy)
Strongly ionising, short range in air (~5 cm), stopped by paper
Example: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂He (Radium-226 decays to Radon-222)
Beta-Minus Decay (β⁻)
Beta-minus decay occurs when a nucleus has too many neutrons. A neutron converts to a proton, emitting an electron and an antineutrino:
¹₀n → ¹₁p + ⁰₋₁e + ν̄_e (antineutrino)
ᴬ_Z X → ᴬ_(Z+1) Y + ⁰₋₁e + ν̄_e
Properties:
Z increases by 1, A unchanged (same nucleon number)
Beta particles have a continuous energy spectrum (0 to E_max)
The continuous spectrum was the key evidence that required the neutrino hypothesis
The antineutrino (ν̄_e) carries away some of the decay energy, which is why beta particles have a range of energies up to a maximum, not a fixed value.
Beta-Plus Decay (β⁺) and Electron Capture
These occur when a nucleus has too many protons.
Beta-plus decay: A proton converts to a neutron, emitting a positron and a neutrino:
¹₁p → ¹₀n + ⁰₊₁e + ν_e (neutrino)
ᴬ_Z X → ᴬ_(Z−1) Y + ⁰₊₁e + ν_e
Electron capture: An inner-shell orbital electron is captured by the nucleus; a proton converts to a neutron:
¹₁p + ⁰₋₁e → ¹₀n + ν_e
Both β⁺ and electron capture produce the same daughter nucleus (Z decreases by 1, A unchanged), but the process and observable products differ.
A positron (β⁺) quickly annihilates with an electron, producing two gamma rays of 0.511 MeV each — this is the basis of PET scanning in medicine.
Gamma Emission (γ)
Gamma emission occurs when a nucleus is in an excited nuclear state after alpha or beta decay. It does not change the proton number or nucleon number:
ᴬ_Z X* → ᴬ_Z X + γ
Properties of gamma radiation:
Electromagnetic radiation, wavelength ~10⁻¹² m
No charge, no mass — only energy is emitted
Weakly ionising, very penetrating — reduced by lead or thick concrete
Gamma photons are monoenergetic (characteristic of the nucleus)
Gamma ray energies are discrete and characteristic of the emitting nucleus, unlike beta particles. This makes gamma spectroscopy a useful tool for identifying radioisotopes.
Example 1: Alpha decay equation
Write the nuclear equation for the alpha decay of Polonium-210 (Po, Z=84).
1 Alpha particle: ⁴₂He. Z decreases by 2 → Z = 84 − 2 = 82 (Lead, Pb). A decreases by 4 → A = 210 − 4 = 206.
2 Write equation: ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He
²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂He (+ energy as kinetic energy of products)
Example 2: Beta-minus decay equation
Carbon-14 undergoes beta-minus decay. Write the full nuclear equation including the antineutrino.
1 β⁻ decay: Z increases by 1 → Z = 6 + 1 = 7 (Nitrogen, N). A unchanged → A = 14.
2 Include antineutrino: ν̄_e
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄_e
Example 3: Identifying decay mode from N and Z
Phosphorus-30 has Z = 15 and N = 15. Stable phosphorus-31 has N = 16. Predict the decay mode of P-30 and write the equation.
1 P-30 has N = 15, Z = 15 — equal N and Z. Stable P-31 has one more neutron (N=16). So P-30 has too few neutrons (too many protons) relative to the stable configuration.
2 Predict: beta-plus decay (or electron capture). For β⁺: Z decreases by 1 → Z = 14 (Silicon, Si). A = 30 unchanged.
³⁰₁₅P → ³⁰₁₄Si + ⁰₊₁e + ν_e
Example 4: Conservation laws in beta decay
Verify that charge, lepton number, and nucleon number are conserved in: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄_e
1 Nucleon number: 14 = 14 + 0 + 0 ✓
2 Charge: 6 = 7 + (−1) + 0 = 6 ✓
3 Lepton number: 0 = 0 + 1 (electron) + (−1) (antineutrino) = 0 ✓ (antineutrino has lepton number −1)
All conservation laws satisfied. The antineutrino is essential to conserve lepton number.
Q1. A nucleus with Z = 90 and N = 142 lies above the N-Z stability curve. Which decay mode would you predict?
Q2. Why do beta particles from a given isotope have a continuous range of energies, while alpha particles are monoenergetic?
Q3. What happens to the atomic number (Z) and mass number (A) in gamma emission?
Q4. In electron capture, what particle is emitted from the nucleus?
Q5. The nuclide ²³⁸₉₂U undergoes alpha decay to Thorium. What is the mass number and atomic number of the Thorium daughter?
Challenge 1. Radon-222 undergoes a decay series: one alpha decay followed by two beta-minus decays. Identify the final nuclide and write out the complete chain of equations.
Start: ²²²₈₆Rn
Alpha decay: ²²²₈₆Rn → ²¹⁸₈₄Po + ⁴₂He
Beta-minus decay 1: ²¹⁸₈₄Po → ²¹⁸₈₅At + ⁰₋₁e + ν̄_e
Beta-minus decay 2: ²¹⁸₈₅At → ²¹⁸₈₆Rn + ⁰₋₁e + ν̄_e
Final nuclide: ²¹⁸₈₆Rn (Radon-218)
Note: each β⁻ decay increases Z by 1, A unchanged.
Challenge 2. Explain why the existence of a continuous beta energy spectrum led physicists to propose the neutrino before any direct detection. What conservation laws demanded a third particle?
In two-body decay (like alpha), all energy and momentum distributions are fixed — products are monoenergetic.
The continuous beta spectrum showed that the electron could carry any energy from 0 to E_max, implying the energy was shared with another particle.
Conservation of energy: the total kinetic energy of electron + undetected particle = Q-value
Conservation of momentum: the recoil of the daughter nucleus was measured and did not match a two-body decay
Conservation of angular momentum (spin): nuclei and electrons have half-integer spin; a two-body decay would violate spin conservation without a third half-spin particle
Pauli proposed the neutrino in 1930; it was detected experimentally in 1956 by Cowan and Reines.
Challenge 3. A nucleus X undergoes beta-plus decay to give ¹³⁶C. Write the equation for the decay and identify nucleus X. Then explain whether X could also have undergone electron capture to reach the same daughter, and write that equation too.
For β⁺ decay: daughter is ¹³₆C (Z=6, A=13). Parent X has Z = 6 + 1 = 7 (Nitrogen), A = 13.
So X = ¹³₇N (Nitrogen-13).
β⁺ equation: ¹³₇N → ¹³₆C + ⁰₊₁e + ν_e
Electron capture to same daughter:
¹³₇N + ⁰₋₁e → ¹³₆C + ν_e
Yes — both processes convert a proton to a neutron (Z: 7→6), producing the same daughter nucleus. β⁺ is favoured when sufficient energy is available (needs minimum 1.022 MeV = 2m_e c²); otherwise electron capture is the only option.