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Radioactive decay is the process by which an unstable atomic nucleus loses energy by emitting radiation. The three main types of nuclear radiation are alpha (α), beta (β), and gamma (γ). This lesson covers the nature, properties, and penetrating power of each type, along with the role of neutrinos and antineutrinos in beta decay. You must be able to write balanced nuclear equations for all decay types. This material is assessed in AQA sections 3.2.1 and 3.8.1.
Spec mapping (AQA 7408): This lesson covers the nature and properties of alpha, beta-minus, beta-plus and gamma radiation, the role of the (anti)neutrino in beta decay, ionising and penetrating power, behaviour in electric and magnetic fields, and the writing of balanced nuclear-decay equations conserving nucleon number A and proton number Z. Mapping is to AQA 7408 sections 3.2.1.2 (stable and unstable nuclei) and 3.8.1.2 (radioactive decay). (Refer to the official AQA specification document for exact wording.)
Synoptic links: (1) Beta decay is mediated by the weak interaction (section 3.2.1.4), and the quark-level transformations d → u + e⁻ + ν̄ₑ and u → d + e⁺ + νₑ are the bridge between this lesson and the particle-physics course. (2) Gamma photons obey the E = hf relation from quantum phenomena (section 3.2.2), so gamma energies in MeV correspond directly to specific frequencies and wavelengths — useful for spectroscopy. (3) Penetrating power and inverse-square behaviour of gamma intensity connect forward to required practical 7 (RP7) and to the radial-field treatments in section 3.7 Fields. The deflection of charged particles in magnetic fields uses the F = BQv result from section 3.7.4 and provides a clean experimental way to distinguish α, β⁻ and β⁺ visually in a cloud chamber.
A nucleus is unstable when the balance of protons and neutrons does not allow the nuclear forces to hold the nucleus together in its lowest energy state. The nucleus can become more stable by emitting particles or electromagnetic radiation. Radioactive decay is:
An alpha particle is a helium-4 nucleus: two protons and two neutrons bound together. It is written as ⁴₂He or ⁴₂α.
| Property | Value |
|---|---|
| Charge | +2e = +3.20 × 10⁻¹⁹ C |
| Mass | 4 u ≈ 6.64 × 10⁻²⁷ kg |
| Speed | Typically 5–10% of the speed of light (~1.5 × 10⁷ m s⁻¹) |
| Ionising ability | Strong — ionises about 10⁴ ion pairs per cm in air |
| Penetrating power | Very low — stopped by a few cm of air or a sheet of paper |
| Deflection in fields | Deflected by electric and magnetic fields (low specific charge, so deflected less than beta) |
In alpha decay, the parent nucleus loses 2 protons and 2 neutrons:
ᴬ_Z X → ᴬ⁻⁴_(Z−2) Y + ⁴₂α
Example — Uranium-238 alpha decay:
²³⁸₉₂U → ²³⁴₉₀Th + ⁴₂α
Check: Nucleon number: 238 = 234 + 4 ✓. Proton number: 92 = 90 + 2 ✓.
Example — Radium-226 alpha decay:
²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α
Check: 226 = 222 + 4 ✓. 88 = 86 + 2 ✓.
Alpha decay is most common in heavy nuclei (A > 200) where the nucleus is too large for the strong nuclear force (which is short-range) to overcome the long-range Coulomb repulsion between all the protons.
Plutonium-239 undergoes alpha decay to uranium-235:
²³⁹₉₄Pu → ²³⁵₉₂U + ⁴₂α
Nuclear masses (in u): m(Pu-239) = 239.05216, m(U-235) = 235.04393, m(α) = 4.00151.
Mass defect: Δm = 239.05216 − (235.04393 + 4.00151) = 0.00672 u.
Energy released: E = 0.00672 × 931.5 = 6.26 MeV.
By conservation of momentum (parent at rest), the alpha and the daughter recoil in opposite directions with equal momenta: p_α = p_U. Their kinetic energies are inversely proportional to their masses:
E_α / E_U = m_U / m_α = 235 / 4 ≈ 58.75.
So E_α = 6.26 × 235/(235+4) ≈ 6.15 MeV, and the U-235 daughter takes the remaining ~0.11 MeV as recoil energy. Note that the daughter recoil is small but non-zero — a 100 keV recoil is enough to displace the U-235 atom from its lattice site in a solid, which contributes to the radiation-damage profile of α emitters in nuclear-fuel cladding.
In beta-minus decay, a neutron in the nucleus transforms into a proton, emitting an electron (the beta-minus particle) and an electron antineutrino (ν̄ₑ).
At the quark level: a down quark changes into an up quark via the weak interaction.
n → p + e⁻ + ν̄ₑ
or in quark terms: d → u + e⁻ + ν̄ₑ
| Property | Value |
|---|---|
| Charge | −1e = −1.60 × 10⁻¹⁹ C |
| Mass | 0.000549 u ≈ 9.11 × 10⁻³¹ kg |
| Speed | Up to ~99% of the speed of light |
| Ionising ability | Moderate — about 100 ion pairs per cm in air |
| Penetrating power | Moderate — stopped by a few mm of aluminium |
| Deflection in fields | Strongly deflected (high specific charge, opposite direction to alpha) |
ᴬ_Z X → ᴬ_(Z+1) Y + ⁰₋₁e + ⁰₀ν̄ₑ
The nucleon number A stays the same (a neutron is replaced by a proton). The proton number Z increases by 1.
Example — Carbon-14 beta-minus decay:
¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ⁰₀ν̄ₑ
Check: A: 14 = 14 + 0 + 0 ✓. Z: 6 = 7 + (−1) + 0 ✓.
Example — Strontium-90 beta-minus decay:
⁹⁰₃₈Sr → ⁹⁰₃₉Y + ⁰₋₁e + ⁰₀ν̄ₑ
When beta decay was first studied, it appeared that energy, momentum, and angular momentum were not conserved. The emitted electrons had a continuous spectrum of energies (from zero up to a maximum), rather than a single fixed energy as expected. In 1930, Wolfgang Pauli proposed that a third, undetected particle was carrying away the missing energy and momentum. Enrico Fermi named this particle the neutrino ("little neutral one"). The neutrino (and its antiparticle, the antineutrino) has:
Exam Tip: The AQA specification requires you to know that the existence of the neutrino was postulated to account for the continuous energy spectrum of beta particles and to conserve energy, momentum, and lepton number. This is a common 2–3 mark question.
In beta-plus decay, a proton in the nucleus transforms into a neutron, emitting a positron (the beta-plus particle, the antiparticle of the electron) and an electron neutrino (νₑ).
At the quark level: an up quark changes into a down quark.
p → n + e⁺ + νₑ
ᴬ_Z X → ᴬ_(Z−1) Y + ⁰₊₁e + ⁰₀νₑ
The nucleon number A stays the same. The proton number Z decreases by 1.
Example — Carbon-11 beta-plus decay:
¹¹₆C → ¹¹₅B + ⁰₊₁e + ⁰₀νₑ
Check: A: 11 = 11 + 0 + 0 ✓. Z: 6 = 5 + 1 + 0 ✓.
Example — Fluorine-18 beta-plus decay (used in PET scans):
¹⁸₉F → ¹⁸₈O + ⁰₊₁e + ⁰₀νₑ
Key Point: Beta-plus decay can only occur inside a nucleus (not for a free proton) because the proton mass is less than the neutron mass — the extra energy comes from the binding energy of the nucleus.
Gamma radiation is high-energy electromagnetic radiation (photons) emitted when a nucleus transitions from an excited state to a lower energy state. It often follows alpha or beta decay, when the daughter nucleus is left in an excited state.
| Property | Value |
|---|---|
| Charge | 0 |
| Mass | 0 |
| Speed | Speed of light, c = 3.00 × 10⁸ m s⁻¹ |
| Ionising ability | Weak — about 1 ion pair per cm in air |
| Penetrating power | Very high — reduced by thick lead or concrete (never fully stopped, intensity follows inverse exponential) |
| Deflection in fields | Not deflected (no charge) |
Gamma emission does not change the proton number or nucleon number — the nucleus just loses energy:
ᴬ_Z X* → ᴬ_Z X + γ
(The asterisk * denotes an excited nuclear state.)
| Property | Alpha (α) | Beta-minus (β⁻) | Beta-plus (β⁺) | Gamma (γ) |
|---|---|---|---|---|
| Identity | ⁴₂He nucleus | Electron (e⁻) | Positron (e⁺) | Photon |
| Charge | +2e | −e | +e | 0 |
| Mass (u) | 4 | 0.000549 | 0.000549 | 0 |
| Typical speed | ~0.05c–0.1c | Up to ~0.99c | Up to ~0.99c | c |
| Ionising power | Strong | Moderate | Moderate | Weak |
| Penetration | Paper / few cm air | Few mm aluminium | Few mm aluminium | Thick lead / concrete |
| Deflected by fields? | Yes (slightly) | Yes (strongly) | Yes (strongly, opposite to β⁻) | No |
| Change in A | −4 | 0 | 0 | 0 |
| Change in Z | −2 | +1 | −1 | 0 |
Exam Tip: When writing nuclear decay equations, always check that both the total nucleon number (A) and the total proton number (Z) are conserved (i.e., the same on both sides of the equation). This is a fundamental conservation law and marks are awarded for showing this check.
A radioactive isotope has nucleon number 241 and proton number 95 (americium-241). It decays to neptunium-237. What type of decay has occurred? Write the nuclear equation.
Change in A: 241 − 237 = 4. Change in Z: 95 − 93 = 2.
Since A decreases by 4 and Z decreases by 2, this is alpha decay.
²⁴¹₉₅Am → ²³⁷₉₃Np + ⁴₂α
The U-238 decay series ends at stable Pb-206 after a sequence of 8 alpha and 6 beta-minus decays. Verify that the changes in A and Z balance.
Initial: ²³⁸₉₂U (A = 238, Z = 92). Final: ²⁰⁶₈₂Pb (A = 206, Z = 82).
Change in A: 238 − 206 = 32. Each α decay reduces A by 4; each β⁻ decay leaves A unchanged. So number of α decays = 32 / 4 = 8 ✓.
Change in Z: 92 − 82 = 10. Eight α decays reduce Z by 16 (each by 2). To finish with Z = 82, we need to increase Z by 6 — and each β⁻ decay increases Z by 1. So number of β⁻ decays = 6 ✓.
The sequence is therefore consistent with the observed 8α + 6β⁻ decay chain. This kind of A and Z bookkeeping is a standard exam-question type and rewards careful conservation-law application.
Consider the β⁺ decay of ¹⁸F to ¹⁸O followed by positron annihilation:
¹⁸₉F → ¹⁸₈O + ⁰₊₁e + νₑ ⁰₊₁e + ⁰₋₁e → 2γ (positron annihilation, 0.511 MeV each)
In the first step, the lepton number of the LHS is 0 (F is a baryon, lepton number 0). On the RHS: the positron has L = −1 and the neutrino has L = +1, summing to 0 ✓.
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