Radioactive Decay

Certain nuclei are unstable. They have too many protons, too many neutrons, or simply too much energy, and they cannot remain in their current state indefinitely. To move towards stability, these nuclei undergo radioactive decay — a spontaneous process in which the nucleus emits radiation and transforms into a different nuclear configuration.

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Types of Radioactive Decay

There are four principal types of radioactive decay you need to know for Edexcel A-Level Physics. Each involves a different emission and a different change to the nucleus.

Alpha (α) Decay

An alpha particle is a helium-4 nucleus: two protons and two neutrons bound together (⁴₂He). When a heavy, proton-rich nucleus undergoes alpha decay, it ejects an alpha particle.

Changes to the nucleus:

• Nucleon number A decreases by 4
• Proton number Z decreases by 2

General equation: ᴬ_Z X → ᴬ⁻⁴_(Z-2) Y + ⁴₂α

Example: ²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂α

Alpha decay typically occurs in heavy nuclei (Z > 82) where the nucleus is too large for the strong nuclear force to hold it together against electromagnetic repulsion.

Why alpha particles? The helium-4 nucleus has an exceptionally high binding energy per nucleon (7.08 MeV), making it an especially stable cluster. This is why the nucleus preferentially ejects an α particle rather than individual nucleons or other clusters — the process releases more energy.

Beta-Minus (β⁻) Decay

In beta-minus decay, a neutron inside the nucleus transforms into a proton. The nucleus emits a high-speed electron (the beta-minus particle) and an antineutrino.

Changes to the nucleus:

• Nucleon number A stays the same
• Proton number Z increases by 1

General equation: ᴬ_Z X → ᴬ_(Z+1) Y + ⁰₋₁β + ν̄ₑ

At the quark level: A down quark in the neutron changes to an up quark: d → u + e⁻ + ν̄ₑ

Example: ¹⁴₆C → ¹⁴₇N + ⁰₋₁β + ν̄ₑ

Beta-minus decay occurs in nuclei with too many neutrons relative to protons (neutron-rich nuclei).

Beta-Plus (β⁺) Decay

In beta-plus decay, a proton transforms into a neutron. The nucleus emits a positron (the antiparticle of the electron) and an electron neutrino.

Changes to the nucleus:

• Nucleon number A stays the same
• Proton number Z decreases by 1

General equation: ᴬ_Z X → ᴬ_(Z-1) Y + ⁰₊₁β + νₑ

Example: ²²₁₁Na → ²²₁₀Ne + ⁰₊₁β + νₑ

Beta-plus decay occurs in proton-rich nuclei. It is less common than beta-minus decay in naturally occurring isotopes but is important in medical PET scanning (the positrons annihilate with electrons to produce detectable gamma rays).

Gamma (γ) Emission

Gamma rays are high-energy photons emitted when a nucleus transitions from an excited energy state to a lower energy state. Gamma emission does not change A or Z — the nucleus remains the same isotope but loses energy.

Gamma emission often follows alpha or beta decay: the daughter nucleus is produced in an excited state and then de-excites by emitting one or more gamma photons.

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Properties of the Radiations

Property	Alpha (α)	Beta-minus (β⁻)	Beta-plus (β⁺)	Gamma (γ)
Nature	⁴₂He nucleus	Electron	Positron	Photon
Charge	+2e	−e	+e	0
Mass	~4 u	~1/1836 u	~1/1836 u	0
Speed	~5% of c	Up to ~99% of c	Up to ~99% of c	c
Ionising power	Very high (~10⁴ ion pairs per cm in air)	Moderate (~10² ion pairs per cm)	Moderate	Low (~1 ion pair per cm)
Penetrating power	Stopped by paper / few cm of air	Stopped by ~3 mm aluminium	Annihilates with electron	Reduced by thick lead / several cm of concrete
Deflection in E/B fields	Deflected (low charge-to-mass ratio)	Deflected opposite to α (high charge-to-mass ratio)	Deflected same direction as α	Not deflected

The key trade-off: the more ionising a radiation is, the less penetrating it is. Alpha particles interact strongly with matter, depositing their energy quickly and stopping within a few centimetres of air. Gamma rays interact weakly, passing through many centimetres of material before being absorbed.

Identifying Radiation in the Lab

A practical method for identifying an unknown source:

1. Paper test: Place paper between source and detector. If count rate drops to background, the source emits alpha only.
2. Aluminium test: If alpha is ruled out, insert 3–5 mm aluminium sheet. If count rate drops to background, the source emits beta.
3. Lead test: If the count rate persists through aluminium, the source emits gamma. A thick lead sheet (several cm) will significantly reduce the count rate.
4. Magnetic field test: Alpha and beta are deflected in opposite directions; gamma is unaffected.

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Decay Chains

Heavy nuclei often undergo a series of decays before reaching a stable configuration. Each decay produces a different daughter nuclide, which may itself be unstable.

Example: Part of the uranium-238 decay chain:

²³⁸U →(α) ²³⁴Th →(β⁻) ²³⁴Pa →(β⁻) ²³⁴U →(α) ²³⁰Th → ... → ²⁰⁶Pb (stable)

The complete U-238 chain involves 8 alpha decays and 6 beta-minus decays before reaching the stable end-product lead-206. You can verify this:

• Total change in A: 238 − 206 = 32 = 8 × 4 (8 alpha decays, each reducing A by 4)
• Total change in Z: 92 − 82 = 10, but 8 alpha decays reduce Z by 16, so 6 beta decays increase Z by 6, giving net change = 16 − 6 = 10 ✔

graph TD
    A["U-238 (Z=92)"] -->|"α decay"| B["Th-234 (Z=90)"]
    B -->|"β⁻ decay"| C["Pa-234 (Z=91)"]
    C -->|"β⁻ decay"| D["U-234 (Z=92)"]
    D -->|"α decay"| E["Th-230 (Z=90)"]
    E -->|"α decay"| F["Ra-226 (Z=88)"]
    F -->|"α decay"| G["Rn-222 (Z=86)"]
    G -->|"... further decays ..."| H["Pb-206 (Z=82, stable)"]

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The Random Nature of Radioactive Decay

Radioactive decay is a random and spontaneous process:

• Random: It is impossible to predict which particular nucleus will decay next, or exactly when any given nucleus will decay. Each unstable nucleus has a fixed probability of decaying in a given time interval, but the actual moment of decay is governed by quantum mechanics and cannot be predicted.
• Spontaneous: The decay is not triggered by external conditions. It is not affected by temperature, pressure, chemical bonding, or any other physical or chemical condition. The process originates within the nucleus itself.

When dealing with large numbers of nuclei (which is always the case in practice — even a tiny sample contains billions of billions of atoms), the statistical behaviour becomes very predictable. We can say with confidence how many nuclei will decay per second on average, even though we cannot say which ones.

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Background Radiation

Background radiation is the low level of ionising radiation that exists everywhere in the environment. It comes from both natural and artificial sources:

Natural sources (~85% of total):

• Radon gas from rocks and soil (largest single source in the UK, ~42%)
• Cosmic rays from space (~12%)
• Rocks and buildings containing radioactive minerals (~12%)
• Food and drink containing naturally radioactive isotopes like K-40 and C-14 (~12%)
• Internal radiation from radioactive atoms within the body (~12%)

Artificial sources (~15% of total):

• Medical procedures (X-rays, CT scans, nuclear medicine)
• Nuclear weapons testing fallout (very small residual)
• Industrial uses
• Nuclear power (tiny contribution under normal operation)

In any experiment involving radioactive sources, the background count rate must be measured and subtracted from the observed count rate to obtain the corrected count rate due to the source alone.

Worked Example: Background Correction

A Geiger counter records 25 counts per minute over 10 minutes with no source present.

Background rate = 25 counts/min.

With a source present, it records 345 counts per minute.

Corrected count rate = 345 − 25 = 320 counts per minute due to the source.

Common exam mistake: forgetting to subtract background, or subtracting it incorrectly (e.g., subtracting the total background count rather than the background rate).

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Inverse Square Law for Gamma Radiation

Gamma radiation obeys an inverse square law because it is emitted uniformly in all directions from a point source:

I = k / r²

where I is intensity, r is distance from source, and k is a constant. This means doubling the distance from a gamma source reduces the intensity to one quarter. Alpha and beta do not follow this law because they are absorbed so quickly that their range is limited.

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Common Exam Mistakes

• Saying the electron in β⁻ decay comes from an electron shell. The electron is created during the decay process itself (a neutron transforms into a proton + electron + antineutrino). It does not come from the atom’s electron cloud.
• Forgetting the neutrino/antineutrino in beta decay equations. The neutrino was postulated by Pauli precisely to conserve energy, momentum, and lepton number.
• Writing that γ rays have mass. Gamma photons have zero rest mass.
• Confusing ionising power with penetrating power. They are inversely related: higher ionising power means lower penetrating power.

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Summary

Radioactive decay is the spontaneous emission of radiation from an unstable nucleus. Alpha decay reduces A by 4 and Z by 2. Beta-minus decay converts a neutron to a proton (Z increases by 1). Beta-plus decay converts a proton to a neutron (Z decreases by 1). Gamma emission releases energy without changing A or Z. Each radiation type has characteristic ionising power, penetration, and behaviour in fields. Decay is random and spontaneous, unaffected by external conditions, and background radiation must always be accounted for in measurements. Heavy nuclei undergo decay chains involving multiple alpha and beta decays before reaching stability.

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A-Level Deep Dive: Radioactive Decay

Spec mapping

Edexcel 9PH0 specification Topic 11 — Nuclear and Particle Physics covers the random and spontaneous nature of radioactive decay; the four principal modes (α, β⁻, β⁺, γ); the use of the activity equation $A = \lambda N$A=λN and the exponential decay laws $N = N_0 e^{-\lambda t}$N=N0​e−λt and $A = A_0 e^{-\lambda t}$A=A0​e−λt; the relationship $\lambda t_{1/2} = \ln 2$λt1/2​=ln2; and applications such as radio-isotope dating and medical tracers (refer to the official specification document for exact wording). Although the topic sits in Paper 2 alongside Topic 12 (Gravitational Fields) and Topic 13 (Oscillations), it is heavily synoptic: examiners routinely link it to capacitor discharge (Topic 7), exponential modelling (mathematics), and inverse-square behaviour (gravitational and electric fields). The Edexcel formula booklet does list $N = N_0 e^{-\lambda t}$N=N0​e−λt and $\lambda = \ln 2 / t_{1/2}$λ=ln2/t1/2​, but candidates are still expected to manipulate them under logarithm operations confidently.

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Worked example with full mark scheme

Question (8 marks):

A wooden artefact recovered from an archaeological dig contains carbon. A reference sample of living wood, of equal mass, has a measured carbon-14 activity of $2.30 \times 10^{-1}$2.30×10−1 Bq. The artefact records an activity of $5.74 \times 10^{-2}$5.74×10−2 Bq, after background subtraction. Carbon-14 has a half-life of $5730$5730 years.

(a) Calculate the decay constant $\lambda$λ for carbon-14, giving your answer in $\text{year}^{-1}$year−1. (2)

(b) Estimate the age of the artefact. (4)

(c) State two assumptions you have made in your dating calculation. (2)

Solution with mark scheme:

(a) Step 1 — apply the half-life relation.

$\lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.6931}{5730 \text{ yr}}$λ=t1/2​ln2​=5730 yr0.6931​

M1 — correct rearrangement of $\lambda t_{1/2} = \ln 2$λt1/2​=ln2, with consistent units. A common slip is to mix seconds and years; here the question explicitly asks for $\text{year}^{-1}$year−1, so the half-life is left in years.

$\lambda = 1.21 \times 10^{-4} \text{ yr}^{-1}$λ=1.21×10−4 yr−1

A1 — correct value to 3 s.f.

(b) Step 1 — set up the decay equation in terms of activities.

Because activity is proportional to the number of undecayed nuclei ($A = \lambda N$A=λN), the same exponential law applies:

$A = A_0 e^{-\lambda t} \implies \dfrac{A}{A_0} = e^{-\lambda t}$A=A0​e−λt⟹A0​A​=e−λt

M1 — recognising that the activity ratio gives the exponential factor; equivalently, identifying $A_0$A0​ as the living-wood reference.

Step 2 — substitute and take logarithms.

$\dfrac{5.74 \times 10^{-2}}{2.30 \times 10^{-1}} = 0.2496$2.30×10−15.74×10−2​=0.2496

$\ln(0.2496) = -\lambda t \implies -1.388 = -(1.21 \times 10^{-4}) t$ln(0.2496)=−λt⟹−1.388=−(1.21×10−4)t

M1 — correct application of the natural logarithm to both sides; sign handled correctly.

Step 3 — solve for $t$t.

$t = \dfrac{1.388}{1.21 \times 10^{-4}} = 1.15 \times 10^{4} \text{ years}$t=1.21×10−41.388​=1.15×104 years

A1 — final value to 3 s.f., correct unit. Approximately $11{,}500$11,500 years.

(c) Any two of:

• The C-14:C-12 ratio in the atmosphere has remained constant over the artefact's lifetime. B1
• The artefact stopped exchanging carbon with its environment at the moment of "death" (i.e. felling/use), so no later contamination or replacement of carbon has occurred. B1
• The half-life of carbon-14 is precisely known and the decay constant has not changed.

Total: 8 marks (M3 A2 B2 with one M-A linked).

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Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A radioactive source contains $8.0 \times 10^{14}$8.0×1014 atoms of an isotope with decay constant $\lambda = 4.0 \times 10^{-3}$λ=4.0×10−3 s$^{-1}$−1.

(a) Calculate the initial activity of the source. (2)

(b) Calculate the activity after $5.0$5.0 minutes, and hence the number of decays that have occurred during this interval (to a reasonable approximation). (4)

Mark scheme decomposition by AO:

(a)

• M1 (AO2) — applying $A = \lambda N$A=λN: $A_0 = (4.0 \times 10^{-3})(8.0 \times 10^{14})$A0​=(4.0×10−3)(8.0×1014).
• A1 (AO1) — $A_0 = 3.2 \times 10^{12}$A0​=3.2×1012 Bq.

(b)

• M1 (AO2) — converting time: $t = 300$t=300 s, computing $\lambda t = 1.2$λt=1.2.
• M1 (AO1) — applying $A = A_0 e^{-\lambda t} = (3.2 \times 10^{12}) e^{-1.2}$A=A0​e−λt=(3.2×1012)e−1.2.
• A1 (AO1) — $A \approx 9.6 \times 10^{11}$A≈9.6×1011 Bq.
• A1 (AO3) — number of decays $\approx N_0 - N \approx N_0(1 - e^{-1.2}) = 8.0 \times 10^{14} \times (1 - 0.301) \approx 5.6 \times 10^{14}$≈N0​−N≈N0​(1−e−1.2)=8.0×1014×(1−0.301)≈5.6×1014 decays. (Equivalent to integrating activity over time.)

Total: 6 marks split AO1 = 3, AO2 = 2, AO3 = 1. This is a classic Paper 2 structure: a procedural opening (AO1), a multi-step substitution (AO2), and a synoptic closer asking the candidate to convert between activity, number, and a derived total.

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Synoptic links

Connects to:

1. Topic 11 itself — alpha, beta-minus, beta-plus and gamma processes: the exponential decay law applies regardless of the specific decay mode. Whether a nucleus emits an α particle (Z drops by 2, A drops by 4), a β⁻ particle (a down quark converts to an up quark, releasing an electron and antineutrino), or a γ photon (no change to A or Z), the underlying probability per nucleus per unit time $\lambda$λ governs the population.

2. Topic 7 — capacitor discharge: the equation $Q = Q_0 e^{-t/RC}$Q=Q0​e−t/RC has identical mathematical structure to $N = N_0 e^{-\lambda t}$N=N0​e−λt, with the time constant $\tau = RC$τ=RC playing the role of $1/\lambda$1/λ. Both arise from a "rate proportional to amount" differential equation: $dQ/dt = -Q/RC$dQ/dt=−Q/RC and $dN/dt = -\lambda N$dN/dt=−λN. Examiners explicitly reward candidates who articulate this analogy.

3. Mathematics — exponential and logarithmic functions: manipulating $N = N_0 e^{-\lambda t}$N=N0​e−λt requires confident use of $\ln(e^x) = x$ln(ex)=x, the constant-ratio property of exponentials, and conversion between forms $e^{-\lambda t}$e−λt and $2^{-t/t_{1/2}}$2−t/t1/2​. The relation $\lambda t_{1/2} = \ln 2$λt1/2​=ln2 is the bridge.