Info: Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4
Explore the simulation above to develop intuition for this topic.
Intuition
Radioactivity is nature’s way of stabilising unstable nuclei: Imagine a crowded room where people are packed too tightly — some will logically move to less crowded areas. Similarly, nuclei with too many or too few neutrons relative to protons are unstable and will spontaneously emit particles to reach a more stable configuration. This emission is radioactivity.
Why it matters: Radioactivity powers the Sun, enables medical imaging and cancer treatment, provides energy through nuclear power, and allows carbon dating of ancient artefacts. Understanding it explains why some materials are dangerous and others are harmless.
The key insight: Radioactive decay is fundamentally random — you cannot predict when a specific nucleus will decay, only the probability that it will decay within a given time. This is why we use half-life (the time for half the nuclei to decay) rather than exact decay times. The randomness is not due to our ignorance — it’s a fundamental property of quantum mechanics.
Definition. Radioactivity is the spontaneous emission of radiation from unstable atomic nuclei.
1. Nuclear Structure
The atom consists of a nucleus containing protons and neutrons (collectively called nucleons), surrounded by electrons.
Property
Proton
Neutron
Electron
Charge
+e
0
−e
Mass (u)
1.00728
1.00867
0.00055
Location
Nucleus
Nucleus
Electron shells
Notation. A nuclide is written as \prescriptAZX where A is the mass number (nucleon number) and Z is the atomic number (proton number). The neutron number is N=A−Z.
Isotopes are atoms of the same element (same Z) with different mass numbers (different N).
2. Types of Radioactive Decay
Alpha Decay (α)
Definition. An alpha particle (α) is a helium nucleus (\prescript42He), Emitted during alpha decay.
\prescriptAZX→\prescriptA−4Z−2Y+\prescript42α
Conservation checks:
Mass number: A=(A−4)+4✓
Atomic number: Z=(Z−2)+2✓
Alpha particles are highly ionising but poorly penetrating (stopped by paper or a few cm of air).
Beta Decay (β−)
Definition. A beta particle (β−) is an electron emitted when a neutron converts to a Proton in beta-minus decay.
A neutron converts to a proton, emitting an electron and an antineutrino:
The antineutrino was postulated by Fermi (1934) to conserve energy and momentum — the observed Electron energy spectrum is continuous, implying a third particle carries away the remaining energy.
Penetration: Beta particles are moderately ionising, stopped by a few mm of aluminium.
Beta-Plus Decay (β+)
Definition. A beta-plus particle (β+) is a positron emitted when a proton converts to a Neutron in beta-plus decay.
A proton converts to a neutron, emitting a positron and a neutrino:
\prescriptAZX→\prescriptAZ−1Y+\prescript0+1β++νe
This only occurs in proton-rich nuclei.
Gamma Decay (γ)
Definition. A gamma ray (γ) is high-energy electromagnetic radiation emitted during Nuclear transitions.
After alpha or beta decay, the daughter nucleus is often in an excited state. It de-excites by Emitting a gamma ray (high-energy photon):
\prescriptAZX∗→\prescriptAZX+γ
No change in A or Z. Gamma rays are weakly ionising but highly penetrating (stopped by thick Lead or concrete).
Definition. The decay constant λ is the probability per unit time of a nucleus decaying.
λ=t1/2ln2
The probability that any single nucleus decays in a small time interval dt is λdt.
If there are N nuclei, the number decaying in dt is:
dN=−λNdt
(The minus sign indicates N decreases.)
dtdN=−λN
Separating variables and integrating:
∫N0NNdN=−∫0tλdt
lnN−lnN0=−λt
ln(N0N)=−λt
N=N0e−λt
Activity
Definition. Activity A is the rate of decay of a radioactive sample: A=λN=−dtdN.
A=λN=−dtdN
Expanding using the exponential decay law:
A=λN0e−λt=A0e−λt
Definition. The becquerel is the SI unit of activity; 1 Bq = 1 decay per second.
1Bq=1decays−1
Half-Life
Definition. The half-life t1/2 is the time taken for half of the radioactive nuclei in a Sample to decay.
N(t)=N0⋅2−t/t1/2
Setting N=N0/2:
2N0=N0e−λt1/2
21=e−λt1/2
ln2=λt1/2
t1/2=λln2
Intuition. After one half-life, half remain. After two, a quarter. After n half-lives, N=N0/2n. The decay is exponential — the activity is always proportional to the number of Remaining nuclei, so it decreases exponentially.
Radiation Detection
Scintillation Detector
A scintillation detector uses a crystal (e.g. Sodium iodide) that emits flashes of light when Radiation passes through it. The light pulses are converted to electrical signals by a Photomultiplier tube. Scintillation detectors are more sensitive than GM tubes and can distinguish Between different types of radiation by the intensity of the light pulse (pulse-height analysis), Making them useful for identifying specific isotopes.
Half-Life Measurement Techniques
Graphical Method
To determine half-life experimentally:
Measure the corrected count rate (subtracting background) at regular time intervals.
Plot a graph of corrected count rate (or activity) against time.
Find the time for the count rate to halve. This is t1/2.
Alternatively, plot lnA vs t; the gradient equals −λ.
The logarithmic method is more accurate because it uses all data points, not just a single halving:
lnA=lnA0−λt
The gradient of the straight line gives −λAnd then t1/2=ln2/λ.
Practical Considerations
For short half-lives (seconds to minutes), readings can be taken in real time with a GM tube and data logger.
For long half-lives (years to millennia), direct measurement is impractical. Instead, measure a known mass of the isotope and use A=λN where N is calculated from the mass and Avogadro’s number.
Problem Set
Problem 1Radium-226 undergoes alpha decay. Write the nuclear equation and identify the daughter nucleus.
Problem 4A sample contains 4.0×1012 nuclei of a radioactive isotope with decay constant 1.2×10−7 s−1. Calculate the initial activity and the number of nuclei remaining After 2.0 hours.
Answer.A0=λN0=1.2×10−7×4.0×1012=4.8×105 Bq.
t=7200 s. N=N0e−λt=4.0×1012×e−1.2×10−7×7200=4.0×1012×e−0.864=4.0×1012×0.421=1.68×1012.
Problem 7Explain why the antineutrino was proposed in beta decay.
Answer. In beta decay, the emitted electron has a continuous energy spectrum rather than a Single energy. This appeared to violate conservation of energy. Pauli proposed (1930) and Fermi Formalised (1934) the existence of an unseen particle (the antineutrino) that carries away the Remaining energy and momentum. The total energy of electron + antineutrino equals the fixed energy Released by the nuclear transition, conserving energy. The antineutrino was experimentally detected In 1956 by Cowan and Reines.
Problem 9Compare and contrast alpha, beta, and gamma radiation in terms of: (a) nature, (b) ionising power, (c) penetrating power, (d) deflection in electric and magnetic fields.
Problem 10A radioactive isotope X with half-life 12 hours decays to a stable daughter Y. A sample initially Contains 1000 atoms of X. How many atoms of Y are present after 36 hours?
Answer. After 36 hours: n=36/12=3 half-lives. NX=1000/23=125. Atoms of Y = 1000−125=875.
Info: Board Coverage AQA Paper 2 | Edexcel CP3 | OCR (A) Paper 2 | CIE P4 Definition. A decay chain (or radioactive series) is a sequence of radioactive decays where each Daughter nuclide is itself radioactive, continuing until a stable nuclide is reached.
The Uranium-238 Decay Series
The most important natural decay chain begins with uranium-238 and ends at stable lead-206 after 14 Decays (8 alpha and 6 beta-minus):
Verification of net change: 8 alpha decays reduce A by 8×4=32 and Z by 8×2=16. 6 beta-minus decays increase Z by 6×1=6 and leave A unchanged. Net: A decreases by 32 (238−206=32✓) and Z decreases by 10 (92−82=10✓).
Other natural decay chains include:
Thorium-232 series→208Pb (6 alpha, 4 beta-minus)
Uranium-235 series→207Pb (7 alpha, 4 beta-minus)
Secular Equilibrium
When a long-lived parent nuclide (very large t1/2) decays through a chain of shorter-lived Daughters, a state of secular equilibrium is established after several daughter half-lives. In Equilibrium:
λ1N1=λ2N2=λ3N3=⋯
Each daughter has the same activity as the parent. This has practical importance in radiological Protection: the activity of radon-222 (a daughter of 238U) in buildings is sustained By the essentially constant uranium in the ground beneath.
5. Applications of Radioactivity
Carbon-14 Dating
6. Case Studies
Chernobyl (1986)
The Chernobyl disaster released approximately 5.2×1018 Bq of radioactive material into The atmosphere. The most significant isotopes released were:
Iodine-131 (t1/2=8.04 days): Caused thyroid cancer, particularly in children who drank milk from cows that had grazed on contaminated grass. The short half-life meant the acute danger passed within months.
Caesium-137 (t1/2=30.2 years): More persistent contaminant. Caesium behaves chemically like potassium and was taken up by plants and animals, entering the food chain. The 30 km exclusion zone around the reactor remains in place partly because of 137Cs contamination.
The different half-lives illustrate a key principle: short-lived isotopes produce intense but brief Radiation hazards, while long-lived isotopes create lower but persistent contamination.
Fukushima Daiichi (2011)
The Fukushima disaster, triggered by a tsunami following the Tohoku earthquake, released significant Quantities of radioactive isotopes including 131I, 134Cs (t1/2=2.06 years), and 137Cs.
Key differences from Chernobyl:
Most radioactive material was released into the Pacific Ocean rather than the atmosphere, leading to rapid dilution of water-soluble isotopes.
134Cs has a short half-life (2.06 years) and decayed rapidly; 137Cs remains the primary long-term concern.
The disaster highlighted the importance of cooling systems for spent fuel and the vulnerability of nuclear plants to extreme natural events.
Both disasters demonstrate that understanding half-lives is critical for predicting contamination Timelines and informing public health responses.
Additional Problems
Problem 11A sample of ancient wood has a 14C activity of 1.5 Bq per gram. A living sample of the Same type of wood has an activity of 12.5 Bq per gram. Calculate the age of the ancient wood. (t1/2 of 14C=5730 years.)
Answer.A/A0=1.5/12.5=0.12. ln0.12=−λt. λ=ln2/5730=1.209×10−4 year−1. t=−ln0.12/(1.209×10−4)=2.12/(1.209×10−4)=17,500 years (2 s.f.).
Problem 12In the uranium-238 decay series, verify that 8 alpha decays and 6 beta-minus decays are consistent With the transformation from 238U to 206Pb.
Answer. After 8 alpha decays: A=238−8×4=206, Z=92−8×2=76. After 6 Beta-minus decays: A unchanged, Z=76+6=82. Final nuclide: A=206, Z=82Which is 206Pb✓.
Problem 13A GM tube records a count rate of 340 counts per minute from a radioactive source. The background Count rate is 25 counts per minute. After 2.5 hours, the count rate (including background) has Fallen to 105 counts per minute. Calculate the half-life of the source.
Problem 14A patient is administered 99mTc with a physical half-life of 6.0 hours for a Diagnostic scan. The biological half-life of 99mTc in the body is 4.8 hours. Calculate The effective half-life and the fraction of the initial activity remaining after 12 hours.
Problem 15Explain why a sample of radon-222 collected from the soil beneath a building continues to produce a Detectable activity even after the original radon-222 has been removed, referring to the concept of Secular equilibrium.
Answer. Radon-222 is part of the 238U decay chain. Its parent, radium-226, has a Half-life of 1600 years — much longer than radon-222’s half-life of 3.82 days. In secular Equilibrium, radon-222 is produced by radium-226 at the same rate it decays, so its activity remains Constant. Removing the radon-222 temporarily reduces the activity, but new radon-222 atoms are Continuously produced by the long-lived parent, causing the radon activity to rebuild towards its Equilibrium value over several radon half-lives (approximately 20-40 days).