How do different isotopes exhibit varying modes and rates of radioactive decay, and what factors influence their stability?
Explore the forces inside the nucleus, the valley of stability, and the remarkable ways atoms transform themselves - with interactive 3D visualisations and real-world applications.
Learning Objectives
Section
Learning Objective
8.1
Explain why most isotopes are unstable
8.2
Compare the strong nuclear force and Coulomb repulsion
8.3
Use the valley of stability to predict decay modes
8.4
Identify the four factors influencing nuclear stability and apply the SEMF
8.5
Describe alpha, beta, and gamma decay mechanisms
8.6
Trace the U-238 decay chain and explain secular equilibrium
8.7
Calculate half-life, activity, and apply radioactive dating
8.8
Evaluate applications of radioactive decay in medicine, dating, and energy
8.1 Why Most Nuclei Are Unstable
Of the approximately 3,300 nuclides known to exist in nature or created in laboratories (as of 2025), only about 279 are truly stable. Every other nucleus is on a slow - or not so slow - journey toward a more stable configuration. Understanding why some nuclei sit peacefully in matter while others spontaneously transform is one of the central problems of nuclear physics.
A nucleus is held together by the strong nuclear force, an attractive force that acts between all nucleons - protons and neutrons alike. At distances of about 1 to 3 femtometres (where 1 fm = 10−15 m), this force is roughly 100 times stronger than the electromagnetic repulsion between two individual nucleons at a distance of 1 fm. However, Coulomb repulsion is infinite-range and scales as Z(Z−1), meaning the total electromagnetic repulsion across all protons in a heavy nucleus can exceed the binding capacity of the strong force. The strong force has a critical limitation: beyond about 3 fm, its influence essentially vanishes.
This tension - short-range attraction versus long-range repulsion - is the root cause of nuclear instability. For a nucleus to be stable, the binding provided by the strong force must outweigh the disruptive Coulomb repulsion. The balance depends on three numbers: the proton count (Z), the neutron count (N), and the total nucleon count, or mass number ($A = Z + N$). When this balance tips too far in either direction, the nucleus decays.
The consequences of nuclear instability extend far beyond the physics laboratory. Radioactive decay provides the clock for radiometric dating, the radiation for cancer therapy, the signatures for medical imaging, and the energy that powers nuclear reactors. Understanding why nuclei decay, how they decay, and at what rate is essential for all of these applications.
8.2 The Forces Inside the Nucleus
The Strong Nuclear Force
The strong nuclear force is one of the four fundamental forces of nature. Unlike electromagnetism, which only acts between charged particles, the strong force acts equally between all nucleon pairs: proton-proton, neutron-neutron, and proton-neutron. This charge independence explains why neutrons, despite having no electric charge, play such a vital role in nuclear stability. A nucleus made only of protons would tear itself apart under Coulomb repulsion; neutrons provide extra binding through the strong force without adding to the repulsion.
At very short distances - below about 0.5 fm - the strong force actually becomes repulsive, creating a "hard core" that prevents nucleons from collapsing into one another. This repulsive core ensures that nuclei have well-defined sizes and do not implode.
Attraction Versus Repulsion
In any nucleus with more than one proton, two competing interactions determine whether the nucleus is stable. The strong nuclear force pulls all nucleons together, while the Coulomb force pushes protons apart. For light nuclei, where protons are packed closely together, the strong force easily dominates. This is why stable light nuclei tend to have roughly equal numbers of protons and neutrons.
As the proton count increases, however, the Coulomb repulsion grows dramatically - it scales roughly as Z(Z−1), meaning the total repulsive force approximately quadruples each time the atomic number doubles. To compensate, more neutrons must be added. The additional neutrons contribute binding energy through the strong force without adding to the Coulomb repulsion. Consequently, heavy stable nuclei require significantly more neutrons than protons. For example, lead-208 contains 82 protons and 126 neutrons, giving an N/Z ratio of about 1.54. Beyond bismuth (Z = 83), no amount of neutron excess can stabilise a nucleus, and all isotopes are radioactive.
Property
Strong Nuclear Force
Electromagnetic Force
Relative strength at 1 fm
~100× stronger than EM
Weaker but long-range
Range
~1–3 fm (very short)
Infinite (falls as 1/r²)
Acts on
All nucleons (p-p, n-n, p-n)
Only charged particles
Effect on stability
Binds nucleons together
Pushes protons apart
Importance in light nuclei
Dominant
Minor
Importance in heavy nuclei
Limited by short range
Becomes increasingly significant
8.3 The Valley of Stability
When every known stable isotope is plotted on a graph of neutron number (N) against proton number (Z), the points form a distinct curved band called the valley of stability. Isotopes lying within this band are generally stable. Those lying outside it are radioactive, and their position relative to the valley predicts how they will decay.
Figure 8.1 - Chart of Nuclides
Interactive Chart
Z = , N = , A =
Main Decay Mode
α decay
EC + β⁺
β⁻
p (proton)
n (neutron)
EC
SF
Stable
Click a nuclide to view details
Drag to pan the chart
Wheel to zoom in/out
Double-click to reset view
-
Click on a nuclide in the chart to view its data.
Figure 8.1 — Interactive chart of nuclides showing the valley of stability with colour-coded decay modes. Click any nuclide to view its half-life and decay properties.
Explore the valley of stability. The diagonal band of stable isotopes (white) is the valley centre. Nuclides above the valley tend toward β⁻ decay; those below tend toward β⁺ or EC. Heavy nuclides (Z > 83) predominantly undergo α decay.
For light elements with Z ≤ 20, the valley follows approximately N = Z. Carbon-12 (6p, 6n), oxygen-16 (8p, 8n), and calcium-40 (20p, 20n) all sit exactly on this line and are exceptionally stable. As Z increases, the valley gradually curves toward the neutron-rich side. By Z = 82 (lead), the stable isotope ²⁰⁸Pb requires 126 neutrons - more than 1.5 neutrons for every proton. No stable isotopes exist beyond bismuth-209 (Z = 83), although some radioactive isotopes such as uranium-238 have half-lives comparable to the age of the Earth.
Predicting Decay from Position
The position of an isotope on the N-Z diagram is a remarkably good predictor of its decay mode:
Position
Excess
Decay mode
Effect on N and Z
Above the valley
Too many neutrons (N/Z too high)
β⁻ decay
N → N−1, Z → Z+1
Below the valley
Too many protons (N/Z too low)
β⁺ decay or electron capture
N → N+1, Z → Z−1
Far below, heavy nuclei
Excess protons + large Z
α decay
N → N−2, Z → Z−2
Any direction, excited state
Excess energy only
γ emission
No change in N or Z
8.4 What Makes a Nucleus Stable?
The N/Z Ratio
The neutron-to-proton ratio is the single most important factor governing nuclear stability. For elements up to Z = 20, the optimal ratio is approximately 1.0. As Z increases, the optimal ratio rises: to about 1.25 for elements in the Z = 21–40 range, and to roughly 1.5 for the heaviest stable elements near lead. This trend is a direct consequence of the Coulomb problem: extra neutrons provide strong-force binding without electromagnetic repulsion. However, this strategy has a limit. Beyond lead-208, the Coulomb repulsion becomes so great that no number of additional neutrons can produce a stable nucleus.
Even-Odd Effects
A striking empirical pattern emerges when stable isotopes are sorted by whether they contain even or odd numbers of protons and neutrons. Nuclei with even numbers of both (even-even nuclei) are far more common than any other type. Only four stable isotopes have odd numbers of both protons and neutrons: ¹H, ⁶Li, ¹⁰B, and ¹⁴N.
Protons (Z)
Neutrons (N)
Stable isotopes
Relative stability
Even
Even
165
Most stable
Even
Odd
57
Moderately stable
Odd
Even
50
Less stable
Odd
Odd
4 (only ¹H, ⁶Li, ¹⁰B, ¹⁴N)
Least stable
This pattern arises from pairing energy: two protons or two neutrons can occupy the same quantum state with opposite spins, lowering their combined energy. The semi-empirical mass formula incorporates this effect through the pairing term, which adds extra binding energy for even-even configurations and reduces it for odd-odd configurations.
Magic Numbers and the Shell Model
Certain specific numbers of protons or neutrons - 2, 8, 20, 28, 50, 82, and 126 - produce nuclei of exceptional stability. These magic numbers were first identified from systematic trends: elements with magic proton numbers have more stable isotopes than their neighbours, and nuclei with magic neutron numbers have more stable isotones (nuclei sharing the same N). Nuclei with both proton and neutron counts at magic values - called doubly magic - are extraordinarily stable. Examples include ⁴He (2, 2), ¹⁶O (8, 8), ⁴⁰Ca (20, 20), and ²⁰⁸Pb (82, 126).
The magic numbers are explained by the nuclear shell model, developed by Maria Goeppert Mayer and others in 1949. In this model, each nucleon moves in an average potential created by all the other nucleons, occupying discrete quantum energy levels much like electrons in an atom. When a shell is completely filled - that is, when the proton or neutron count equals a magic number - the nucleus gains exceptional stability, analogous to the chemical inertness of noble gases.
The shell model explains many observations that the simpler liquid drop model cannot: the existence of magic numbers, the spins and parities of nuclear ground states, the energies of excited states, and the enhanced stability of doubly magic nuclei. Tin (Z = 50), for instance, has ten stable isotopes - more than any other element - because 50 is a magic proton number. Its neighbours indium (Z = 49) and antimony (Z = 51) have only two stable isotopes each.
Binding Energy and the Semi-Empirical Mass Formula
The binding energy of a nucleus is the energy required to separate it into its constituent nucleons. It corresponds to the mass difference between the separated nucleons and the actual nucleus, converted to energy via $E = mc^2$. The binding energy per nucleon (BE/A) is a direct measure of stability: higher values mean each nucleon is more tightly bound.
Figure 8.2 - Binding Energy per Nucleon
Interactive Chart
BE/A curve ⁵⁶Fe peak (8.79 MeV) Fusion region Fission region
Figure 8.2 — Binding energy per nucleon curve showing the peak at iron-56, the turning point between fusion and fission energy release.
The curve peaks at iron-56, the most stable nucleus in existence. Light nuclei release energy through fusion; heavy nuclei release energy through fission.
The binding energy per nucleon curve rises rapidly for light nuclei, reaches a maximum at iron-56 (8.79 MeV per nucleon), and then gradually falls for heavier elements. This curve explains why both nuclear fusion (combining light nuclei to climb the curve) and nuclear fission (splitting heavy nuclei to move toward the peak) release energy. Iron-56 sits at the very top of this curve, making it the most stable nucleus in existence.
The semi-empirical mass formula (SEMF), also called the Weizsäcker formula, provides a quantitative framework for understanding binding energy by combining five physically motivated terms:
Each nucleon gains constant binding from nearest neighbours
15.75 MeV
Surface (−asA2/3)
Surface nucleons have fewer neighbours
17.8 MeV
Coulomb (−acZ(Z−1)/A1/3)
Electrostatic repulsion between all proton pairs
0.711 MeV
Asymmetry (−aa(N−Z)²/A)
Energy penalty when N ≠ Z (Pauli exclusion principle)
23.7 MeV
Pairing (±δ)
Extra binding for even-even; reduced for odd-odd
±11.18/√A MeV
The SEMF reproduces experimental binding energies within 1–2 MeV for most nuclei but systematically underestimates binding near magic numbers, where the shell model must be invoked to account for the extra stability conferred by closed shells.
Using the SEMF - A Worked Example
Let's calculate the binding energy per nucleon for iron-56 (Z=26, N=30, A=56) using the SEMF.
Term
Calculation
Value (MeV)
Volume
15.75 × 56
+882.00
Surface
−17.8 × 562/3 = −17.8 × 15.02
−267.36
Coulomb
−0.711 × 26×25 / 561/3 = −0.711 × 650 / 3.83
−120.61
Asymmetry
−23.7 × (30−26)² / 56 = −23.7 × 16 / 56
−6.77
Pairing
+11.18 / √56 = +11.18 / 7.48
+1.49
Total BE
~489 MeV
BE/A
489 / 56
~8.73 MeV/nucleon
This is within 1% of the experimental value of 8.79 MeV/nucleon, demonstrating the predictive power of the SEMF. The slight underestimate near magic numbers (N=28 is close to a magic number) is expected.
8.5 How Unstable Nuclei Decay
Alpha Decay
Alpha decay is the emission of a helium-4 nucleus: two protons and two neutrons bound together as a single particle. It is the dominant decay mode for very heavy nuclei (typically Z > 83, A > 210) because the alpha particle is itself an exceptionally stable cluster, and its emission simultaneously reduces both the proton count and the total mass.
Alpha decay equation. Q is the energy released, appearing as kinetic energy shared between the alpha particle and the recoiling daughter nucleus.
The Q-value - the energy released - is calculated from the mass difference between the parent nucleus and the products: $Q = (m_X - m_Y - m_\alpha)c^2$. This energy appears as kinetic energy shared between the alpha particle and the recoiling daughter nucleus. Because the alpha particle is much lighter, it carries away the majority of the energy, typically 4–9 MeV.
Alpha decay is fundamentally a quantum tunnelling process. The alpha particle exists as a pre-formed cluster inside the nucleus, bouncing against the potential wall created by the strong force. Each collision with the nuclear surface offers a small probability of tunnelling through the Coulomb barrier - the electrostatic potential that would classically trap the particle inside. This tunnelling mechanism, first explained by George Gamow in 1928, was one of the earliest compelling demonstrations of quantum mechanics.
The relationship between alpha decay energy and half-life is described by the Geiger-Nuttall law:
Geiger-Nuttall law. A modest change in decay energy produces an enormous change in half-life.
A modest change in decay energy produces an enormous change in half-life. Polonium-212 decays with $Q_\alpha = 8.95$ MeV and a half-life of only 0.3 microseconds, whereas uranium-238 has $Q_\alpha = 4.27$ MeV and a half-life of 4.47 billion years - a difference of twenty orders of magnitude in half-life arising from a factor of only two in decay energy.
Figure 8.3 - Alpha Decay Visualisation
Interactive 3D
Proton Neutron Alpha particle
Figure 8.3 — Three-dimensional visualisation of alpha decay showing the emission of a helium-4 nucleus from a heavy parent nucleus via quantum tunnelling.
Uranium-238 → Thorium-234 + α - Drag to rotate. Scroll to zoom. Click "Trigger Alpha Decay" to watch the alpha particle escape via quantum tunnelling.
Beta-Minus Decay
Beta-minus decay occurs in neutron-rich nuclei and converts a neutron into a proton, emitting an electron (the beta particle) and an electron antineutrino:
The mass number A is unchanged, but the atomic number Z increases by one, transforming the nucleus into a different element.
The Q-value is shared as kinetic energy among the electron, antineutrino, and recoiling daughter nucleus. Unlike alpha particles, which are emitted with discrete energies, beta particles display a continuous energy spectrum from near zero up to a maximum equal to the Q-value. This continuous spectrum was historically puzzling and led Wolfgang Pauli to postulate the existence of the neutrino in 1930.
Beta decay is mediated by the weak nuclear force, operating through the exchange of W and Z bosons. It is the mechanism by which neutron-rich nuclei adjust their N/Z ratio and move toward the valley of stability.
Figure 8.4 - Beta-Minus Decay Visualisation
Interactive 3D
Proton Neutron Electron (β⁻) Antineutrino
Figure 8.4 — Interactive 3D model of beta-minus decay, showing a neutron converting into a proton with emission of an electron and antineutrino.
Carbon-14 → Nitrogen-14 + e⁻ + ν̄ₑ - Watch a neutron convert to a proton while the electron and antineutrino streak away in opposite directions.
Beta-Plus Decay and Electron Capture
For proton-rich nuclei, the reverse process occurs. A proton transforms into a neutron, emitting a positron and an electron neutrino:
$$p \rightarrow n + e^+ + \nu_e$$
Proton conversion via the weak nuclear force
Because a positron has the same rest mass as an electron (0.511 MeV/c²), β⁺ decay is only energetically possible when the Q-value exceeds 1.022 MeV. For nuclei where the Q-value is positive but below this threshold, electron capture provides an alternative pathway. The nucleus captures one of its own orbital electrons - most likely from the inner K or L shell - converting a proton into a neutron and emitting only a neutrino:
Electron capture. After capture, the vacancy in an inner shell produces characteristic X-rays unique to the daughter element.
After electron capture, the resulting vacancy in an inner electron shell is filled by electrons from outer shells, producing characteristic X-rays unique to the daughter element. Both β⁺ decay and electron capture reduce Z by one while keeping A constant, shifting proton-rich nuclei toward the valley of stability.
Figure 8.5 - Beta-Plus Decay Visualisation
Interactive 3D
Proton Neutron Positron (β⁺) Neutrino
Figure 8.5 — Visualisation of beta-plus decay (positron emission), the process underlying PET scanning in nuclear medicine.
Fluorine-18 → Oxygen-18 + e⁺ + νₑ - Positron emission is the basis of PET scans in medicine.
Gamma Emission
Gamma decay involves the emission of a high-energy photon from an excited nucleus. It changes neither the proton count, the neutron count, nor the mass number - it simply allows the nucleus to shed excess energy and transition to a lower energy state:
Gamma emission typically follows alpha or beta decay, as the daughter nucleus is often left in an excited state.
Gamma emission typically follows alpha or beta decay, as the daughter nucleus is often left in an excited state. The excess energy is released almost instantaneously, usually within 10−12 seconds, as one or more gamma photons. Gamma rays are the most penetrating form of nuclear radiation.
Figure 8.6 - Gamma Decay Visualisation
Interactive 3D
Proton Neutron Gamma photon
Figure 8.6 — Three-dimensional model of gamma emission, where an excited nucleus releases a high-energy photon as it transitions to its ground state.
⁹⁹ᵐTc → ⁹⁹Tc + γ - The excited nucleus vibrates, then releases a gamma photon as it relaxes to ground state.
Build an Atom - Interactive Simulation
Use this custom interactive tool to build your own atoms by adding protons, neutrons, and electrons. Observe how the element identity, stability, and charge change as you adjust the particle counts.
Figure 8.7 - Build an Atom
Interactive Simulation
Particles
-
Empty
Add a proton to start building
Neutral
Figure 8.7 — Interactive simulation for constructing atoms by adding protons, neutrons, and electrons, with real-time validation against known nuclear data.
Summary of Decay Modes
Decay mode
Emission
ΔZ
ΔA
ΔN
Typical Q-value
Occurs when
α
⁴He nucleus
−2
−4
−2
4–9 MeV
Heavy nuclei (Z > 83, A > 210)
β⁻
Electron + antineutrino
+1
0
−1
0.01–10 MeV
Neutron-rich nuclei
β⁺
Positron + neutrino
−1
0
+1
> 1.022 MeV
Proton-rich nuclei (light)
EC
Neutrino only
−1
0
+1
0–1.022 MeV
Proton-rich nuclei (alternative)
γ
Photon
0
0
0
keV–MeV
Excited nucleus relaxing
Conservation Laws in Nuclear Decay
Quantity
α decay
β⁻ decay
β⁺ decay
Electron Capture
γ emission
Mass number A
Conserved
Conserved
Conserved
Conserved
Conserved
Charge (Z)
−2
+1
−1
−1
0
Lepton number
0
+1 (e⁻) + (−1) (ν̄ₑ) = 0
−1 (e⁺) + (+1) (νₑ) = 0
+1 (captured e⁻) + (−1) (νₑ) = 0
0
Baryon number
Conserved
Conserved
Conserved
Conserved
Conserved
Energy
Released as Q
Released as Q
Released as Q (if >1.022 MeV)
Released as Q
Released as Eᵧ
The neutrino's existence was postulated precisely because beta decay appears to violate energy conservation without it. Including the neutrino ensures conservation of energy, momentum, and lepton number simultaneously.
Beyond the Big Five: Spontaneous Fission
For very heavy nuclei (Z ≥ 100), the nucleus can split into two roughly equal fragments plus several neutrons. This is spontaneous fission, and it competes with alpha decay as the primary decay mode. For elements beyond fermium (Z = 100), spontaneous fission typically dominates. This process is the basis of nuclear power reactors, where induced (not spontaneous) fission of U-235 is controlled to release energy.
Physics History: Chien-Shiung Wu (1956)
Beta decay doesn't just convert neutrons to protons - it also violates parity, one of the fundamental symmetries of nature. In 1956, physicists Tsung-Dao Lee and Chen-Ning Yang theorised that the weak force might not respect mirror symmetry. Chien-Shiung Wu, an experimental physicist at Columbia University, designed a brilliant experiment using cobalt-60 at near-absolute-zero temperatures. Her results confirmed that beta particles are emitted preferentially in one direction - definitively proving that parity is violated in weak interactions. This was one of the most profound discoveries in 20th-century physics.
8.6 Decay Chains
Many heavy radioactive isotopes cannot reach stability through a single decay. Instead they undergo a decay chain - a sequence of successive alpha and beta decays, each producing a new radioactive daughter, until a stable isotope is finally reached.
The uranium-238 series is the most extensively studied decay chain. It consists of 14 steps - eight alpha decays and six beta-minus decays - that transform U-238 into stable lead-206. The half-lives within this single chain vary extraordinarily, from 4.47 billion years for U-238 down to 164 microseconds for polonium-214. This variation demonstrates how sensitive radioactive decay rates are to the specific nuclear structure and Q-value of each isotope.
Figure 8.8 - Uranium-238 Decay Chain
Interactive Diagram
U-238
4.47 × 10⁹ y
α
Th-234
24.1 d
β⁻
Pa-234
6.7 h
β⁻
U-234
2.45 × 10⁵ y
α
Th-230
7.54 × 10⁴ y
α
Ra-226
1600 y
α
Rn-222
3.82 d
α
Po-218
3.05 m
α
Pb-214
26.8 m
β⁻
Bi-214
19.7 m
β⁻
Po-214
164 μs
α
Pb-210
22.2 y
β⁻
Bi-210
5.01 d
β⁻
Po-210
138 d
α
Pb-206
STABLE
α decay β⁻ decay Stable
Figure 8.8 — Complete decay chain of uranium-238 to stable lead-206, showing all 14 steps with decay modes and half-lives.
Step
Isotope
Decay
Half-life
1
U-238
α
4.47 × 10⁹ y
2
Th-234
β⁻
24.1 d
3
Pa-234
β⁻
6.7 h
4
U-234
α
2.45 × 10⁵ y
5
Th-230
α
7.54 × 10⁴ y
6
Ra-226
α
1600 y
7
Rn-222
α
3.82 d
8
Po-218
α
3.05 m
9
Pb-214
β⁻
26.8 m
10
Bi-214
β⁻
19.7 m
11
Po-214
α
164 μs
12
Pb-210
β⁻
22.2 y
13
Bi-210
β⁻
5.01 d
14
Po-210
α
138 d → Pb-206 (stable)
This decay chain is the primary natural source of background radiation and is responsible for the presence of radon-222 gas in buildings, which is a significant public health concern.
Why Doesn't the Decay Chain Run Away?
After a long time (many times the longest daughter half-life), the decay chain reaches secular equilibrium: the activity of every daughter equals the activity of the parent. This means 1 gram of U-238 produces radon-222 at exactly the same rate that the U-238 decays. Mathematically: $A_{\text{daughter}} = A_{\text{parent}}$ when $t \gg t_{1/2}(\text{daughter})$. This is why radon gas in buildings is a constant health concern - it's continuously produced by uranium in the ground.
8.7 Half-Life and Activity
The Mathematics of Decay
Radioactive decay is a first-order kinetic process: the rate of decay at any instant is directly proportional to the number of radioactive nuclei present:
$$\frac{dN}{dt} = -\lambda N$$
N = number of radioactive nuclei, t = time, λ = decay constant
Here N is the number of radioactive nuclei, t is time, and λ is the decay constant - a characteristic property of each isotope. Integrating this equation gives the exponential decay law:
$$N(t) = N_0 e^{-\lambda t}$$
Exponential decay law
The half-life (t₁/₂) is the time required for half the nuclei in a sample to decay. Setting N = N₀/2 in the equation above yields the fundamental relationship:
Half-life is an intrinsic property of each radioisotope, independent of temperature, pressure, chemical environment, and the age of the sample.
Half-life is an intrinsic property of each radioisotope. It is independent of temperature, pressure, chemical environment, and the age of the sample - a unique feature of nuclear processes that distinguishes radioactive decay from chemical reactions.
Mean Lifetime
Related to half-life is the mean lifetime (τ), which represents the average time an individual nucleus survives before decaying. The relationship is:
Mean lifetime is the natural time constant in the exponential decay equation: $N = N_0 e^{-t/\tau}$
While half-life is more commonly used, mean lifetime appears in many theoretical derivations and is the natural time constant in the exponential decay equation.
Activity and Its Units
The activity of a sample is its rate of decay - the number of disintegrations per unit time:
$$A = \lambda N = \frac{0.693 \cdot N}{t_{1/2}}$$
Activity equation
The SI unit of activity is the becquerel (Bq), defined as one disintegration per second. An older unit still in common use is the curie (Ci), originally defined as the activity of one gram of radium-226 (3.70 × 10¹⁰ Bq). Activity, like the number of nuclei, decays exponentially: $A(t) = A_0 e^{-\lambda t}$. This equation is the foundation of all radioactive dating techniques.
Figure 8.9 - Half-Life Simulator
100
Remaining
0
Decayed
0.0
Half-lives
100%
Percent Left
Figure 8.9 — Interactive half-life simulator showing probabilistic decay of individual nuclei alongside the exponential decay curve.
Decay Curve
$N = N_0 e^{-\lambda t}$
Exponential decay law
Isotope
Half-life
Decay mode
Application
¹⁴C
5,730 years
β⁻
Radiocarbon dating
²³⁸U
4.47 × 10⁹ years
α
Geological dating, nuclear fuel
⁶⁰Co
5.27 years
β⁻ + γ
Cancer radiotherapy
⁹⁹ᵐTc
6.0 hours
γ (isomeric)
Diagnostic nuclear medicine
¹³¹I
8.0 days
β⁻ + γ
Thyroid diagnosis and treatment
²¹⁰Po
138 days
α
Thermoelectric power (space probes)
8.8 Applications of Radioactive Decay
Nuclear Medicine
Radioactive isotopes have transformed modern medicine. The ideal diagnostic radioisotope emits gamma rays detectable outside the body and has a half-life long enough for the procedure but short enough to minimise patient dose. Technetium-99m is the most widely used diagnostic isotope, employed in approximately 80% of all nuclear medicine scans worldwide. Its 6-hour half-life, pure gamma emission, and versatile chemistry make it suitable for imaging bones, heart muscle, kidneys, and the thyroid.
Physics in Action
How technetium is produced
Technetium-99m is produced in small nuclear generators located in hospitals. Molybdenum-99 decays by beta emission to form technetium-99m:
Technetium-99m is flushed from the generator using saline solution, then attached to an appropriate chemical compound before being administered to the patient.
Radiometric Dating
The predictable mathematics of radioactive decay provides a precise clock for measuring the age of materials. Carbon-14 dating relies on the constant production of ¹⁴C in the atmosphere by cosmic rays and its incorporation into living organisms. After death, the ¹⁴C content decreases with its 5,730-year half-life, allowing archaeologists to date organic materials up to about 50,000 years old. For geological timescales, uranium-lead dating uses the ²³⁸U → ²⁰⁶Pb chain (half-life 4.47 billion years) to date the oldest rocks and meteorites in the solar system.
PET Imaging
Positron emission tomography (PET) directly exploits beta-plus decay. A patient is injected with a glucose analog labelled with fluorine-18 (¹⁸F), which undergoes β⁺ decay with a 110-minute half-life. The emitted positron annihilates with an electron, producing two 511 keV gamma rays travelling in opposite directions. Detecting these coincident photons allows PET scanners to build 3D images of metabolic activity - particularly useful for detecting cancers, which consume more glucose than healthy tissue.
How Radiometric Dating Actually Works
Radiometric dating doesn't just measure activity - it measures parent-to-daughter isotope ratios. In uranium-lead dating, scientists measure the ratio of ²⁰⁶Pb to ²³⁸U in a mineral. Because each U-238 decay produces exactly one Pb-206 atom, this ratio increases predictably over time. The age equation is: $t = \frac{1}{\lambda} \ln\left(1 + \frac{N_{\text{daughter}}}{N_{\text{parent}}}\right)$. A critical assumption is that the system remained closed - no lead or uranium entered or left the mineral since it formed.
Why U-235, Not U-238?
Nuclear reactors use uranium-235 as fuel, not the more abundant U-238. This is because U-235 is fissile - its nucleus splits when struck by a slow (thermal) neutron, releasing energy and more neutrons to sustain a chain reaction. U-238 absorbs neutrons without fissioning, acting as a "poison" that would stop the chain reaction if concentrated. Natural uranium contains only 0.7% U-235, so enrichment (increasing the U-235 fraction to 3–5%) is required for most reactor designs.
Energy and Industry
Radioactive decay and induced fission of heavy elements power nuclear reactors, which provide roughly 10% of global electricity. On a smaller scale, americium-241 in smoke detectors uses alpha particles to ionise air, while industrial thickness gauges employ beta radiation to monitor paper, metal, and plastic production lines.
Physics in Action
Smoke detectors
Domestic smoke detectors contain a tiny amount of americium-241, an alpha emitter. The alpha particles ionise air molecules between two electrodes, creating a small electric current. When smoke enters the chamber, it absorbs the alpha particles, reducing the current and triggering the alarm. Because alpha particles cannot penetrate the plastic casing, the device is completely safe in normal use.
Common Misconceptions - Don't Fall For These!
Misconception
The Reality
"All radiation is dangerous"
Not all radiation is equal. Alpha particles are stopped by paper; gamma rays are the most penetrating. The danger depends on type, dose, and exposure time.
"Half-life means exactly half the atoms decay at t₁/₂"
Half-life is a statistical average. For a large sample, approximately half decay in one half-life. Individual atoms decay randomly.
"Neutrons don't matter because they have no charge"
Neutrons are essential! They provide strong-force binding without adding Coulomb repulsion. Without neutrons, no nucleus heavier than hydrogen-1 could exist.
"Nuclear waste stays radioactive forever"
Most nuclear waste decays to safe levels within centuries. Only a small fraction (transuranic elements) remains hazardous for millennia.
"Fusion is always better than fission"
Fusion produces less long-lived waste but requires temperatures of ~100 million °C. We haven't yet achieved sustained net-energy fusion on Earth.
8.9 Chapter Summary
Nuclear instability arises from the competition between the short-range strong nuclear force, which binds nucleons together, and the long-range electromagnetic repulsion between protons.
The stability of a nucleus depends on four interconnected factors: the neutron-to-proton ratio, which increases from approximately 1.0 for light elements to 1.5 for heavy elements; the even-odd configuration of nucleons, with even-even nuclei being the most stable; magic numbers (2, 8, 20, 28, 50, 82, 126) that confer exceptional stability through the nuclear shell model; and the overall binding energy per nucleon, which peaks at ⁵⁶Fe.
When a nucleus is unstable, it decays via one of several mechanisms. Alpha decay dominates for heavy nuclei and is governed by quantum tunnelling through the Coulomb barrier, with the Geiger-Nuttall law relating decay energy to half-life.
Beta-minus decay converts neutrons to protons and moves neutron-rich nuclei toward the valley of stability, while beta-plus decay and electron capture convert protons to neutrons and move proton-rich nuclei toward stability.
Gamma emission releases excess energy without changing the nucleus's composition. These processes often occur in decay chains, such as the 14-step uranium-238 series that terminates at stable lead-206.
All radioactive decay follows first-order kinetics described by $N = N_0 e^{-\lambda t}$, with a characteristic half-life that is independent of external conditions. Activity is measured in becquerels or curies.
The predictable nature of radioactive decay underpins crucial applications in nuclear medicine, radiometric dating, and nuclear energy production.
Interactive Check Questions
Question 8.1
A nucleus located above the valley of stability has an excess of neutrons. Which decay mode will most likely restore it toward stability?
Correct! β⁻ decay converts a neutron into a proton, reducing N and increasing Z, moving the nucleus diagonally toward the valley of stability.
Question 8.2
Technetium-99m has a half-life of 6.0 hours. A sample has an initial activity of 800 MBq. How many hours will pass before the activity falls below 100 MBq? Show your working.
Answer: 18 hours
After one half-life (6 h): 800 → 400 MBq
After two half-lives (12 h): 400 → 200 MBq
After three half-lives (18 h): 200 → 100 MBq
So 18 hours must pass before the activity falls below 100 MBq.
Question 8.3
Using the N/Z ratio and even-odd rules, predict whether phosphorus-30 (Z = 15, N = 15) is likely to be stable. If radioactive, which decay mode would you expect? Explain your reasoning.
Model Answer:
³⁰P has Z = 15 (odd) and N = 15 (odd), making it an odd-odd nucleus - only four odd-odd nuclei are stable (¹H, ⁶Li, ¹⁰B, ¹⁴N), and ³⁰P is not one of them.
For Z = 15, the valley of stability prefers slightly more neutrons than protons (N/Z ≈ 1.1–1.2), so N/Z = 1.0 is proton-rich. You would expect β⁺ decay or electron capture to convert a proton to a neutron, moving the nucleus toward the valley of stability.
Question 8.4
Calculate the Q-value for alpha decay of polonium-210 (mass = 209.9829 u) to lead-206 (mass = 205.9745 u). The alpha particle mass is 4.0026 u. Express your answer in MeV (1 u = 931.5 MeV/c²).
Explain why polonium-212 has a half-life of 0.3 microseconds while uranium-238 has a half-life of 4.47 billion years, despite both undergoing alpha decay.
Model Answer:
The Geiger-Nuttall law relates alpha decay energy to half-life: higher Q-value means a thinner Coulomb barrier and much higher tunnelling probability. Po-212 has Q = 8.95 MeV, while U-238 has Q = 4.27 MeV. A factor of only two in energy produces a difference of twenty orders of magnitude in half-life.
Question 8.6
Use the SEMF to explain why ⁵⁶Fe is more stable than ⁵⁵Fe.
Model Answer:
⁵⁶Fe is even-even (Z=26 even, N=30 even), giving it an extra pairing term energy bonus. ⁵⁵Fe is odd-A (no pairing bonus). The pairing term adds ~+1.5 MeV of extra binding for even-even nuclei, making ⁵⁶Fe significantly more stable.
Question 8.7
A sample of carbon-14 has an activity of 320 Bq. How old is the sample if the initial activity (when the organism was alive) was 640 Bq? The half-life of ¹⁴C is 5730 years.
Answer: 5730 years
The activity has halved from 640 Bq to 320 Bq, which means exactly one half-life has elapsed. Age = 5730 years.
Question 8.8
The binding energy per nucleon curve peaks at ⁵⁶Fe. Explain why (a) fusion of light nuclei releases energy, and (b) fission of heavy nuclei releases energy, using the concept of binding energy per nucleon.
Model Answer:
(a) Light nuclei have low BE/A; fusing them into heavier nuclei increases BE/A toward the Fe-56 peak → energy is released.
(b) Heavy nuclei have lower BE/A than Fe-56; splitting them into medium-mass fragments moves the products toward the peak → energy is released.
Question 8.9 - Data Interpretation
The graph below shows the binding energy per nucleon versus mass number. Use it to answer the following:
(a) Identify the most stable nucleus and its BE/A value.
(b) Explain why fusion of hydrogen into helium releases energy.
(c) A nucleus with A = 240 undergoes fission into two fragments with A = 120 and A = 120. Estimate the energy released per fission event.
(d) Why can't ⁵⁶Fe release energy through either fusion or fission?
Model Answer:
(a) ⁵⁶Fe at ~8.79 MeV/nucleon.
(b) Hydrogen has very low BE/A (~1 MeV). Helium-4 has BE/A ≈ 7 MeV. The product is more tightly bound, so energy is released.
(c) A≈240 has BE/A ≈ 7.6 MeV. A≈120 has BE/A ≈ 8.5 MeV. Energy released ≈ 240 × (8.5 − 7.6) ≈ ~200 MeV.
(d) ⁵⁶Fe is at the peak of the curve. Neither combining it with another nucleus (fusion) nor splitting it (fission) can increase BE/A further.
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Flashcards
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Strong nuclear force
The fundamental force that binds protons and neutrons together in the nucleus. It is ~100× stronger than electromagnetism at nuclear distances but has a very short range (~1–3 fm).