The key idea
IGCSE radioactivity plus the two ideas that were hiding underneath: mass-energy equivalence, which explains where nuclear energy comes from, and the exponential decay law, which turns half-life from a counting trick into an equation.
Understand it · 01
Separate a random event from a predictable population
An unstable nucleus has a probability of decaying in a given time interval. The moment of decay of one particular nucleus cannot be predicted, but a large collection has a reproducible average exponential trend. The decay constant λ is the probability of decay per unit time in the usual small-interval interpretation. Activity is the expected decay rate, A = λN, measured in becquerels. A detector’s count rate is not automatically equal to activity because it detects only a fraction of emissions.
Half-life is the time for the expected number of undecayed nuclei, or the activity, to fall to one half. It is linked to the decay constant by t½ = ln 2/λ. Equal half-life intervals produce equal fractional reductions, not equal numbers of decays. Subtract background count rate before using measured counts to estimate a half-life. Statistical fluctuations become relatively more conspicuous as the source count rate falls towards the background.
Understand it · 02
Apply conservation rules to the actual process
Alpha emission reduces nucleon number by four and proton number by two. In beta-minus decay, a neutron changes into a proton with an electron and an electron antineutrino emitted; nucleon number stays unchanged. Beta-plus decay changes a proton into a neutron with a positron and an electron neutrino emitted. Gamma emission changes the nucleus’s energy state without changing its proton or nucleon number. These different processes must not be described as interchangeable ways of losing electrons.
Particle reactions also require conservation of quantities such as charge, energy and momentum. At the relevant IAL level, baryon number and lepton number help identify whether a proposed reaction is possible. A proton and neutron are baryons composed of quarks; an electron is a lepton and is not composed of quarks in this model. Antiparticles have opposite additive quantum numbers to their corresponding particles. Do not decide whether a reaction is allowed by charge balance alone.
Understand it · 03
Compare the energy of initial and final states
The mass of a bound nucleus is less than the total mass of its separate nucleons. The difference corresponds to binding energy through E = Δmc²: energy must be supplied to separate the nucleus completely. Binding energy per nucleon measures how strongly the nucleons are bound on average. A nuclear process can release energy when its products are more tightly bound overall, so that their total rest mass is lower than that of the initial system.
Fission of very heavy nuclei and fusion of sufficiently light nuclei can both release energy by moving towards more tightly bound products. This does not mean that every fission or fusion reaction releases energy; compare the actual masses or binding energies. For a decay experiment, use approved laboratory procedures and minimise exposure by time, distance and shielding appropriate to the source. Record background for a suitable interval and keep source–detector geometry fixed so changing detection efficiency is not mistaken for radioactive decay.
Key definitions
- Mass defect
- The difference between the total mass of the separate nucleons and the mass of the nucleus they form.
- Binding energy
- The energy required to separate a nucleus into its individual nucleons — equivalently, the energy released when the nucleus forms from them: E = Δmc².
- Decay constant (λ)
- The probability per unit time that a given nucleus will decay: A = λN.
- Activity
- The number of decays per unit time of a sample, measured in becquerels (Bq).
Equations, units & conditions
Mass-energy
E = m c²
Units: J
Activity
A = λ N
Units: Bq
Decay law
N = N₀ e^(−λt)
Half-life
t½ = ln 2 ÷ λ
Worked examples, step by step
Example 1
A radioactive isotope has a half-life of 8.0 days. What fraction of a sample remains after 20 days?
Follow the method
- λ = ln 2 ÷ t½ = 0.693 ÷ 8.0 = 0.0866 per day.
- N/N₀ = e^(−λt) = e^(−0.0866 × 20) = e^(−1.73).
N/N₀ ≈ 0.18 (about 18% remains)
Practice questions with worked solutions
Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.
Build · 3 suggested marks
1. Background-corrected half-life
A detector measures 850 counts per minute initially and 250 counts per minute 12 minutes later. Background is 50 counts per minute. Estimate the half-life.
Show solution and self-check: question 1
- Correct the readings: 800 and 200 counts per minute.
- 200/800 = 1/4, so two half-lives have elapsed.
- Half-life = 12/2.
6.0 minutes.
Self-check · one mark per point
- Subtract background from both readings.
- Recognise the source rate has fallen through two half-lives.
- Obtain 6.0 minutes.
Build · 3 suggested marks
2. Energy from a mass change
A nuclear reaction has a rest-mass decrease of 3.0 × 10⁻²⁹ kg. Find the released energy in joules and MeV, using c = 3.0 × 10⁸ m s⁻¹ and 1 MeV = 1.60 × 10⁻¹³ J.
Show solution and self-check: question 2
- E = Δmc² = 3.0 × 10⁻²⁹ × (3.0 × 10⁸)² = 2.7 × 10⁻¹² J.
- Energy in MeV = 2.7 × 10⁻¹²/(1.60 × 10⁻¹³).
2.7 × 10⁻¹² J, approximately 17 MeV.
Self-check · one mark per point
- Use the square of c.
- Calculate the energy in joules.
- Convert to approximately 16.9 MeV before rounding.
Common mistakes & how to avoid them
- Reading the binding energy curve backwards. The peak of the binding-energy-per-nucleon curve is in the iron/nickel region. Fusion moves light nuclei up the curve towards it; fission moves heavy nuclei up towards it from the other side. Both release energy because binding energy per nucleon INCREASES.
- Mass units chaos: masses given in atomic mass units (u) must become kilograms (1 u = 1.66 × 10⁻²⁷ kg) before E = mc² — or use the energy equivalent 1 u × c² ≈ 931.5 MeV. A mass unit is not itself an energy unit.
- Confusing activity with count rate. A detector catches only a fraction of the decays; count rate is proportional to activity, not equal to it, and background must be subtracted first.
- Decay-constant logic inverted: a SHORT half-life means a LARGE decay constant (λ = ln 2 ÷ t½) — the nucleus is more likely to decay each second, not less.
Understand the why
Questions, explained
Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.
Where does the energy in fission and fusion come from?
An energy-releasing nuclear reaction has products with less total rest mass than the initial system. The mass difference corresponds to released energy ΔE = Δmc², often because the products are more tightly bound overall.
Binding energy is energy required to separate a nucleus into its free nucleons. A more tightly bound state has a lower total mass for the same constituent content. Suitable fusion among light nuclei and fission of heavy nuclei can both move towards more tightly bound products, but not every proposed reaction releases energy. Compare the actual initial and final masses, accounting consistently for whether atomic or nuclear masses are supplied. Do not treat mass as disappearing without an energy account: rest energy is converted into other forms, including kinetic energy and emitted radiation.
Energy from a mass decrease
A reaction has mass decrease 0.020 u. Use 1 u × c² = 931.5 MeV to find the released energy.
- ΔE = 0.020 × 931.5 MeV.
- The conversion factor is an energy equivalent of mass, not an equality between mass and energy units.
18.63 MeV, approximately 19 MeV.
What is the relationship between half-life and the decay constant?
Half-life and decay constant are related by t½ = ln 2/λ. A larger decay constant means a greater decay probability per short time interval and therefore a shorter time for the population’s expected activity to halve.
The time unit used for λ must match the time unit used in the exponential. A value per day cannot be combined directly with time in seconds. The half-life concerns the statistical behaviour of many nuclei, not a deadline at which an individual nucleus must decay. A nucleus that has survived several half-lives does not become old and more likely to decay under the constant-λ model. Its future short-interval probability is unchanged, which is why successive equal intervals reduce the expected population by the same fraction.
Decay constant from half-life
An isotope has half-life 6.0 hours. Find decay constant in s⁻¹.
- Convert half-life: 6.0 × 3600 = 21,600 s.
- λ = ln 2/21,600.
3.21 × 10⁻⁵ s⁻¹.
Why is radioactive decay exponential?
Radioactive decay is exponential because the expected decay rate is proportional to the number of undecayed nuclei remaining. As that population shrinks, fewer decays occur per unit time, while the same fraction is lost in equal intervals.
The probability for one nucleus is not reduced just because the sample is smaller. Instead, there are fewer nuclei available to decay. This distinction explains why the curve becomes less steep without reaching zero at a fixed finite time in the continuous model. Real counts fluctuate around that trend because the events are random. A tiny sample can depart noticeably from a smooth curve, while a large population gives a more stable average. Background counts add an approximately constant component and must be removed before testing the source’s exponential relationship.
Why is detector count rate different from activity?
Activity is the source’s decay rate, while detector count rate records only events detected by a particular setup. Geometry, absorption and detector efficiency affect that fraction, and background adds counts unrelated to the source.
A detector cannot usually register every emitted particle or photon. Moving it closer may raise count rate even though the source activity has not changed, so keep source–detector geometry fixed in a decay experiment. Measure background separately over a suitable interval and subtract rates, not raw counts taken over unequal times. Dead time can also make a detector miss events at high rates. These effects explain why a count-rate measurement requires calibration or modelling before it can be interpreted as an absolute activity, while relative changes can still be useful under stable conditions.
Why use binding energy per nucleon instead of total binding energy?
Total binding energy measures the energy needed to separate the entire nucleus, while binding energy per nucleon divides that total by nucleon number. The per-nucleon value enables comparisons between nuclei of very different sizes.
A heavy nucleus can have a large total binding energy simply because it contains many nucleons, yet be less tightly bound on average than a smaller nucleus. The familiar curve peaks in the iron–nickel region rather than increasing indefinitely with mass number. In a reaction involving different products, compare the full energy balance rather than assuming that one product’s per-nucleon value alone determines the released energy. The graph is a useful guide to suitable fission and fusion processes, but precise calculations require the masses or binding energies of all participants.
Why is an antineutrino included in beta-minus decay?
In beta-minus decay, a neutron becomes a proton while an electron and an electron antineutrino are emitted. Including the antineutrino allows the reaction to satisfy lepton-number, energy and momentum conservation as well as charge conservation.
The electron is produced in the decay process rather than being a pre-existing orbital electron ejected from the atom. Nucleon number stays constant because one nucleon changes type, while proton number increases by one. The emitted electron’s energy is not always a single fixed value because the available energy is shared with the antineutrino and recoiling nucleus. For beta-plus decay, a proton becomes a neutron with a positron and electron neutrino. Use the correct neutrino or antineutrino symbol in particle equations; balancing charge alone does not check all the required conservation laws.
Your learning checklist
Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.
- Define mass defect and binding energy; use E = mc²
- Sketch binding energy per nucleon against nucleon number; explain fission and fusion
- Define activity and the decay constant: A = λN
- Use the exponential decay law: N = N₀e^(−λt)
- Relate half-life and decay constant: t½ = ln 2 ÷ λ
Sources & how to cite these notes
Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.
Reference this page
Dr Desouky Physics Academy. Edexcel IAL Physics: Nuclear physics. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/nuclear-physics
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