E.3 Radioactive decay

Syllabus
First assessment 2025
Topic
—
Level
SL

Learning objectives

Identify Isotopes

Define an isotope

Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons. They share chemical identity but can have different physical properties.

Read the numbers

The proton number ZZ stays fixed within an element. Different isotopes have different nucleon numbers AA, so their neutron numbers N=A−ZN=A-Z differ.

Common trap

Do not define isotopes only as atoms with different A and Z. The same proton number is essential; otherwise the atoms are different elements.

E.3.1 Exam Analysis

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Calculate Mass Defect

Define mass defect

A bound nucleus has less mass than the separated protons and neutrons that form it. The missing mass is the mass defect Δm\Delta m, associated with the energy released when the nucleus forms.

Convert mass to binding energy

Use Eb=Δmc2E_b=\Delta mc^2. If Δm\Delta m is in unified atomic mass units, the convenient conversion is approximately 931.5 MeV/c2931.5\,\mathrm{MeV}/c^2 per u, giving energy directly in MeV.

Worked example — mass defect

If separated nucleons have total mass 4.0320 u4.0320\,\mathrm{u} and the nucleus has mass 4.0015 u4.0015\,\mathrm{u}, then Δm=0.0305 u\Delta m=0.0305\,\mathrm{u}. Hence Eb=(0.0305)(931.5)=28.4 MeVE_b=(0.0305)(931.5)=28.4\,\mathrm{MeV}. The positive result is the energy needed to separate the nucleus, and the same energy magnitude was released when it formed.

Interpret the sign

Binding energy is the energy required to separate the nucleons completely, and the same amount is released when the bound nucleus forms. It is positive as a required or released energy magnitude.

Common trap

Do not multiply a mass difference in u by c² again after using 931.5 MeV per u; that conversion already includes the mass–energy relation.

E.3.2 Exam Analysis

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Read Binding Energy Curve

Read the curve

Binding energy per nucleon rises for light nuclei, reaches a broad maximum for medium-mass nuclei, then decreases gradually for very heavy nuclei. The curve compares average nuclear stability per nucleon, not total binding energy.

Predict energy release

Fusion of light nuclei can move products upward toward the maximum. Fission of very heavy nuclei can also move products upward. In either case, the increase in binding energy per nucleon corresponds to released energy.

Sketch the trend

Show a rise from the light-nucleus region, a maximum between roughly A=50 and A=100, and a slow decline for larger A. Exact numerical values are not required for the qualitative graph.

Common trap

Do not claim that the heaviest nucleus is most stable simply because it has the largest total binding energy. Use binding energy per nucleon to compare stability.

E.3.3 Exam Analysis

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Apply Mass-Energy Equivalence

Use E=mc²

A change in rest mass corresponds to energy through E=mc2E=mc^2. In a nuclear reaction, compare the total mass before and after to find the mass converted into released or absorbed energy.

\Delta E=\Delta mc^2

Worked example — energy from a mass decrease

For Δm=2.0×10−12 kg\Delta m=2.0\times10^{-12}\,\mathrm{kg}, ΔE=(2.0×10−12)(3.00×108)2=1.8×105 J\Delta E=(2.0\times10^{-12})(3.00\times10^8)^2=1.8\times10^5\,\mathrm{J}. A smaller total rest mass of the products means this energy is released.

Compare energy yields

Energy released per reaction is proportional to mass converted. Energy released per unit mass also depends on the converted fraction: divide the energy from one reaction by the mass of fuel involved.

Track the system

Mass–energy equivalence applies to the mass difference of the defined reaction system. Do not compare only the total mass of the reactants without accounting for products.

Common trap

Do not confuse a large energy per reaction with a large energy per unit mass. The question’s denominator determines the comparison.

E.3.4 Exam Analysis

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Model Strong Nuclear Force

Describe the force

The strong nuclear force is attractive between nucleons at nuclear separations and has a very short range. It can bind protons and neutrons despite the electrostatic repulsion between protons.

Explain stability

At short distances the strong force can dominate, while the electromagnetic force is repulsive and long range. A stable nucleus requires the attractive nuclear interaction to overcome proton repulsion within the nucleus.

Keep the range distinction

The strong force does not act as a long-range force between separated nuclei. Its short range is why increasing nuclear size makes stability more difficult.

Common trap

Do not call the strong force repulsive between nucleons in the binding explanation, and do not confuse it with the weak nuclear interaction.

E.3.5 Exam Analysis

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Model Random Decay

Treat each nucleus independently

Radioactive decay is spontaneous and random: the exact nucleus and instant of decay cannot be predicted. For a large sample, however, the fraction decaying per unit time follows a stable statistical law.

Separate random from law-like

Random decay does not mean the activity is random noise. The expected number of decays is predictable from the number of undecayed nuclei and the decay constant.

Apply the statistical model

You cannot identify which nucleus will decay next. But if two large samples contain the same nuclide and the second has twice as many undecayed nuclei, its expected activity is twice as large. Individual unpredictability and ensemble predictability coexist.

Common trap

Do not claim that randomness prevents prediction of half-life or activity. It prevents prediction of an individual decay, not the ensemble behaviour.

E.3.6 Exam Analysis

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Compare Nuclear Decays

Alpha decay

Alpha decay emits a 24He{}^{4}_{2}\mathrm{He} nucleus. The parent’s nucleon number decreases by 4 and proton number decreases by 2.

Beta decay

In beta-minus decay, a neutron becomes a proton and an electron is emitted, so AA is unchanged and ZZ increases by 1. In beta-plus decay, a proton becomes a neutron and a positron is emitted, so AA is unchanged and ZZ decreases by 1.

Gamma decay

Gamma emission changes the nucleus from an excited state to a lower energy state. Neither AA nor ZZ changes.

Common trap

Do not change A during beta decay, and do not treat gamma emission as a change of element.

E.3.7 Exam Analysis

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Write Decay Equations

Balance alpha decay

Write ZAX→Z−2A−4Y+24He{}^{A}_{Z}X\rightarrow{}^{A-4}_{Z-2}Y+{}^{4}_{2}\mathrm{He}. Check both A and Z on the two sides.

Balance beta decay

For beta-minus use ZAX→Z+1AY+−10e+νˉe^{A}_{Z}X\rightarrow{}^{A}_{Z+1}Y+{}^{0}_{-1}e+\bar{\nu}_e. For beta-plus use ZAX→Z−1AY++10e+νe^{A}_{Z}X\rightarrow{}^{A}_{Z-1}Y+{}^{0}_{+1}e+\nu_e. Gamma emission adds 00γ^{0}_{0}\gamma after an excited daughter.

Balance a reaction

Conserve total nucleon number and charge. For uranium-235 absorbing a neutron and producing xenon-140 and strontium-94, the remaining nucleon number identifies the emitted neutrons.

Common trap

Do not omit the neutrino or antineutrino when the syllabus asks for a complete beta-decay equation, and do not balance A while leaving charge unbalanced.

E.3.8 Exam Analysis

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Track Neutrinos in Beta Decay

Identify the neutral leptons

A neutrino νe\nu_e and an antineutrino νˉe\bar{\nu}_e are neutral, extremely low-mass leptons. They interact very weakly with matter, so they are difficult to detect directly.

Choose the correct particle

Beta-minus decay emits an electron and an electron antineutrino: n→p+e−+νˉen\rightarrow p+e^-+\bar{\nu}_e. Beta-plus decay emits a positron and an electron neutrino: p→n+e++νep\rightarrow n+e^++\nu_e (inside a nucleus).

Check lepton number

An electron has lepton number +1+1, so the accompanying antineutrino has −1-1. A positron has −1-1, so the accompanying neutrino has +1+1. Each beta reaction therefore keeps the initial total lepton number at zero.

Common trap

Do not swap the beta partners: β−\beta^- pairs with νˉe\bar{\nu}_e, while β+\beta^+ pairs with νe\nu_e. The continuous beta spectrum is treated separately in the HL objective E.3.18.

E.3.9 Exam Analysis

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Compare Radiation Types

Alpha radiation

Alpha particles are heavy and doubly charged. They interact strongly with matter, so they are highly ionizing but have low penetration and a short range in air.

Beta radiation

Beta particles are much lighter and singly charged. They are moderately ionizing and more penetrating than alpha particles, but can be deflected by electric and magnetic fields.

Gamma radiation

Gamma photons are neutral and travel at the speed of light in vacuum. They are weakly ionizing compared with alpha and beta, but have the greatest penetration.

Common trap

Do not rank penetration and ionization in the same order. The usual qualitative order is alpha > beta > gamma for ionization and gamma > beta > alpha for penetration.

E.3.10 Exam Analysis

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Track Activity and Half-Life

Define activity

Activity is the number of nuclear decays per unit time, measured in becquerels: one Bq is one decay per second. As the number of undecayed nuclei falls, activity falls.

Use half-life steps

After each half-life, half of the remaining nuclei survive: N=N0(1/2)nN=N_0(1/2)^n, where n=t/T1/2n=t/T_{1/2} is the number of half-lives elapsed.

Track count rate

If detector efficiency and background are unchanged, count rate is proportional to activity. Apply the same half-life scaling to the net count rate.

Common trap

Do not halve the original amount repeatedly without using the remaining amount, and do not confuse count rate with the number of nuclei when background is present.

E.3.11 Exam Analysis

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Calculate Half-Life Changes

Use integer half-lives

If activity changes from A0A_0 to AA, use A/A0=(1/2)nA/A_0=(1/2)^n to find the number of half-lives nn. For example, a fall to one-eighth means three half-lives.

Worked example — integer half-lives

A net count rate falls from 640 s−1640\,\mathrm{s^{-1}} to 80 s−180\,\mathrm{s^{-1}} in 18 h. Since 80/640=1/8=(1/2)380/640=1/8=(1/2)^3, three half-lives elapsed. Therefore T1/2=18/3=6.0 hT_{1/2}=18/3=6.0\,\mathrm{h}.

Find the half-life

Once nn is known, divide the elapsed time by nn: T1/2=t/nT_{1/2}=t/n. This is often quicker and clearer than starting with the exponential form.

Check the direction

A decay interval must reduce activity or count rate. If the calculated half-life or number of half-lives implies growth, revisit the ratio.

Common trap

Do not call a drop to one-eighth “one half-life”; half-life is the time for one factor of one-half.

E.3.12 Exam Analysis

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Correct for Background

Separate sample and background

A detector count rate can include decays from the sample plus background radiation. The measured rate is Rmeasured=Rsample+RbackgroundR_{measured}=R_{sample}+R_{background}.

Subtract before analysing

Estimate the background count rate with the source absent or from the long-time plateau, then calculate Rnet=Rmeasured−RbackgroundR_{net}=R_{measured}-R_{background}. Use the net rate for half-life comparisons.

Worked example — subtract, decay, restore

A detector reads 260 Bq260\,\mathrm{Bq} with a 20 Bq20\,\mathrm{Bq} background. The initial net rate is 240 Bq240\,\mathrm{Bq}. After four half-lives it is 240/16=15 Bq240/16=15\,\mathrm{Bq}, so the detector reads 15+20=35 Bq15+20=35\,\mathrm{Bq}.

Interpret a non-zero limit

If the measured rate approaches a non-zero constant, the remaining signal may be background radiation or a systematic detector contribution. The sample activity itself may have continued toward zero.

Common trap

Do not fit a half-life directly to a count rate that still contains background; the offset distorts the decay curve.

E.3.13 Exam Analysis

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Retrieve the SL Nuclear Model

Retrieve the nuclear structure

Isotopes differ in neutrons; mass defect becomes binding energy; the binding-energy curve explains why fusion and fission can release energy; and the strong force competes with electromagnetic repulsion.

Retrieve the decay model

Alpha, beta and gamma decays change A and Z differently. Radioactive decay is random but statistically predictable; use half-life, count-rate scaling and background correction carefully.