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E.3

Radioactive decay

Theme E · Nuclear and quantum physics · SL and HL, with additional HL material marked

🎯What you need to be able to do

  • Describe the strong nuclear force and what makes a nucleus stable or unstable.
  • Describe alpha, beta and gamma radiation and compare their properties.
  • Write balanced decay equations.
  • Use half-life and activity, and describe background radiation.
  • HLUse the decay constant and the exponential decay law.

📚The physics

Why any nucleus holds together is the right place to start. Protons repel one another electrically, so something stronger must act at short range: the strong nuclear force, attractive between nucleons, roughly a hundred times stronger than the electric force but effective only over about \( 10^{-15} \) m. Because it is short-ranged while the electric repulsion is not, large nuclei become progressively harder to hold together — which is why every nucleus beyond bismuth is unstable.

The three radiations. An alpha particle is a helium nucleus: heavily ionising, stopped by paper or a few centimetres of air, deflected slightly in a magnetic field. A beta-minus particle is an electron from the nucleus: moderately ionising, stopped by a few millimetres of aluminium, deflected strongly and in the opposite direction to alpha. Gamma is a high-energy photon: weakly ionising, reduced but never fully stopped by lead, undeflected by fields because it is uncharged. The ordering is consistent — the more strongly ionising, the less penetrating, because ionising is how it loses energy.

Decay equations must balance nucleon number and proton number on both sides. Alpha decay drops \(A\) by 4 and \(Z\) by 2. Beta-minus decay leaves \(A\) unchanged and raises \(Z\) by 1, because a neutron becomes a proton plus an electron plus an antineutrino. Gamma emission changes neither — it simply carries away surplus energy from an excited nucleus.

The neutrino was predicted before it was found, and the reason is worth telling. Beta particles emerge with a continuous range of energies rather than the single value conservation of energy demands. Rather than abandon conservation of energy, Pauli proposed an unseen third particle sharing the energy. It took twenty-six years to detect. That is a better illustration of how physics actually works than most textbook set-pieces.

Half-life \( t_{1/2} \) is the time for half the undecayed nuclei to decay, or equivalently for the activity to halve. It is constant for a given isotope and completely unaffected by temperature, pressure or chemical state. Decay is a random process: we cannot say when a particular nucleus will decay, only what fraction of a large number will.

Activity is the number of decays per second, measured in becquerels.

HLThe decay law

The rate of decay is proportional to how many nuclei remain, \( \dfrac{dN}{dt} = -\lambda N \), which integrates to

\( N = N_0 e^{-\lambda t} \)
\( A = \lambda N \)
\( \lambda = \dfrac{\ln 2}{t_{1/2}} \)

On a log-linear plot the decay becomes a straight line of gradient \(-\lambda\), which is how half-lives are measured in practice.

Background radiation must be subtracted before any measurement is used. It comes mostly from natural sources — radon gas, rocks and soil, cosmic rays, and the potassium-40 in our own bodies — with a small medical and artificial contribution.

✏️Worked example

A sample of iodine-131, half-life 8.0 days, has an initial activity of \( 4.8 \times 10^{6} \) Bq.

(a) Activity after 24 days. That is exactly three half-lives, so the activity halves three times: \( 4.8 \rightarrow 2.4 \rightarrow 1.2 \rightarrow 0.60 \times 10^{6} \) Bq. Whole numbers of half-lives never need the exponential.

(b) HLDecay constant. \( \lambda = \ln 2/t_{1/2} = 0.693/(8.0 \times 24 \times 3600) = 0.693/691200 = 1.0 \times 10^{-6} \) s\(^{-1}\). Note the conversion of days to seconds, so that the activity in becquerels is per second.

(c) HLNumber of undecayed nuclei initially. From \( A = \lambda N \), \( N = A/\lambda = (4.8 \times 10^{6})/(1.0 \times 10^{-6}) = 4.8 \times 10^{12} \) nuclei.

(d) HLActivity after 20 days. Not a whole number of half-lives, so use the exponential: \( t = 20 \times 86400 = 1.728 \times 10^{6} \) s, and \( A = 4.8 \times 10^{6} \times e^{-(1.0 \times 10^{-6} \times 1.728 \times 10^{6})} = 4.8 \times 10^{6} \times e^{-1.73} = 8.5 \times 10^{5} \) Bq.

Sanity check. Twenty days is between two and three half-lives, so the answer must lie between \( 1.2 \times 10^{6} \) and \( 0.60 \times 10^{6} \). It does. Always bracket an exponential answer with the neighbouring half-lives before trusting it.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. What fraction of a radioactive sample remains undecayed after four half-lives?
The amount halves each time: \( \tfrac{1}{2} \to \tfrac{1}{4} \to \tfrac{1}{8} \to \tfrac{1}{16} \). So \( 1/16 \), or 6.25%.
2. A source of half-life 5.0 minutes has an initial activity of \( 8.0 \times 10^{5} \) Bq. Find its activity after 15 minutes.
15 minutes is exactly three half-lives: \( 8.0 \to 4.0 \to 2.0 \to 1.0 \times 10^{5} \) Bq. A whole number of half-lives never needs the exponential.
3. HLFind the decay constant of that source, in s\(^{-1}\).
\( \lambda = \dfrac{\ln 2}{t_{1/2}} = \dfrac{0.693}{300} = 2.3 \times 10^{-3} \) s\(^{-1}\). The half-life must be in seconds for the activity to come out in becquerels.
4. Complete the alpha decay of uranium-238: \( ^{238}_{92}\mathrm{U} \rightarrow\ ? \)
Alpha decay drops \(A\) by 4 and \(Z\) by 2: \( ^{238}_{92}\mathrm{U} \rightarrow\ ^{234}_{90}\mathrm{Th} + ^{4}_{2}\mathrm{He} \). Both nucleon number and proton number balance.
5. Complete the beta-minus decay of carbon-14, including all emitted particles.
\( ^{14}_{6}\mathrm{C} \rightarrow\ ^{14}_{7}\mathrm{N} + ^{0}_{-1}\mathrm{e} + \bar{\nu} \). A neutron becomes a proton plus an electron plus an antineutrino, so \(A\) is unchanged and \(Z\) rises by one. The antineutrino must be included — it was proposed precisely because beta particles emerge with a continuous range of energies.
6. HLA sample starts with \( 1.0 \times 10^{12} \) undecayed nuclei and has \( \lambda = 2.31 \times 10^{-3} \) s\(^{-1}\). How many remain after 600 s?
\( N = N_0e^{-\lambda t} = 1.0 \times 10^{12} \times e^{-(2.31 \times 10^{-3} \times 600)} = 1.0 \times 10^{12} \times e^{-1.386} = 2.5 \times 10^{11} \). Check: 600 s is exactly two half-lives, and a quarter remains, as it should.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • HyperPhysics — radioactivity, decay modes and half-life
  • xkcd — the radiation dose chart, for a sense of scale of real exposures