Physics › Nuclear physics › Radioactive decay and half-life
Radioactive decay and half-life
A nucleus decays at random, with a fixed probability per second that heating, pressure and chemical state do not change. One constant, the decay constant λ, sets that probability, and from it come the exponential decay law, the half-life, and a log-linear plot of the same shape you met in capacitor discharge.
Pick your board and the few notes written for the other boards quietly fold away, here and in the practice players. Nothing is deleted: every folded piece reopens on a tap.
Builds on The time constant and exponential decay and Stable and unstable nuclei.
The maths behind it: Solving differential equations on InkMaths.
IN THIS TOPIC
- Give the experimental evidence that decay is random, and say why an activity is a statistical quantity.
- Use λ, A = λN and the exponential decay equations.
- Convert a mass into a number of nuclei with molar mass and the Avogadro constant.
- Determine a half-life from decay curves and from log graphs.
- Model a decay step by step from ΔN/Δt = −λN, and say why the model runs low and how to fix it.
- Predict the decay mode of a nuclide from its position on the N against Z graph.
- Write balanced decay equations, and account for gamma emission from an excited state.
COMMON MISCONCEPTION
After two half-lives, a sample has all gone.
Random for one, dependable for a mole
Radioactive decay is random. No measurement can tell you when a given nucleus will go, and nothing you do to it, heat, pressure or chemistry, alters its chances. What each nucleus carries instead is a fixed decay constant λ, the probability of decaying per second. Model it with dice. Each die has no memory at all, yet a large enough handful loses very nearly one sixth of its number per throw.
None of that has to be taken on trust, because a counter shows it to you in a minute. Point a Geiger tube at a steady source and record the number of counts in one ten-second interval after another. The readings do not repeat: 412, then 380, then 401, then 435. Nothing about the experiment has changed between them, so every one of those differences is the randomness of individual decays showing through. Those fluctuations in count rate are the experimental evidence that decay is random. Were each nucleus keeping to a schedule, every interval would return the same number.
The scatter has a predictable size even though no single reading does. A count of N carries an uncertainty of about √N, so a mean of 400 scatters by about 20, one part in twenty, while a mean of 10 000 scatters by about 100, one part in a hundred. Count for longer and the fractional scatter shrinks, which is why a serious measurement counts for minutes rather than seconds. It also settles what kind of quantity an activity is. It is a statistical average from the outset, and quoting one as though it were an exact reading claims a precision the physics does not offer.
With N undecayed nuclei each carrying probability λ per second, the expected rate of loss is
and the size of that rate is the activity A, measured in becquerels, one decay per second.
Counting nuclei is chemistry's job. A mass m of a nuclide with molar mass M holds m/M moles, so N = (m/M) × NA nuclei, with the Avogadro constant from the data booklet. Exam questions lean on this chain often.
The exponential and its half-life
A rate of loss proportional to the amount remaining is the recipe you met with capacitor discharge, and it has the same solution.
The half-life is the time for N to halve. Set N = N0/2 and take logs. From e-λT = 1/2 comes λT = ln 2, so
A big decay constant means a short half-life and a fierce source. Because activity is proportional to N, it decays with the same clock,
which is the form the detector actually sees. Two half-lives leave a quarter and three leave an eighth, and halving forever never reaches nothing.
WORKED EXAMPLE
Iodine-131, by the ladder and the law
Iodine-131 has a half-life of 8.0 days. Find its decay constant, and the fraction of a sample remaining after 24 days.
λ = ln 2/T½ = 0.693/(8.0 × 86 400) = 1.0 × 10−6 s−1. The conversion into seconds is needed here because the answer feeds an activity in becquerels, which are per second. λ itself may sit in any reciprocal-time unit, days included, so long as t is measured in the matching unit.
Twenty-four days is exactly three half-lives, so the ladder answers without the exponential: (½)3 = one eighth.
The exponential agrees, as it must; the ladder is just the exponential sampled at its own rungs, and choosing whichever tool fits the numbers is the skill.
Reading decay data
From a decay curve, read the half-life directly. Find the time for the activity to halve, then check it again from a different starting point, because a genuine exponential halves in the same time everywhere along its length.
For scattered data the better tool is the promised twin of the capacitor practical. Taking natural logs gives ln N = ln N0 − λt, a straight line of ln N against t with gradient −λ. Straightness is the evidence the decay is exponential, and the gradient gives λ, from which T½ = ln 2/λ follows. Symbol for symbol, this is the RP9 analysis with λt in place of t/RC.
The half-life sets the applications. Radioactive waste must be stored securely for many half-lives of its longest-lived components, which for some isotopes means centuries. Dating runs the clock backwards. Living things hold a known fraction of carbon-14, so the fraction still left in a bone, with T½ = 5730 years, dates its death.
GUIDED PRACTICE
A half-life from a table
A counter records corrected activities of 1200, then 300 counts per minute, sixteen minutes apart. Find the half-life.
Show the working
1200 down to 300 is a fall to one quarter, so two half-lives have passed.
So T½ = 16/2 = 8.0 minutes. Counting halvings before reaching for logarithms turns many decay questions into arithmetic.
INDEPENDENT PRACTICE
Dating a fire
Charcoal from an ancient hearth shows a carbon-14 activity one quarter that of living wood. Estimate the age of the fire. (T½ = 5730 years.)
Show the working
One quarter remaining means two half-lives since the wood stopped exchanging carbon.
Age ≈ 2 × 5730 = 11 000 years, a hearth from the end of the last ice age. The clock was set by death and read by decay.
Running the decay forwards
ΔN/Δt = −λN is a statement about the next instant rather than about the whole decay, and a spreadsheet can march it forwards without ever solving it. Choose an interval Δt short enough that only a small fraction of the sample goes in it. Take the undecayed count at the start of the interval, work out the loss as ΔN = −λNΔt, subtract it, and repeat down the column. The exponential appears as a result rather than an assumption, and a half-life read off the modelled curve agrees with ln 2/λ.
The model carries the same built-in error as the capacitor version of it, for the same reason. Across each step the decay rate is held at the value it had at the beginning, while the real sample's rate falls throughout the step, so every step removes slightly too many nuclei and the modelled curve runs below the true exponential.
WORKED EXAMPLE
A modelled sample against the equation
A source has λ = 0.050 s−1. Model it with steps of 1.0 s and compare the fraction left after 20 s with the exponential's answer, then repeat with steps of 0.10 s.
Each 1.0 s step removes λNΔt, which is 5.0% of whatever is present, so the model multiplies the count by 0.95 once per step. Twenty steps give 0.9520 = 0.359 of the sample.
The equation gives e−λt = e−1.0 = 0.368, so the model is about 2.5% low.
At Δt = 0.10 s each step takes 0.50%, and 0.995200 = 0.367, within a third of a per cent. The quantity to watch is λΔt, the fraction lost per step: keep it below about 0.01 and the model tracks the equation more closely than any counter could tell.
Two things the model cannot do are worth saying out loud before anyone offers a spreadsheet as evidence. It cannot fluctuate: its column is the expected count, smooth to as many decimal places as you keep, while a real sample scatters about that expectation by roughly √N. And it will report a third of a nucleus, because the arithmetic goes on working long after the physics has stopped. To check the model against the analysis, log the modelled counts and plot them against time; the line should be straight with a gradient of magnitude λ.
Which nuclei decay, and how
Plot every stable nuclide as a point with proton number Z across and neutron number N up, and they hug a narrow stable band, with N roughly equal to Z for light nuclei and bending neutron-rich as Z grows. Extra neutrons add strong-force glue without adding proton repulsion, and that is the reason for the bend.
Position on the map predicts the decay. Neutron-rich nuclei, above the band, undergo β⁻ decay, in which a neutron becomes a proton, so N falls by 1 and Z rises by 1. ⁹⁰Sr becomes ⁹⁰Y, Z climbing from 38 to 39, plus e⁻ and an antineutrino e. Proton-rich nuclei, below the band, travel the other way by β⁺ decay or by electron capture, each turning a proton into a neutron. The heaviest nuclei shed bulk instead. In α decay, ²²⁶Ra becomes ²²²Rn with Z falling from 88 to 86, plus ⁴He, so A drops by 4 and Z by 2. In every equation, A and Z balance across the arrow.
Decay often leaves the daughter in an excited state. Nuclei have energy levels much as atoms do, drawn as nuclear energy level diagrams, and the drop to the ground state emits a γ photon. Medicine exploits one such state deliberately. Technetium-99m is a long-lived excited state that emits gamma alone, with a six hour half-life. Injected as a tracer, its photons escape the body to a camera, and within a couple of days the activity has fallen to almost nothing.
ASSESSMENT FOCUS
- The word random earns its mark only with a definition attached. Constant probability of decay per nucleus per unit time, unaffected by external conditions. Say probability, not chance.
- Asked for evidence of randomness rather than a definition, give the measurement: repeated counts over equal intervals from an unchanging source scatter about a mean instead of repeating. The phrase that scores is scatter about a mean, and adding that the scatter is about √N explains why long counts are worth the wait.
- Convert mass to nuclei through moles. Divide by the molar mass, then multiply by the Avogadro constant. Grams and kilograms trip this chain, and the booklet's Nₐ is per mole.
- For a half-life from a log graph the gradient is −λ, so read its magnitude and use T½ = ln 2/λ. Quote the straightness of the line as your evidence that the decay is exponential.
- Asked to model ΔN/Δt = −λN a step at a time, give the recipe and its flaw together. Each row loses λNΔt of the count the row above it held, and holding the rate fixed across a step in which it is really falling makes the model decay a little too fast, which is cured by shrinking Δt until λΔt is small.
- Decay equations balance twice. Nucleon numbers across the top, proton numbers across the bottom, and the antineutrino written in for β⁻.
- Technetium-99m answers want three properties. Gamma only, so it leaves the body without heavy ionisation. A six hour half-life, long enough to image and short enough to clear. Emission from an excited nuclear state.
- Every measured count rate includes background. Subtract the background, measured with the source removed, before any half-life or activity calculation, because markers look for that correction line in the working.
CHECK YOURSELF
Caesium-137 has a half-life of 30 years. A sealed source contains 3.0 μg of caesium-137 (molar mass 137 g mol⁻¹). Find the decay constant and the activity of the source. Take 1 year = 3.15 × 10⁷ s.
Show a hint
Half-life to λ first; then mass to moles to nuclei; then A = λN.
Show the answer
T½ = 30 × 3.15 × 107 = 9.45 × 108 s, so λ = ln 2 / (9.45 × 108) = 7.3 × 10-10 s-1.
N = (3.0 × 10-6 / 137) × 6.02 × 1023 = 1.3 × 1016 nuclei.
A = λN = 7.3 × 10-10 × 1.3 × 1016 = 9.7 × 106 Bq. Ten million decays a second, out of three millionths of a gram.
One nucleus is a coin toss; a mole of them is a clock.
Every half-life keeps the same fraction, so the count halves forever and never reaches zero.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their mark schemes on the radioactive decay and half-life questions page.
WHERE TO GO NEXT
- Required practical 12: the inverse-square law for gamma radiation puts this topic in the lab, and the written papers ask about it.
- Exponentials and logarithms is the maths this lesson leans on, worked through from GCSE.
CHECK YOUR PROGRESS
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- Give the experimental evidence that decay is random, and say why an activity is a statistical quantity.
- Use λ, A = λN and the exponential decay equations.
- Convert a mass into a number of nuclei with molar mass and the Avogadro constant.
- Determine a half-life from decay curves and from log graphs.
- Model a decay step by step from ΔN/Δt = −λN, and say why the model runs low and how to fix it.
- Predict the decay mode of a nuclide from its position on the N against Z graph.
- Write balanced decay equations, and account for gamma emission from an excited state.
Open the full revision checklist to track your progress across the whole unit.