Physics › Nuclear physics › Radioactive decay and the decay constant
Radioactive decay and the decay constant
Each unstable nucleus carries a fixed probability of decay per second, unchanged by heating, pressure or chemical state, so decay is random. That probability is the decay constant λ, fixing the activity A = λN, the decay law N = N₀e^(−λt) and the half-life ln 2 over λ. A spreadsheet builds the same decay a step at a time.
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Radioactive decay and half-life, part 1 of 2. Part 2 is Half-life and nuclear instability.
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.
- Model a decay step by step from ΔN/Δt = −λN, and say why the model runs low and how to fix it.
COMMON MISCONCEPTION
After two half-lives, a sample has all gone.
Each half-life halves what remains rather than removing it all: after two half-lives a quarter of the nuclei are still undecayed, and the exponential model approaches zero without reaching it.
Randomness and the decay constant
Radioactive decay is random. Each unstable nucleus has a constant probability of decay per unit time, independent of its age, and nothing done to it (heat, pressure or chemistry) alters that probability. The probability per second is the decay constant λ. 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.
Watch Stable nuclei and photoelectric graphs, a worked question on the InkPhysics YouTube channel
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 larger decay constant corresponds to a shorter half-life. Because activity is proportional to N, it falls with the same half-life,
which is the form the detector actually sees. Two half-lives leave a quarter and three leave an eighth. The exponential model approaches zero asymptotically, while a real finite sample contains a discrete number of nuclei and may eventually contain no undecayed nuclei.
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, since halving at each half-life is the same equation evaluated at whole numbers of half-lives. Use whichever form suits the numbers given.
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.
The model has two limitations to state before a spreadsheet is offered 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.
Reading decay data has moved to the next part of this lesson.
Which nuclei decay, and how has moved to the next part of this lesson.
ASSESSMENT FOCUS
- The word random needs its definition attached: a 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. Add that the scatter is about √N to explain 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.
- 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.
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.
A larger decay constant corresponds to a shorter half-life. The decay time of a single nucleus is unpredictable, but a large sample has a statistically predictable exponential decay rate.
The exponential model approaches zero asymptotically. A real finite sample contains a discrete number of nuclei and may eventually contain no undecayed nuclei.
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 the decay constant 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 step by step.
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.
- Model a decay step by step from ΔN/Δt = −λN, and say why the model runs low and how to fix it.
Open the full revision checklist to track your progress across the whole unit.