University Physics IV · Nuclear Physics · 13.7
Decay Chains, Equilibrium & Counting Statistics
Every source you will ever count sits somewhere in a chain, and what the detector reports is the daughter as much as the parent. Here is how to solve the coupled equations, read off which equilibrium you are standing in, and fit a half-life to counts that scatter because they are Poisson, not because you were careless.
Build the model
Connect the measurement to the mechanism.
Radioactive decay has one physical input: a nucleus keeps no record of how long it has already existed, so its chance of decaying in the next dt is λ dt with λ constant. Separating variables in dN/dt = −λN turns that single assumption into the whole exponential law, and since a detector registers decays rather than atoms, the measured quantity is the activity A = λN, which falls with the same λ. A chain repeats the statement one level down: the daughter is fed at λ₁N₁ and drained at λ₂N₂, so dN₂/dt = λ₁N₁ − λ₂N₂, and one integrating factor gives the Bateman solution — a difference of two exponentials, the faster building the daughter up, the slower carrying it away.
Which lifetime is longer then decides everything: a short-lived daughter is dragged onto the parent's clock and settles at bλ₂/(λ₂ − λ₁) times the parent's activity, where b is the fraction of parent decays that feed it, while a long-lived daughter never settles at all. What the model costs is that every one of these curves is an ensemble mean. A single nucleus does not decay exponentially; it waits, then decays once, completely.
Counts in a fixed window are Poisson with standard deviation √N, so an honest half-life comes from a weighted fit to background-subtracted rates, and residuals scattering by a percent on ten thousand counts are the physics behaving, not the apparatus failing.
- Simple definition
- A decay chain is a sequence of nuclides in which each member is created by the decay of the one before it and removed by its own decay, so its population obeys a gain-minus-loss equation rather than a simple exponential.
- Example
- In ⁹⁹Mo (t½ = 65.9 h) → ⁹⁹ᵐTc (t½ = 6.01 h), the technetium activity climbs from zero, peaks 22.8 h after the generator is eluted, then holds at 0.963 of the molybdenum activity while both fall with the 65.9 h half-life.
λ is a probability per unit time, identical for a nucleus made a second ago and one that has already waited ten half-lives.
λ in s⁻¹, N a pure count, A in becquerel (1 Bq = 1 decay s⁻¹)
One mean life leaves 1/e = 37% and one half-life 50%, so τ is the later mark; ten half-lives leave 2⁻¹⁰ = 1/1024, the practical rule that a source is three orders down.
τ is the mean waiting time of one nucleus, t½ its median, so τ > t½
The faster exponential builds the daughter up and the slower one carries it away — which is which depends on the half-lives, not the order in the chain.
for N₂(0) = 0; multiply by the branching fraction b if only part of the parent feeds this daughter
This is the milking interval of a radionuclide generator: 22.8 h for ⁹⁹Mo → ⁹⁹ᵐTc, which is why such a column is eluted about once a day.
the peak is where feeding balances loss, bλ₁N₁ = λ₂N₂; b cancels out of tₘₐₓ
What goes constant is the ratio, not the activities: in secular equilibrium both members still decay, on the parent's clock.
holds for λ₂ > λ₁ once t ≫ 1/(λ₂ − λ₁); tends to b when λ₂ ≫ λ₁
Counting hands you its own error bar: 10 000 counts is 1.0%, and getting to 0.5% costs four times the counting time.
Ng gross counts in live time tg, Nb background counts in tb, rates in s⁻¹
One assumption, and the exponential is forced
A nucleus has no internal clock. Whatever it has been doing for the last ten half-lives, its chance of decaying in the next interval dt is λ dt, with λ a constant fixed by the nuclide alone — not by temperature, pressure, or chemical state, bar narrow exceptions that need the atomic electrons, such as electron capture, which a fully stripped ion cannot do at all. Applied to N identical nuclei this gives dN/dt = −λN, and separating variables, dN/N = −λ dt, integrates in one line to ln(N/N₀) = −λt, or N(t) = N₀e(−λt). Two time constants describe the same curve: the mean life τ = 1/λ is the average waiting time of a single nucleus, while the half-life t½ = ln2/λ = 0.693τ is its median, so τ = 1.443 t½ is always the longer. For ¹³¹I with t½ = 8.02 d, λ = 0.6931/8.02 d = 0.0864 d⁻¹ = 1.00 × 10⁻⁶ s⁻¹ and τ = 11.6 d. Nothing in the derivation mentions the sample size, which is why the same λ governs a milligram and a handful of atoms — even though only the milligram will trace a smooth exponential.
Activity is what the detector actually counts
You cannot count atoms; you count decays. The activity A = λN, in becquerel, is that decay rate, and because N falls exponentially so does A, with the same λ: A(t) = A₀e(−λt). Two consequences are worth carrying. First, activity and mass are only loosely related. A 4.5 MBq ¹³¹I capsule holds A/λ = 4.5 × 10¹² atoms, about 1 ng; the same 4.5 MBq of ²³⁸U needs 9.2 × 10²³ atoms, which is 362 g. The atom count is larger by exactly the ratio of decay constants, 2.0 × 10¹¹, and the mass by a further factor 238/131 for the heavier atoms — 3.7 × 10¹¹ in all. Second, no detector sees A. It records C = εA, where ε lumps together solid angle, intrinsic efficiency, self-absorption and the window — hard to pin down better than a few per cent, and usually not worth measuring, because as a constant multiplier it cancels from every ratio of counts. That is why a half-life comes out of raw count rates with no calibration at all, while an absolute activity in becquerel does not.
Two coupled equations and one integrating factor
A chain applies the same statement twice. The parent is unfed, so N₁(t) = N₁(0)e(−λ₁t). The daughter is fed by the parent at λ₁N₁ and drained by its own decay at λ₂N₂: dN₂/dt = λ₁N₁ − λ₂N₂. Move the loss term across, dN₂/dt + λ₂N₂ = λ₁N₁(0)e(−λ₁t), and multiply by the integrating factor e(λ₂t). The left side collapses to d(N₂e(λ₂t))/dt and the right becomes λ₁N₁(0)e((λ₂−λ₁)t), so integrating from 0 with N₂(0) = 0 and dividing back out gives N₂(t) = [λ₁N₁(0)/(λ₂ − λ₁)](e(−λ₁t) − e(−λ₂t)). Read the structure instead of memorising it. At small t the bracket rises linearly and the daughter accumulates at rate λ₁N₁(0), the parent's decay rate. At large t whichever exponential dies more slowly is left standing alone. And if only a fraction b of parent decays feed this particular daughter, every term picks up the same b, scaling the curve without shifting its peak.
Which lifetime is longer decides the regime
Divide the Bateman result by the parent to get A₂/A₁ = [bλ₂/(λ₂ − λ₁)](1 − e(−(λ₂−λ₁)t)), and three cases fall out. If λ₂ ≫ λ₁ — ²²⁶Ra at 1600 y feeding ²²²Rn at 3.82 d, with b = 1 — the bracket saturates within a few weeks and the prefactor is 1/(1 − t₂/t₁) = 1.0000065, so the radon activity matches the radium activity to seven parts in a million. That is secular equilibrium, and both members are still decaying, on the parent's 1600-year clock. If λ₂ > λ₁ but only modestly — ⁹⁹Mo at 65.9 h feeding ⁹⁹ᵐTc at 6.01 h — the same expression gives λ₂/(λ₂ − λ₁) = 1/(1 − 6.01/65.9) = 1.100: transient equilibrium, in which the daughter would run 10% above the parent if every parent decay fed it. Only 87.5% do, so the measured ⁹⁹ᵐTc activity settles at 0.875 × 1.100 = 0.963 of the ⁹⁹Mo activity — below its parent, and still falling on the parent's 65.9 h clock. And if λ₂ < λ₁ the prefactor is negative while the bracket runs negative too, their product growing without bound; there is no equilibrium, because the parent burns away and leaves the daughter decaying on its own longer clock. One line covers it: equilibrium exists only when the daughter is the shorter-lived member.
Fitting λ to counts that are Poisson
Counting is a Poisson process: a run yielding N counts has variance N and standard deviation √N, so the relative uncertainty is 1/√N — 1.0% at 10 000 counts, and 0.5% only after four times the counting time. Three things follow for a half-life measurement. Subtract the background first, as a rate rather than raw counts: a constant background flattens the tail and, fitted blind, biases λ low. Propagate both terms — for gross counts Ng in tg and background Nb in tb, the net rate is Ng/tg − Nb/tb with uncertainty √(Ng/tg² + Nb/tb²). Then weight the fit. Taking logs straightens the exponential, but it also converts an absolute uncertainty into a relative one, σ(ln r) = σᵣ/r ≈ 1/√N, so the late, weak points carry far the largest error bars. An unweighted straight line through ln(counts) treats them as equals and lets the noisiest data set the slope; weights w = 1/σ², or a Poisson maximum-likelihood fit, put that right. Curvature surviving all of this is usually physics — a second chain member, or dead time at high rates.
The exponential is a mean, not a trajectory
N(t) = N₀e(−λt) is an expectation value. A single nucleus does not follow it: it waits for a time drawn from the exponential distribution of mean τ = 1/λ, then decays once, completely. What varies smoothly is the count of a large ensemble. If each of N₀ nuclei survives independently with probability p = e(−λt), the number left is binomial with mean N₀p and standard deviation √(N₀p(1 − p)), which for small p is the familiar Poisson √N. The consequence is quantitative. A sample of 10¹² atoms fluctuates by of order one part in a million and traces a curve indistinguishable from the formula; twenty atoms of a superheavy nuclide give a jagged staircase whose half-life can only be got by maximum likelihood on the individual decay times. Read in reverse, this is the useful statement: when residuals scatter by 1% on 10 000 counts, the apparatus is behaving, and tightening the fit further is chasing noise.
Change one variable at a time
Make the relationship visible.
With b = 1, raise the half-life ratio r from 1.2 to 12 and watch the daughter's peak slide left while the late-time ratio falls from 6 towards 12/11 = 1.09 — transient equilibrium tightening into secular. Lowering b scales the whole daughter curve down without moving its peak. Below r = 1 the daughter outlives the parent and the ratio never settles at all.
PARENT A₁/A₀35.4 %
DAUGHTER A₂/A₀34.1 %
RATIO A₂/A₁0.964
DAUGHTER PEAKS AT0.35 t½
Live interpretationPARENT A₁/A₀: 35.4 %. DAUGHTER A₂/A₀: 34.1 %. RATIO A₂/A₁: 0.964. DAUGHTER PEAKS AT: 0.35 t½
Catch the common trap
Explain before calculating.
A freshly separated source contains only a parent of half-life 60 h, which decays entirely to a daughter of half-life 6.0 h. Once the initial transient has died away, what does the counting show?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA thyroid-therapy capsule of ¹³¹I (t½ = 8.02 d) is calibrated at 4.5 MBq. Find the decay constant in s⁻¹, the mean life, the number of ¹³¹I atoms present, and the activity 30 days later.
- λ = ln2/t½ = 0.6931/8.02 d = 0.08643 d⁻¹; dividing by 86 400 s per day gives λ = 1.00 × 10⁻⁶ s⁻¹.
- Mean life τ = 1/λ = 1.443 × 8.02 d = 11.6 d, longer than the half-life because the few long survivors drag the mean above the median.
- Activity fixes the population: N = A/λ = 4.5 × 10⁶ s⁻¹ ÷ 1.00 × 10⁻⁶ s⁻¹ = 4.5 × 10¹² atoms. At 131 g mol⁻¹ that is 4.5 × 10¹² × 131 ÷ 6.022 × 10²³ = 9.8 × 10⁻¹⁰ g, about one nanogram.
- After 30 d: λt = 0.08643 × 30 = 2.593, so A = 4.5 MBq × e(−2.593) = 4.5 × 0.0748 = 0.34 MBq — 7.5% of the calibrated activity, since 30 d is 3.74 half-lives.
Answerλ = 1.00 × 10⁻⁶ s⁻¹, τ = 11.6 d, N = 4.5 × 10¹² atoms (≈ 1 ng), and A(30 d) = 0.34 MBq.
MediumA ⁹⁹Mo generator (t½ = 65.9 h) is eluted at t = 0, leaving 20.0 GBq of ⁹⁹Mo and no ⁹⁹ᵐTc (t½ = 6.01 h). Only 87.5% of ⁹⁹Mo decays feed the metastable state. Find when the ⁹⁹ᵐTc activity peaks, how large it is then, and the ratio it settles to.
- λ₁ = 0.6931/65.9 h = 0.010518 h⁻¹ and λ₂ = 0.6931/6.01 h = 0.115332 h⁻¹, so λ₂/λ₁ = 10.965 and λ₂ − λ₁ = 0.104814 h⁻¹.
- tₘₐₓ = ln(λ₂/λ₁)/(λ₂ − λ₁) = ln(10.965)/0.104814 = 2.3947/0.104814 = 22.8 h. The branching fraction cancels out of this: b scales the daughter curve without moving its peak.
- At the peak the feed balances the loss, bλ₁N₁ = λ₂N₂, i.e. A₂ = bA₁. Here A₁(22.8 h) = 20.0 × e(−0.010518 × 22.8) = 20.0 × e(−0.2398) = 20.0 × 0.7868 = 15.74 GBq, so A₂ = 0.875 × 15.74 = 13.8 GBq.
- Long after the transient, A₂/A₁ → bλ₂/(λ₂ − λ₁) = 0.875 × 0.115332/0.104814 = 0.875 × 1.100 = 0.963, so the ⁹⁹ᵐTc stays a little below its parent and then falls with the parent's 65.9 h half-life, not its own 6.01 h one.
Answer⁹⁹ᵐTc peaks 22.8 h after elution at 13.8 GBq, then tracks the parent at A₂/A₁ = 0.963 — which is why the column is milked about once a day.
HardA detector, left untouched, records 8420 counts in 60 s at t = 0 and 2115 counts in 60 s at t = 120 min. A separate 600 s run with the source removed gives 1200 background counts. Find the half-life with its Poisson uncertainty, and say which measurement limits it.
- Background rate: b = 1200/600 = 2.000 s⁻¹, with ub = √1200/600 = 0.058 s⁻¹. Subtract it before anything else — a background left in flattens the tail and biases λ low.
- Net rates: n₁ = 8420/60 − 2.000 = 138.33 s⁻¹ and n₂ = 2115/60 − 2.000 = 33.25 s⁻¹, with u₁ = √(8420/60² + 1200/600²) = 1.530 s⁻¹ and u₂ = √(2115/60² + 1200/600²) = 0.769 s⁻¹.
- Efficiency and geometry are identical for both readings, so they cancel in the ratio: λ = ln(n₁/n₂)/Δt = ln(138.33/33.25)/7200 s = ln(4.1604)/7200 = 1.4256/7200 = 1.980 × 10⁻⁴ s⁻¹.
- Half-life: t½ = ln2/λ = 0.6931/1.980 × 10⁻⁴ = 3.50 × 10³ s = 58.3 min.
- Propagate: uλ/λ = √((u₁/n₁)² + (u₂/n₂)²)/ln(n₁/n₂) = √(0.01106² + 0.02312²)/1.4256 = 0.02563/1.4256 = 1.80%, so u(t½) = 0.0180 × 58.3 min = 1.0 min.
- The weak late point carries (0.02312/0.02563)² = 81% of the variance, so the cheapest improvement is a longer count at t = 120 min, not at t = 0.
Answert½ = 58.3 ± 1.0 min from Poisson statistics alone. Four fifths of the variance sits in the late, weak reading — count longer there.