The decay law, activity, and decay chains

Let P(t) be the probability that one particular nucleus is still undecayed at time t, and let its chance of decaying in the next interval dt be λ dt. Surviving to t + dt means surviving to t and then not decaying, so P(t + dt) = P(t)(1 − λ dt), which gives dP/dt = −λP and, with P(0) = 1, P(t) = e^(−λt). Read the other way, the conditional survival P(t + s)/P(t) = e^(−λs) does not depend on t: a nucleus that has lasted a week has exactly the odds of a fresh one. The only continuous functions with P(t + s) = P(t)P(s) are exponentials, so memorylessness and the exponential are one statement. Lifetimes then have the probability density λe^(−λt), normalised because ∫₀^∞ λe^(−λt) dt = 1, with mean τ = ∫₀^∞ t λe^(−λt) dt = 1/λ. A population that wears out behaves differently: adult human mortality roughly doubles every eight years, which is why nobody quotes a half-life for people.

With N₀ independent nuclei, the expected number left is N(t) = N₀P(t). The rate equation gives the same result directly: in dt the expected number of decays is λN dt, so dN = −λN dt. Separate the variables, dN/N = −λ dt, and integrate from (0, N₀) to (t, N): ln(N/N₀) = −λt, so N = N₀e^(−λt). Setting N = N₀/2 gives t½ = ln 2/λ = 0.693τ, the same whenever the clock starts. Check the result three ways. Dimensions: λt must be a pure number, so λ is in s⁻¹. Limits: λ → 0 gives a stable nucleus, N = N₀ for ever, and small t gives N ≈ N₀(1 − λt). Shape: ln N = ln N₀ − λt is a straight line, so a semi-log plot turns the decay constant into a slope. For ²²⁶Ra, t½ = 1600 years gives λ = 1.37 × 10⁻¹¹ s⁻¹: a given radium nucleus has about one chance in 7 × 10¹⁰ of decaying in the next second.

A detector responds to the decay rate, the activity A = −dN/dt = λN, measured in becquerels (1 Bq = 1 s⁻¹). Because A carries the factor λ, equal activities can hide wildly different amounts of material. One megabecquerel of ⁹⁹ᵐTc (t½ = 6.01 h) is N = A/λ = 3.12 × 10¹⁰ nuclei, about 5 pg; one megabecquerel of ²³⁸U (t½ = 4.468 × 10⁹ years) is 2.03 × 10²³ nuclei, about 80 g. Specific activity, the activity per unit mass λN_A/M, makes the same point: for ²²⁶Ra it is 3.66 × 10¹⁰ Bq g⁻¹, which is why the curie, 3.7 × 10¹⁰ Bq, began as the activity of one gram of radium. A counter never records A itself, only a count rate R = εA, where the efficiency ε folds in geometry and detector response. Because ε is the same at every time, a ratio of count rates is a ratio of activities, and λ can be measured without ever knowing ε.

Each nucleus in a large sample decays within a counting window with some small probability p, independently of the rest, so the count is binomial with mean Np and, for huge N and small p, Poisson: its variance equals its mean, and a count n carries σ = √n. Subtracting a background adds variances. Two equal windows a time Δt apart give λ = ln(n₁/n₂)/Δt, and since var(ln n) ≈ 1/n, σ_λ ≈ √(1/n₁ + 1/n₂)/Δt, dominated by the smaller, later count. That sets the design. Wait too briefly and ln(n₁/n₂) is lost in noise; wait too long and n₂ starves. For a fixed first count σ_λ is least at λΔt = 2.2, about 3.2 half-lives, and anywhere from 1.5 to 3 mean lives costs under 10% more. With many windows, fit ln n against t with weights n. The same statistics explain why the law looks deterministic: the survivors are binomial, with relative scatter √((1 − P)/(N₀P)), which vanishes as N₀ grows.

In a chain 1 → 2 → …, the daughter gains nuclei at the parent's decay rate and loses them at its own: dN₂/dt = λ₁N₁ − λ₂N₂, with N₁ = N₁(0)e^(−λ₁t). The equation is first-order and linear, so multiply by the integrating factor e^(λ₂t): d(N₂e^(λ₂t))/dt = λ₁N₁(0)e^{((λ₂ − λ₁)t)}. Integrating from 0 gives N₂(t) = N₁(0)λ₁[e^(−λ₁t) − e^(−λ₂t)]/(λ₂ − λ₁) + N₂(0)e^(−λ₂t), the last term being any daughter already present, decaying on its own. Check it: at t = 0 it returns N₂(0); for a pure parent the initial slope is λ₁N₁(0), feeding with no drain yet; and as λ₂ → λ₁ the bracket over (λ₂ − λ₁) tends to te^(−λ₁t), so N₂ = N₁(0)λ₁te^(−λ₁t) stays finite. For a pure parent the daughter's activity is greatest where dN₂/dt = 0, that is where λ₁N₁ = λ₂N₂. At its peak, then, the daughter is exactly as active as its parent, at tₘₐₓ = ln(λ₂/λ₁)/(λ₂ − λ₁).

Which exponential in N₂ dies first decides the late-time behaviour. If the daughter is the shorter-lived, λ₂ > λ₁, then e^(−λ₂t) goes first and A₂/A₁ → λ₂/(λ₂ − λ₁), a constant above 1. This is transient equilibrium: the pair decays together at the parent's rate with the daughter the more active, because its stock was built when the parent was more active than it is now. The fission products ¹³²Te (3.20 d) and ¹³²I (2.30 h) come within 0.1% of their limiting ratio, 1.031, in a day, although the number ratio N₂/N₁ = λ₁/(λ₂ − λ₁) is only 0.031. When λ₂ ≫ λ₁ the ratio tends to 1, secular equilibrium: radon in a sealed radium source reaches 99.6% of the radium activity in 30 days, and since every member of such a chain has the same activity, each population is proportional to its half-life. If instead λ₂ < λ₁, the parent dies first, A₂/A₁ grows without limit, and the daughter ends by decaying at its own rate. None of this predicts the λs or branching ratios: those are data.