---
title: Serial Decay, the Bateman Equations, and Radioactive Equilibrium
module: Radioactive Decay
moduleNumber: 4
lessonNumber: 2
order: 402
summary: >
  A radioactive parent that decays into a radioactive daughter obeys a coupled
  pair of rate equations whose solution is the Bateman formula. Depending on the
  half-life ordering the chain settles into secular equilibrium (equal
  activities), transient equilibrium (a fixed activity ratio), or no equilibrium.
  Constant production under irradiation drives the activity toward a saturation
  value equal to the production rate, competing decay modes split the total
  decay constant into partial constants, and the natural decay series in secular
  equilibrium underpin radiometric dating.
topics: [Radioactive Decay]
sources:
  - book: Krane
    ref: "Ch. 6 — Radioactive Decay; §6.3 Production and Decay, §6.4 Growth of Daughter Activities, §6.5 Types of Decays"
  - book: Wong
    ref: "Ch. 3 — Nuclear Reactions and Decays (decay kinetics context)"
draft: false
---

The exponential law $N(t) = N_0 e^{-\lambda t}$ describes a single unstable species
in isolation. Most radioactive nuclei are not isolated: the daughter of one decay
is frequently unstable itself, so activity flows through a chain, and a nucleus
under a neutron beam is produced and destroyed at the same time. Each of these
situations is a linear rate equation with a source term, and the family of
solutions organizes the observable activities of every decay series.

## Activity and its units

The **activity** of a sample is the number of disintegrations per unit time,

$$
A(t) = \lambda N(t) = -\frac{\d N}{\d t},
$$

not the number of nuclei present. A detector counts decays, so activity, not
$N$, is the measured quantity. For a single species $A(t) = A_0 e^{-\lambda t}$
with the same decay constant as the population.

> **Definition (Units of activity).** The SI unit is the **becquerel**,
> $1\,\mathrm{Bq} = 1\ \text{decay/s}$. The older **curie** is
> $1\,\mathrm{Ci} = 3.7 \times 10^{10}\,\mathrm{Bq}$, originally the activity of
> one gram of $^{226}\mathrm{Ra}$. The **specific activity** is the activity per
> unit mass, $a = \lambda N_A / M$ for a pure sample of molar mass $M$, and rises
> as the half-life falls.

A short half-life packs a large activity into few atoms: $1\,\mu\mathrm{g}$ of
$^{131}\mathrm{I}$ ($t_{1/2} = 8.02\,\mathrm{d}$) carries about $4.6\,\mathrm{GBq}$,
while $1\,\mathrm{g}$ of $^{238}\mathrm{U}$ ($t_{1/2} = 4.47 \times 10^9\,\mathrm{y}$)
carries only $12\,\mathrm{kBq}$.

## The two-member chain

Let a parent (species 1) decay to a daughter (species 2) that itself decays to a
stable end product. Writing $N_1, N_2$ for the populations and $\lambda_1,
\lambda_2$ for the decay constants,

$$
\frac{\d N_1}{\d t} = -\lambda_1 N_1,
\qquad
\frac{\d N_2}{\d t} = \lambda_1 N_1 - \lambda_2 N_2.
$$

The daughter is fed at the rate the parent decays, $\lambda_1 N_1$, and removed at
its own rate $\lambda_2 N_2$. The first equation gives $N_1(t) = N_1(0)\,
e^{-\lambda_1 t}$. Substituting into the second yields a linear first-order
equation with an exponential source,

$$
\frac{\d N_2}{\d t} + \lambda_2 N_2 = \lambda_1 N_1(0)\,e^{-\lambda_1 t}.
$$

Multiplying by the integrating factor $e^{\lambda_2 t}$ makes the left side an
exact derivative, $\frac{\d}{\d t}\!\left(N_2 e^{\lambda_2 t}\right) =
\lambda_1 N_1(0)\, e^{(\lambda_2 - \lambda_1)t}$. Integrating from $0$ to $t$ with
the initial condition $N_2(0) = 0$ gives the two-member result,

$$
N_2(t) = N_1(0)\,\frac{\lambda_1}{\lambda_2 - \lambda_1}
\left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right).
$$

The daughter activity follows as $A_2 = \lambda_2 N_2$,

$$
A_2(t) = A_1(0)\,\frac{\lambda_2}{\lambda_2 - \lambda_1}
\left(e^{-\lambda_1 t} - e^{-\lambda_2 t}\right),
\qquad
A_1(0) = \lambda_1 N_1(0).
$$

Starting from zero, the daughter activity rises as its stock builds, reaches a
maximum, and then falls once the parent supply thins. The maximum sits where
$\d N_2/\d t = 0$, that is where production balances loss, $\lambda_1 N_1 =
\lambda_2 N_2$:

$$
t_{\max} = \frac{\ln(\lambda_2/\lambda_1)}{\lambda_2 - \lambda_1}.
$$

At $t_{\max}$ the two activities are momentarily equal, $A_1(t_{\max}) =
A_2(t_{\max})$.

$$
% caption: The daughter population starts at zero, climbs while production
% lambda1 N1 exceeds loss lambda2 N2, and peaks at t-max where the two rates
% balance, after which it tracks the decaying parent.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.8,0) node[right, black!70] {$t$};
  \draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {activity};
  % parent: exponential decline
  \draw[black, very thick] (0,3.9)
    .. controls (1.2,2.9) and (2.4,2.25) .. (3.6,1.85)
    .. controls (5.0,1.4) and (6.2,1.15) .. (7.4,0.98);
  \node[black, anchor=west, font=\scriptsize] at (6.0,1.55) {parent};
  % daughter: rise then fall, peak at t_max touching parent
  \draw[acc, very thick] (0,0.05)
    .. controls (0.8,1.55) and (1.6,2.3) .. (2.6,2.4)
    .. controls (3.8,2.5) and (5.2,1.5) .. (7.4,0.9);
  \node[acc, anchor=south west, font=\scriptsize] at (2.7,2.42) {daughter};
  % t_max marker
  \draw[black, dashed] (2.6,0) -- (2.6,2.4);
  \node[black!70, anchor=north, font=\scriptsize] at (2.6,0) {$t_{\max}$};
\end{tikzpicture}
$$

## The Bateman equations

For a straight chain $1 \to 2 \to \cdots \to n$ with only the head populated at
$t = 0$, the same integrating-factor method applied down the chain produces the
**Bateman solution**,[^bateman]

$$
N_n(t) = N_1(0)\left(\prod_{i=1}^{n-1}\lambda_i\right)
\sum_{i=1}^{n}
\frac{e^{-\lambda_i t}}{\displaystyle\prod_{\substack{j=1 \\ j\neq i}}^{n}(\lambda_j - \lambda_i)}.
$$

Each member contributes a decaying exponential in every downstream population,
weighted so that the initial conditions $N_k(0) = 0$ for $k \geq 2$ hold. The
formula assumes distinct decay constants; equal constants are handled by a
limiting form. The two-member result above is the $n = 2$ case.

$$
% caption: A serial chain: each species is produced by the decay of the one
% before it and removed by its own decay, so the source term of stage k is the
% loss term of stage k minus one.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \tikzset{spec/.style={draw=acc, fill=acc!10, minimum width=1.5cm, minimum height=0.8cm, font=\footnotesize}}
  \node[spec] (a) at (0,0) {parent};
  \node[spec] (b) at (2.6,0) {daughter};
  \node[spec] (c) at (5.2,0) {grand};
  \node[spec] (d) at (7.8,0) {stable};
  \draw[->, acc, thick] (a) -- (b);
  \draw[->, acc, thick] (b) -- (c);
  \draw[->, acc, thick] (c) -- (d);
  \node[black, font=\scriptsize, anchor=south] at (1.3,0.42) {decay 1};
  \node[black, font=\scriptsize, anchor=south] at (3.9,0.42) {decay 2};
  \node[black, font=\scriptsize, anchor=south] at (6.5,0.42) {decay 3};
\end{tikzpicture}
$$

## Radioactive equilibrium

The long-time behavior of a two-member chain depends entirely on which member
lives longer. For $t \gg 1/\lambda_2$ the term $e^{-\lambda_2 t}$ vanishes and the
activity ratio approaches a limit set by the two decay constants:

$$
\frac{A_2}{A_1} = \frac{\lambda_2}{\lambda_2 - \lambda_1}
\left(1 - e^{-(\lambda_2 - \lambda_1)t}\right)
\;\xrightarrow{\;t\to\infty\;}\;
\frac{\lambda_2}{\lambda_2 - \lambda_1}
\quad(\lambda_2 > \lambda_1).
$$

Three regimes follow from the sign and size of $\lambda_2 - \lambda_1$.

- **Secular equilibrium** ($\lambda_1 \ll \lambda_2$, parent effectively stable on
  the daughter's timescale). The ratio tends to $\lambda_2/(\lambda_2-\lambda_1)
  \approx 1$, so $A_2 \to A_1$: the daughter decays exactly as fast as it is
  produced, $\lambda_1 N_1 = \lambda_2 N_2$. Every member of a long chain headed by
  a very long-lived parent reaches the same activity.
- **Transient equilibrium** ($\lambda_1 < \lambda_2$, comparable). The ratio tends
  to the constant $\lambda_2/(\lambda_2-\lambda_1) > 1$, so the daughter activity
  settles above the parent's and both then fall with the parent's slow constant
  $\lambda_1$.
- **No equilibrium** ($\lambda_1 > \lambda_2$, parent shorter-lived). No constant
  ratio exists. The parent disappears first, and the accumulated daughter
  afterward decays with its own constant $\lambda_2$.

$$
% caption: Secular equilibrium: with a nearly stable parent the daughter activity
% climbs to match the parent, and the two curves coincide once the daughter is
% destroyed as fast as it is made.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.8,0) node[right, black!70] {$t$};
  \draw[->, black] (0,0) -- (0,4.0) node[above, black!70] {activity};
  % parent nearly flat (very long-lived), slight decline
  \draw[black, very thick] (0,3.2) -- (7.4,3.05);
  \node[black, anchor=south east, font=\scriptsize] at (7.4,3.05) {parent};
  % daughter rising toward parent
  \draw[acc, very thick] (0,0.05)
    .. controls (0.9,2.15) and (2.0,2.95) .. (3.4,3.08)
    .. controls (4.6,3.12) and (6.0,3.08) .. (7.4,3.02);
  \node[acc, anchor=north west, font=\scriptsize] at (2.4,2.7) {daughter};
  % equal-activity annotation
  \draw[black, dashed] (5.4,0) -- (5.4,3.05);
  \node[black!70, anchor=north, font=\scriptsize] at (5.4,0) {equal activities};
\end{tikzpicture}
$$

Secular equilibrium is the working principle of the natural series: in an ancient
uranium ore every daughter down to the stable lead end has an activity equal to
that of $^{238}\mathrm{U}$, so counting one member measures the whole chain. A
transient-equilibrium pair is exploited in the $^{99}\mathrm{Mo}/^{99\mathrm{m}}
\mathrm{Tc}$ generator: the $66\,\mathrm{h}$ molybdenum parent feeds the
$6.0\,\mathrm{h}$ technetium daughter, which is eluted for imaging while its
activity tracks the parent.

$$
% caption: Transient equilibrium on a logarithmic activity axis: the daughter
% rises, peaks at t-max where it equals the parent, and thereafter runs parallel
% to and above the parent, both falling with the parent constant.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.8,0) node[right, black!70] {$t$};
  \draw[->, black] (0,0) -- (0,4.4) node[above, black!70] {log activity};
  % parent straight declining line (semilog)
  \draw[black, very thick] (0.2,4.0) -- (7.4,1.15);
  \node[black, anchor=north east, font=\scriptsize] at (7.2,1.35) {parent};
  % daughter rises, peaks touching parent at t_max, then parallel above
  \draw[acc, very thick] (0.2,0.5)
    .. controls (1.0,2.7) and (1.8,3.2) .. (2.4,3.22)
    .. controls (3.6,3.1) and (5.4,2.3) .. (7.4,1.55);
  \node[acc, anchor=south, font=\scriptsize] at (3.6,3.0) {daughter};
  % t_max marker
  \draw[black, dashed] (2.4,0) -- (2.4,3.22);
  \node[black!70, anchor=north, font=\scriptsize] at (2.4,0) {$t_{\max}$};
\end{tikzpicture}
$$

## Production and decay under irradiation

A target held in a reactor or accelerator beam is a source with a constant
production rate. Let $R$ be the number of daughter nuclei formed per unit time
(for a thin target, $R = \phi\,\sigma\,N_{\text{target}}$ with flux $\phi$ and
cross section $\sigma$), and let the product decay with constant $\lambda$. The
population obeys

$$
\frac{\d N}{\d t} = R - \lambda N,
\qquad N(0) = 0,
$$

whose solution is $N(t) = \frac{R}{\lambda}\!\left(1 - e^{-\lambda t}\right)$. The
activity is

$$
A(t) = \lambda N(t) = R\left(1 - e^{-\lambda t}\right).
$$

Activity climbs toward the **saturation value** $A_\infty = R$, the point where
formation and decay balance. Saturation is independent of the decay constant: a
short-lived product saturates quickly at the same ceiling a long-lived one
approaches slowly. After $n$ half-lives the activity has reached $1 - 2^{-n}$ of
saturation, so $50\%$ after one, $75\%$ after two, and $87.5\%$ after three;
irradiating far beyond a few half-lives yields diminishing returns.

$$
% caption: Under constant production the activity rises as one minus e to the
% minus lambda t toward the saturation ceiling R, reaching 50, 75, and 87.5
% percent of it after one, two, and three half-lives.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \draw[->, black] (0,0) -- (7.8,0) node[right, black!70] {$t$ (half-lives)};
  \draw[->, black] (0,0) -- (0,4.2) node[above, black!70] {$A(t)$};
  % saturation ceiling
  \draw[black, dashed] (0,3.5) -- (7.4,3.5);
  \node[black, anchor=south east, font=\scriptsize] at (7.4,3.5) {saturation $= R$};
  % rising saturation curve through (1.5,1.75),(3,2.625),(4.5,3.06)
  \draw[acc, very thick] (0,0)
    .. controls (0.7,1.15) and (1.1,1.55) .. (1.5,1.75)
    .. controls (2.3,2.15) and (2.6,2.45) .. (3.0,2.63)
    .. controls (3.9,2.95) and (4.6,3.12) .. (5.4,3.28)
    .. controls (6.2,3.4) and (6.8,3.45) .. (7.4,3.48);
  % half-life ticks
  \foreach \x/\lab in {1.5/1,3.0/2,4.5/3} {
    \draw[black] (\x,0.06) -- (\x,-0.06);
    \node[black, anchor=north, font=\scriptsize] at (\x,-0.05) {\lab};
  }
  % 50% marker
  \draw[black, dotted] (0,1.75) -- (1.5,1.75);
  \node[black, anchor=east, font=\scriptsize] at (0,1.75) {$\tfrac{1}{2}R$};
\end{tikzpicture}
$$

## Branching and partial decay constants

A nucleus that can decay by more than one route has a **partial decay constant**
$\lambda_i$ for each mode. The modes are independent competing processes, so the
total decay probability per unit time is their sum,

$$
\lambda = \sum_i \lambda_i,
\qquad
t_{1/2} = \frac{\ln 2}{\lambda},
$$

and the population still falls as $e^{-\lambda t}$ with the total constant. The
fraction of decays through mode $i$ is the **branching ratio**

$$
b_i = \frac{\lambda_i}{\lambda},
\qquad \sum_i b_i = 1.
$$

Each partial constant defines a **partial half-life** $t_{1/2,i} = \ln 2/
\lambda_i = t_{1/2}/b_i$, the half-life the nucleus would have if that mode acted
alone. Partial half-lives are always longer than the observed half-life. For
$^{40}\mathrm{K}$, electron capture and $\beta^+$ to $^{40}\mathrm{Ar}$ take about
$b_{\mathrm{EC}} = 0.11$ of decays and $\beta^-$ to $^{40}\mathrm{Ca}$ the
remaining $0.89$; the argon branch is what makes potassium-argon dating possible.

$$
% caption: Competing decay modes split the total constant: a parent with partial
% constants for two channels decays with their sum, and the branching ratio of a
% channel is its partial constant over the total.
\begin{tikzpicture}[>=stealth, font=\footnotesize]
  \definecolor{acc}{HTML}{4A6FA5}
  \tikzset{spec/.style={draw=acc, fill=acc!10, minimum width=1.6cm, minimum height=0.8cm, font=\footnotesize}}
  \node[spec] (p) at (0,2.4) {parent};
  \node[spec] (a) at (2.8,3.4) {product A};
  \node[spec] (b) at (2.8,1.4) {product B};
  \draw[->, acc, thick] (p) -- (a) node[midway, above, black, font=\scriptsize] {mode A};
  \draw[->, black, thick] (p) -- (b) node[midway, below, black, font=\scriptsize] {mode B};
\end{tikzpicture}
$$

## Natural decay series and radiometric dating

Alpha decay lowers the mass number by four and beta decay leaves it unchanged, so
every heavy chain conserves $A \bmod 4$. Three of the four series survive in
nature because their heads outlive the age of the Earth.

| Series | $A$ | Head | $t_{1/2}$ (yr) | Stable end |
| --- | --- | --- | --- | --- |
| Thorium | $4n$ | $^{232}\mathrm{Th}$ | $1.40\times 10^{10}$ | $^{208}\mathrm{Pb}$ |
| Neptunium | $4n{+}1$ | $^{237}\mathrm{Np}$ | $2.14\times 10^{6}$ | $^{209}\mathrm{Bi}$ |
| Uranium | $4n{+}2$ | $^{238}\mathrm{U}$ | $4.47\times 10^{9}$ | $^{206}\mathrm{Pb}$ |
| Actinium | $4n{+}3$ | $^{235}\mathrm{U}$ | $7.04\times 10^{8}$ | $^{207}\mathrm{Pb}$ |

The neptunium series is extinct: its head decays in a few million years, short
against the $4.5$-billion-year age of the solar system, so it long ago reached its
stable bismuth end. In an undisturbed mineral the surviving series sit in secular
equilibrium, every intermediate at the activity of its long-lived head.

Radiometric dating reads the clock set by a long-lived parent decaying to a stable
daughter. If a mineral begins with $N_0$ parent atoms and no daughter, then at
time $t$ the parent count is $N = N_0 e^{-\lambda t}$ and the accumulated stable
daughter is $D = N_0 - N = N(e^{\lambda t} - 1)$. Solving for the age,

$$
t = \frac{1}{\lambda}\ln\!\left(1 + \frac{D}{N}\right).
$$

The ratio $D/N$ of daughter to remaining parent, both measured by mass
spectrometry, fixes the age without knowing the original amount. Uranium-lead
($^{238}\mathrm{U}\to{}^{206}\mathrm{Pb}$), rubidium-strontium, and
potassium-argon clocks date rocks over billions of years.

> **Example (Carbon-14 dating).** Cosmic-ray neutrons keep atmospheric
> $^{14}\mathrm{C}$ in production-decay balance with living tissue, at a specific
> activity of $0.25\,\mathrm{Bq}$ per gram of carbon. After death, uptake stops
> and the $^{14}\mathrm{C}$ decays with $t_{1/2} = 5730\,\mathrm{yr}$
> ($\lambda = 1.21\times 10^{-4}\,\mathrm{yr^{-1}}$). A sample measured at
> $0.125\,\mathrm{Bq/g}$, half the living value, has aged one half-life,
> $t = 5730\,\mathrm{yr}$; a sample at $0.031\,\mathrm{Bq/g}$, one-eighth, has aged
> three half-lives, $t \approx 1.72\times 10^{4}\,\mathrm{yr}$. The method reaches
> back about ten half-lives, near $5\times 10^{4}\,\mathrm{yr}$, before the
> activity is too low to measure.

[^bateman]: Krane, §6.4; the closed-form serial-decay solution is due to Bateman.