Radioactive Decay and Bateman Equations

Derivation

Radioactive Decay and Bateman Equations

D-272 Home PU-304 Threads chance · matter Depends on —

Statement

For a radioactive nuclide decaying with constant per-nucleus probability rate \(\lambda\), the population obeys \(N(t)=N_0 e^{-\lambda t}\); for a linear chain \(1\to 2\to\cdots\to n\) in which each species \(i\) decays with rate \(\lambda_i\) and feeds only species \(i+1\), starting from a pure parent \(N_1(0)=N_1^0\), \(N_i(0)=0\) for \(i>1\), the daughter populations are given by the Bateman solution \(\displaystyle N_n(t)=N_1^0\left(\prod_{j=1}^{n-1}\lambda_j\right)\sum_{i=1}^{n}\frac{e^{-\lambda_i t}}{\displaystyle\prod_{\substack{k=1\\k\neq i}}^{n}(\lambda_k-\lambda_i)}\), from which secular and transient equilibrium follow as limits.

Why it matters

Almost every naturally occurring heavy nuclide sits in a decay chain — the uranium, thorium and actinium series each pass through a dozen intermediate radionuclides before reaching stable lead. Predicting the activity of any member (radon in\-growth in a basement, \(^{99\text{m}}\)Tc eluted from a \(^{99}\)Mo generator, \(^{140}\)La following \(^{140}\)Ba in fallout) requires exactly the Bateman machinery derived here.

The same linear first\-order kinetics govern neutron\-activation product build\-up in reactors, radiopharmaceutical dosimetry, and radiometric dating. Understanding when a daughter tracks its parent (equilibrium) versus when it decouples is the difference between a usable calibration and a systematic error of orders of magnitude.

Assumptions

Decay is a first\-order stochastic process with a constant rate constant \(\lambda\).If \(\lambda\) depended on the number of nuclei or on time, the governing equation would be nonlinear or non\-autonomous and the exponential/Bateman closed forms would not hold.

The number of nuclei is large enough to treat \(N(t)\) as a smooth deterministic variable.For small \(N\) the discreteness and Poisson fluctuations dominate; \(N(t)=N_0 e^{-\lambda t}\) then describes only the expectation value, not any single realisation.

Each species has a single decay mode feeding exactly one daughter (linear chain).Branching (e.g. \(^{212}\)Bi to \(^{212}\)Po and \(^{208}\)Tl) requires branching ratios and parallel terms; the simple product\-of\-rates prefactor is replaced by a sum over paths.

All decay constants are distinct, \(\lambda_i\neq\lambda_j\) for \(i\neq j\).If two rates coincide the Bateman denominator \(\prod(\lambda_k-\lambda_i)\) develops a \(0/0\); the correct limit produces \(t\,e^{-\lambda t}\) (resonant/degenerate) terms via l'Hôpital.

The system is closed: no external production, no loss channels other than decay.With continuous production \(R(t)\) (reactor irradiation, cosmogenic input) each equation gains a source term and the homogeneous Bateman solution must be convolved with \(R\).

Derivation

\[ \text{Prob}(\text{decay in } dt)=\lambda\,dt \quad\Rightarrow\quad -dN=\lambda N\,dt \]

Constant per\-nucleus decay probability; expected number of decays in \(dt\) is \(\lambda N\,dt\). Sign is negative because decays remove parent nuclei. A

\[ \frac{dN}{dt}=-\lambda N \]

Divide by \(dt\); this is the linear first\-order ODE for a single species. A

\[ \frac{dN}{N}=-\lambda\,dt \;\Rightarrow\; \int_{N_0}^{N}\frac{dN'}{N'}=-\lambda\int_0^t dt' \]

Separation of variables is legal since \(N>0\) and \(\lambda\) is constant. A

\[ \ln\!\frac{N}{N_0}=-\lambda t \;\Rightarrow\; N(t)=N_0\,e^{-\lambda t} \]

Integrate and exponentiate; applies the initial condition \(N(0)=N_0\). Defines mean life \(\tau=1/\lambda\) and half\-life \(t_{1/2}=\ln 2/\lambda\). A

\[ \frac{dN_1}{dt}=-\lambda_1 N_1,\qquad \frac{dN_i}{dt}=\lambda_{i-1}N_{i-1}-\lambda_i N_i\ \ (i\ge 2) \]

Balance equations for the chain: species \(i\) is produced by decay of \(i-1\) at rate \(\lambda_{i-1}N_{i-1}\) and lost at rate \(\lambda_i N_i\). This is a coupled linear system. B

\[ N_1(t)=N_1^0\,e^{-\lambda_1 t} \]

The parent equation decouples and integrates as in step 4. A

\[ \frac{dN_i}{dt}+\lambda_i N_i=\lambda_{i-1}N_{i-1} \;\Rightarrow\; \frac{d}{dt}\!\left(e^{\lambda_i t}N_i\right)=\lambda_{i-1}e^{\lambda_i t}N_{i-1} \]

Multiply by the integrating factor \(e^{\lambda_i t}\); the left side becomes an exact derivative. Solve the chain recursively, each \(N_{i-1}\) already known as a sum of exponentials. C

\[ N_i(t)=\lambda_{i-1}e^{-\lambda_i t}\!\int_0^t e^{\lambda_i t'}N_{i-1}(t')\,dt' \]

Integrate step 7 with \(N_i(0)=0\). Substituting \(N_{i-1}\) as a sum of exponentials, each integral \(\int_0^t e^{(\lambda_i-\lambda_m)t'}dt'\) is elementary and reproduces the exponential structure with a shifted denominator. C

\[ \int_0^t e^{(\lambda_i-\lambda_m)t'}\,dt'=\frac{e^{(\lambda_i-\lambda_m)t}-1}{\lambda_i-\lambda_m} \]

Each recursion step introduces one new denominator factor \((\lambda_i-\lambda_m)\) and one new exponential \(e^{-\lambda_m t}\); induction on the chain length assembles them into a single sum. C

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

Collecting all recursion levels by induction gives the Bateman closed form. The numerator carries the product of all upstream production rates; the denominator, for the \(i\)\-th residue, is the product of rate differences — precisely the partial\-fraction residue of \(\prod_j\lambda_j/\prod_k(s+\lambda_k)\) at \(s=-\lambda_i\), which is the Laplace\-transform derivation. C

\[ A_n(t)=\lambda_n N_n(t) \]

The measurable quantity is activity (decays per second), not population; multiply the number by the species' own decay constant. A

Result

\[ N(t)=N_0 e^{-\lambda t},\qquad N_n(t)=N_1^0\left(\prod_{j=1}^{n-1}\lambda_j\right)\sum_{i=1}^{n}\frac{e^{-\lambda_i t}}{\displaystyle\prod_{\substack{k=1\\k\neq i}}^{n}(\lambda_k-\lambda_i)} \]

Reading. A single species falls off as a pure exponential with mean life \(1/\lambda\). Any daughter in a chain is a signed superposition of the exponentials of every ancestor and of itself: each mode \(e^{-\lambda_i t}\) is weighted by the total upstream production rate divided by how far that mode's rate sits from all the others. Fast\-decaying intermediates contribute short transients; the slowest ancestor dominates the long\-time tail, so a short\-lived daughter ends up "tracking" a long\-lived parent.

Units check. Each \(\lambda_j\) has units \(\text{s}^{-1}\). The prefactor \(\prod_{j=1}^{n-1}\lambda_j\) has units \(\text{s}^{-(n-1)}\); the denominator \(\prod_{k\neq i}(\lambda_k-\lambda_i)\) runs over \(n-1\) factors, units \(\text{s}^{-(n-1)}\). Their ratio is dimensionless, the exponential is dimensionless, so \(N_n\) inherits the units of \(N_1^0\) (a pure number of nuclei). Activity \(A=\lambda N\) then carries \(\text{s}^{-1}\) (becquerel), as required.

Limiting cases

Single species (\(n=1\)): the sum has one term with an empty product denominator \(=1\), recovering \(N_1=N_1^0 e^{-\lambda_1 t}\).
Short times (\(\lambda_i t\ll 1\)): expand \(e^{-\lambda_i t}\approx 1-\lambda_i t+\tfrac12\lambda_i^2 t^2\); the daughter grows linearly, \(N_2\approx N_1^0\lambda_1 t\), i.e. initial in\-growth at the parent's decay rate.
Secular equilibrium (\(\lambda_1\ll\lambda_2\), long\-lived parent): after \(t\gg 1/\lambda_2\), \(A_2\to A_1\), so \(\lambda_2 N_2=\lambda_1 N_1\) and every daughter carries the same activity as the parent.
Transient equilibrium (\(\lambda_1<\lambda_2\), comparable): the daughter\-to\-parent activity ratio stabilises at \(A_2/A_1=\lambda_2/(\lambda_2-\lambda_1)>1\); daughter activity exceeds parent then decays with the parent's rate.
No equilibrium (\(\lambda_1>\lambda_2\), short\-lived parent): the parent vanishes first; the daughter rises, peaks, then decays with its own constant \(\lambda_2\) — no fixed ratio is reached.

Breaks when

Degenerate rates \(\lambda_i=\lambda_j\). The denominator \(\prod_{k\neq i}(\lambda_k-\lambda_i)\) vanishes and the closed form is singular. The physical solution is finite: taking the limit produces polynomial\-in\-\(t\) prefactors, e.g. for two equal rates \(N_2=N_1^0\lambda t\,e^{-\lambda t}\). Must use the confluent (l'Hôpital) form.
Branching decay. When a nuclide has multiple exit channels (partial constants \(\lambda_i^{(a)}\), \(\lambda_i^{(b)}\)), the feed into any one daughter is scaled by a branching ratio \(b=\lambda_i^{(a)}/\lambda_i\); the simple \(\prod\lambda_j\) numerator is wrong and one must sum over all chain paths.
Small populations. Near the end of a chain or for trace nuclides, \(N\sim\mathcal{O}(1)\); the deterministic ODE only gives the mean, and actual counts follow a Poisson/multinomial branching process with fractional standard deviation \(\sim 1/\sqrt{N}\).
Open systems / continuous production. Reactor irradiation, accelerator targets or cosmogenic input add a source term \(+R_i(t)\); the homogeneous Bateman result must be convolved with the production history and no longer applies unmodified.

Failure modes

Confusing \(\lambda\) with \(t_{1/2}\): writing \(N=N_0 e^{-t/t_{1/2}}\) instead of \(e^{-\lambda t}\) with \(\lambda=\ln 2/t_{1/2}\), losing a factor of \(\ln 2\approx 0.693\).
Reporting population instead of activity: a Geiger counter measures \(A=\lambda N\), not \(N\). In secular equilibrium the activities are equal but the populations differ by the ratio \(\lambda_1/\lambda_2\) (huge for a long\-lived parent).
Using the parent's \(\lambda\) in the daughter's activity: the daughter's activity is \(\lambda_2 N_2\), never \(\lambda_1 N_2\).
Assuming equilibrium too early: secular/transient ratios are asymptotic; applying them before \(\sim 5\text{–}10\) daughter half\-lives have elapsed gives large errors.
Dropping the initial\-condition term: forgetting that daughters start at \(N_i(0)=0\), and thus omitting the negative residue that enforces the in\-growth from zero.
Sign error in denominators: writing \((\lambda_i-\lambda_k)\) instead of \((\lambda_k-\lambda_i)\), flipping the sign of every daughter term.

Discussion

The decay law is the archetype of a Markovian, memoryless process: an atom that has "survived" for a time \(t\) has exactly the same probability of decaying in the next instant as a freshly created one. This memorylessness is what forces the survival probability to be exponential — it is the only continuous distribution with a constant hazard rate. Radioactivity was the first clear demonstration in nature of a genuinely irreducible quantum probability, with no hidden clock inside the nucleus counting down.

The Bateman equations are the physical incarnation of a linear compartmental system: mass flows one way through a cascade of "tanks" each draining at its own rate. The same mathematics reappears in pharmacokinetics (drug metabolite chains), chemical reaction kinetics, and heat\-exchanger cascades. The structure of the solution — a sum of the system's natural exponential modes weighted by residues at the eigenvalues \(-\lambda_i\) — is exactly the Laplace/partial\-fraction picture, making the eigenvalues of the tridiagonal rate matrix the decay constants themselves.

Viewing the chain as \(\dot{\mathbf N}=\mathbf{M}\mathbf N\) with \(\mathbf M\) lower\-bidiagonal, the eigenvalues are the diagonal entries \(-\lambda_i\) and the Bateman coefficients are the entries of the matrix exponential \(e^{\mathbf M t}\). Because \(\mathbf M\) is defective exactly when two \(\lambda_i\) coincide, the appearance of \(t\,e^{-\lambda t}\) secular terms in the degenerate case is not a coincidence but the generalized\-eigenvector (Jordan block) structure surfacing physically. This unifies the "resonant" polynomial terms with the theory of repeated roots in linear ODE systems.

Common misconceptions. Secular equilibrium does not mean the daughter stops decaying — it decays vigorously, but is replenished at exactly the rate it is lost, so its population is stationary while its activity equals the parent's. Equally, "equilibrium" is a dynamic steady state that itself slowly erodes on the parent's timescale; over geological time even a secular\-equilibrium daughter follows the parent's decay. And half\-life is a property of the ensemble, not a countdown for any individual atom.

Worked examples

\[ \textbf{Radon in-growth from radium.}\quad ^{226}\text{Ra}\ (t_{1/2}=1600\,\text{yr}) \to\ ^{222}\text{Rn}\ (t_{1/2}=3.82\,\text{d}) \]

Two\-species chain, long\-lived parent — expect secular equilibrium. Compute the daughter activity as a fraction of parent activity after 30 days. B

\[ \lambda_2=\frac{\ln 2}{3.82\,\text{d}}=0.1815\,\text{d}^{-1},\qquad \lambda_1=\frac{\ln 2}{1600\times365.25\,\text{d}}=1.19\times10^{-6}\,\text{d}^{-1} \]

Convert half\-lives to rate constants in consistent units (per day). A

\[ \frac{A_2(t)}{A_1(t)}=\frac{\lambda_2}{\lambda_2-\lambda_1}\left(1-e^{-(\lambda_2-\lambda_1)t}\right)\approx 1-e^{-\lambda_2 t} \]

Since \(\lambda_1\ll\lambda_2\), the parent is essentially constant; the two\-species Bateman activity ratio reduces to the in\-growth factor. B

\[ \frac{A_2}{A_1}=1-e^{-0.1815\times 30}=1-e^{-5.44}=1-0.00432=0.9957 \]

Evaluate at \(t=30\) days. A

\[ \frac{A_2}{A_1}\approx 0.996 \]

Reading. After 30 days (\(\approx 7.9\) radon half\-lives) the sealed radium sample's radon activity has reached 99.6% of the parent activity — essentially full secular equilibrium. This is why a radium source must be sealed for a few weeks before its radon activity is used as a standard.

Units check. \(\lambda_2 t=(0.1815\,\text{d}^{-1})(30\,\text{d})\) is dimensionless; the ratio of activities is dimensionless. Consistent.

\[ \textbf{Technetium generator.}\quad ^{99}\text{Mo}\ (t_{1/2}=65.9\,\text{h}) \to\ ^{99\text{m}}\text{Tc}\ (t_{1/2}=6.01\,\text{h}) \]

Transient equilibrium: parent longer\-lived but comparable. Find the time at which the daughter activity peaks after a fresh elution (\(N_2(0)=0\)). Take the branching to \(^{99\text{m}}\)Tc as 100% for this estimate. B

\[ \lambda_1=\frac{\ln 2}{65.9}=0.01052\,\text{h}^{-1},\qquad \lambda_2=\frac{\ln 2}{6.01}=0.1153\,\text{h}^{-1} \]

Rate constants in per\-hour. A

\[ A_2(t)=\frac{\lambda_2\lambda_1 N_1^0}{\lambda_2-\lambda_1}\left(e^{-\lambda_1 t}-e^{-\lambda_2 t}\right) \]

Two\-species Bateman activity with \(N_2(0)=0\). The peak occurs where \(dA_2/dt=0\). B

\[ \frac{dA_2}{dt}=0\;\Rightarrow\;\lambda_1 e^{-\lambda_1 t}=\lambda_2 e^{-\lambda_2 t}\;\Rightarrow\; t_{\max}=\frac{\ln(\lambda_2/\lambda_1)}{\lambda_2-\lambda_1} \]

Set derivative to zero and solve for \(t\); take logs. B

\[ t_{\max}=\frac{\ln(0.1153/0.01052)}{0.1153-0.01052}=\frac{\ln(10.96)}{0.1048}=\frac{2.394}{0.1048}=22.8\,\text{h} \]

Numerical evaluation. A

\[ t_{\max}=\frac{\ln(\lambda_2/\lambda_1)}{\lambda_2-\lambda_1}\approx 22.8\,\text{h} \]

Reading. Following a fresh elution the \(^{99\text{m}}\)Tc activity rebuilds and peaks about 23 hours later, which is why hospital generators are typically eluted once per day. Thereafter parent and daughter enter transient equilibrium with \(A_2/A_1=\lambda_2/(\lambda_2-\lambda_1)=1.10\).

Units check. \(\ln(\lambda_2/\lambda_1)\) is dimensionless; dividing by \((\lambda_2-\lambda_1)\) with units \(\text{h}^{-1}\) yields hours. Correct.

Problems

A pure sample of \(^{131}\)I (\(t_{1/2}=8.02\) d) has initial activity 500 MBq. What is its activity after 3 weeks?
Solution
\(\lambda=\ln2/8.02=0.0864\,\text{d}^{-1}\). \(t=21\,\text{d}\): \(A=500\,e^{-0.0864\times21}=500\,e^{-1.815}=500\times0.1628=81.4\,\text{MBq}.\)
Show that at exactly one half\-life the number of undecayed nuclei is \(N_0/2\), and that the fraction decaying between \(t_{1/2}\) and \(2t_{1/2}\) is \(1/4\) of the original.
Solution
\(N(t_{1/2})=N_0 e^{-\lambda\cdot\ln2/\lambda}=N_0 e^{-\ln2}=N_0/2\). At \(2t_{1/2}\): \(N=N_0/4\). Fraction decaying in the interval \(= (N_0/2 - N_0/4)/N_0 = 1/4.\)
For the two\-species chain \(1\to2\) with \(N_2(0)=0\), derive \(N_2(t)=\dfrac{\lambda_1 N_1^0}{\lambda_2-\lambda_1}\big(e^{-\lambda_1 t}-e^{-\lambda_2 t}\big)\) directly from the Bateman formula.
Solution
For \(n=2\): \(N_2=N_1^0\lambda_1\Big[\dfrac{e^{-\lambda_1 t}}{\lambda_2-\lambda_1}+\dfrac{e^{-\lambda_2 t}}{\lambda_1-\lambda_2}\Big] =\dfrac{\lambda_1 N_1^0}{\lambda_2-\lambda_1}\big(e^{-\lambda_1 t}-e^{-\lambda_2 t}\big).\) The two residue denominators are \((\lambda_2-\lambda_1)\) for \(i=1\) and \((\lambda_1-\lambda_2)=-(\lambda_2-\lambda_1)\) for \(i=2\), giving the difference of exponentials.
A \(^{140}\)Ba (\(t_{1/2}=12.75\) d) parent feeds \(^{140}\)La (\(t_{1/2}=1.68\) d). Starting from pure barium, estimate the daughter\-to\-parent activity ratio at long times and the time to reach transient equilibrium (say 5 daughter half\-lives).
Solution
\(\lambda_1=\ln2/12.75=0.0544\,\text{d}^{-1}\), \(\lambda_2=\ln2/1.68=0.4126\,\text{d}^{-1}\). Transient ratio \(A_2/A_1=\lambda_2/(\lambda_2-\lambda_1)=0.4126/0.3582=1.152\). Equilibrium after \(\sim5\times1.68=8.4\) d. So after \(\approx\)8–9 days the lanthanum activity settles to about 1.15 times the barium activity and then decays with the 12.75\-d barium half\-life.
Two nuclides in a chain happen to have equal decay constants \(\lambda_1=\lambda_2=\lambda\). Find \(N_2(t)\) for \(N_2(0)=0\) and locate its maximum.
Solution
The Bateman denominator vanishes, so take the limit \(\lambda_2\to\lambda_1\) of \(N_2=\dfrac{\lambda_1 N_1^0}{\lambda_2-\lambda_1}(e^{-\lambda_1 t}-e^{-\lambda_2 t})\). Writing \(\lambda_2=\lambda_1+\epsilon\), \(e^{-\lambda_2 t}\approx e^{-\lambda_1 t}(1-\epsilon t)\), so \(N_2\to \lambda_1 N_1^0\, t\, e^{-\lambda_1 t}\), i.e. \(N_2(t)=N_1^0\lambda t\,e^{-\lambda t}\). Maximum: \(dN_2/dt=N_1^0\lambda e^{-\lambda t}(1-\lambda t)=0\Rightarrow t_{\max}=1/\lambda=\tau\), the mean life.