Parent–Daughter Kinetics and the Bateman Equations

Introduction

Many radionuclides decay into daughters that are themselves radioactive. Uranium-238 starts a chain of fourteen decays ending at lead-206; molybdenum-99 decays to technetium-99m, which then decays to technetium-99. To predict how much of an intermediate is present at any time, we need a pair of coupled rate equations. Their solution, first written generally by Harry Bateman in 1910, is the basis of generator design, dating methods and the handling of fission-product mixtures.

Core explanation

Setting up the equations. Consider a chain 1 → 2 → 3, where nuclide 3 is stable. Parent 1 simply decays:

dN₁/dt = −λ₁N₁, so N₁ = N₁⁰e^(−λ₁t)

Daughter 2 is formed at the rate at which the parent decays and destroyed by its own decay:

dN₂/dt = λ₁N₁ − λ₂N₂

This is the same mathematics as consecutive first-order reactions A → B → C in chemical kinetics.

Solution for the daughter. Substituting N₁ and solving the linear differential equation (for example with an integrating factor e^(λ₂t)) gives

N₂ = λ₁N₁⁰/(λ₂ − λ₁) − e^(−λ₂t)) + N₂⁰e^(−λ₂t)

The first term is the daughter grown in from the parent; the second is any daughter present at the start, decaying on its own. Multiplying by λ₂ gives the daughter activity A₂.

Shape of the curve. Starting from pure parent (N₂⁰ = 0), N₂ rises from zero, passes through a maximum, and then decays. At the maximum dN₂/dt = 0, so λ₁N₁ = λ₂N₂: the daughter activity equals the parent activity at that instant. Differentiating gives the time of maximum:

t max = ln(λ₂/λ₁)/(λ₂ − λ₁)

Branching. If only a fraction b of parent decays produce the daughter, multiply the ingrowth term by b. For molybdenum-99, about 88% of decays populate technetium-99m.

Special case λ₁ = λ₂. The general formula becomes 0/0; taking the limit gives N₂ = λN₁⁰t e^(−λt).

Longer chains. For a chain 1 → 2 → … → n starting with pure parent, Bateman's general solution is a sum of exponentials:

Nₙ = N₁⁰(λ₁λ₂…λₙ₋₁) Σᵢ e^(−λᵢt)/Πⱼ≠ᵢ(λⱼ − λᵢ)

In practice, such chains are solved numerically, but the structure — one exponential per member — explains why complex mixtures show multi-component decay curves.

Formulae

dN₂/dt = λ₁N₁ − λ₂N₂; N₂ = λ₁N₁⁰/(λ₂ − λ₁) − e^(−λ₂t)) + N₂⁰e^(−λ₂t); t max = ln(λ₂/λ₁)/(λ₂ − λ₁); A₂ = λ₂N₂.

Step-by-step reasoning

To find the daughter activity at time t after separating a pure parent:

1. Calculate λ₁ and λ₂ from the two half-lives, in the same unit. 2. Note that N₂⁰ = 0, so only the ingrowth term is needed. 3. Evaluate A₂ = bλ₂λ₁N₁⁰(e^(−λ₁t) − e^(−λ₂t))/(λ₂ − λ₁), or equivalently A₂ = bA₁⁰λ₂(e^(−λ₁t) − e^(−λ₂t))/(λ₂ − λ₁). 4. Check the limits: A₂ = 0 at t = 0, and A₂ → 0 again at very long times.

Visual explanation

Plot three curves on the same axes: the parent falling exponentially, the daughter rising from zero to a peak and then falling, and the stable end-product climbing steadily towards N₁⁰. The daughter peak lies exactly where its curve crosses the parent activity curve. The simulation draws random decays and the exact Bateman curves together.

Real-world analogy

Think of a bath with a tap and an open plug-hole. Water flows in from a draining tank (the parent) and leaves through the plug-hole at a rate proportional to the depth (the daughter decay). The level rises while inflow exceeds outflow, peaks when the two are equal, and then falls as the tank empties.

Real-world example

Hospital technetium generators are loaded with molybdenum-99 (t½ = 66.0 h). After each elution removes the technetium-99m (t½ = 6.01 h), the daughter grows back, reaching its maximum after about 23 hours. This is why generators are commonly eluted once a day.

Why?

Why does the maximum occur exactly when A₁ = A₂? The daughter number stops rising when its rate of formation (λ₁N₁, the parent activity) equals its rate of loss (λ₂N₂, its own activity). Before that, formation wins; afterwards, decay wins.

Common misconception

"The daughter always ends up with more atoms than the parent." The ratio depends on the half-lives. A short-lived daughter never accumulates many atoms; at its peak it has the same activity as the parent but far fewer nuclei, because N₂ = A/λ₂ is small when λ₂ is large.

Worked example

Question: Calculate the time at which technetium-99m activity is greatest after a molybdenum-99 generator is eluted. Take t½(⁹⁹Mo) = 66.0 h and t½(⁹⁹ᵐTc) = 6.01 h.

Reasoning: λ₁ = 0.693/66.0 = 0.01050 h⁻¹; λ₂ = 0.693/6.01 = 0.1153 h⁻¹. t max = ln(0.1153/0.01050)/(0.1153 − 0.01050) = ln(10.98)/0.1048 = 2.396/0.1048.

Answer: t max ≈ 22.9 h, so about 23 hours.

Quick check

1. At the instant the daughter population is at its maximum, how does its activity compare with the parent's activity? Answer: They are equal (for 100% branching), because the formation and decay rates of the daughter balance.

Exam focus

Be able to write dN₂/dt = λ₁N₁ − λ₂N₂, quote or derive the solution, and use t max = ln(λ₂/λ₁)/(λ₂ − λ₁). Examiners often ask you to sketch parent, daughter and end-product curves and to identify where the daughter curve peaks.

Advanced insight

Real chains can branch and recombine: bismuth-212 decays both by β⁻ to polonium-212 and by α to thallium-208, and both routes end at lead-208. Nuclear data codes solve thousands of coupled Bateman equations, including neutron-induced production terms, to predict the inventory of spent fuel. Matrix-exponential methods replace the explicit formula because nearly equal decay constants make the sum of exponentials numerically unstable.

Summary

For a chain 1 → 2 → 3, the daughter obeys dN₂/dt = λ₁N₁ − λ₂N₂, whose solution is the Bateman equation. From pure parent, the daughter grows in, peaks at t max = ln(λ₂/λ₁)/(λ₂ − λ₁) when A₂ = A₁, and then decays. Longer chains give sums of exponentials, one per member, and branching fractions scale the ingrowth.

Practice questions

1. Write the differential equation for the number of daughter nuclei and explain each term. Answer: dN₂/dt = λ₁N₁ − λ₂N₂; λ₁N₁ is the rate of formation from the parent and λ₂N₂ is the rate of loss by the daughter's own decay. 2. What happens to the general Bateman expression if λ₁ = λ₂, and what is the correct result? Answer: It becomes 0/0 and is undefined; taking the limit gives N₂ = λN₁⁰t e^(−λt). 3. A parent has t½ = 10.0 h and its daughter t½ = 2.00 h. Calculate the time of maximum daughter activity. Answer: λ₁ = 0.0693 h⁻¹, λ₂ = 0.3466 h⁻¹; t max = ln 5/(0.2773) = 1.609/0.2773 ≈ 5.80 h. 4. Why is the stable end-product curve always rising when starting from pure parent? Answer: It is only formed, never destroyed, so its number increases continuously and approaches N₁⁰ at long times.