Chemistry / Nuclear & Radiochemistry Decay Kinetics, Fission, Fusion & Radiometrics 100% Free Open Access
Chapter 3 β€’ Theory & Derivations

Unit 3: Decay Kinetics of Unstable Nuclei: Half-Life, Successive Decay & Bateman Equilibria

Comprehensive mathematical exposition of radioactive decay kinetics: the differential decay rate law, mean life derivations, Poisson counting statistics, branching decay channels, the general solution of successive multi-nuclide decay chains via the Bateman equations, and quantitative criteria for secular, transient, and non-equilibrium regimes.

Β§3.1 The Fundamental Radioactive Decay Differential Rate Law & Integral Kinetics

Radioactive decay is a first-order unimolecular kinetic transformation governed by the fundamental quantum decay probability $\lambda$. Consider an ensemble of $N(t)$ identical, non-interacting unstable radioactive nuclei. In any infinitesimal time interval $dt$, the probability $dp$ that an individual nucleus disintegrates is:

$$dp = \lambda \, dt$$

where $\lambda$ is the decay constant ($\text{s}^{-1}$), an intrinsic nuclear property independent of chemical bonding, physical state, temperature, or pressure.

The Differential Rate Law

The expected number of nuclear disintegrations $dN$ occurring in the sample within time interval $dt$ is:

$$dN = -N(t) \cdot dp = -\lambda N(t) dt \implies \frac{dN}{dt} = -\lambda N(t)$$

The negative sign signifies that the population of parent nuclei decreases monotonically over time.

Integral Form of the Decay Law

Separating variables and integrating from initial boundary condition $N(0) = N_0$ at $t = 0$:

$$\int_{N_0}^{N(t)} \frac{dN'}{N'} = -\lambda \int_0^t dt'$$
$$\ln\left(\frac{N(t)}{N_0}\right) = -\lambda t \implies N(t) = N_0 e^{-\lambda t}$$

The number of disintegrated daughter nuclei $N_D(t)$ formed (assuming a stable daughter and $N_D(0) = 0$) is:

$$N_D(t) = N_0 - N(t) = N_0 (1 - e^{-\lambda t})$$

``` Number of Nuclei Nβ‚€ β–² β”‚\ Parent N(t) = Nβ‚€ e^(-Ξ»t) Daughter N_D(t) = Nβ‚€ (1 - e^(-Ξ»t)) β”‚ \ / Nβ‚€/2β”‚--\------------------------------/-- (Half-Life T₁/β‚‚) β”‚ \ / β”‚ \ / β”‚ \________________________/ └──────┴────────────────────────► Time t T₁/β‚‚ ```

Radioactive Activity Formulation

The absolute rate of disintegration is defined as activity $A(t)$:

$$A(t) \equiv -\frac{dN}{dt} = \lambda N(t) = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}$$

Activity decays with the identical exponential rate constant $\lambda$ as the number of atoms.

Logarithmic Linearization

Taking the natural logarithm of both sides:

$$\ln A(t) = \ln A_0 - \lambda t$$

Plotting $\ln A(t)$ versus time $t$ yields a straight line with:

  • Vertical intercept: $\ln A_0$
  • Slope: $-\lambda$

In base-10 logarithms:

$$\log_{10} A(t) = \log_{10} A_0 - \frac{\lambda}{2.302585} t$$

Β§3.2 Half-Life, Mean Life & Decay Rate Parameter Interconversions

The kinetics of radionuclide disappearance can be characterized using three mathematically related parameters:

1. Decay Constant $\lambda$ (dimensions: $[\text{T}]^{-1}$, typically $\text{s}^{-1}$, $\text{h}^{-1}$, or $\text{yr}^{-1}$)

2. Half-Life $T_{1/2}$ (dimensions: $[\text{T}]$)

3. Mean Life $\tau$ (dimensions: $[\text{T}]$)

Detailed Derivation of Half-Life ($T_{1/2}$)

The half-life is the duration required for the active population (or activity) to decline to exactly half its initial value:

$$\frac{N(T_{1/2})}{N_0} = \frac{1}{2} = e^{-\lambda T_{1/2}}$$

Taking the natural logarithm:

$$\ln\left(\frac{1}{2}\right) = -\ln 2 = -\lambda T_{1/2} \implies T_{1/2} = \frac{\ln 2}{\lambda}$$

Using $\ln 2 \approx 0.69314718056$:

$$T_{1/2} \approx \frac{0.693147}{\lambda} \quad \text{or} \quad \lambda = \frac{0.693147}{T_{1/2}}$$

The surviving fraction after an arbitrary number of half-lives $n = t / T_{1/2}$ is:

$$\frac{N(t)}{N_0} = e^{-\lambda t} = e^{-(\ln 2 / T_{1/2}) t} = \left(e^{-\ln 2}\right)^{t / T_{1/2}} = \left(\frac{1}{2}\right)^{t / T_{1/2}} = 2^{-n}$$

| Half-Lives Elapsed ($n$) | Surviving Fraction ($N/N_0$) | Percentage Surviving | Decayed Fraction ($1 - N/N_0$) | | :--- | :--- | :--- | :--- | | 0 | $1$ | $100.00\%$ | $0.00\%$ | | 1 | $1/2$ | $50.00\%$ | $50.00\%$ | | 2 | $1/4$ | $25.00\%$ | $75.00\%$ | | 3 | $1/8$ | $12.50\%$ | $87.50\%$ | | 5 | $1/32$ | $3.125\%$ | $96.875\%$ | | 7 | $1/128$ | $0.781\%$ | $99.219\%$ | | 10 | $1/1024$ | $0.0977\%$ | $99.9023\%$ |

After 10 half-lives, less than $0.1\%$ ($1/1024$) of the initial radioactivity remains. In radioprotection and waste management, $10 T_{1/2}$ is the standard operational threshold for total decay of short-lived isotopes.

Detailed Derivation of Mean Life ($\tau$)

The mean lifetime $\tau$ is the arithmetic average lifespan of all nuclei in an ensemble. The probability distribution function $P(t) dt$ representing the probability that a nucleus decays between time $t$ and $t + dt$ is:

$$P(t) dt = \frac{|dN|}{N_0} = \lambda e^{-\lambda t} dt$$

Note that this distribution is properly normalized:

$$\int_0^\infty P(t) dt = \lambda \int_0^\infty e^{-\lambda t} dt = \lambda \left[ -\frac{1}{\lambda} e^{-\lambda t} \right]_0^\infty = 1$$

The expectation value $\langle t \rangle = \tau$ is:

$$\tau = \int_0^\infty t P(t) dt = \int_0^\infty t (\lambda e^{-\lambda t}) dt = \lambda \int_0^\infty t e^{-\lambda t} dt$$

Using the standard definite integral $\int_0^\infty t^n e^{-\lambda t} dt = \frac{n!}{\lambda^{n+1}}$ with $n = 1$:

$$\tau = \lambda \left(\frac{1!}{\lambda^2}\right) = \frac{1}{\lambda}$$

Converting between mean life and half-life:

$$\tau = \frac{T_{1/2}}{\ln 2} \approx 1.442695 \cdot T_{1/2}$$
$$T_{1/2} = \tau \ln 2 \approx 0.693147 \cdot \tau$$

Total Number of Decays Over All Time

The total cumulative disintegrations $N_{\text{tot}}$ that occur between $t = 0$ and $t = \infty$ is:

$$N_{\text{tot}} = \int_0^\infty A(t) dt = \int_0^\infty A_0 e^{-\lambda t} dt = A_0 \left[ -\frac{1}{\lambda} e^{-\lambda t} \right]_0^\infty = \frac{A_0}{\lambda} = A_0 \cdot \tau$$

The total number of atoms initially present is equal to the initial activity multiplied by the mean life:

$$N_0 = A_0 \tau$$

Β§3.3 Stochastic Foundations of Nuclear Counting: Poisson Statistics & Variance Limits

Because radioactive decay is a completely random stochastic process, repeated counts recorded over identical time intervals fluctuate. Understanding these statistical fluctuations is essential for determining experimental uncertainties, detection limits, and confidence intervals in radiation measurements.

The Binomial Distribution Formulation

Consider a sample containing $N_0$ radioactive atoms. Over an observation interval $\Delta t$, the probability that any specific nucleus decays is:

$$p = 1 - e^{-\lambda \Delta t}$$

The probability that it survives is $q = 1 - p = e^{-\lambda \Delta t}$. The probability $P(n)$ of observing exactly $n$ disintegrations out of $N_0$ atoms is governed by the Binomial distribution:

$$P(n) = \frac{N_0!}{n! (N_0 - n)!} p^n (1 - p)^{N_0 - n}$$

The Poisson Approximation

In real radiochemical counting experiments:

  1. The total number of atoms is extraordinarily large ($N_0 \sim 10^{12} - 10^{20}$).
  2. The decay probability over typical counting intervals is tiny ($p \ll 1$).
  3. The expected mean count $\mu = N_0 p$ remains finite.

Taking the mathematical limit $N_0 \to \infty$ and $p \to 0$ while holding $\mu = N_0 p$ constant, the Binomial distribution converges rigorously to the Poisson distribution:

$$P(n) = \frac{\mu^n e^{-\mu}}{n!}$$

where $\mu$ is the true mean count.

``` P(n) Probability β–² β”‚ Poisson Distribution (ΞΌ = 10) β”‚ ___ β”‚ / \ β”‚ / \ β”‚ / \ β”‚ / \ β”‚_______/ \________ └───────┴──────┴───────┴───────► Recorded Counts n 0 ΞΌ=10 20 ```

Variance and Standard Deviation of Poisson Counting

For a Poisson distribution:

  • Expectation Value: $\langle n \rangle = \mu$
  • Variance: $\sigma^2 = \mu$
  • Standard Deviation: $\sigma = \sqrt{\mu}$

In an experimental determination, the true mean $\mu$ is unknown and is estimated by the observed single count $N$:

$$\sigma \approx \sqrt{N}$$

The result of any radiation measurement is expressed with its one-sigma standard uncertainty ($68.3\%$ confidence interval) as:

$$\text{Measured Count} = N \pm \sqrt{N}$$

Relative Standard Deviation (RSD) and Counting Precision

The relative standard uncertainty (coefficient of variation) is:

$$\text{RSD} = \frac{\sigma}{N} = \frac{\sqrt{N}}{N} = \frac{1}{\sqrt{N}}$$
$$\% \text{RSD} = \frac{100\%}{\sqrt{N}}$$

| Observed Counts ($N$) | Standard Deviation ($\sigma = \sqrt{N}$) | Relative Standard Deviation (% RSD) | Metrological Level | | :--- | :--- | :--- | :--- | | 100 | $\pm 10$ | $10.0\%$ | Screening survey | | 1,000 | $\pm 31.6$ | $3.16\%$ | Routine monitoring | | 10,000 | $\pm 100$ | $1.00\%$ | Analytical standard | | 100,000 | $\pm 316.2$ | $0.316\%$ | High-precision metrology | | 1,000,000 | $\pm 1000$ | $0.100\%$ | Absolute activity calibration |

To improve counting precision by a factor of 10, the counting duration (or accumulated count total) must be increased by a factor of $10^2 = 100$.

Net Count Rate and Background Subtraction Uncertainty

Every radiation detector records ambient background radiation counts $N_b$ during background counting time $t_b$, in addition to gross sample counts $N_g$ accumulated over sample counting time $t_g$. The net count rate $R_n$ is:

$$R_n = R_g - R_b = \frac{N_g}{t_g} - \frac{N_b}{t_b}$$

By propagation of independent random errors:

$$\sigma_{R_n}^2 = \sigma_{R_g}^2 + \sigma_{R_b}^2 = \frac{\sigma_{N_g}^2}{t_g^2} + \frac{\sigma_{N_b}^2}{t_b^2} = \frac{N_g}{t_g^2} + \frac{N_b}{t_b^2} = \frac{R_g}{t_g} + \frac{R_b}{t_b}$$
$$\sigma_{R_n} = \sqrt{\frac{R_g}{t_g} + \frac{R_b}{t_b}}$$

Β§3.4 Branching Decay Mechanics, Partial Decay Constants & Competing Transition Ratios

Many unstable nuclei can decay through multiple competing pathways. For example, potassium-40 ($^{40}\text{K}$) undergoes both negative beta decay ($\beta^-$) into calcium-40 and electron capture (EC) into argon-40: ``` β”Œβ”€β”€β”€β–Ί ⁴⁰Ca + e⁻ + Ξ½Μ„_e (β⁻ decay: 89.28%) ⁴⁰K ──────── └───► ⁴⁰Ar + Ξ½_e + Ξ³ (EC decay: 10.72%) ``` Similarly, copper-64 ($^{64}\text{Cu}$) decays via three competing modes: $\beta^-$ ($39.0\%$), $\beta^+$ ($17.6\%$), and electron capture ($43.4\%$).

Partial Decay Constants

Let an unstable nuclide decay through $m$ independent competing branches, each characterized by its own partial decay constant $\lambda_i$:

$$-\left(\frac{dN}{dt}\right)_i = \lambda_i N$$

The total rate of disappearance of the parent nucleus is the sum of the disappearance rates across all individual decay modes:

$$-\frac{dN}{dt} = \sum_{i=1}^m -\left(\frac{dN}{dt}\right)_i = \sum_{i=1}^m \lambda_i N = \left(\sum_{i=1}^m \lambda_i\right) N$$

Defining the total decay constant $\lambda_{\text{tot}}$:

$$\lambda_{\text{tot}} = \sum_{i=1}^m \lambda_i = \lambda_1 + \lambda_2 + \dots + \lambda_m$$

The population of the parent decays exponentially according to the total decay constant:

$$N(t) = N_0 e^{-\lambda_{\text{tot}} t}$$

Effective Total Half-Life

Because $T_{1/2} = \ln 2 / \lambda$, the total half-life $T_{1/2,\text{tot}}$ satisfies:

$$\frac{\ln 2}{T_{1/2,\text{tot}}} = \frac{\ln 2}{T_{1/2,1}} + \frac{\ln 2}{T_{1/2,2}} + \dots + \frac{\ln 2}{T_{1/2,m}}$$

Dividing through by $\ln 2$:

$$\frac{1}{T_{1/2,\text{tot}}} = \sum_{i=1}^m \frac{1}{T_{1/2,i}} = \frac{1}{T_{1/2,1}} + \frac{1}{T_{1/2,2}} + \dots + \frac{1}{T_{1/2,m}}$$

Notice that this is formally identical to the formula for resistors connected in parallel. For two competing branches:

$$T_{1/2,\text{tot}} = \frac{T_{1/2,1} \cdot T_{1/2,2}}{T_{1/2,1} + T_{1/2,2}}$$

The total half-life is always strictly shorter than the shortest partial half-life.

Branching Ratio (Branching Fraction)

The branching ratio $BR_i$ (or branching fraction $f_i$) of the $i$-th decay channel is defined as the fraction of all decays that proceed through that channel:

$$BR_i = \frac{\lambda_i}{\lambda_{\text{tot}}} = \frac{\lambda_i}{\sum \lambda_k} = \frac{T_{1/2,\text{tot}}}{T_{1/2,i}}$$

By definition, the sum of all branching ratios equals unity:

$$\sum_{i=1}^m BR_i = 1.00 \quad (100\%)$$

Daughter Growth Kinetics in Branching Decay

The production rate of daughter nucleus $D_i$ formed via the $i$-th branch is:

$$\frac{d N_{D,i}}{dt} = \lambda_i N(t) = \lambda_i N_0 e^{-\lambda_{\text{tot}} t}$$

Assuming $N_{D,i}(0) = 0$ and that the daughter is stable:

$$N_{D,i}(t) = \int_0^t \lambda_i N_0 e^{-\lambda_{\text{tot}} t'} dt' = \frac{\lambda_i}{\lambda_{\text{tot}}} N_0 (1 - e^{-\lambda_{\text{tot}} t}) = BR_i \cdot N_0 (1 - e^{-\lambda_{\text{tot}} t})$$

At any time $t$, the ratio of the numbers of atoms of two alternative stable daughters produced is:

$$\frac{N_{D,1}(t)}{N_{D,2}(t)} = \frac{BR_1}{BR_2} = \frac{\lambda_1}{\lambda_2} = \text{constant}$$

This constant ratio forms the physical foundation of K-Ar (potassium-argon) geological dating, where the accumulated $^{40}\text{Ar}$ relative to $^{40}\text{Ca}$ determines rock crystallization age over billions of years.

Β§3.5 Successive Decay Series & Rigorous Derivation of the Master Bateman Equations

In many radioactive decay chains, an unstable parent radionuclide decays into a daughter which is itself radioactive, giving rise to a cascade of successive decays:

$$N_1 \xrightarrow{\lambda_1} N_2 \xrightarrow{\lambda_2} N_3 \xrightarrow{\lambda_3} \dots \xrightarrow{\lambda_{n-1}} N_n \xrightarrow{\lambda_n} \dots$$

The System of Coupled Differential Rate Equations

The time evolution of the population of each nuclide in the chain is governed by a balance between production from its immediate parent and disappearance via its own decay:

$$\frac{dN_1}{dt} = -\lambda_1 N_1$$
$$\frac{dN_2}{dt} = \lambda_1 N_1 - \lambda_2 N_2$$
$$\frac{dN_3}{dt} = \lambda_2 N_2 - \lambda_3 N_3$$
$$\vdots$$
$$\frac{dN_i}{dt} = \lambda_{i-1} N_{i-1} - \lambda_i N_i$$

Derivation for a Two-Step Chain ($N_1 \to N_2 \to N_3$)

Let the initial conditions at $t = 0$ be $N_1(0) = N_1^0$ and $N_2(0) = 0, N_3(0) = 0$.

1. Parent Solution:

$$N_1(t) = N_1^0 e^{-\lambda_1 t}$$

2. Daughter Solution:

Substitute $N_1(t)$ into the daughter differential equation:

$$\frac{dN_2}{dt} + \lambda_2 N_2 = \lambda_1 N_1^0 e^{-\lambda_1 t}$$

This is a first-order linear ordinary differential equation. Multiplying by the integrating factor $e^{\lambda_2 t}$:

$$e^{\lambda_2 t}\left(\frac{dN_2}{dt} + \lambda_2 N_2\right) = \frac{d}{dt}\left(N_2 e^{\lambda_2 t}\right) = \lambda_1 N_1^0 e^{(\lambda_2 - \lambda_1) t}$$

Integrating both sides from $0$ to $t$:

$$N_2(t) e^{\lambda_2 t} - N_2(0) = \lambda_1 N_1^0 \int_0^t e^{(\lambda_2 - \lambda_1) t'} dt'$$

Assuming $\lambda_1 \ne \lambda_2$:

$$N_2(t) e^{\lambda_2 t} = \frac{\lambda_1 N_1^0}{\lambda_2 - \lambda_1} \left( e^{(\lambda_2 - \lambda_1) t} - 1 \right)$$

Multiplying by $e^{-\lambda_2 t}$:

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

The daughter activity $A_2(t) = \lambda_2 N_2(t)$ is:

$$A_2(t) = \frac{\lambda_2 \lambda_1}{\lambda_2 - \lambda_1} N_1^0 \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right) = \frac{\lambda_2}{\lambda_2 - \lambda_1} A_1(t) \left( 1 - e^{-(\lambda_2 - \lambda_1) t} \right)$$

General Harry Bateman Solution (1910)

For an arbitrary linear chain of $n$ nuclides with initial condition $N_1(0) = N_1^0$ and $N_2(0) = N_3(0) = \dots = N_n(0) = 0$, Harry Bateman proved via Laplace transforms that the population of the $n$-th nuclide is:

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

Expanded explicitly:

$$N_n(t) = C_1 e^{-\lambda_1 t} + C_2 e^{-\lambda_2 t} + \dots + C_n e^{-\lambda_n t}$$

where each coefficient $C_j$ is given by:

$$C_j = \frac{\lambda_1 \lambda_2 \cdots \lambda_{n-1}}{(\lambda_1 - \lambda_j)(\lambda_2 - \lambda_j)\cdots(\lambda_{j-1} - \lambda_j)(\lambda_{j+1} - \lambda_j)\cdots(\lambda_n - \lambda_j)} N_1^0$$

This elegant theorem solves the isotopic inventory of any nuclear decay chain, from uranium decay series to complex reactor core transmutations.

Β§3.6 Secular and Transient Radioactive Equilibria: Mathematical Conditions & Time-to-Peak

In a successive decay chain $N_1 \xrightarrow{\lambda_1} N_2 \xrightarrow{\lambda_2} N_3$, the temporal relationship between parent and daughter activities depends fundamentally on the relative magnitudes of the decay constants $\lambda_1$ and $\lambda_2$. Three distinct kinetic regimes emerge:

1. Secular Equilibrium ($\lambda_1 \ll \lambda_2$, i.e., $T_{1/2,1} \gg T_{1/2,2}$)

When the parent half-life is thousands or millions of times longer than the daughter half-life, the parent activity remains essentially constant over experimental timescales:

$$\lambda_1 \approx 0 \implies e^{-\lambda_1 t} \approx 1 \quad \text{and} \quad \lambda_2 - \lambda_1 \approx \lambda_2$$

Substituting into the daughter activity equation:

$$A_2(t) = \frac{\lambda_2 \lambda_1}{\lambda_2 - \lambda_1} N_1^0 (e^{-\lambda_1 t} - e^{-\lambda_2 t}) \approx \frac{\lambda_2 \lambda_1}{\lambda_2} N_1^0 (1 - e^{-\lambda_2 t}) = \lambda_1 N_1^0 (1 - e^{-\lambda_2 t})$$
$$A_2(t) = A_1^0 (1 - e^{-\lambda_2 t})$$

``` Activity A(t) A₁ ▲──────────────────────────────────── Parent A₁(t) = A₁⁰ (Constant) β”‚ /──────────────────────── Daughter Aβ‚‚(t) in Secular Eq: Aβ‚‚ = A₁ β”‚ / β”‚ / β”‚ / β”‚_______/ └───────┴────────────────────────────► Time t ~ 7 T₁/β‚‚,β‚‚ ```

As $t \gg T_{1/2,2}$ ($t \ge 7 T_{1/2,2}$), $e^{-\lambda_2 t} \to 0$:

$$A_2(t) \longrightarrow A_1(t) \implies \lambda_1 N_1 = \lambda_2 N_2$$
$$\frac{N_1}{N_2} = \frac{\lambda_2}{\lambda_1} = \frac{T_{1/2,1}}{T_{1/2,2}}$$

In secular equilibrium, the daughter activity becomes exactly equal to the parent activity, and the abundance ratio of parent to daughter is directly proportional to their half-lives! Classic natural examples:

  • $^{226}\text{Ra}$ ($T_{1/2} = 1600\text{ yr}$) $\xrightarrow{\alpha} \, ^{222}\text{Rn}$ ($T_{1/2} = 3.82\text{ days}$)
  • $^{238}\text{U}$ ($T_{1/2} = 4.47 \times 10^9\text{ yr}$) $\xrightarrow{\alpha} \, ^{234}\text{Th}$ ($T_{1/2} = 24.1\text{ days}$)

2. Transient Equilibrium ($\lambda_1 < \lambda_2$, but parent decays noticeably, $T_{1/2,1} \approx 10 T_{1/2,2}$)

When the parent half-life is moderately longer than the daughter, the daughter activity rises, reaches a maximum, and then decays with the apparent half-life of the parent. For large $t$ such that $e^{-\lambda_2 t} \ll e^{-\lambda_1 t}$:

$$A_2(t) \approx \frac{\lambda_2}{\lambda_2 - \lambda_1} A_1^0 e^{-\lambda_1 t} = \frac{\lambda_2}{\lambda_2 - \lambda_1} A_1(t)$$

Notice that in transient equilibrium, the daughter activity exceeds the parent activity by the constant factor:

$$\frac{A_2(t)}{A_1(t)} = \frac{\lambda_2}{\lambda_2 - \lambda_1} > 1$$

Derivation of Time to Maximum Daughter Activity ($t_{\max}$)

To determine when the daughter activity reaches its peak, differentiate $N_2(t)$ with respect to $t$ and set the derivative to zero:

$$\frac{dN_2}{dt} = \frac{\lambda_1 N_1^0}{\lambda_2 - \lambda_1} \left( -\lambda_1 e^{-\lambda_1 t} + \lambda_2 e^{-\lambda_2 t} \right) = 0$$
$$\lambda_1 e^{-\lambda_1 t_{\max}} = \lambda_2 e^{-\lambda_2 t_{\max}}$$

Taking natural logarithms:

$$\ln\lambda_1 - \lambda_1 t_{\max} = \ln\lambda_2 - \lambda_2 t_{\max}$$
$$(\lambda_2 - \lambda_1) t_{\max} = \ln\left(\frac{\lambda_2}{\lambda_1}\right)$$

Yielding the Time-to-Peak Equation:

$$t_{\max} = \frac{\ln(\lambda_2 / \lambda_1)}{\lambda_2 - \lambda_1} = \frac{\ln(T_{1/2,1} / T_{1/2,2})}{\lambda_2 - \lambda_1}$$

At exactly $t = t_{\max}$, substituting $\lambda_1 e^{-\lambda_1 t} = \lambda_2 e^{-\lambda_2 t}$ into $A_2(t)$:

$$A_2(t_{\max}) = A_1(t_{\max})$$

At the peak, the parent and daughter activity curves cross each other! The premier radiopharmaceutical example is the $^{99}\text{Mo} \to ^{99m}\text{Tc}$ generator:

  • $^{99}\text{Mo}$: $T_{1/2} = 66.0\text{ hours}$
  • $^{99m}\text{Tc}$: $T_{1/2} = 6.01\text{ hours}$
  • $t_{\max} \approx 22.9\text{ hours}$ post-elution, determining the optimal clinical elution schedule.

Β§3.7 Non-Equilibrium Decay Regimes & Radionuclide Inventory Evolution

The third kinetic regime occurs when the parent decays faster than the daughter.

No Equilibrium ($\lambda_1 > \lambda_2$, i.e., $T_{1/2,1} < T_{1/2,2}$)

When the parent has a shorter half-life than the daughter, no equilibrium is ever established. As time progresses:

$$e^{-\lambda_1 t} \ll e^{-\lambda_2 t}$$

The exponential term of the parent vanishes rapidly, leaving only the daughter's decay term:

$$N_2(t) \approx \frac{\lambda_1}{\lambda_1 - \lambda_2} N_1^0 e^{-\lambda_2 t}$$

All initial parent nuclei disintegrate into daughter nuclei, which then decay away at their own leisurely characteristic rate $\lambda_2$. Examples:

  • $^{140}\text{Ba}$ ($T_{1/2} = 12.75\text{ days}$) $\xrightarrow{\beta^-} \, ^{140}\text{La}$ ($T_{1/2} = 1.678\text{ days}$)
  • $^{218}\text{Po}$ ($T_{1/2} = 3.10\text{ min}$) $\xrightarrow{\alpha} \, ^{214}\text{Pb}$ ($T_{1/2} = 26.8\text{ min}$)

``` Activity A(t) β–² β”‚ Parent A₁(t) (Rapid Fall) β”‚ \ β”‚ \ Daughter Growth and Slow Decay Aβ‚‚(t) β”‚ \ /──────────────\ β”‚ \____/ \_________________ └────────────────────────────────────────────────► Time t ```

Comprehensive Comparison of Equilibrium Regimes

| Equilibrium Type | Half-Life Relationship | Ratio of Decay Constants | Activity Ratio at Late Times | Cross-Over at $t_{\max}$? | Representative Real-World Example | | :--- | :--- | :--- | :--- | :--- | :--- | | Secular | $T_{1/2,1} \gg T_{1/2,2}$ ($>10^4\times$) | $\lambda_1 \ll \lambda_2$ | $\frac{A_2(t)}{A_1(t)} \to 1.000$ | No ($A_2$ approaches $A_1$ asymptotically) | $^{226}\text{Ra} \xrightarrow{\alpha} ^{222}\text{Rn}$ | | Transient | $T_{1/2,1} > T_{1/2,2}$ ($\sim 10\times$) | $\lambda_1 < \lambda_2$ | $\frac{A_2(t)}{A_1(t)} \to \frac{\lambda_2}{\lambda_2 - \lambda_1} > 1$ | Yes ($A_2 = A_1$ at $t = t_{\max}$) | $^{99}\text{Mo} \xrightarrow{\beta^-} ^{99m}\text{Tc}$ | | No Equilibrium | $T_{1/2,1} < T_{1/2,2}$ | $\lambda_1 > \lambda_2$ | $\frac{A_2(t)}{A_1(t)} \to \infty$ ($A_1 \to 0$) | Yes ($A_2$ overtakes $A_1$, parent vanishes) | $^{218}\text{Po} \xrightarrow{\alpha} ^{214}\text{Pb}$ |

Radionuclide Inventory in Nuclear Reactors and Spent Fuel

Understanding these three regimes is the core principle of nuclear fuel inventory codes (e.g., ORIGEN). In freshly discharged spent nuclear fuel:

  1. Short-lived fission products ($T_{1/2} \sim \text{seconds to days}$) undergo rapid non-equilibrium cascades into longer-lived daughters.
  2. In intermediate storage ponds ($1 - 5\text{ years}$), isotopes like $^{137}\text{Cs} \to ^{137m}\text{Ba}$ ($T_{1/2} = 2.55\text{ min}$) and $^{90}\text{Sr} \to ^{90}\text{Y}$ ($T_{1/2} = 64.1\text{ hours}$) exist in strict secular equilibrium, where the daughter activity matches the 30-year parent activity.
  3. In deep geological repositories ($>10^4\text{ years}$), actinide decay chains ($^{241}\text{Pu} \to ^{241}\text{Am} \to ^{237}\text{Np}$) govern long-term radiotoxicity.

Β§3.8 Matrix Exponential & Laplace Transform Methods in Bateman Decay Networks

While the classical Bateman equations provide elegant solutions for simple linear chains ($N_1 \to N_2 \to N_3$), real-world nuclear fuel cycles and reactor transmutation networks involve branched, cyclic, and cross-feeding pathways (e.g., $(n, \gamma)$ activation competing with simultaneous beta and alpha decays). For such complex networks, modern nuclear inventory codes (e.g., ORIGEN, CINDER) formulate the problem via matrix differential equations.

Matrix Formulation of Transmutation Chains

Let the column vector $\vec{N}(t) = [N_1(t), N_2(t), \dots, N_k(t)]^T$ denote the time-dependent population inventory of $k$ distinct nuclides. The governing system of coupled first-order differential equations is:

$$\frac{d\vec{N}(t)}{dt} = \mathbf{A} \vec{N}(t)$$

where $\mathbf{A}$ is the $k \times k$ transition rate matrix (or burnup matrix):

  • Diagonal elements represent the total destruction rate (decay constant plus neutron absorption):
$$A_{ii} = -(\lambda_i + \sigma_{a,i} \Phi)$$
  • Off-diagonal elements represent the production rate of nuclide $i$ from precursor $j$:
$$A_{ij} = \lambda_{j \to i} + \sigma_{j \to i} \Phi \quad (i \ne j)$$

Formal Solution via the Matrix Exponential

The exact analytical solution of this autonomous linear system with initial inventory $\vec{N}(0)$ is:

$$\vec{N}(t) = \exp(\mathbf{A} t) \vec{N}(0)$$

where the matrix exponential is defined by the convergent Taylor series:

$$\exp(\mathbf{A} t) = \mathbf{I} + \mathbf{A} t + \frac{(\mathbf{A} t)^2}{2!} + \frac{(\mathbf{A} t)^3}{3!} + \dots = \sum_{m=0}^\infty \frac{(\mathbf{A} t)^m}{m!}$$

Numerical Solution via the Chebyshev Rational Approximation Method (CRAM)

Direct Taylor summation of $\exp(\mathbf{A} t)$ suffers from numerical instability when the spectrum of eigenvalues $\lambda_i$ spans dozens of orders of magnitude (from fractions of a second to billions of years). Modern reactor burnup solvers utilize the Chebyshev Rational Approximation Method (CRAM), approximating $\exp(\mathbf{A} t)$ by a rational function $\hat{r}_k(z) = p_k(z) / q_k(z)$ on the negative real axis:

$$\exp(\mathbf{A} t) \approx \alpha_0 \mathbf{I} + 2 \text{Re}\left( \sum_{j=1}^{k/2} \alpha_j (\mathbf{A} t - \theta_j \mathbf{I})^{-1} \right)$$

where $\alpha_j$ and $\theta_j$ are pre-computed complex poles and residues. CRAM evaluates the complete isotopic inventory of thousands of isotopes across an entire nuclear reactor core in milliseconds with precision exceeding $10^{-14}$!

Mathematical Taxonomy of Radioactive Decay Kinetics & Equilibrium Regimes

``` SUCCESSIVE DECAY KINETICS N₁ ──(λ₁)──► Nβ‚‚ ──(Ξ»β‚‚)──► N₃

REGIME CONDITION ACTIVITY RATIO (Late Times) ───────────── ─────────────────────── ─────────────────────────────── Secular Eq λ₁ β‰ͺ Ξ»β‚‚ (T₁/β‚‚ ≫ 10⁴×) Aβ‚‚(t) / A₁(t) ──► 1.000 Transient Eq λ₁ < Ξ»β‚‚ (T₁/β‚‚ β‰ˆ 10Γ—) Aβ‚‚(t) / A₁(t) ──► Ξ»β‚‚ / (Ξ»β‚‚ - λ₁) > 1 No Equilibrium λ₁ > Ξ»β‚‚ (T₁/β‚‚ < T₁/β‚‚,β‚‚) Aβ‚‚(t) / A₁(t) ──► ∞ (A₁ vanishes) ```

In the secular regime, the abundance of all intermediate daughters in an undisturbed rock is directly proportional to their half-lives:

$$\frac{N_1}{T_{1/2,1}} = \frac{N_2}{T_{1/2,2}} = \frac{N_3}{T_{1/2,3}} = \dots = \frac{N_n}{T_{1/2,n}}$$

This simple relation allows geochemists to determine the age of continents, meteorites, and lunar samples.

Intermediate Example 3.1: Bateman Transient Equilibrium Peak Time for Mo-99 / Tc-99m Generator

A molybdenum-99 / technetium-99m medical generator utilizes fission-produced $^{99}\text{Mo}$ ($T_{1/2,1} = 66.00\text{ hours}$) which beta-decays with a branching fraction $BR = 0.875$ into metastable $^{99m}\text{Tc}$ ($T_{1/2,2} = 6.007\text{ hours}$). Immediately following complete saline elution (milking) at $t = 0$, the column contains zero $^{99m}\text{Tc}$ ($N_2(0) = 0$).

  1. Calculate the decay constants $\lambda_1$ and $\lambda_2$ in $\text{h}^{-1}$.
  2. Derive and calculate the exact time $t_{\max}$ (in hours) at which the $^{99m}\text{Tc}$ activity on the column reaches its theoretical maximum.
  3. Calculate the ratio of $^{99m}\text{Tc}$ activity to $^{99}\text{Mo}$ activity on the column at transient equilibrium ($t \gg t_{\max}$).

Step 1: Decay Constants

$$\lambda_1 = \frac{\ln 2}{66.00\text{ h}} = \frac{0.693147}{66.00} \approx 0.010502\text{ h}^{-1}$$
$$\lambda_2 = \frac{\ln 2}{6.007\text{ h}} = \frac{0.693147}{6.007} \approx 0.115390\text{ h}^{-1}$$

Step 2: Time to Maximum Activity $t_{\max}$

Because branching fraction $BR$ is a constant multiplier, it does not alter the location of the extremum:

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

Compute the ratio and difference:

$$\frac{\lambda_2}{\lambda_1} = \frac{0.115390}{0.010502} \approx 10.9874$$
$$\ln\left(\frac{\lambda_2}{\lambda_1}\right) = \ln(10.9874) \approx 2.39675$$
$$\lambda_2 - \lambda_1 = 0.115390 - 0.010502 = 0.104888\text{ h}^{-1}$$
$$t_{\max} = \frac{2.39675}{0.104888\text{ h}^{-1}} \approx 22.85\text{ hours}$$

The technetium activity peaks at $22.85\text{ hours}$ post-elution (explaining why radiopharmacies elute generators once every 24 hours).

Step 3: Activity Ratio at Transient Equilibrium

Accounting for the branching ratio $BR = 0.875$:

$$\frac{A_2(t)}{A_1(t)} = BR \cdot \frac{\lambda_2}{\lambda_2 - \lambda_1} = 0.875 \times \frac{0.115390}{0.104888} = 0.875 \times 1.10012 \approx 0.9626$$

At transient equilibrium, the $^{99m}\text{Tc}$ activity tracks at $96.3\%$ of the $^{99}\text{Mo}$ parent activity.

Easy Example 3.2: Secular Equilibrium Activity and Mass in Natural Uranium Ores

In an undisturbed geological mineral specimen of pitchblende ($\text{U}_3\text{O}_8$) containing $1.000\text{ kg}$ of natural uranium ($99.274\%$ $^{238}\text{U}$, $T_{1/2} = 4.468 \times 10^9\text{ yr}$), all decay chain daughters reside in strict secular equilibrium. Radium-226 ($^{226}\text{Ra}$, $T_{1/2} = 1600\text{ yr}$) is one of the intermediate alpha-emitting daughters in the uranium series.

  1. Calculate the absolute activity of $^{238}\text{U}$ in the rock in $\text{MBq}$.
  2. State the activity of $^{226}\text{Ra}$ in the rock.
  3. Calculate the mass of $^{226}\text{Ra}$ present in milligrams ($\text{mg}$).

Step 1: Activity of $^{238}\text{U}$

Mass of $^{238}\text{U}$:

$$m = 1000\text{ g} \times 0.99274 = 992.74\text{ g}$$

Number of atoms:

$$N_{238} = \frac{992.74\text{ g}}{238.05\text{ g/mol}} \times 6.02214 \times 10^{23}\text{ mol}^{-1} \approx 2.5113 \times 10^{24}\text{ atoms}$$

Decay constant in $\text{s}^{-1}$:

$$T_{1/2} = 4.468 \times 10^9\text{ yr} \times 3.15576 \times 10^7\text{ s/yr} = 1.40999 \times 10^{17}\text{ s}$$
$$\lambda_{238} = \frac{\ln 2}{1.40999 \times 10^{17}\text{ s}} \approx 4.91597 \times 10^{-18}\text{ s}^{-1}$$

Activity:

$$A_{238} = \lambda_{238} N_{238} = (4.91597 \times 10^{-18}\text{ s}^{-1})(2.5113 \times 10^{24}) \approx 1.2345 \times 10^7\text{ Bq} = 12.35\text{ MBq}$$

Step 2: Activity of $^{226}\text{Ra}$

In secular equilibrium, the activity of every radioactive member in an unbranched chain is identical:

$$A_{\text{Ra}} = A_{238} = 12.35\text{ MBq} \quad (0.334\text{ mCi})$$

Step 3: Mass of $^{226}\text{Ra}$

From secular equilibrium relation:

$$\lambda_{\text{Ra}} N_{\text{Ra}} = \lambda_{238} N_{238} \implies N_{\text{Ra}} = N_{238} \frac{\lambda_{238}}{\lambda_{\text{Ra}}} = N_{238} \frac{T_{1/2,\text{Ra}}}{T_{1/2,238}}$$
$$N_{\text{Ra}} = (2.5113 \times 10^{24}) \times \frac{1600\text{ yr}}{4.468 \times 10^9\text{ yr}} \approx 8.993 \times 10^{17}\text{ atoms}$$

Mass of radium:

$$m_{\text{Ra}} = \frac{N_{\text{Ra}}}{N_A} \times M_{\text{Ra}} = \frac{8.993 \times 10^{17}}{6.02214 \times 10^{23}} \times 226.03\text{ g/mol} \approx 3.376 \times 10^{-4}\text{ g} = 0.338\text{ mg}$$

One metric ton ($1000\text{ kg}$) of uranium ore yields only $\approx 338\text{ milligrams}$ of radium (illustrating why the Curies had to process tons of pitchblende).

Intermediate Example 3.3: Poisson Counting Statistics and Minimum Detectable Activity

A low-background proportional counter records gross counts from an environmental water sample over a counting period $t_g = 60.0\text{ min}$, registering $N_g = 1,440\text{ counts}$. A blank background count accumulated over $t_b = 120.0\text{ min}$ registered $N_b = 1,800\text{ counts}$.

  1. Calculate the gross count rate $R_g$ and background count rate $R_b$ in counts per minute (cpm).
  2. Calculate the net count rate $R_n$ and its standard uncertainty $\sigma_{R_n}$ in cpm.
  3. Compute the $95\%$ confidence interval ($1.96 \sigma$) for the net count rate.
  4. If detector efficiency is $\epsilon = 0.350$ ($35\%$), calculate the sample activity in Becquerels (Bq).

Step 1: Gross and Background Count Rates

Gross count rate:

$$R_g = \frac{N_g}{t_g} = \frac{1440\text{ counts}}{60.0\text{ min}} = 24.00\text{ cpm}$$

Background count rate:

$$R_b = \frac{N_b}{t_b} = \frac{1800\text{ counts}}{120.0\text{ min}} = 15.00\text{ cpm}$$

Step 2: Net Count Rate and Uncertainty

Net count rate:

$$R_n = R_g - R_b = 24.00 - 15.00 = 9.00\text{ cpm}$$

Uncertainty propagation:

$$\sigma_{R_n} = \sqrt{\frac{R_g}{t_g} + \frac{R_b}{t_b}} = \sqrt{\frac{24.00}{60.0} + \frac{15.00}{120.0}} = \sqrt{0.400 + 0.125} = \sqrt{0.525} \approx 0.7246\text{ cpm}$$

Result: $R_n = 9.00 \pm 0.72\text{ cpm}$.

Step 3: 95% Confidence Interval

$$95\%\text{ CI} = R_n \pm 1.96 \sigma_{R_n} = 9.00 \pm (1.96 \times 0.7246) = 9.00 \pm 1.42\text{ cpm} \quad [7.58\text{ to }10.42\text{ cpm}]$$

Step 4: Net Activity in Becquerels

Convert net rate to counts per second (cps):

$$R_n = \frac{9.00\text{ cpm}}{60\text{ s/min}} = 0.150\text{ cps}$$

Sample activity:

$$A = \frac{R_n}{\epsilon} = \frac{0.150\text{ cps}}{0.350\text{ cps/Bq}} \approx 0.4286\text{ Bq} \pm 0.035\text{ Bq}$$
Advanced Example 3.4: Potassium-40 Dual Branching Decay and K-Ar Geochronology

Potassium-40 ($^{40}\text{K}$) undergoes branched decay with a total half-life $T_{1/2,\text{tot}} = 1.248 \times 10^9\text{ yr}$:

  • $89.28\%$ via $\beta^-$ to $^{40}\text{Ca}$ ($\lambda_\beta = 4.962 \times 10^{-10}\text{ yr}^{-1}$)
  • $10.72\%$ via Electron Capture / $\beta^+$ to $^{40}\text{Ar}$ ($\lambda_{\text{EC}} = 5.960 \times 10^{-11}\text{ yr}^{-1}$)

A volcanic rock specimen contains $N_{\text{Ar}} = 3.50 \times 10^{15}\text{ atoms}$ of radiogenic $^{40}\text{Ar}$ and $N_{\text{K}} = 2.80 \times 10^{16}\text{ atoms}$ of $^{40}\text{K}$. Assuming zero initial argon was trapped at solidification:

  1. Derive the K-Ar age equation for $t$ in terms of $N_{\text{Ar}} / N_{\text{K}}$.
  2. Calculate the geological age $t$ of the volcanic formation in millions of years (Ma).

Step 1: Derivation of K-Ar Age Equation

The accumulation rate of $^{40}\text{Ar}$ is:

$$N_{\text{Ar}}(t) = \frac{\lambda_{\text{EC}}}{\lambda_{\text{tot}}} N_{\text{K}}(0) (1 - e^{-\lambda_{\text{tot}} t})$$

The remaining potassium-40 atoms is:

$$N_{\text{K}}(t) = N_{\text{K}}(0) e^{-\lambda_{\text{tot}} t} \implies N_{\text{K}}(0) = N_{\text{K}}(t) e^{\lambda_{\text{tot}} t}$$

Substituting $N_{\text{K}}(0)$:

$$N_{\text{Ar}}(t) = \frac{\lambda_{\text{EC}}}{\lambda_{\text{tot}}} N_{\text{K}}(t) e^{\lambda_{\text{tot}} t} (1 - e^{-\lambda_{\text{tot}} t}) = \frac{\lambda_{\text{EC}}}{\lambda_{\text{tot}}} N_{\text{K}}(t) (e^{\lambda_{\text{tot}} t} - 1)$$

Dividing by $N_{\text{K}}(t)$:

$$\frac{N_{\text{Ar}}}{N_{\text{K}}} = \frac{\lambda_{\text{EC}}}{\lambda_{\text{tot}}} (e^{\lambda_{\text{tot}} t} - 1)$$

Solving for age $t$:

$$e^{\lambda_{\text{tot}} t} = 1 + \frac{\lambda_{\text{tot}}}{\lambda_{\text{EC}}} \left(\frac{N_{\text{Ar}}}{N_{\text{K}}}\right)$$
$$t = \frac{1}{\lambda_{\text{tot}}} \ln\left[ 1 + \frac{\lambda_{\text{tot}}}{\lambda_{\text{EC}}} \left(\frac{N_{\text{Ar}}}{N_{\text{K}}}\right) \right]$$

Step 2: Numerical Age Calculation

Given parameters:

$$\lambda_{\text{tot}} = \lambda_\beta + \lambda_{\text{EC}} = 4.962 \times 10^{-10} + 5.960 \times 10^{-11} = 5.558 \times 10^{-10}\text{ yr}^{-1}$$
$$\frac{\lambda_{\text{tot}}}{\lambda_{\text{EC}}} = \frac{5.558 \times 10^{-10}}{5.960 \times 10^{-11}} \approx 9.3255$$

Atomic ratio:

$$\frac{N_{\text{Ar}}}{N_{\text{K}}} = \frac{3.50 \times 10^{15}}{2.80 \times 10^{16}} = 0.1250$$

Argument of logarithm:

$$1 + 9.3255 \times 0.1250 = 1 + 1.16569 = 2.16569$$

Calculate age $t$:

$$t = \frac{\ln(2.16569)}{5.558 \times 10^{-10}\text{ yr}^{-1}} = \frac{0.77274}{5.558 \times 10^{-10}} \approx 1.3903 \times 10^9\text{ yr} \approx 1,390\text{ Ma}$$

The rock solidified $1.39\text{ billion years}$ ago.

Advanced Example 3.5: Three-Nuclide Bateman Chain Activity Inventory

Consider a three-member decay series:

$$N_1 \xrightarrow{\lambda_1 = 0.20\text{ h}^{-1}} N_2 \xrightarrow{\lambda_2 = 0.50\text{ h}^{-1}} N_3 \xrightarrow{\lambda_3 = 0.80\text{ h}^{-1}} N_4 \text{ (Stable)}$$

Initially at $t = 0$, pure parent nuclide is prepared with $N_1(0) = 1.00 \times 10^{12}\text{ atoms}$ and $N_2(0) = N_3(0) = 0$.

  1. Write down the explicit Bateman expansion for $N_3(t)$.
  2. Calculate the number of atoms $N_3$ and activity $A_3$ at $t = 2.0\text{ hours}$.

Step 1: Bateman Coefficients for $N_3(t)$

For $n = 3$, Bateman formula gives:

$$N_3(t) = N_1(0) \lambda_1 \lambda_2 \left[ \frac{e^{-\lambda_1 t}}{(\lambda_2 - \lambda_1)(\lambda_3 - \lambda_1)} + \frac{e^{-\lambda_2 t}}{(\lambda_1 - \lambda_2)(\lambda_3 - \lambda_2)} + \frac{e^{-\lambda_3 t}}{(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3)} \right]$$

Pre-factor:

$$N_1(0) \lambda_1 \lambda_2 = (1.00 \times 10^{12})(0.20)(0.50) = 1.00 \times 10^{11}$$

Denominators:

  1. Term 1 ($j = 1$):
$$(\lambda_2 - \lambda_1)(\lambda_3 - \lambda_1) = (0.50 - 0.20)(0.80 - 0.20) = (0.30)(0.60) = 0.18$$
$$C_1 = \frac{1.00 \times 10^{11}}{0.18} \approx 5.5556 \times 10^{11}$$
  1. Term 2 ($j = 2$):
$$(\lambda_1 - \lambda_2)(\lambda_3 - \lambda_2) = (0.20 - 0.50)(0.80 - 0.50) = (-0.30)(0.30) = -0.09$$
$$C_2 = \frac{1.00 \times 10^{11}}{-0.09} \approx -1.1111 \times 10^{12}$$
  1. Term 3 ($j = 3$):
$$(\lambda_1 - \lambda_3)(\lambda_2 - \lambda_3) = (0.20 - 0.80)(0.50 - 0.80) = (-0.60)(-0.30) = +0.18$$
$$C_3 = \frac{1.00 \times 10^{11}}{0.18} \approx 5.5556 \times 10^{11}$$

Notice $C_1 + C_2 + C_3 = 5.5556 \times 10^{11} - 11.1111 \times 10^{11} + 5.5556 \times 10^{11} = 0$, satisfying $N_3(0) = 0$.

Step 2: Evaluation at $t = 2.0\text{ h}$

$$e^{-\lambda_1 t} = e^{-0.20 \times 2} = e^{-0.40} \approx 0.67032$$
$$e^{-\lambda_2 t} = e^{-0.50 \times 2} = e^{-1.00} \approx 0.36788$$
$$e^{-\lambda_3 t} = e^{-0.80 \times 2} = e^{-1.60} \approx 0.20190$$

Substitute:

$$N_3(2) = 10^{11} \left[ \frac{0.67032}{0.18} - \frac{0.36788}{0.09} + \frac{0.20190}{0.18} \right] = 10^{11} [3.7240 - 4.0876 + 1.1217] = 10^{11} [0.7581] \approx 7.581 \times 10^{10}\text{ atoms}$$

Activity $A_3$ in Becquerels:

$$\lambda_3 = 0.80\text{ h}^{-1} = \frac{0.80}{3600\text{ s}} \approx 2.222 \times 10^{-4}\text{ s}^{-1}$$
$$A_3 = \lambda_3 N_3 = (2.222 \times 10^{-4}\text{ s}^{-1})(7.581 \times 10^{10}) \approx 1.685 \times 10^7\text{ Bq} = 16.85\text{ MBq}$$
Intermediate Example 3.6: Continuous Radionuclide Production Kinetics in Cyclotron Beam

A target is irradiated in a biomedical cyclotron to produce fluorine-18 ($^{18}\text{F}$, $T_{1/2} = 109.77\text{ min}$) via the $^{18}\text{O}(p, n)^{18}\text{F}$ nuclear reaction at a constant production rate $R = 2.50 \times 10^9\text{ atoms/s}$.

  1. Formulate the differential rate equation for $^{18}\text{F}$ inventory during irradiation and derive $N(t)$ and $A(t)$.
  2. Calculate the saturation activity $A_{\text{sat}}$ in $\text{GBq}$.
  3. Calculate the activity produced after an irradiation time of $t = 2.00\text{ hours}$.
  4. Determine the percentage of saturation achieved.

Step 1: Production Differential Rate Equation

During irradiation, atoms are generated at constant rate $R$ and simultaneously decay at rate $\lambda N$:

$$\frac{dN}{dt} = R - \lambda N \implies \frac{dN}{dt} + \lambda N = R$$

Using integrating factor $e^{\lambda t}$:

$$\frac{d}{dt}(N e^{\lambda t}) = R e^{\lambda t}$$

Integrating with $N(0) = 0$:

$$N(t) e^{\lambda t} = \frac{R}{\lambda}(e^{\lambda t} - 1) \implies N(t) = \frac{R}{\lambda}(1 - e^{-\lambda t})$$

The activity $A(t) = \lambda N(t)$ is:

$$A(t) = R (1 - e^{-\lambda t})$$

Step 2: Saturation Activity $A_{\text{sat}}$

As $t \to \infty$, the decay rate exactly equals the production rate ($e^{-\lambda t} \to 0$):

$$A_{\text{sat}} = R = 2.50 \times 10^9\text{ Bq} = 2.50\text{ GBq} \quad (\sim 67.6\text{ mCi})$$

Step 3: Activity After $t = 2.00\text{ hours}$

Convert half-life to hours:

$$T_{1/2} = \frac{109.77\text{ min}}{60\text{ min/h}} \approx 1.8295\text{ h}$$
$$\lambda = \frac{\ln 2}{1.8295\text{ h}} \approx 0.37887\text{ h}^{-1}$$

Saturation factor:

$$1 - e^{-\lambda t} = 1 - e^{-(0.37887 \times 2.00)} = 1 - e^{-0.75774} = 1 - 0.46872 = 0.53128$$

Activity:

$$A(2\text{ h}) = 2.50\text{ GBq} \times 0.53128 \approx 1.328\text{ GBq}$$

Step 4: Percentage of Saturation

$$\% \text{ Saturation} = 53.13\%$$

After $\approx 1.1$ half-lives, over half of the maximum achievable activity has already been reached.

Intermediate Example 3.7: Decay Chain Daughter Growth with Non-Zero Initial Daughter Activity

A radionuclide generator container arrives at a hospital containing $^{131m}\text{Xe}$ ($T_{1/2,1} = 11.9\text{ days}$) which decays into ground-state $^{131}\text{Xe}$ or similar chain. Consider an idealized chain where parent $N_1$ ($T_{1/2,1} = 20.0\text{ h}$) decays to daughter $N_2$ ($T_{1/2,2} = 4.0\text{ h}$). Due to incomplete prior separation, at $t = 0$ the daughter activity is not zero, with initial values: $A_1(0) = 50.0\text{ MBq}$ and $A_2(0) = 15.0\text{ MBq}$.

  1. Derive the general Bateman expression for $N_2(t)$ and $A_2(t)$ incorporating $N_2(0) \ne 0$.
  2. Calculate the daughter activity $A_2$ at $t = 6.0\text{ hours}$.

Step 1: Derivation with Initial Daughter Population

The differential equation is:

$$\frac{dN_2}{dt} + \lambda_2 N_2 = \lambda_1 N_1(0) e^{-\lambda_1 t}$$

Integrating with integrating factor $e^{\lambda_2 t}$ from $N_2(0)$ to $N_2(t)$:

$$N_2(t) e^{\lambda_2 t} - N_2(0) = \frac{\lambda_1 N_1(0)}{\lambda_2 - \lambda_1}(e^{(\lambda_2 - \lambda_1) t} - 1)$$
$$N_2(t) = N_2(0) e^{-\lambda_2 t} + \frac{\lambda_1 N_1(0)}{\lambda_2 - \lambda_1}(e^{-\lambda_1 t} - e^{-\lambda_2 t})$$

Multiplying through by $\lambda_2$:

$$A_2(t) = A_2(0) e^{-\lambda_2 t} + \frac{\lambda_2}{\lambda_2 - \lambda_1} A_1(0) (e^{-\lambda_1 t} - e^{-\lambda_2 t})$$

This reflects linear superposition: independent exponential decay of initial daughter plus in-growth from parent decay.

Step 2: Numerical Calculation at $t = 6.0\text{ h}$

Decay constants:

$$\lambda_1 = \frac{\ln 2}{20.0\text{ h}} = 0.034657\text{ h}^{-1}$$
$$\lambda_2 = \frac{\ln 2}{4.0\text{ h}} = 0.173287\text{ h}^{-1}$$
$$\lambda_2 - \lambda_1 = 0.173287 - 0.034657 = 0.138630\text{ h}^{-1}$$
$$\frac{\lambda_2}{\lambda_2 - \lambda_1} = \frac{0.173287}{0.138630} = 1.250$$

Exponentials at $t = 6.0\text{ h}$:

$$e^{-\lambda_1 t} = e^{-0.034657 \times 6} = e^{-0.20794} \approx 0.81225$$
$$e^{-\lambda_2 t} = e^{-0.173287 \times 6} = e^{-1.03972} \approx 0.35355$$

Compute components:

  1. Residual initial daughter:
$$A_{2,\text{init}} = 15.0\text{ MBq} \times 0.35355 \approx 5.303\text{ MBq}$$
  1. Growth from parent:
$$A_{2,\text{growth}} = 1.250 \times 50.0\text{ MBq} \times (0.81225 - 0.35355) = 62.5 \times 0.45870 \approx 28.669\text{ MBq}$$

Total daughter activity:

$$A_2(6\text{ h}) = 5.303 + 28.669 = 33.97\text{ MBq}$$
Intermediate Example 3.8: Actinium Natural Series Branching Decay Kinetics at Bismuth-211

In the natural actinium ($4n + 3$) radioactive decay series, bismuth-211 ($^{211}_{83}\text{Bi}$, $T_{1/2} = 2.14\text{ min}$) undergoes branched decay:

  • Alpha decay ($99.72\%$, $\lambda_\alpha$) to thallium-207 ($^{207}_{81}\text{Tl}$, $T_{1/2} = 4.77\text{ min}$)
  • Beta decay ($0.28\%$, $\lambda_\beta$) to polonium-211 ($^{211}_{84}\text{Po}$, $T_{1/2} = 0.516\text{ s}$)

Both branches terminate at stable lead-207 ($^{207}_{82}\text{Pb}$).

  1. Calculate the total decay constant $\lambda_{\text{tot}}$ and partial decay constants $\lambda_\alpha$ and $\lambda_\beta$ in $\text{s}^{-1}$.
  2. In a sample containing initially $N_0 = 1.00 \times 10^9\text{ atoms}$ of pure $^{211}\text{Bi}$ at $t = 0$, compute the total number of $^{207}\text{Tl}$ atoms and $^{211}\text{Po}$ atoms formed after complete decay ($t \to \infty$).

Step 1: Decay Constants

$$T_{1/2} = 2.14\text{ min} = 128.4\text{ s}$$
$$\lambda_{\text{tot}} = \frac{\ln 2}{128.4\text{ s}} \approx 0.0053983\text{ s}^{-1}$$

Partial constants:

$$\lambda_\alpha = BR_\alpha \cdot \lambda_{\text{tot}} = 0.9972 \times 0.0053983\text{ s}^{-1} \approx 0.0053832\text{ s}^{-1}$$
$$\lambda_\beta = BR_\beta \cdot \lambda_{\text{tot}} = 0.0028 \times 0.0053983\text{ s}^{-1} \approx 1.5115 \times 10^{-5}\text{ s}^{-1}$$

Step 2: Cumulative Atom Yields

As $t \to \infty$, all initial $^{211}\text{Bi}$ atoms disintegrate. The total number of atoms produced through each branch equals the branching fraction multiplied by $N_0$:

$$N(^{207}\text{Tl}) = BR_\alpha \cdot N_0 = 0.9972 \times 1.00 \times 10^9 = 9.972 \times 10^8\text{ atoms}$$
$$N(^{211}\text{Po}) = BR_\beta \cdot N_0 = 0.0028 \times 1.00 \times 10^9 = 2.80 \times 10^6\text{ atoms}$$

Both branches terminate at stable $^{207}\text{Pb}$, so exactly $1.00 \times 10^9$ lead-207 atoms are created.

Intermediate Example 3.9: Laplace Transform Solution of Two-Step Successive Decay Chain

Solve the successive radioactive decay chain $N_1 \xrightarrow{\lambda_1} N_2 \xrightarrow{\lambda_2} N_3$ using Laplace transforms. Initial conditions: $N_1(0) = N_1^0$ and $N_2(0) = 0$.

  1. Transform the coupled differential equations into the algebraic $s$-domain:
$$\mathcal{L}\left\{\frac{dN_1}{dt}\right\} = s \tilde{N}_1(s) - N_1^0 = -\lambda_1 \tilde{N}_1(s)$$
$$\mathcal{L}\left\{\frac{dN_2}{dt}\right\} = s \tilde{N}_2(s) - 0 = \lambda_1 \tilde{N}_1(s) - \lambda_2 \tilde{N}_2(s)$$
  1. Solve algebraically for $\tilde{N}_2(s)$ in the $s$-domain.
  2. Use partial fraction expansion and inverse Laplace transformation $\mathcal{L}^{-1}\{1/(s+a)\} = e^{-at}$ to recover the Bateman daughter equation.

Step 1: Solve for $\tilde{N}_1(s)$ in the $s$-Domain

$$(s + \lambda_1) \tilde{N}_1(s) = N_1^0 \implies \tilde{N}_1(s) = \frac{N_1^0}{s + \lambda_1}$$

Step 2: Solve for $\tilde{N}_2(s)$

$$(s + \lambda_2) \tilde{N}_2(s) = \lambda_1 \tilde{N}_1(s) = \frac{\lambda_1 N_1^0}{s + \lambda_1}$$
$$\tilde{N}_2(s) = \frac{\lambda_1 N_1^0}{(s + \lambda_1)(s + \lambda_2)}$$

Step 3: Partial Fraction Decomposition and Inversion

Expand $\tilde{N}_2(s)$ into partial fractions:

$$\frac{1}{(s + \lambda_1)(s + \lambda_2)} = \frac{A}{s + \lambda_1} + \frac{B}{s + \lambda_2}$$

Multiply by $(s + \lambda_1)(s + \lambda_2)$:

$$1 = A(s + \lambda_2) + B(s + \lambda_1)$$
  • Setting $s = -\lambda_1$: $1 = A(\lambda_2 - \lambda_1) \implies A = \frac{1}{\lambda_2 - \lambda_1}$
  • Setting $s = -\lambda_2$: $1 = B(\lambda_1 - \lambda_2) \implies B = -\frac{1}{\lambda_2 - \lambda_1}$

Substitute coefficients:

$$\tilde{N}_2(s) = \frac{\lambda_1 N_1^0}{\lambda_2 - \lambda_1} \left[ \frac{1}{s + \lambda_1} - \frac{1}{s + \lambda_2} \right]$$

Taking the inverse Laplace transform:

$$\mathcal{L}^{-1}\left\{\frac{1}{s + \lambda}\right\} = e^{-\lambda t}$$
$$N_2(t) = \frac{\lambda_1 N_1^0}{\lambda_2 - \lambda_1} \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right)$$

The Laplace transform method derives the Bateman equation directly without guessing integrating factors!

Solved Honors Problems & Derivations

Step-by-step rigorous solutions with full physical, thermodynamic, and nuclear kinematic validation.