Physics / Nuclear & Particle Nuclear Physics I 100% Free Open Access
Chapter 2 • Theory & Derivations

Radioactivity, Decay Kinetics & Radiometric Dating

Rigorous mathematical formulation of nuclear decay: Segre chart of stability, differential and integral radioactive decay laws, half-life, mean lifetime, activity units (Becquerel, Curie), Bateman equations for successive multi-step chains, secular and transient radioactive equilibria, and radiometric dating (Carbon-14, Potassium-Argon, Uranium-Lead isochrons).

§2.1 Nuclear Stability, the Segre Chart & The Radioactive Decay Law

1. The Segrè Chart of Nuclides ($N$ vs. $Z$)

Nuclear stability is dictated by the delicate quantum balance between attractive short-range nucleon-nucleon strong forces and long-range disruptive Coulomb electrostatic repulsion between protons. Plotting neutron number $N$ against atomic number $Z$ for all known nuclides yields the Segrè Chart:

  • Light Stable Nuclei ($Z \le 20$): Follow the $N = Z$ line of exact symmetry ($N/Z = 1.0$), as in $^4_2\text{He}, {^{12}_6\text{C}}, {^{16}_8\text{O}}, {^{40}_{20}\text{Ca}}$.
  • Heavy Stable Nuclei ($Z > 20$): Curve progressively toward the neutron axis to provide additional strong-force binding without adding disruptive Coulomb charges, reaching $N/Z \approx 1.54$ at $^{208}_{82}\text{Pb}_{126}$.
  • Proton Drip Line & Neutron Drip Line: Boundaries beyond which proton or neutron separation energies vanish ($S_p \le 0$ or $S_n \le 0$); nucleons drip spontaneously from the nucleus on strong-interaction timescales ($\sim 10^{-21}\text{ s}$).
  • Upper Limit of Natural Stability: Terminated at Bismuth-209 ($Z = 83$). All heavier elements ($Z \ge 84$) are naturally radioactive.

2. The Fundamental Radioactive Decay Law

Radioactive decay is an intrinsically stochastic, memoryless quantum process governed by the uncertainty principle. For a population of $N(t)$ identical, unstable radioactive parent nuclei at time $t$, the probability of any given nucleus disintegrating per unit time is a fundamental constant, the decay constant $\lambda$ ($\text{s}^{-1}$):

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

Integrating with the initial condition $N(0) = N_0$ at $t = 0$ yields the exponential Rutherford-Soddy Radioactive Decay Law:

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

3. Half-Life $T_{1/2}$ and Mean Lifetime $\tau$

  • Half-Life ($T_{1/2}$): The elapsed time required for exactly one-half of the initial radioactive nuclei to decay ($N(T_{1/2}) = N_0 / 2$):
    $$\frac{N_0}{2} = N_0 e^{-\lambda T_{1/2}} \implies T_{1/2} = \frac{\ln(2)}{\lambda} = \frac{0.693147}{\lambda}$$
  • Mean Lifetime ($\tau$): The statistical average lifetime of a radioactive nucleus:
    $$\tau = \langle t \rangle = \frac{\int_0^\infty t \left( \lambda N_0 e^{-\lambda t} \right) dt}{N_0} = \frac{1}{\lambda} = \frac{T_{1/2}}{\ln(2)} \approx 1.4427 T_{1/2}$$

§2.2 Activity, Radiation Units & Counting Statistics

1. Radioactivity and Decay Rate

The activity $A(t)$ of a radioactive sample is defined as the number of disintegrations occurring per unit time:

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

2. Units of Radioactivity

  • Becquerel ($\text{Bq}$): The SI unit of activity, defined as precisely one nuclear disintegration per second:
    $$1\text{ Bq} \equiv 1\text{ disintegration/s}$$
  • Curie ($\text{Ci}$): The historical unit, historically based on the activity of 1 gram of pure Radium-226:
    $$1\text{ Ci} \equiv 3.700 \times 10^{10}\text{ Bq} = 37\text{ GBq}$$
  • Specific Activity ($a$): Activity per unit mass of the pure radioisotope:
    $$a_{\text{spec}} = \frac{A}{m} = \frac{\lambda N_A}{M} = \frac{\ln(2) N_A}{T_{1/2} M}$$

3. Counting Statistics: Poisson and Gaussian Distributions

Because nuclear disintegrations are independent random events occurring with small probability $p \ll 1$ in a large population $N \gg 1$, the probability of recording exactly $n$ counts in a time interval $\Delta t$ follows the Poisson distribution:

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

where $\mu = \langle n \rangle$ is the mean count. The variance equals the mean: $\sigma^2 = \mu$. The standard deviation in any single nuclear counting measurement of $N_{\text{counts}}$ is:

$$\sigma = \sqrt{N_{\text{counts}}}$$

The fractional statistical counting uncertainty is $\frac{\sigma}{N} = \frac{1}{\sqrt{N_{\text{counts}}}}$. To achieve a precision of $1\%$, one must accumulate at least $N_{\text{counts}} = (1/0.01)^2 = 10,000$ counts.

§2.3 Successive Radioactive Transformations & Bateman Equations

1. Multi-Step Radioactive Decay Series

In nature, heavy radioactive isotopes decay through successive chains (e.g., the Uranium-238, Thorium-232, and Actinium-235 series). Consider a general radioactive cascade:

$$N_1 \xrightarrow{\lambda_1} N_2 \xrightarrow{\lambda_2} N_3 \xrightarrow{\lambda_3} \dots \xrightarrow{\lambda_m} N_m \text{ (stable)}$$

The system of coupled linear differential equations governing the populations is:

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

2. Derivation of the Daughter Population $N_2(t)$

Assuming initially pure parent at $t = 0$ ($N_1(0) = N_0$ and $N_2(0) = 0$):

$$\frac{dN_2}{dt} + \lambda_2 N_2 = \lambda_1 N_0 e^{-\lambda_1 t}$$

Multiplying by the integrating factor $e^{\lambda_2 t}$:

$$\frac{d}{dt} \left( N_2 e^{\lambda_2 t} \right) = \lambda_1 N_0 e^{(\lambda_2 - \lambda_1)t} \implies N_2 e^{\lambda_2 t} = \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1} e^{(\lambda_2 - \lambda_1)t} + C$$

Applying initial condition $N_2(0) = 0 \implies C = - \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1}$ yields:

$$N_2(t) = \frac{\lambda_1 N_0}{\lambda_2 - \lambda_1} \left( e^{-\lambda_1 t} - e^{-\lambda_2 t} \right)$$

The daughter activity is $A_2(t) = \lambda_2 N_2(t) = \frac{\lambda_1 \lambda_2 N_0}{\lambda_2 - \lambda_1} (e^{-\lambda_1 t} - e^{-\lambda_2 t})$.

3. General Bateman Solution for the $n$-th Isotope

Harry Bateman (1910) solved the general $n$-step system in closed analytical form:

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

§2.4 Radioactive Equilibria: Secular, Transient & Non-Equilibrium

1. Classification of Radioactive Equilibria

The physical behavior of a parent-daughter system depends on the ratio of parent half-life $T_{1/2, 1}$ to daughter half-life $T_{1/2, 2}$ (or decay constants $\lambda_1$ vs. $\lambda_2$):

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

When the parent is extremely long-lived compared to the daughter (e.g., $^{226}\text{Ra}$ with $T_{1/2} = 1600\text{ y}$ decaying to $^{222}\text{Rn}$ with $T_{1/2} = 3.82\text{ d}$, or $^{238}\text{U}$ with $4.47 \times 10^9\text{ y}$):

  • $\lambda_2 - \lambda_1 \approx \lambda_2$ and $e^{-\lambda_1 t} \approx 1$.
  • For times $t \gg \tau_2 = 1/\lambda_2$, the transient exponential vanishes ($e^{-\lambda_2 t} \to 0$):
$$N_2(t) \to \frac{\lambda_1}{\lambda_2} N_1 \implies \lambda_1 N_1 = \lambda_2 N_2 \implies A_1 = A_2$$

In secular equilibrium, the activity of the daughter equals the activity of the parent. Across an entire undisturbed natural decay series:

$$A_1 = A_2 = A_3 = \dots = A_n \implies \lambda_1 N_1 = \lambda_2 N_2 = \dots = \lambda_n N_n$$

3. Transient Equilibrium ($\lambda_1 < \lambda_2$, comparable orders of magnitude)

When the parent is longer-lived than the daughter, but parent decay is non-negligible (e.g., $^{99}\text{Mo}$ with $T_{1/2} = 66\text{ h}$ decaying to $^{99m}\text{Tc}$ with $T_{1/2} = 6.0\text{ h}$, widely used in nuclear medicine generators):

For $t \gg 1/\lambda_2$, $e^{-\lambda_2 t} \ll e^{-\lambda_1 t}$, yielding:

$$N_2(t) \approx \frac{\lambda_1}{\lambda_2 - \lambda_1} N_1(t) \implies \frac{A_2(t)}{A_1(t)} = \frac{\lambda_2}{\lambda_2 - \lambda_1} > 1$$

In transient equilibrium, the daughter activity decays with the half-life of the parent, but exceeds the parent activity by the constant factor $\frac{\lambda_2}{\lambda_2 - \lambda_1}$.

4. Non-Equilibrium ($\lambda_1 > \lambda_2$)

If the parent is shorter-lived than the daughter, no equilibrium can ever be established. The parent decays rapidly, leaving the daughter to decay independently with its own characteristic decay constant $\lambda_2$.

§2.5 Radiometric Dating Principles: Carbon-14, K-Ar & U-Pb Isochrons

1. Radiocarbon ($^{14}\text{C}$) Dating

Willard Libby (1949, Nobel Prize 1960) developed radiocarbon dating based on cosmic-ray production of $^{14}\text{C}$ in the upper atmosphere via neutron capture on nitrogen:

$$^1_0 n + {^{14}_7\text{N}} \to {^{14}_6\text{C}} + {^1_1 p}$$

The radioactive $^{14}\text{C}$ ($T_{1/2} = 5730\pm40\text{ y}$, $\beta^-$ emitter) oxidizes to $^{14}\text{CO}_2$ and mixes into the biosphere via photosynthesis and the food chain, establishing an equilibrium specific activity in living organic tissue:

$$A_0 \approx 15.3\text{ disintegrations per minute per gram of Carbon} \approx 0.255\text{ Bq/g}$$

Upon death, biological carbon exchange terminates, and $^{14}\text{C}$ decays exponentially without replenishment: $A(t) = A_0 e^{-\lambda t}$. The age of the archaeological specimen is calculated as:

$$t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A(t)}\right) = \frac{T_{1/2}}{\ln(2)} \ln\left(\frac{A_0}{A(t)}\right) = 8267 \ln\left(\frac{A_0}{A(t)}\right) \text{ years}$$

2. Potassium-Argon ($^{40}\text{K}$-$^{40}\text{Ar}$) Dating

Used for geological samples ($10^5$ to $4.5 \times 10^9$ years). Natural Potassium contains $0.0117\%$ $^{40}\text{K}$ ($T_{1/2} = 1.248 \times 10^9\text{ y}$), which exhibits branched decay:

  • $89.3\%$ decays via $\beta^-$ to $^{40}\text{Ca}$.
  • $10.7\%$ decays via electron capture to noble gas $^{40}\text{Ar}$.

When volcanic rock crystallizes from magma, molten lava outgasses all volatile Argon ($^{40}\text{Ar}_{\text{initial}} = 0$). Trapped radiogenic $^{40}\text{Ar}$ subsequently accumulates in the solid mineral lattice:

$$N(^{40}\text{Ar}) = \frac{\lambda_{EC}}{\lambda_{\text{tot}}} N(^{40}\text{K}) \left( e^{\lambda_{\text{tot}} t} - 1 \right) \implies t = \frac{1}{\lambda_{\text{tot}}} \ln\left( 1 + \frac{\lambda_{\text{tot}}}{\lambda_{EC}} \frac{N(^{40}\text{Ar})}{N(^{40}\text{K})} \right)$$

3. Uranium-Lead ($^{238}\text{U}$-$^{206}\text{Pb}$) Isochron Dating

In zircon crystals ($\text{ZrSiO}_4$), initial lead contamination is eliminated during crystallization. For mineral samples containing non-radiogenic common lead ($^{204}\text{Pb}$), measuring the isotopic ratios yields the isochron equation:

$$\left( \frac{^{206}\text{Pb}}{^{204}\text{Pb}} \right)_{\text{today}} = \left( \frac{^{206}\text{Pb}}{^{204}\text{Pb}} \right)_0 + \left( \frac{^{238}\text{U}}{^{204}\text{Pb}} \right)_{\text{today}} \left( e^{\lambda_{238} t} - 1 \right)$$

Plotting $^{206}\text{Pb}/^{204}\text{Pb}$ vs. $^{238}\text{U}/^{204}\text{Pb}$ for different co-genetic mineral grains yields a straight line whose slope $m = e^{\lambda_{238} t} - 1$ determines the geological age $t$ independently of the initial lead content.

Solved Problem Example 2.1: Activity and Specific Activity of Cobalt-60 Radiation Source

Cobalt-60 ($^{60}_{27}\text{Co}$) is a widely used industrial and medical gamma source with a half-life of $T_{1/2} = 5.271 \text{ years}$ ($1.663 \times 10^8 \text{ s}$) and atomic mass $M = 59.9338 \text{ g/mol}$. (a) Calculate the decay constant $\lambda$ in $\text{s}^{-1}$. (b) Determine the specific activity of pure $^{60}\text{Co}$ in $\text{Bq/g}$ and in $\text{Ci/g}$. (c) What mass of $^{60}\text{Co}$ is required to fabricate a $5000 \text{ Ci}$ radiotherapy source?

Step 1: Calculate the Decay Constant lambda
$$\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693147}{1.6635 \times 10^8\text{ s}} \approx 4.1668 \times 10^{-9}\text{ s}^{-1}$$

Evaluate lambda = ln(2) / T_1/2 in SI units.

Step 2: Calculate the Specific Activity per Gram
$$a_{\text{spec}} = \frac{\lambda N_A}{M} = \frac{(4.1668 \times 10^{-9}\text{ s}^{-1})(6.0221 \times 10^{23}\text{ mol}^{-1})}{59.9338\text{ g/mol}} = \frac{2.5093 \times 10^{15}}{59.9338}\text{ Bq/g} \approx 4.1868 \times 10^{13}\text{ Bq/g} = 41.87\text{ TBq/g}$$

Compute activity per gram of pure isotope.

Step 3: Convert Specific Activity to Curies per Gram
$$a_{\text{spec}} = \frac{4.1868 \times 10^{13}\text{ Bq/g}}{3.700 \times 10^{10}\text{ Bq/Ci}} \approx 1131.6\text{ Ci/g}$$

Divide by 3.7 x 10^10 Bq per Curie.

Step 4: Calculate the Mass Required for a 5000 Ci Source
$$m = \frac{A_{\text{target}}}{a_{\text{spec}}} = \frac{5000\text{ Ci}}{1131.6\text{ Ci/g}} \approx 4.419\text{ g}$$

Divide target activity by specific activity.

Final Answer & Physical Insight

\lambda = 4.17 \times 10^{-9} \text{ s}^{-1}, \quad a_{\text{spec}} = 4.19 \times 10^{13} \text{ Bq/g} \quad (1132 \text{ Ci/g}), \quad m = 4.42 \text{ g}

Solved Problem Example 2.2: Transient Equilibrium and Maximum Activity in Technetium-99m Generator

In a $^{99}\text{Mo}$-$^{99m}\text{Tc}$ medical isotope generator, parent $^{99}\text{Mo}$ ($T_{1/2, 1} = 66.0 \text{ h}$, $\lambda_1 = 0.01050 \text{ h}^{-1}$) decays to daughter $^{99m}\text{Tc}$ ($T_{1/2, 2} = 6.01 \text{ h}$, $\lambda_2 = 0.11533 \text{ h}^{-1}$) with a branching ratio of $87.5\%$. Starting with fresh pure $^{99}\text{Mo}$ ($N_2(0) = 0$): (a) Calculate the time $t_{\text{max}}$ at which the daughter $^{99m}\text{Tc}$ activity reaches its maximum value. (b) Calculate the ratio of daughter activity to parent activity at transient equilibrium.

Step 1: Formulate the Condition for Maximum Daughter Activity
$$\frac{dN_2}{dt} = 0 \implies \lambda_1 N_1(t_{\text{max}}) = \lambda_2 N_2(t_{\text{max}}) \implies t_{\text{max}} = \frac{\ln(\lambda_2 / \lambda_1)}{\lambda_2 - \lambda_1}$$

Setting the derivative of N_2(t) in the Bateman equation to zero yields the maximum population time.

Step 2: Calculate the Numerical Value of t_max
$$t_{\text{max}} = \frac{\ln(0.11533 / 0.01050)}{0.11533 - 0.01050} = \frac{\ln(10.984)}{0.10483}\text{ h} = \frac{2.3964}{0.10483}\text{ h} \approx 22.86\text{ hours}$$

Evaluate t_max in hours. The generator reaches peak activity at approx 22.9 hours after elution.

Step 3: Calculate the Equilibrium Activity Ratio
$$\frac{A_2}{A_1} = \text{Branching Ratio} \times \frac{\lambda_2}{\lambda_2 - \lambda_1} = 0.875 \times \frac{0.11533}{0.11533 - 0.01050} = 0.875 \times \frac{0.11533}{0.10483} \approx 0.875 \times 1.1001 \approx 0.9626$$

Multiply the transient equilibrium activity factor by the 87.5% branching fraction.

Final Answer & Physical Insight

t_{\text{max}} = 22.86 \text{ hours}, \quad \frac{A_2}{A_1} = 0.963 \quad (\text{Transient Equilibrium})

Solved Problem Example 2.3: Archaeological Radiocarbon Dating of an Ancient Wooden Artifact

An ancient charcoal sample excavated from an archaeological site has a measured $^{14}\text{C}$ activity of $3.82 \text{ disintegrations per minute per gram of Carbon}$ ($\text{dpm/g}$). A living modern reference standard exhibits an activity of $15.30 \text{ dpm/g}$. The half-life of $^{14}\text{C}$ is $5730 \text{ years}$. (a) Calculate the decay constant $\lambda$ of $^{14}\text{C}$ in $\text{year}^{-1}$. (b) Determine the calendar age of the archaeological sample.

Step 1: Calculate the Decay Constant lambda
$$\lambda = \frac{\ln(2)}{T_{1/2}} = \frac{0.693147}{5730\text{ y}} \approx 1.2097 \times 10^{-4}\text{ y}^{-1}$$

Compute the decay constant per year.

Step 2: Apply the Radiocarbon Age Equation
$$t = \frac{1}{\lambda} \ln\left(\frac{A_0}{A(t)}\right) = \frac{1}{1.2097 \times 10^{-4}\text{ y}^{-1}} \ln\left(\frac{15.30}{3.82}\right) = (8266.5\text{ y}) \ln(4.0052)$$

Substitute modern initial activity A_0 and measured activity A(t).

Step 3: Evaluate Natural Logarithm and Final Age
$$t = (8266.5\text{ y}) \times (1.3876) \approx 11470\text{ years}$$

Multiply to find the age in years.

Final Answer & Physical Insight

\lambda = 1.21 \times 10^{-4} \text{ y}^{-1}, \quad t = 11470 \text{ years} \quad (\sim 9450 \text{ BCE})

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