3.8 Decay Kinetics, Half-Life & Medical Radiation Applications

3.8 Decay Kinetics, Half-Life & Medical Radiation Applications

4. Radioactive Decay Kinetics and Half-Life Calculations

Radioactive decay is a strictly stochastic process that obeys first-order rate kinetics. The rate of decay (Activity, $A$) is directly proportional to the number of radioactive nuclei present ($N$):

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

Where $\lambda$ is the first-order decay constant (units: $\text{s}^{-1}$ or $\text{hr}^{-1}$). Integrating yields the exponential decay law:

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

Half-Life ($t_{1/2}$)

The half-life ($t_{1/2}$) is the time required for half of the initial radioactive nuclei to decay:

$$t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda} \iff \lambda = \frac{0.693}{t_{1/2}}$$

Fraction Remaining Shortcuts for Test Day

For an elapsed time $t$ equal to $n$ integer half-lives ($n = t / t_{1/2}$):

$$\text{Fraction Remaining} = \left(\frac{1}{2}\right)^n, \quad \text{Fraction Decayed} = 1 - \left(\frac{1}{2}\right)^n$$

• $n = 1$: $1/2 = 50.0\%$ remaining ($50.0\%$ decayed)
• $n = 2$: $1/4 = 25.0\%$ remaining ($75.0\%$ decayed)
• $n = 3$: $1/8 = 12.5\%$ remaining ($87.5\%$ decayed)
• $n = 4$: $1/16 = 6.25\%$ remaining ($93.75\%$ decayed)
• $n = 5$: $1/32 = 3.125\%$ remaining ($96.875\%$ decayed)

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5. Clinical Applications in Diagnostic Imaging and Radiotherapy

1. Positron Emission Tomography (PET) and Pair Annihilation

PET imaging exploits the biochemistry of malignant cells and the physics of positron-electron annihilation:

1. Biochemical Trapping: The radiotracer 2-deoxy-2-[18F]fluoro-D-glucose ($^{18}\text{F}$-FDG) is injected intravenously. Malignant tumors exhibit aerobic glycolysis with up-regulated GLUT1/GLUT3 transporters and high hexokinase activity (the Warburg effect). Hexokinase phosphorylates FDG to FDG-6-phosphate; lacking a 2'-hydroxyl group, FDG-6-phosphate cannot undergo isomerization by phosphoglucose isomerase and becomes trapped inside tumor cells.
2. Positron Emission: Fluorine-18 decays via positron emission: $$^{18}_9\text{F} \rightarrow ^{18}_8\text{O} + ^0_{+1}\text{e} + \nu_e$$
3. Pair Annihilation: The emitted positron ($e^+$) traverses $1-2\text{ mm}$ through tissue, losing kinetic energy until it collides with an ambient tissue electron ($e^-$). The two particles annihilate each other, converting their entire rest masses into pure electromagnetic energy: $$e^+ + e^- \rightarrow 2\gamma$$ To conserve both energy and linear momentum from near-stationary reactants, annihilation produces two identical $511\text{ keV}$ gamma photons emitted in exactly opposite directions ($180^\circ$ apart).
4. Coincidence Detection: The circular ring of scintillation detectors surrounding the patient registers simultaneous arrival of two $511\text{ keV}$ photons along a "line of response" (coincidence window $\sim 5-10\text{ ns}$), mathematically reconstructing 3D metabolic tumor maps.

2. Technetium-99m ($^{99m}\text{Tc}$) in Diagnostic Scintigraphy

Technetium-99m is the workhorse of hospital nuclear medicine:

• Decays purely via isomeric gamma de-excitation ($^{99m}_{43}\text{Tc} \rightarrow ^{99}_{43}\text{Tc} + \gamma$) with $t_{1/2} = 6.0\text{ hours}$.
• Its $140\text{ keV}$ gamma photon easily exits the human body to strike gamma cameras without depositing excessive ionizing dose in healthy tissues (absence of high-LET alpha or beta particles).

3. Iodine-131 ($^{131}\text{I}$) Targeted Radiotherapy

The thyroid gland concentrates iodide ions via the basolateral sodium-iodide symporter (NIS). Administering oral sodium iodide $^{131}\text{I}$ achieves selective targeted thyroid destruction:

• Emits destructive $\beta^-$ particles that penetrate only $1-2\text{ mm}$ of tissue, selectively destroying hyperplastic follicular cells or thyroid carcinoma metastases while sparing adjacent parathyroid glands and larynx.

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6. Worked Numerical Examples

Example 1: Multistep Nuclear Decay Balancing

Uranium-238 ($^{238}_{92}\text{U}$) decays through a series of radioactive steps before stabilizing as Lead-206 ($^{206}_{82}\text{Pb}$). How many total alpha ($\alpha$) decays and beta-minus ($\beta^-$) decays occur in this decay chain?

Solution:

1. Let $x$ represent the number of alpha decays ($^4_2\alpha$) and $y$ represent the number of beta-minus decays ($^0_{-1}\text{e}$): $$^{238}_{92}\text{U} \rightarrow ^{206}_{82}\text{Pb} + x(^4_2\alpha) + y(^0_{-1}\text{e}) + y(\bar{\nu}_e)$$
2. Balance mass numbers ($A$): $$238 = 206 + 4x + 0y \implies 4x = 238 - 206 = 32 \implies x = 8\text{ alpha decays}$$
3. Balance atomic numbers ($Z$) using $x = 8$: $$92 = 82 + 2x - 1y \implies 92 = 82 + 2(8) - y$$ $$92 = 82 + 16 - y = 98 - y \implies y = 98 - 92 = 6\text{ beta-minus decays}$$ Result: The decay chain comprises 8 alpha decays and 6 beta-minus decays.

Example 2: Binding Energy of Helium-4

The measured atomic mass of a Helium-4 nucleus ($^4_2\text{He}$) is $4.0015\text{ amu}$. Calculate its mass defect ($\Delta m$), total nuclear binding energy ($E_b$), and binding energy per nucleon ($E_b / A$). Use $m_p = 1.00728\text{ amu}$ and $m_n = 1.00866\text{ amu}$.

Solution:

1. Helium-4 contains $Z = 2$ protons and $N = 4 - 2 = 2$ neutrons.
2. Sum the masses of the individual unbound nucleons: $$\text{Mass of constituents} = 2(1.00728\text{ amu}) + 2(1.00866\text{ amu}) = 2.01456 + 2.01732 = 4.03188\text{ amu}$$
3. Compute mass defect ($\Delta m$): $$\Delta m = 4.03188\text{ amu} - 4.00150\text{ amu} = 0.03038\text{ amu}$$
4. Convert mass defect into binding energy using $1\text{ amu} = 931.5\text{ MeV}$: $$E_b = 0.03038\text{ amu} \times 931.5\text{ MeV}/\text{amu} \approx 28.30\text{ MeV}$$
5. Compute binding energy per nucleon ($A = 4$): $$\frac{E_b}{A} = \frac{28.30\text{ MeV}}{4\text{ nucleons}} \approx 7.08\text{ MeV per nucleon}$$