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3.8 Decay Kinetics, Half-Life & Medical Radiation Applications
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3.8 Decay Kinetics, Half-Life & Medical Radiation Applications
Key Takeaways Explain This Topic AI
Radioactive decay follows first-order exponential kinetics N(t) = N0 (1/2)^(t / t1/2) with decay constant lambda = ln(2) / t1/2 approx 0.693 / t1/2.
Positron emission tomography (PET) utilizes fluorine-18 (FDG) positron decay and subsequent electron pair annihilation to generate coincident 511 keV gamma rays at 180 degrees for high-resolution metabolic imaging.
The three classical radiations differ sharply in penetrating power and ionising density: alpha particles are stopped by paper or the stratum corneum yet are devastating if internalised, beta particles penetrate millimetres of tissue and are stopped by thin aluminium, and gamma photons require dense lead or concrete shielding.
Last updated: September 2026
- 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 A A ) is directly proportional to the number of radioactive nuclei present ( N N N ): Activity A ( t ) = − d N d t = λ N ( t ) \text{Activity } A(t) = -\frac{dN}{dt} = \lambda N(t) Activity A ( t ) = − d t d N = λ N ( t )
Where λ \lambda λ is the first-order decay constant (units: s − 1 \text{s}^{-1} s − 1 or hr − 1 \text{hr}^{-1} hr − 1 ). Integrating yields the exponential decay law: N ( t ) = N 0 e − λ t = N 0 ( 1 2 ) t / t 1 / 2 N(t) = N_0 e^{-\lambda t} = N_0 \left(\frac{1}{2}\right)^{t / t_{1/2}} N ( t ) = N 0 e − λ t = N 0 ( 2 1 ) t / t 1/2
Half-Life ( t 1 / 2 t_{1/2} t 1/2 )
The half-life ( t 1 / 2 t_{1/2} t 1/2 ) is the time required for half of the initial radioactive nuclei to decay: t 1 / 2 = ln 2 λ ≈ 0.693 λ ⟺ λ = 0.693 t 1 / 2 t_{1/2} = \frac{\ln 2}{\lambda} \approx \frac{0.693}{\lambda} \iff \lambda = \frac{0.693}{t_{1/2}} t 1/2 = λ ln 2 ≈ λ 0.693 ⟺ λ = t 1/2 0.693
Fraction Remaining Shortcuts for Test Day
For an elapsed time t t t equal to n n n integer half-lives ( n = t / t 1 / 2 n = t / t_{1/2} n = t / t 1/2 ): Fraction Remaining = ( 1 2 ) n , Fraction Decayed = 1 − ( 1 2 ) n \text{Fraction Remaining} = \left(\frac{1}{2}\right)^n, \quad \text{Fraction Decayed} = 1 - \left(\frac{1}{2}\right)^n Fraction Remaining = ( 2 1 ) n , Fraction Decayed = 1 − ( 2 1 ) n
- n = 1 n = 1 n = 1 : 1 / 2 = 50.0 % 1/2 = 50.0\% 1/2 = 50.0% remaining ( 50.0 % 50.0\% 50.0% decayed)
- n = 2 n = 2 n = 2 : 1 / 4 = 25.0 % 1/4 = 25.0\% 1/4 = 25.0% remaining ( 75.0 % 75.0\% 75.0% decayed)
- n = 3 n = 3 n = 3 : 1 / 8 = 12.5 % 1/8 = 12.5\% 1/8 = 12.5% remaining ( 87.5 % 87.5\% 87.5% decayed)
- n = 4 n = 4 n = 4 : 1 / 16 = 6.25 % 1/16 = 6.25\% 1/16 = 6.25% remaining ( 93.75 % 93.75\% 93.75% decayed)
n = 5 n = 5 n = 5 : 1 / 32 = 3.125 % 1/32 = 3.125\% 1/32 = 3.125% remaining ( 96.875 % 96.875\% 96.875% decayed)
5. Clinical Applications in Diagnostic Imaging and Radiotherapy
Radioisotope Primary Decay Mode Half-Life ( t 1 / 2 t_{1/2} t 1/2 ) Emitted Radiation Primary Clinical Application
Fluorine-18 ( 18 F ^{18}\text{F} 18 F ) β + \beta^+ β + (Positron) 110 minutes 110\text{ minutes} 110 minutes e + e^+ e + followed by 2 × 511 keV γ 2 \times 511\text{ keV } \gamma 2 × 511 keV γ Oncologic PET imaging ( 18 F ^{18}\text{F} 18 F -FDG)
Technetium-99m ( 99 m Tc ^{99m}\text{Tc} 99 m Tc ) Isomeric γ \gamma γ decay 6.0 hours 6.0\text{ hours} 6.0 hours 140 keV γ 140\text{ keV } \gamma 140 keV γ photon SPECT cardiac perfusion, bone scintigraphy
Iodine-131 ( 131 I ^{131}\text{I} 131 I ) β − \beta^- β − ( 90 % 90\% 90% ) and γ \gamma γ ( 10 % 10\% 10% ) 8.02 days 8.02\text{ days} 8.02 days β − \beta^- β − electron + γ \gamma γ photon Targeted thyroid ablation (Graves' disease, thyroid cancer)
Iodine-123 ( 123 I ^{123}\text{I} 123 I ) Electron Capture (EC) 13.2 hours 13.2\text{ hours} 13.2 hours 159 keV γ 159\text{ keV } \gamma 159 keV γ photon Diagnostic thyroid uptake imaging (non-ablative)
1. Positron Emission Tomography (PET) and Pair Annihilation
PET imaging exploits the biochemistry of malignant cells and the physics of positron-electron annihilation:
Biochemical Trapping : The radiotracer 2-deoxy-2-[18F]fluoro-D-glucose ( 18 F ^{18}\text{F} 18 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.
Positron Emission : Fluorine-18 decays via positron emission: 9 18 F → 8 18 O + + 1 0 e + ν e ^{18}_9\text{F} \rightarrow ^{18}_8\text{O} + ^0_{+1}\text{e} + \nu_e 9 18 F → 8 18 O + + 1 0 e + ν e
Pair Annihilation : The emitted positron ( e + e^+ e + ) traverses 1 − 2 mm 1-2\text{ mm} 1 − 2 mm through tissue, losing kinetic energy until it collides with an ambient tissue electron ( e − e^- e − ). The two particles annihilate each other, converting their entire rest masses into pure electromagnetic energy: e + + e − → 2 γ e^+ + e^- \rightarrow 2\gamma e + + e − → 2 γ To conserve both energy and linear momentum from near-stationary reactants, annihilation produces two identical 511 keV 511\text{ keV} 511 keV gamma photons emitted in exactly opposite directions ( 180 ∘ 180^\circ 18 0 ∘ apart) .
Coincidence Detection : The circular ring of scintillation detectors surrounding the patient registers simultaneous arrival of two 511 keV 511\text{ keV} 511 keV photons along a "line of response" (coincidence window ∼ 5 − 10 ns \sim 5-10\text{ ns} ∼ 5 − 10 ns ), mathematically reconstructing 3D metabolic tumor maps.
2. Technetium-99m ( 99 m Tc ^{99m}\text{Tc} 99 m Tc ) in Diagnostic Scintigraphy
Technetium-99m is the workhorse of hospital nuclear medicine:
Decays purely via isomeric gamma de-excitation ( 43 99 m Tc → 43 99 Tc + γ ^{99m}_{43}\text{Tc} \rightarrow ^{99}_{43}\text{Tc} + \gamma 43 99 m Tc → 43 99 Tc + γ ) with t 1 / 2 = 6.0 hours t_{1/2} = 6.0\text{ hours} t 1/2 = 6.0 hours .
Its 140 keV 140\text{ keV} 140 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 I ^{131}\text{I} 131 I ) Targeted Radiotherapy
The thyroid gland concentrates iodide ions via the basolateral sodium-iodide symporter (NIS). Administering oral sodium iodide 131 I ^{131}\text{I} 131 I achieves selective targeted thyroid destruction:
Emits destructive β − \beta^- β − particles that penetrate only 1 − 2 mm 1-2\text{ mm} 1 − 2 mm of tissue, selectively destroying hyperplastic follicular cells or thyroid carcinoma metastases while sparing adjacent parathyroid glands and larynx.
- 6. Worked Numerical Examples
- Example 1: Multistep Nuclear Decay Balancing
Uranium-238 ( 92 238 U ^{238}_{92}\text{U} 92 238 U ) decays through a series of radioactive steps before stabilizing as Lead-206 ( 82 206 Pb ^{206}_{82}\text{Pb} 82 206 Pb ). How many total alpha ( α \alpha α ) decays and beta-minus ( β − \beta^- β − ) decays occur in this decay chain?
Solution :
Let x x x represent the number of alpha decays ( 2 4 α ^4_2\alpha 2 4 α ) and y y y represent the number of beta-minus decays ( − 1 0 e ^0_{-1}\text{e} − 1 0 e ): 92 238 U → 82 206 Pb + x ( 2 4 α ) + y ( − 1 0 e ) + y ( ν ˉ e ) ^{238}_{92}\text{U} \rightarrow ^{206}_{82}\text{Pb} + x(^4_2\alpha) + y(^0_{-1}\text{e}) + y(\bar{\nu}_e) 92 238 U → 82 206 Pb + x ( 2 4 α ) + y ( − 1 0 e ) + y ( ν ˉ e )
Balance mass numbers ( A A A ): 238 = 206 + 4 x + 0 y ⟹ 4 x = 238 − 206 = 32 ⟹ x = 8 alpha decays 238 = 206 + 4x + 0y \implies 4x = 238 - 206 = 32 \implies x = 8\text{ alpha decays} 238 = 206 + 4 x + 0 y ⟹ 4 x = 238 − 206 = 32 ⟹ x = 8 alpha decays
Balance atomic numbers ( Z Z Z ) using x = 8 x = 8 x = 8 : 92 = 82 + 2 x − 1 y ⟹ 92 = 82 + 2 ( 8 ) − y 92 = 82 + 2x - 1y \implies 92 = 82 + 2(8) - y 92 = 82 + 2 x − 1 y ⟹ 92 = 82 + 2 ( 8 ) − y 92 = 82 + 16 − y = 98 − y ⟹ y = 98 − 92 = 6 beta-minus decays 92 = 82 + 16 - y = 98 - y \implies y = 98 - 92 = 6\text{ beta-minus decays} 92 = 82 + 16 − y = 98 − y ⟹ y = 98 − 92 = 6 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 ( 2 4 He ^4_2\text{He} 2 4 He ) is 4.0015 amu 4.0015\text{ amu} 4.0015 amu . Calculate its mass defect ( Δ m \Delta m Δ m ), total nuclear binding energy ( E b E_b E b ), and binding energy per nucleon ( E b / A E_b / A E b / A ). Use m p = 1.00728 amu m_p = 1.00728\text{ amu} m p = 1.00728 amu and m n = 1.00866 amu m_n = 1.00866\text{ amu} m n = 1.00866 amu .
Solution :
Helium-4 contains Z = 2 Z = 2 Z = 2 protons and N = 4 − 2 = 2 N = 4 - 2 = 2 N = 4 − 2 = 2 neutrons.
Sum the masses of the individual unbound nucleons: Mass of constituents = 2 ( 1.00728 amu ) + 2 ( 1.00866 amu ) = 2.01456 + 2.01732 = 4.03188 amu \text{Mass of constituents} = 2(1.00728\text{ amu}) + 2(1.00866\text{ amu}) = 2.01456 + 2.01732 = 4.03188\text{ amu} Mass of constituents = 2 ( 1.00728 amu ) + 2 ( 1.00866 amu ) = 2.01456 + 2.01732 = 4.03188 amu
Compute mass defect ( Δ m \Delta m Δ m ): Δ m = 4.03188 amu − 4.00150 amu = 0.03038 amu \Delta m = 4.03188\text{ amu} - 4.00150\text{ amu} = 0.03038\text{ amu} Δ m = 4.03188 amu − 4.00150 amu = 0.03038 amu
Convert mass defect into binding energy using 1 amu = 931.5 MeV 1\text{ amu} = 931.5\text{ MeV} 1 amu = 931.5 MeV : E b = 0.03038 amu × 931.5 MeV / amu ≈ 28.30 MeV E_b = 0.03038\text{ amu} \times 931.5\text{ MeV}/\text{amu} \approx 28.30\text{ MeV} E b = 0.03038 amu × 931.5 MeV / amu ≈ 28.30 MeV
Compute binding energy per nucleon ( A = 4 A = 4 A = 4 ): E b A = 28.30 MeV 4 nucleons ≈ 7.08 MeV per nucleon \frac{E_b}{A} = \frac{28.30\text{ MeV}}{4\text{ nucleons}} \approx 7.08\text{ MeV per nucleon} A E b = 4 nucleons 28.30 MeV ≈ 7.08 MeV per nucleon