Radioactivity, Dosimetry — I-131 Thyroid Treatment: 17-Phys-B1 Radiation Physics, December 2013, Q6

Question 6: Radioactivity, Dosimetry — I-131 Thyroid Treatment (9 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Given data
Quantity	Symbol	Value
Administered activity	$A_0$	100 MBq
Physical (radiological) half-life	$T_{\text{phys}}$	8 days
Biological half-life	$T_{\text{bio}}$	2 days
Thyroid uptake fraction (immediate)	$f$	60%
Mean beta / gamma energy per decay	$\bar{E}_\beta,\bar{E}_\gamma$	192 keV / 370 keV
Thyroid mass	$m$	20 g = 0.020 kg

Find. (b) time for the body activity to fall to one-quarter of its initial value; (c) the cumulative (time-integrated) activity absorbed in the body.

Approach. Combine physical decay and biological elimination into a single effective half-life, since both processes remove ${}^{131}$I from the body simultaneously; use that effective decay constant for (b), and integrate the resulting exponential activity-vs-time curve over all time to get the cumulated activity for (c).

Part (a) — why iodine treats thyroid disease. The thyroid gland naturally and selectively absorbs iodine from the bloodstream to synthesize the thyroid hormones T3 and T4; because ${}^{131}$I is chemically indistinguishable from stable iodine, the thyroid concentrates it just as avidly, delivering a locally targeted internal radiation dose that irradiates (and shrinks/ablates) overactive thyroid tissue while largely sparing other organs that do not take up iodine.
Part (b) — time to one-quarter activity. Physical decay and biological clearance act simultaneously, so their rate constants add, giving an effective half-life: $$\frac{1}{T_{\text{eff}}} = \frac{1}{T_{\text{phys}}} + \frac{1}{T_{\text{bio}}} = \frac{1}{8} + \frac{1}{2} = 0.625\text{ day}^{-1} \;\Longrightarrow\; T_{\text{eff}} = 1.6\text{ days}$$ One-quarter of the initial value is exactly two effective half-lives ($\tfrac14 = (\tfrac12)^2$): $$\boxed{t = 2\,T_{\text{eff}} = 2(1.6) = 3.2\text{ days}}$$
Part (c) — cumulative activity. The activity actually incorporated in the body is the thyroid's 60% uptake share of the administered dose, $A_{0,\text{thyroid}} = 0.60 \times 100\text{ MBq} = 60\text{ MBq}$, decaying with the effective decay constant $\lambda_{\text{eff}}=\ln2/T_{\text{eff}}$. The time-integrated (cumulated) activity is the area under the exponential decay curve from uptake to infinity: $$\tilde{A} = \int_0^\infty A_{0,\text{thyroid}}\,e^{-\lambda_{\text{eff}}t}\,dt = \frac{A_{0,\text{thyroid}}}{\lambda_{\text{eff}}} = A_{0,\text{thyroid}}\times\frac{T_{\text{eff}}}{\ln 2}$$ With $T_{\text{eff}}=1.6\text{ d} = 138{,}240\text{ s}$: $$\boxed{\tilde{A} = (60\times10^6\text{ Bq})\times\frac{138{,}240\text{ s}}{0.6931} \approx 1.20\times10^{13}\text{ Bq}\cdot\text{s}}$$
Part (d) — beta vs. gamma for treatment. Beta radiation is the better therapeutic choice. Beta particles at these energies have a range of only a few millimetres in tissue, so essentially all of their energy is deposited locally within the thyroid itself — concentrating the dose exactly where ablation is wanted. Gamma rays (370 keV mean here) travel centimetres to metres through tissue, depositing most of their energy well outside the gland (and outside the patient), which delivers unwanted whole-body dose without contributing efficiently to the therapeutic effect; gamma emission is instead what makes ${}^{131}$I (and lower-dose ${}^{123}$I) useful for diagnostic imaging rather than treatment.

Question 6 — results
Quantity	Value
Effective half-life	$T_{\text{eff}} = 1.6$ days
(b) Time to one-quarter activity	3.2 days
(c) Cumulative activity absorbed in the body	≈ 1.20×10¹³ Bq·s
(d) Preferred radiation type for treatment	Beta (short range, local dose)

Topic. Effective half-life for a radiopharmaceutical undergoing both physical decay and biological clearance, and cumulated (time-integrated) activity as the basis of internal dosimetry.

Key relations. $1/T_{\text{eff}} = 1/T_{\text{phys}} + 1/T_{\text{bio}}$; $\tilde{A}=\int_0^\infty A(t)\,dt = A_0\,T_{\text{eff}}/\ln2$; MIRD-style dose $=\tilde{A}\times S$-factor.

Why this works. Physical decay removes radioactive atoms regardless of where they sit in the body, while biological clearance (excretion/turnover) removes all iodine atoms, radioactive or not, at its own independent rate; because both are first-order processes acting on the same population simultaneously, their rate constants (not their half-lives) add directly, which is why the reciprocal-half-life formula is used rather than averaging the two half-lives. The cumulated activity $\tilde{A}$ is exactly the quantity MIRD-style internal dosimetry multiplies by a tissue-specific dose-per-decay "S-factor" (the role played by the two dose-rate coefficients given in this problem) to obtain absorbed dose.

Common pitfall. Using the physical half-life alone (ignoring biological clearance) when a patient's body actively excretes the isotope — this overestimates $T_{\text{eff}}$ and hence the cumulated activity and dose; also forgetting to apply the 60% uptake fraction before integrating, which would overstate the cumulated activity in the thyroid compartment specifically by a factor of $1/0.6$.

Source. J. R. Cember and T. E. Johnson, Introduction to Health Physics, Ch. 8 (internal dosimetry, effective half-life, cumulated activity).

Question 6 of 7: Radioactivity, Dosimetry — I-131 Thyroid Treatment

Question 6: Radioactivity, Dosimetry — I-131 Thyroid Treatment (9 marks)