Question 5 of 7: Internal Dosimetry of Iodine-131 in the Thyroid
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 98-Phys-B1 Radiation Physics, National Examination
May 2015 — a three-hour open-book examination in which any
non-communicating calculator is permitted. The cover page states the exam has 7
questions worth a total of 100 points, of which only 80 points' worth need be answered
for full marks; every question and sub-part is nonetheless answered in full below so the paper
remains a complete study resource. The cover page also invites the candidate to submit a
written statement of any assumptions made where a question is open to interpretation —
this licence is used in Question 1(a) (proton-mass vs. hydrogen-atom-mass convention), Question
3(d) (single effective attenuation coefficient for the polychromatic X-ray beam) and Question
5(c) (reading "the body" as the thyroid uptake compartment, consistent with the biological
half-life given). The exam's own page-6 marking-scheme summary is internally inconsistent for
two questions (it prints "15 points" for both Q2 and Q3, but their own per-sub-part marks,
given beside each sub-part on the question pages, sum to 18 and 17 respectively); the
sub-part marks are used below since only that reading makes the paper's own stated 100-point
total add up exactly ($12+18+17+18+18+10+7=100$).
Reference texts. K. S. Krane, Introductory Nuclear Physics
(nuclear masses and binding energy, radioactive decay); F. H. Attix, Introduction to
Radiological Physics and Radiation Dosimetry (X-ray production, photon attenuation);
J. R. Cember and T. E. Johnson, Introduction to Health Physics, 5th ed. (internal
dosimetry, effective half-life, shielding, radiation survey practice); J. E. Turner,
Atoms, Radiation, and Radiation Protection, 3rd ed. (radiation interactions,
detectors, health-physics standards).
Question 5: Internal Dosimetry of Iodine-131 in the Thyroid (18 marks)
Given. Administered activity $A_0=100$ MBq; physical half-life
$T_{\text{phys}}=8$ d; biological half-life $T_{\text{bio}}=2$ d; thyroid uptake $f=60\%$,
immediate; mean beta energy 192 keV, mean gamma energy 370 keV; infinite-medium dose-rate
constants $S_\beta=27.72\times10^{-15}$ and $S_\gamma=48.55\times10^{-15}$
kg·Sv/(Bq·s); thyroid mass $m=16$ g; per-gram absorbed-fraction table above.
Find. (a) the rationale for iodine as the therapeutic radionuclide;
(b) time to reach $1/8$ activity; (c) cumulated activity in the thyroid; (d) the total
absorbed fraction of 370 keV gammas in the 16 g thyroid; (e)–(f) cumulative gamma and
beta absorbed dose to the thyroid; (g) which emission is therapeutically dominant.
Approach. Combine physical and biological clearance into an effective
half-life, use it to get both the $1/8$-activity time and the time-integrated (cumulated)
activity in the thyroid, interpolate the absorbed-fraction table to 370 keV and scale it by
thyroid mass, then apply the given infinite-medium dose-rate constants — used directly
for beta (whose short range makes the thyroid effectively "infinite" for local absorption) and
scaled down by the finite absorbed fraction for gamma (which is not locally absorbed in a
16 g organ).
Part (a) — why iodine. The thyroid gland actively and selectively
concentrates iodine from the bloodstream because iodine is the essential raw material for
synthesizing thyroid hormones (T3, T4). Radioactive iodine administered systemically is
therefore preferentially taken up and concentrated in thyroid tissue by the gland's own normal
physiology, delivering a targeted internal radiation dose to (over)active thyroid tissue with
comparatively little uptake elsewhere in the body.
Part (b) — time to $1/8$ activity. Physical decay and biological
clearance act in parallel, so their rate constants add, giving an effective half-life
$$\frac{1}{T_{\text{eff}}} = \frac{1}{T_{\text{phys}}}+\frac{1}{T_{\text{bio}}}
= \frac{1}{8}+\frac{1}{2} = \frac{5}{8}\text{ d}^{-1}
\quad\Rightarrow\quad T_{\text{eff}} = 1.6\text{ d}$$
Since $1/8=(1/2)^3$, reaching one-eighth of the initial activity takes exactly three effective
half-lives:
$$\boxed{t_{1/8} = 3\,T_{\text{eff}} = 3(1.6) = 4.8\text{ days}}$$
Part (c) — cumulated activity. With immediate 60% thyroid uptake,
the thyroid's initial activity is $A_{\text{thy},0}=0.60\times100=60$ MBq $=6.0\times10^7$ Bq.
Clearance from the thyroid follows the same effective half-life found in (b), so the
time-integral of activity (the "cumulated activity," $\tilde A=\int_0^\infty A(t)\,dt$) is
$$\tilde A = \frac{A_{\text{thy},0}}{\lambda_{\text{eff}}}
= A_{\text{thy},0}\,\frac{T_{\text{eff}}}{\ln2}, \qquad
T_{\text{eff}}=1.6\text{ d} = 1.6\times86{,}400\text{ s} = 1.3824\times10^5\text{ s}$$
$$\boxed{\tilde A = 6.0\times10^7 \times \frac{1.3824\times10^5}{0.6931}
= 1.197\times10^{13}\text{ Bq}\!\cdot\!\text{s}}$$
Part (d) — absorbed fraction at 370 keV. The table gives absorbed
fraction per gram at 100/200/500 keV; interpolating linearly between the 200 keV and 500 keV
points for the given 370 keV mean gamma energy:
$$\phi(370\text{ keV}) = 1.55\times10^{-3} + \frac{370-200}{500-200}\big(1.66\times10^{-3}-1.55\times10^{-3}\big)
= 1.612\times10^{-3}\text{ g}^{-1}$$
Scaling by the 16 g thyroid mass gives the total (dimensionless) absorbed fraction of the
370 keV gammas actually stopped in the thyroid itself:
$$\boxed{\text{AF}_\gamma = 1.612\times10^{-3}\times16 = 0.0258\ \ (2.6\%)}$$
i.e. roughly 97.4% of the thyroid's own gamma emissions escape the 16 g gland without
depositing energy in it.
Part (e) — cumulative gamma dose. The given $S_\gamma$ is defined
for an infinite medium (absorbed fraction $=1$ by construction); for the real,
small thyroid the dose must be scaled down by the actual absorbed fraction found in (d). Using
the cumulated activity concentration $\tilde A/m$ ($m=16$ g $=0.016$ kg):
$$\frac{\tilde A}{m} = \frac{1.197\times10^{13}}{0.016} = 7.479\times10^{14}\text{ Bq}\!\cdot\!\text{s/kg}$$
$$D_{\gamma,\infty} = S_\gamma\,\frac{\tilde A}{m}
= 48.55\times10^{-15}\times7.479\times10^{14} = 36.31\text{ Sv (infinite-medium value)}$$
$$\boxed{D_\gamma = D_{\gamma,\infty}\times\text{AF}_\gamma = 36.31\times0.0258 = 0.937\text{ Sv}}$$
Part (f) — cumulative beta dose. Beta particles from ${}^{131}$I
have a range of only a few millimetres in tissue — far smaller than the thyroid's own
dimensions (a compact 16 g gland spans several centimetres) — so essentially all beta
energy is absorbed locally and the infinite-medium constant applies directly, with no
absorbed-fraction correction:
$$\boxed{D_\beta = S_\beta\,\frac{\tilde A}{m}
= 27.72\times10^{-15}\times7.479\times10^{14} = 20.73\text{ Sv}}$$
Part (g) — beta vs. gamma for therapy. Beta delivers roughly
22 times the local thyroid dose that gamma does (20.7 Sv vs. 0.94 Sv) precisely because its
short range confines essentially all its energy to the gland itself, while gamma's much longer
range lets 97% of its energy escape and irradiate the rest of the body instead. For treating
Graves' disease the goal is to destroy overactive thyroid tissue while minimizing dose to
everything else, so the therapeutically effective (and dose-efficient) emission is
beta. Gamma still has clinical value — it is exactly what makes the
uptake and biodistribution externally imageable/measurable — but from a pure
dose-delivery standpoint it is mostly "wasted" dose to the rest of the body rather than useful
therapeutic dose to the target.
Check: part (c) asks for the cumulated activity "absorbed in the body,"
which is read here as the thyroid uptake compartment (the 60% that is retained with the stated
2-day biological half-life), since every subsequent sub-part (d)–(g) builds on exactly
this quantity for thyroid dose. This is flagged per the exam's own note on stating
interpretation assumptions.