PET/CT & Nuclear Medicine Facility Shielding (AAPM TG-108)

PET/CT & Nuclear Medicine Facility Shielding (AAPM TG-108)

Positron emission tomography brings a shielding problem that ordinary nuclear medicine does not: every positron annihilation releases two 511 keV photons, far more penetrating than the 140 keV photons of technetium-99m. The patient is also a moving source, injected in one room, resting in an uptake room, imaged in another, and sometimes held in a recovery area. AAPM Task Group 108 (PET and PET/CT Shielding Requirements, Medical Physics 2006) gives the methods most US physicists use, and its numbers come up on board examinations.

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Source Terms: Point Source versus Patient

The most common PET agent, $^{18}\text{F}$-FDG, has a physical half-life of about $110\text{ minutes}$. TG-108 reviewed the published "dose-rate constants" for F-18, which range from about $0.134$ (air kerma rate constant) to $0.188\ \mu\text{Sv}\cdot\text{m}^2/(\text{MBq}\cdot\text{h})$ (an older maximum-dose value), and recommends the effective-dose-equivalent value because regulatory limits are written in effective dose:

The patient value is the mean of published measurements around patients, corrected back to the time of administration.

Decay During Occupancy

Because the activity decays while the patient sits in a room, TG-108 replaces the initial dose rate times the time with the initial dose rate times the time times a reduction factor:

$R_t = 1.443 \left(\frac{T_{1/2}}{t}\right)\left[1 - e^{-0.693\,t/T_{1/2}}\right]$

For F-18, $R_t = 0.91$, $0.83$, and $0.76$ for $t = 30$, $60$, and $90\text{ minutes}$.

Uptake-Room Equation

With $N$ patients per week, administered activity $A_0$ (MBq), uptake time $t_U$ (h), and distance $d$ (m), the weekly dose at the protected point is $0.092 \times N A_0 t_U R_{tU}/d^2$ (in µSv). Setting this equal to $P/T$ gives the required transmission:

$B = \frac{10.9\,P\,d^2}{T\,N\,A_0\,t_U\,R_{tU}}$

With the US design goals of $P = 20\ \mu\text{Sv}$ per week for uncontrolled areas and $100\ \mu\text{Sv}$ per week for controlled areas, the constant $10.9P$ becomes 218 and 1,090. TG-108's own example (40 patients per week, $555\text{ MBq}$, $1\text{ h}$ uptake, an uncontrolled fully occupied point $4\text{ m}$ away) needs $B = 0.189$, met by about $1.2\text{ cm}$ of lead or $15\text{ cm}$ of concrete.

Imaging-Room Equation

By the time the patient reaches the scanner the activity has decayed by $F_U = e^{-0.693\,t_U/T_{1/2}}$ (0.68 after a 60-minute uptake), and voiding removes roughly 15% of the activity, so TG-108 multiplies by 0.85. The constant becomes 256 (uncontrolled) or 1,280 (controlled), and $t_I$ and $R_{tI}$ for the imaging time replace the uptake values. The gantry itself shields some walls, and TG-108 notes that for a five-bed-position scan the scanner conservatively reduces the dose by about 20%; manufacturer data can be used where available.

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Transmission at 511 keV

Narrow-beam ("good geometry") values give an HVL of $4.1\text{ mm}$ of lead and $3.4\text{ cm}$ of concrete, but calculations with these values underestimate the shielding needed because they ignore scatter buildup. TG-108 instead provides Monte Carlo broad-beam transmission factors:

Thickness in the first column is millimetres for lead and centimetres for concrete (density $2.35\text{ g/cm}^3$) and iron.

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Worked Example: Uptake Room Next to an Office

An uptake room handles $N = 25$ patients per week, each given $A_0 = 370\text{ MBq}$ of $^{18}\text{F}$-FDG and resting $t_U = 1\text{ h}$ ($R_{tU} = 0.83$). A fully occupied office (uncontrolled, $T = 1$) lies $3.0\text{ m}$ from the patient chair.

$B = \frac{218 \times (3.0)^2}{1 \times 25 \times 370 \times 1.0 \times 0.83} = \frac{1{,}962}{7{,}678} = 0.256$

From the TG-108 lead data, $9\text{ mm}$ transmits $0.289$ and $10\text{ mm}$ transmits $0.249$, so about $10\text{ mm}$ of lead is needed (about $13\text{ cm}$ of concrete would also work). A $3/8$-inch ($9.5\text{ mm}$) sheet falls just short, so the designer would specify $1/2$-inch ($12.7\text{ mm}$) lead or verify the existing wall's concrete. Without the reduction factor, or using narrow-beam HVLs, the answer would be wrong in opposite directions, which is why each step matters.

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PET/CT and the Rest of the Department

• Add the CT: The CT of a PET/CT is shielded with the NCRP 147 dose-length-product method, and its secondary air kerma is added to the PET contribution at each point. Lead sized for 511 keV usually covers the CT scatter as well, but the sum must still meet the design goal.
• Other rooms: Hot labs, dose-calibrator areas, injection rooms, patient toilets, and recovery bays are all sources; a single wall may receive contributions from several rooms.
• Floors and ceilings: Concrete slabs attenuate 511 keV photons much less than diagnostic x-rays, so rooms above and below uptake rooms need explicit calculation.

Pitfall

• Using the point-source constant for patients, or omitting decay: Using $0.143$ instead of $0.092\ \mu\text{Sv}\cdot\text{m}^2/(\text{MBq}\cdot\text{h})$ around patients and ignoring $R_t$ overestimates the dose, while using narrow-beam HVLs underestimates the lead; TG-108 corrects both errors.