I am a physician who thought she was good at math, but apparently am not as I cannot figure out this mathematical/physics question. (My background is obviously NOT nuclear medicine!) A family friend recently asked for my professional opinion on the possible radiation exposure to the child in the following encounter. This family friend had had a PET scan with 18FDG (18-fluorodeoxyglucose). She received a total dose of 11.6 mCi of tracer. Approximately 5 hours after the injection she had a brief contact with a 2-3 year-old child (a girl). She was within 10 feet of the child for approximately 5-10 minutes. (She had direct contact with the child for less time.) She had voided perhaps once (or twice) after the injection and before the above contact, so a fair amount of the 18FDG had presumably been removed from her body via her kidneys. From the literature I have read online, I don't believe there was any significant radiation exposure to be concerned about. But I am unable to arrive at an approximation of the dose equivalent (rems) the child may have received. (I am not sophisticated enough to convert radiation activity (mCi) to dose absorbed (rads).) I would like to know if such a calculation is, indeed, possible, as a raw number (even if only an approximation) would be much more reassuring. The child can be assumed to weigh 16 kg and the adult weighs 72 kg. Thank you for your help!


1 Answer 1


You can get an upper limit by simply treating the case of

  • The radioisotope structured as a point source.
  • The whole dose still present, corrected for half-life.
  • No shielding.
  • A simple $\frac{\text{presented area}}{4\pi r^2}$ for the acceptance (in this case the radiation is emitted in all directions, acceptance represents the fraction that hits the target, the formula I've chosen here is a consequence of the point source assumption above). For so young a child I would use roughly a quarter square meter for the presented area (generous estimate because we want an upper limit).
  • All impinging dose fully stopped in the subject.

The acceptance and shielding issues are the ones that would be difficult to do correctly.


  • The isotope in the tracer is $^{18}\mathrm{F}$ which has a half-life of about 110 minutes and a primary line (97%) of $633.5 \text{ keV}$ positrons for a total energy per decay of $1656 \text{ ke}$V. The decay product is $^{18}\mathrm{O}$ which is stable so there are no significant subsequent decays. Starting with $11.6 \text{ mCi} \approx 4.6 \times 10^8 \text{ Bq}$ and allowing for 2.5 half-lives we get a present activity of $8.2 \times 10^7 \text{ Bq}$.
  • Given a range of about 3 meters we get an acceptance of $\frac{0.25 \text{ m}^2}{4 \pi (3\text{ m}^2)} = 2.2 \times 10^{-4}$

Allowing 600 seconds, that gives us an exposure of $$(600\text{ s}) (8.2 \times 10^7 \text{ Bq}) (2.2 \times 10^{-4}) (1656 \text{ keV}) = 1.8 \times 10^{13} \text{ eV} = 2.9 \times 10^{-6} \text{ J} \, .$$ Allowing a mass of 16 kilograms that gives us a upper limit does of $1.8 \times 10^{-7} \text{ Sv}$ excluding the brief contact interaction and assuming no loss of radioisotope. That's less than a dental x-ray.

In short this interaction does not represent a significant risk to the child.

But it gets better. I assumed the KE of the positron in the dose, but that would be absorbed in the patient, so the child would only see the annihilation gammas. So the expectation should be reduced by about a third, and gammas will be significantly shielded by the patient and the intervening air.

  • 4
    $\begingroup$ also less than the dose absorbed during a transatlantic flight ($7\cdot10^{-5} $ Sievert) according to this source $\endgroup$ Feb 16, 2013 at 19:34

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