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Category: Radiation Fundamentals — Doses and Dose Calculations

The following question was answered by an expert in the appropriate field:

Q

I traveled in a closed compartment of a vehicle with about 50 passengers. During the trip, an M1A1 collimator was damaged and leaked 370 GBq of tritium (3H) in the compartment. Fourteen days after the event I submitted a urine sample which showed 110 Bq in my system. What was the approximate dose of exposure on the day of the event?

A

From what I understand, the device you cite uses tritium gas as the radioactive source to stimulate luminescence as a light source for use of the collimator.

As has been discussed in International Commission on Radiological Protection (ICRP) Publication 68, when tritium gas, HT, is inhaled, as would be the presumed mechanism of exposure for the case you describe, a small fraction, on the order of 0.01% of the gas inhaled gets absorbed and converted to tritiated water, HTO. It is this component that dominates the dose, which is several times greater than the dose contribution from the inhaled gas that irradiates lung tissue. The ICRP effective dose coefficient for tritium gas is 1.8 × 10-15 Sv Bq-1.

You state that 14 days after your exposure, results of a urine analysis confirmed the presence of about 110 Bq of tritium in your body, which we shall assume was tritiated water. Considering that tritiated water has about a 10-day effective half-life in the body, the likely amount of tritiated water that would have been present 14 days earlier would have been, which may be back-calculated as (110 Bq)e(ln2/10)(14) = 290 Bq. If the tritiated water represented 0.01% of the tritiated gas inhaled, the tritium intake would have been 290 Bq/0.0001 = 2.9 × 106 Bq. If the total tritium activity in the collimator originally was 3.7 × 1011 Bq, your intake would have represented a tiny fraction, 0.0000078, of the original amount.

This intake is inconsequential from a dose impact point of view. In this case the projected effective dose would be (2.9 × 106 Bq)(1.8 × 10-15 Sv Bq-1) = 5.2 × 10-9 Sv or 5.2 × 10-3 µSv, which is about the dose one might receive from natural external background radiation in about a three minute period.

I hope this satisfies your needs and provides assurance that the dose consequences of your exposure were negligible, based on the information you provided and the assumptions made.

George Chabot, PhD, CHP

Ask the Experts is posting answers using only SI (the International System of Units) in accordance with international practice. To convert these to traditional units we have prepared a conversion table. You can also view a diagram to help put the radiation information presented in this question and answer in perspective. Explanations of radiation terms can be found here.
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