I-125 dose calc formulae? -Reply

[FRAMEP@ORAU.GOV]

>Anyone have I-125 dose calculation math they'd like to share?

John, as was pointed out, a nice and easy way is to derive the Sv/Bq
or rem/uCi factors from the ALIs. Its even easier to look up the dose
conversion factors in Fed Guidance Report 11. But, as Tina Turner
would say: we never like to do anything nice and easy. On the other
hand, we're not maroons so let's just deal with the thyroid.

The first step is to identify all the places the iodine goes in the
body and how long it stays there. Then we calculate the number of
decays that will occur in the various tissues over the next 50 years.
Figure out how much of the energy released in all the decays 
deposits in the thyroid, divide by the mass of the thyroid, multiply
by a quality factor, play some hocus pocus with a weighting factor
and voila.

If 1 Bq of iodine is ingested, ICRP 30 assumes 0.3 Bq goes to the
thyroid. Thats a little high, lets use Mike's 0.25 which might also
be high. Not knowing what actually happens with your RIA compound
puts a big question mark in our calcs. The rest of the ingested
iodine goes straight to the potty (what happens then has been a
subject of considerable interest lately). Whats in the thyroid leaves
with a 120 d biological half life and takes a uniform distribution
the rest of the body.

In the thyroid, the effective half life(T) is 40 days i.e. 120 x
60/(120 + 60) which in seconds is 3.456E6. If A is the activity that
reaches the thyroid, the number of I-125 decays in 50 years in the
thyroid is:
            1.44 T A  = 1.44 x 3.456E6 x 0.25 
                      = 1.244E6 decays

BTW 1.44 T is the inverse of the effective decay constant.

For reasons that will become obvious we can ignore the iodine outside
the thyroid.

We then take multiply the energy of each of I-125's auger and
conversion electrons by its fractional intensity (yeild) and add up
the products. The sum for the decay data I have is 16.93 keV per
decay. All of this energy is assumed to be deposited in the
thyroidi.e. the absorbed fraction is 1.

e.g. for the k-auger: 22.7 keV x 0.2 = 4.54 keV.

We then multiply the energy of each of I-125's x and gamma rays by
its fractional intensity and (because some of this energy escapes the
thyroid) the absorbed fraction. The latter is the fraction of the
energy at a given energy released in the thyroid that is absorbed in
the thyroid. These absorbed fractions are obtained from MIRD Pamphlet
5. I used an absorbed fraction of 1 for the 3.8 keV x-ray and 0.15
for the other photons which hover around 30 keV. Then we sum the
absorbed energies. 

e.g. 27.47 keV x 0.73 x 0.15 = 3 keV absorbed

The total photon (x and gamma ray) energy absorbed in the thyroid per
decay is 6.65 keV

Add this to the 16.93 keV for the auger and conversion electrons and
we have 23.58 keV of energy absorbed in the thyroid per decay in the
thyroid. We can ignore the iodine outside the thyroid (boy are we in
luck here) because the absorbed fraction for augers and conversion
electrons will be zero and the absorbed fraction for I-125's low
energy photons is next to zero.

So the ingestion of 1 Bq results in the 20 g (0.02 kg) thyroid
absorbing:

23.58 keV/decay x 1.244E6 decays = 2.93E7 keV
                                 = 2.93E7 x 1.6 E-16J/keV
                                 = 4.69 E-9 joules

This is 2.347 E-7 J/kg = 2.347 E-7 Gy per Bq ingested. This becomes
2.347 E-4 mGy/Bq or 2.347 E2 mGy/MBq ingested. This is pretty close
to  Mike's calculated value and whats in Fed Guid Rep 11. Multiply by
the weighting factor of 0.03 for the thyroid and we have an committed
effective dose equivalent of 7 E-9 Sv per ingested Bq, also very
close to whats in FGR 11 (the dose to everything besides the thyroid
is small potatos). 

Fun stuff:-)

Paul Frame
Professional Training Programs
ORISE
framep@orau.gov