Oe2)( 82 , hr B, = fea 2,5 x 107 at 365;" ant R(t) = 265 x 19°(0,0169) = 42p0T4fapy initial uptake Then from equation 3 De 2528 (Sta? (00987) - 2601és (02352) = 3.48 x 10° +87 & rep ae din approximately 100 days. As a check, let us see how much of this rather high concentration of activity would remain by t = 106d (March 24 - dune 15). Qe 25210 0,0205(1 x10") - 0 0.0855 9 0.06 pe in total For a 11 gm thyroid, this 1s approxiuately 5.5 x 107? e/g, or about 5efx? (2,22010/) = 1e2x10” counts per minute. For an average 40% efficient scintillation counter with well, about 5000 opm would be present due to thia first doge, at the time of counting. Sinee this is 0.06 yo out of a total of 0.506, ahout 11% of the activity counted would be due to the Ghot 2 exposure. This would reduce hy 11% the estimates of i, for Shot 9, hence also the value of q ami thus those of R, ani D for Shot 2. A dose of about 3200 rep might then be mre accurate. However, it should be noted that on the “high spot" hypothesis the total dose from shot 2 might be much less as pointed out in ref(1), also that irregular distribution of activity in the thyroid might alter the dose estinates based on it. In the other direction, p may be dloser te 0,30 than 0.2, ard hence more retention and higher dosages would have cecurred. -20- Sy Now ARCHIVES