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