As a check, one may ostimate a half thickness value for a given energy spectrun
from the absorption curve in Al. (See Table). Using the Feather Value, it is
seen that for d = 1 om, the value of pd for these intervals exceeds 6 and the
cylinder behaves like an infinite slab with respect to self absorption, for
which ( ref. 6
):
Te Ap(/bpe
sees
eee
ce
eee
ec
ecco
se
(12)
If one takes all the betas to be absorbed within a 3mm shell of intestinal
mucosa whose density is about 1 gn/cmJ, the total mass of irradiated wall will
be 69017 ga. The total area will be a = 2nrl = 2000 cm’, and the total flux will
be Ia, or
Qa Ia = S2iqit)
20001 2an't) 2 21%
betas/seo x 3,7 x 10+ .
PB Ag(t)
kp 2000
all of which is absorbed in ths tissue. Hence equation (3) will apply. Thus
Bo552_ aw ap(t)
6907
Bp
=
107 = ap(t)
BR.
©
rep/day. sce ce ec ccee
(13)
Where Eis in Mey, pin cat, aq(t) in po/tt? .
Since it has been assumed that the sheep turns over 2000 gn/day, this
amount is always in the intestine, with ite activity falling off as A(t) = Ar,
Then the total dose is
Ds LZap tpt] Etdt... ee ee eee cee ew we (A)
pe
‘4 )
The integral will be divided into three periods over which n, Es and thus p,
are taken as constant to a first appraximation. Then for the shot 2 case, where
t, = 0.15 day and At, = 2300 po/ft”,
3
ny ( pith t4)
D = 147 (2300) 2, t°°
=PE
i=l
Pa
a- 2)
cee
eee
ve
(15)
-19-
DO’ ARCHIVES