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