Dose Commitment = C k If exposure (inhalation) should stop at time t, the additional dose (Day) Dp = RD.Q, 0.6 a TeexpChyt) t- An * og at NF (6) delivered to subcompartment k by the radionuclide present at time t would be T-t Da (RfDQ A) l-exp(-a tt) P . exp(-h, (T-t) de Pare Ryfgy Ay) CL-exp (A, #)) [T-exp (a, (Tot) ) Ag f C, = RD.Q) k =D, k + D l-exp(-A, t) t - —* exp(-A, (T-t) ) : aut | x where T is any time subsequent to t, i.e., T>t Cc, 0.6 ( x a £ Ak THORACIC LYMPH NODES 1-exp(-A, €) Oo RE D/A Lee TO exp (A (T-E))] k (3) The thoracic lymph nodes (L) receive radionuclide from Ph at the biological elimination rate, ”, For class Y compounds, there are two subcompartments, Li and Lr. EQUATIONS USED FOR NTS PLUTONIUM MODEL. 1.4 receives 90 percent of the radionuclide in qh and transfers it to B at the biological elimination rate, ve The remaining 10 percent is retained in Lr from which it is lost by radioactive decay, Mae only. Pulmonary Region = Lungs Q L = Qa, Li + Q Lr Qp = Dez (E/A,) C-exp(-agt)) © (fp/0 ) (L-exp(-ht)) + (£ fh_)Cl-exp(-A,t)) + go 8 8 (£/\_)(L-exp(-2,t))1 hoh h As he = he = Aw f, + ft + fi, = 0.6, and fe. = 0.4, Eq. Q, = Da, L-exp(-A, t) (4) (4) reduces to dQ . — Qe reece? | dQ aH ena = ichch Po SHE = cee ih abn dt ioLi dt rLr - =f Fh nQ/ s tas; —_ Qo = b- S CL-£ (1-f AL, / (Sea) 0.6 —, + 0.4 he Qa. £,D,9,/5(S+a,) For t 2 10, L-exp(-) ,t) - 1, and . _ Q =- D.o, [0.6 (1-exp( AEA, + 0.4/4] is an accurate approximation, 696 (5) = _,b fy Vf —_— Q = Af D 25S retvery + —————+__ S(SA, SA) (8)