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)