Answer
1.Radioactive or chemical decay
2. X(t) = Xoe^-kt
lnY(t) = -t + lnYo
4. 6.07mg
Explanation:
Let the liver and and blood compartment be represented by the symbol L and B respectively
For the liver
Suppose a first Order removal process started with an amount X, in which amount b disappeared in time t, the process is decay process which can be represented as follows,
∫dX/dt = -K1X
By rearrangement and integration;
∫dX/X = -K1t
ln X = -K1t + C
Since At t= 0, X = Xo then
C = lnXo
The equation becomes:
lnX = -K1t + lnXo
lnX - lnXo = -K1t
ln(X/Xo) = -K1t
X/Xo = e^-kt
X(t) = Xoe^-kt
X(t) = Xoe^-0.5t........(2)
Similarly for B (blood), suppose a first order flow flow of the dye move from the blood to the liver, let Y be the initial concentration, and amount b that has flown to the liver in time t
B---------> B(t)
t=0 Yo 0
t=t. Y-b. b
dY/dt = -K12(Y-b)...........(3)
Let Y-b = Y(t)
∫dY/dt = -∫K12t
By rearrangement and integration;
∫dy/Y(t) = ∫-K12dt
lnY(t) = -K12t + C1
at t= 0, C1 = ln(Yo)
Therefore ln Y(t) = -K12t + lnYo
But K12 =1
ln Y(t) = -t + lnYo.............(4)
(3) The assumptions used here
is that of a decay for the liver . The amount remaining taking as the amount of a substance
(4) using the equation 2,
X(t)= Xo e^-K1t........(2)
For time t = 1hour, and an initial amount X = 10mg, K = 0.5
X(t) = 10× e^-0.5
X(t) = 10 × 0.607
X(t) = 6.07mg
(5) within the scope of information presented, I have no data to make this judgment.
