The decay rate (dA/dt) of a radioactive isotope is directly proportional to the

Question

The decay rate (dA/dt) of a radioactive isotope is directly proportional to the amount(A) of the isotope present. A nuclear reactor, operating at a constant power level, produces isotope "X" at a constant rate, Q. Isotope "X" decays with a half life of 30 days. The rate of change in the amount of "X" per unit time, (dA/dt), is equal to the production of "X" less the removal (decay) rate.

Find an expression for the amount of "X" present in the reactor after 60 days. What will be the steady state amount "X".

Explanation / Answer

dAin/dt = Q
dAout/dt = k*A

The half life is 30 days, so,
A(t) = A0ekt
A(30) = A0/2 =  A0e30k
1/2 =  e30k
ln(0.5) = 30k
k = ln(0.5)/30 = -0.0231
(We'll just call this k for the time being)
dA/dt = Q + kA (there's a plus sign here because k is negative)
A' - kA = Q

Homogenous solution:
A' - kA = 0
Use characteristic equation:
r - k = 0
r = k
A(t) = Bekt

Particular solution:
A' - kA = Q
A constant will work:
A = -Q/k

Total solution:
A(t) = -Q/k + Bekt
Say that at t=0, there was A0 of isotope "X" available:
A(0) = A0 = -Q/k + B
B = A0 + Q/k

A(60) = -Q/k + Be60k
A(60) = -Q/k + Be60*ln(0.5)/30
A(60) = -Q/k + Be2*ln(0.5)
A(60) = -Q/k + B(0.5)2
A(60) = -Q/k + B/4
A(60) = -Q/k + (A0 + Q/k)/4
A(60) = -Q/k + A0/4 - Q/(4k)
A(60) = A0/4 - 5Q/(4k)
A(60) = A0/4 - 5Q/(4*ln(0.5)/30)  
A(60) = A0/4 - 75Q/(2*ln(0.5))

The steady state is the amount remaining as t approaches infinity.
A(t) = -Q/k + Bekt
Since k is negative, the exponential term will fall to zero. So, the steady state amount is
-Q/k = 43.3*Q

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