SOLUTION: The amount of a radioactive tracer remaining after t days is given by A = A0 e�0.058t, where A0 is the starting amount at the beginning of the time period. How many days will it

Question 72299: The amount of a radioactive tracer remaining after t days is given
by A = A0 e�0.058t, where A0 is the starting amount at the beginning of
the time period. How many days will it take for one half of the
original amount to decay?
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The amount of a radioactive tracer remaining after t days is given
by A = A0 e�0.058t, where A0 is the starting amount at the beginning of
the time period. How many days will it take for one half of the
original amount to decay?
---------
If half decays that leaves (1/2)Ao.
EQUATION:
(1/2)Ao = Ao e^(-0.058t)
Divide both sides by Ao to get:
1/2 = e^(-0.058t)
Take the natural log of both sides to get:
ln(1/2) = -0.058t
t= ln*1/2) / -0.058
t=11.95 days
The half-life of the tracer is 11.95 days.
-----------
Cheers,
Stan H.

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