Question 463878: I have a question regarding radioactive decay. I am trying to solve the following problem:
A certain amount of radioactive material decays according to the function A(t)=Ao e^-3t, where time is measured in hours. What is the half-life of the material in hours?
So I know that I'm trying to find when A(t) equals half the original amount (Ao), but I'm not sure how to solve. I know that the solution is supposed to be ln(2)/3, just wondering how to get there.
Thanks,
Jessica
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A certain amount of radioactive material decays according to the function
A(t)= Ao e^-3t, where time is measured in hours. What is the half-life of the material in hours?
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So I know that I'm trying to find when A(t) equals half the original amount (Ao), but I'm not sure how to solve. I know that the solution is supposed to be ln(2)/3, just wondering how to get there.
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Let A(t) = (1/2)Ao
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Solve: (1/2)Ao = Ao*e^-3t
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e^-3t = (1/2)
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Take the ln of both sides:
-3t = ln(1/2)
-3t = -ln(2)
t = [ln(2)]/3
t = 0.2318 years
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Cheers,
Stan H.
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