SOLUTION: The amount of grams "A" of a certain radioactive substance dumped in the Craney Island Disposal Area at time "t" is given by the formula "A=(Ao)e^(-0.1t)"
where "t" is the time m
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Question 832377: The amount of grams "A" of a certain radioactive substance dumped in the Craney Island Disposal Area at time "t" is given by the formula "A=(Ao)e^(-0.1t)"
where "t" is the time measured in days and "Ao" is the original amount of material. Find the half-life of this substance.
Would really appreciate any help with this. I have tried everything I could think of but because there are no amounts to work with I'm not sure what to do with it.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The amount of grams "A" of a certain radioactive substance dumped in the Craney Island Disposal Area at time "t" is given by the formula "A=(Ao)e^(-0.1t)"
where "t" is the time measured in days and "Ao" is the original amount of material. Find the half-life of this substance.
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A(t) = Ao*e^(-0.1t)
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Ao is the amount you start with.
A(t) is the amount after time = t.
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If A(t) is (1/2)Ao, e^(-0.1t) must be 1/2
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Solve: e^(-0.1t) = 1/2
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Take the natural log of both sides to get:
-0.1t = ln(0.5) = -0.6931
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t = -0.6931/-0.1 = 6.93 days.
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Cheers,
Stan H.
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