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Math Help - Radioactive

  1. #1
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    Question Radioactive

    The amount of a radioactive tracer remaining after t days is given by A=Ao e^-0.058t, where Ao 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? My answer is 12 days is that right. Thanks for looking at my answer.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by kwtolley
    The amount of a radioactive tracer remaining after t days is given by A=Ao e^-0.058t, where Ao 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? My answer is 12 days is that right. Thanks for looking at my answer.
    Can you tell us how you arrived at your answer?

    RonL
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  3. #3
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    Post missing table

    Looks like I'm missing a table I need to use with this question. I need to find half - life Bismuth isotope Ao. Then maybe I can get a better answer. Thanks for your help.
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  4. #4
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    Quote Originally Posted by kwtolley
    Looks like I'm missing a table I need to use with this question. I need to find half - life Bismuth isotope Ao. Then maybe I can get a better answer. Thanks for your help.
    You don't need a value for A0. Here's a hint: What is the definition of the term "half life?"

    -Dan
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  5. #5
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    Hello, kwtolley!

    You probably did it correctly . . . it's hard to see your work from here.


    The amount of a radioactive tracer remaining after t days is given by: A = A_o e^{-0.058t}
    where A_o 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?

    The question is: When is: A = \frac{1}{2}A_o ?

    The equation is: . A_oe^{-0.058t} \:=\:\frac{1}{2}A_o\quad\Rightarrow\quad e^{-0.058t}\:=\:\frac{1}{2}

    Take logs: . \ln\left(e^{-0.058t}\right) \:= \:\ln(0.5)\quad\Rightarrow\quad (-0.058t)\ln e\:=\:\ln(0.5)

    Therefore: . t \:= \:\frac{\ln(0.5)}{-0.058} \;=\;11.95081346\:\approx\;12

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  6. #6
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    Smile thank you

    Thanks for checking my problem.
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