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Math Help - Another Growth/Decay Question

  1. #1
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    Another Growth/Decay Question

    Please let me know if I am on the right track or way off tangent!

    Plutonium 241: A(t)=Aoe^-.053t

    Q: The Half Life of Plutonium 241 is approximately 13 years.

    A: How much of a sample weighing 4g will remain after 100 years?

    B: How much time is necessary for a sample weighing 4g to decay to .1g?


    So for question A, this is what I thought:

    1/2=4e^-.053 (100)

    On the left hand side, I used 1/2 to represent the half life and on the RS, I plugged in 4 for Ao which is my initial amount? For T, I used 100 years for time. This is where I think I goofed up. I don't think I set up the equation correctly.



    TIA
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  2. #2
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    Quote Originally Posted by mvho View Post
    Please let me know if I am on the right track or way off tangent!

    Plutonium 241: A(t)=Aoe^-.053t

    Q: The Half Life of Plutonium 241 is approximately 13 years.

    A: How much of a sample weighing 4g will remain after 100 years?

    B: How much time is necessary for a sample weighing 4g to decay to .1g?


    So for question A, this is what I thought:

    1/2=4e^-.053 (100)

    On the left hand side, I used 1/2 to represent the half life and on the RS, I plugged in 4 for Ao which is my initial amount? For T, I used 100 years for time. This is where I think I goofed up. I don't think I set up the equation correctly.



    TIA
    (a) A(100) = 4e^{-0.053 \cdot 100}

    A(100) = 0.019 grams


    (b) 0.1 = 4e^{-0.053t}

    solve for t using logarithms
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  3. #3
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    Thanks a lot ( How is it that everyone is able to type in the exponents?). To clarify, I would only set up the left hand side as 1/2 if they were looking for the half life correct?

    And would you mind, breaking down the Right hand side after taking the natural ln?

    Would it be:

    -.053*100 ln4e?
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  4. #4
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    Quote Originally Posted by mvho View Post
    Thanks a lot ( How is it that everyone is able to type in the exponents?). To clarify, I would only set up the left hand side as 1/2 if they were looking for the half life correct?

    And would you mind, breaking down the Right hand side after taking the natural ln?

    Would it be:

    -.053*100 ln4e?
    Yes, the LHS is only 0.5 when you want to measure the time it will take for a substance to decay to half it's original amount which is the definition of half-life.

    A) Skeeter solved this one directly

    B) Set
    • A(t) = 0.1
    • A_0 =


    As A(t) = A_0e^{-0.053t}

    Divide by A_0: \frac{A}{A_0} = e^-{0.053t}

    Take the log of both sides: ln(\frac{A}{A_0}) = -0.053t

    Divide through by -0.053: t = -\frac{1}{0.053} \times ln(\frac{A}{A_0}) = -\frac{ln(A) - ln(A_0)}{0.053}

    Put in your values for A and A_0
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  5. #5
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    Thanks everyone.
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