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Math Help - Plutonium Decay Problem

  1. #1
    Junior Member xterminal01's Avatar
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    Plutonium Decay Problem

    Plutonium,
    239Pu, has a half-life of 24,100 years. What percent of a given amount remains after 100,000 years?
    Is this done correctly?


    1/24100ln(1/2)

    k=0.00002876

    e^(0.00002876(100000))
    = 56.35% after 100,000 years
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  2. #2
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    Nope

    If it has a halflife of 24.000years how can there be more then half (over 50%) left after more then four half lifes?
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  3. #3
    MHF Contributor
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    Quote Originally Posted by xterminal01 View Post
    Plutonium,
    239Pu, has a half-life of 24,100 years. What percent of a given amount remains after 100,000 years?
    Is this done correctly?


    1/24100ln(1/2)

    k=0.00002876

    e^(0.00002876(100000))
    = 56.35% after 100,000 years

    let P = amount of plutonium at any time t

    P_0 = amount of plutonium at t = 0

    P = P_0 e^{kt}

    \frac{P}{P_0} = e^{kt}

    \frac{1}{2} = e^{24100k}

    \ln\left(\frac{1}{2}\right) = 24100k

    k = -\frac{\ln{2}}{24100} \approx -2.876

    at t = 100000 years ...

    \frac{P}{P_0} = e^{100000k} \approx 0.056 = 5.6%
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  4. #4
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    I belive this is correct

    Half-life = ln(2)/lambda

    This gives lambda = 2.876*10^(-5)

    Then N = N0*e^(-lambda*t)

    Set N0 = 100% gives you:

    100*e^(-2.876*10^-5*100000) = 5.9%
    Sounds right 50;25;12.5;6.25 = 4 halflifes and gives you about 6% left


    ops: Too slow, evenwithout this latex =)
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