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Math Help - Half-Life

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    Half-Life

    The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
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    Quote Originally Posted by lilikoipssn View Post
    The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
    Let L(0) denote the initial level of radiation and t the passed time. Then the level after t years is calculated by:

    L(t)=L(0)\cdot e^{k\cdot t} with k a constant typical for the decaying process.

    From L(0) = 1 and L(10) = 0.8 you can derive k:

    0.8 = 1\cdot e^{k\cdot 10}~\implies~ k=\frac1{10} \cdot \ln(0.8)~\approx~-0.02231...

    Now calculate the time which is necessary to reach only the half of the radioactivity:

    0.5 = e^{\frac1{10} \cdot \ln(0.8) \cdot t}~\implies~ t\approx 31.06\ yrs

    Btw: That is a very optimistical result
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  3. #3
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    Quote Originally Posted by lilikoipssn View Post
    The level of radioactivity on the site of a nuclear explosion is decaying exponentially. The level measured in 1980 was found to be 0.8 times the level measured in 1970. What is the half-life.
    A(t)=A_0\left( \frac{1}{2}\right)^{\frac{t}{h}}

    let 1970 be time t=0

    Then

    A(10)=.8A_0=A_0\left( \frac{1}{2}\right)^{\frac{10}{h}}

    .8=\left( \frac{1}{2}\right)^{\frac{10}{h}}

    \ln(.8)=\frac{10}{h}\ln\left( \frac{1}{2}\right)

    h=\frac{10 \ln \left( \frac{1}{2}\right)}{\ln(.8)} \approx 31.06

    I hope this helps.
    Last edited by TheEmptySet; May 12th 2008 at 09:54 PM. Reason: typo
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