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anonymous

  • 5 years ago

the half-life of plutonium-239 is 24,100 years what is its initial amount if after 1000 years its amount is 2.1 grams?

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  1. anonymous
    • 5 years ago
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    2.4g I'll explain in a minute.

  2. anonymous
    • 5 years ago
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    Sorry, 2.04g

  3. anonymous
    • 5 years ago
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    This is exponential decay. The rate of growth or decay in something undergoing this kind of change is in proportion to itself. That is,\[\frac{dN}{dt} =n^2N\]where I've written n^2 as the constant of proportionality to emphasize it's magnitude (we can worry about signs later). You solve this type of equation as\[\frac{dN}{N}=n^2dt \rightarrow \int\limits_{}^{}\frac{dN}{N}=\int\limits_{}{}n^2dt \rightarrow \ln N=n^2t + c\]

  4. anonymous
    • 5 years ago
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    Exponentiate both sides to get\[N=e^{n^2t+c}=e^ce^{n^2t}=Ce^{n^2t}\]

  5. anonymous
    • 5 years ago
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    When t=0, the equation is N(0)=C, so C is the initial amount. Also, when t=24100 years, half of the initial substance is gone, so at this time,\[\frac{1}{2}N(0)=N(0)e^{n^2(24,100)} \rightarrow \frac{1}{2}=e^{n^2(24,100)}\]take the natural log of both sides to solve for n^2:\[\ln \frac{1}{2}=n^2(24100)yr \rightarrow n^2=\frac{\ln 1/2}{24100yr}=-\frac{\ln 2}{24000}\]

  6. anonymous
    • 5 years ago
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    So your equation for the amount of plutonium-239 at time t is:\[N(t)=2.1 e^{-\frac{\ln 2}{24100yr}t} grams\]where t is in years.

  7. anonymous
    • 5 years ago
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    For t=1000 years,\[N(1000)=2.1 e^{-\ln 2/24100yr \times 1000yr}g \approx 2.04g\]

  8. anonymous
    • 5 years ago
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    Okay?

  9. anonymous
    • 5 years ago
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    There's a typo. above -> 24000 should be 24100.

  10. anonymous
    • 5 years ago
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    thanks

  11. anonymous
    • 5 years ago
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    np

  12. anonymous
    • 5 years ago
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    Is this your first time here?

  13. anonymous
    • 5 years ago
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    yes

  14. anonymous
    • 5 years ago
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    We like to be thanked in fan points ;) There should be a link next to my name, "Become a fan".

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