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Thread: Radioactive decay

  1. #1
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    Radioactive decay

    A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by GreenandGold View Post
    A sample of a radioactive substance decayed to 91.5% of its original amount after a year. What is the half life of the substance? Also have to find the amount of time for it to decay to 30% of the original amount.
    Radioactivity can be defined using this model:

    $\displaystyle N=N_0e^{-kt}$, where

    $\displaystyle N$ is the current amount of the substance
    $\displaystyle N_0$ is the initial amount of the substance
    $\displaystyle k$ is the decay constant
    $\displaystyle t$ is the length of time

    Now, we are told that we have $\displaystyle .915N_0$ after 1 year.

    Thus, $\displaystyle .915N_0=N_0e^{-k}$

    Solving for k, we see that $\displaystyle k=-\ln(.915)\approx 0.0888$

    Now, we can find the half life:

    Half life is defined as $\displaystyle \lambda=\frac{\ln 2}{k}$, where

    $\displaystyle \lambda$ is the half life
    $\displaystyle k$ is the decay constant

    Thus, $\displaystyle \lambda=\frac{\ln 2}{.0888}\approx \color{red}\boxed{7.803\text{ years}}$

    Now, our decay model is $\displaystyle N=N_0e^{-0.0888t}$

    Can you find $\displaystyle t$ when $\displaystyle N=.3N_0$?

    --Chris
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    take the natural log of both sides?
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    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by GreenandGold View Post
    take the natural log of both sides?
    Yup. Now what do you get for your answer?

    --Chris
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    I got -13 but i did something wrong. is it 7.803(ln(.3)/ln(2))
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    ok i added a negative to the decay since its negative. so i got 1.35535
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by GreenandGold View Post
    ok i added a negative to the decay since its negative. so i got 1.35535
    Hmmm...

    $\displaystyle N=N_0e^{-0.0888t}\implies.3N_0=N_0e^{-0.0888t}$ $\displaystyle \implies \ln(.3)=-0.0888t\implies t=-\frac{\ln(.3)}{0.0888}\approx\color{red}\boxed{13. 558\text{ years}}$

    --Chris
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  8. #8
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    ok i see now!!! thanks...
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