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Thread: Exponential Decay-Radioactivity

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    Exponential Decay-Radioactivity

    A sample of a radioactive substance decayed to 95.5% of its original amount after a year. What is the half-life of the substance in years, and how long would it take the sample to decay to 25% of its original amount in years.

    My roommate just asked me this and I don't remember how to do it! Please help!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by acg716 View Post
    A sample of a radioactive substance decayed to 95.5% of its original amount after a year. What is the half-life of the substance in years, and how long would it take the sample to decay to 25% of its original amount in years.

    My roommate just asked me this and I don't remember how to do it! Please help!
    Let $\displaystyle N_0$ be the original amount of the substance and $\displaystyle N$ be the current amount of the substance.

    Now, radioactive decay takes on the form $\displaystyle N=N_0e^{-kt}$, where $\displaystyle k$ is the decay constant. We need to know what $\displaystyle k$ is, because the half life of this substance is defined by $\displaystyle \lambda=\frac{\ln 2}{k}$, where $\displaystyle \lambda$ is the half life.

    -------------------------------------------------------------------------

    The first bit of information gives us enough info to find k.

    It takes a year for the substance to decay to 95.5% of the original amount.

    Thus, we see that $\displaystyle N=N_0e^{-kt}\implies .955N_0=N_0e^{-k}$, where we replaced $\displaystyle N$ with $\displaystyle .955N_0$.

    This now simplifies to $\displaystyle .955=e^{-kt}\implies \ln(.955)=-k\implies \color{red}\boxed{k\approx 0.046}$

    Now we can find the half life:

    $\displaystyle \lambda=\frac{\ln 2}{0.046}\implies\color{red}\boxed{\lambda\approx 15.05~\text{years}}$

    -------------------------------------------------------------------------

    Now, can you find $\displaystyle t$, such that $\displaystyle .25=e^{-0.046t}$?? [this was what the second part of the question was asking for]

    Does this make sense?

    --Chris
    Last edited by Chris L T521; Oct 14th 2008 at 07:03 PM. Reason: wrong percentage...
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    thank you so much!!
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