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Math Help - DE:simple chemical conversion

  1. #1
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    DE:simple chemical conversion

    a certain radioactive material has a half-life of 38 hours.find how long it takes for 90% of the radioactivity to be dissipated?
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  2. #2
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    Hello, alderon!

    A certain radioactive material has a half-life of 38 hours.
    Find how long it takes for 90% of the radioactivity to be dissipated?
    The "decay" formula is: . A \;=\;A_oe^{-kt}

    . . where A_o = initial amount, A = final amount,
    . . t = time, and k is a constant to be determined.


    We are told that when t = 38,\;A \,= \,\frac{1}{2}A_o

    So we have: . \frac{1}{2}A_o \;=\;A_oe^{-38k}\quad\Rightarrow\quad \frac{1}{2}\;=\;e^{-38k}

    Multiply by 2e^{38k}\!:\;\;e^{38t} \:=\:2\quad\Rightarrow\quad 38k \:=\:\ln(2)\quad\Rightarrow\quad k \:=\:\frac{\ln(2)}{38} \:\approx\: 0.01824

    Hence, we have: . A \;=\;A_oe^{-0.01824t}


    The problem asks: .When is A \,=\,\frac{1}{10}A_o ?

    We have: . \frac{1}{10}A_o \;=\;A_oe^{-0.01824t}\quad\Rightarrow\quad \frac{1}{10} \;=\;e^{-0.01824t}

    Multiply by 10e^{0.01824}\!:\;\;e^{0.01824t} \:=\:10\quad\Rightarrow\quad 0.01824t \:=\:\ln10\quad\Rightarrow\quad t \:=\:\frac{\ln10}{0.01824}


    Therefore: . t \:=\:126.2382178 \:\approx\:126.24\text{ hours}

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