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Thread: Need help logarithmic exc.

  1. #1
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    Need help logarithmic exc.

    Im stuck on this study question.

    It says.

    Iodine-135 has a half life of 8 days. How long would it take for 28kg of iodine-135 to decay to 200g?

    The answer s 57.03 days.

    Could someone explain how this answer was gotten? The unit is logarithmic funtions so I assume that they are involved, but how?
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  2. #2
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    Quote Originally Posted by Skor View Post
    Im stuck on this study question.

    It says.

    Iodine-135 has a half life of 8 days. How long would it take for 28kg of iodine-135 to decay to 200g?

    The answer s 57.03 days.

    Could someone explain how this answer was gotten? The unit is logarithmic funtions so I assume that they are involved, but how?
    This an exponential decay problem, so the formula is:
    $\displaystyle m(t) = m_0e^{-kt}$

    where $\displaystyle m_0$ is the original amount of material and k is a constant equal to $\displaystyle \frac{ln(2)}{t_{1/2}}$ where $\displaystyle t_{1/2}$ is the half-life.

    So:
    $\displaystyle 0.200 = 28 e^{-ln(2)t/(8)}$ where t is in days.

    $\displaystyle \frac{0.200}{28} = e^{-0.086643t}$

    $\displaystyle 0.007143 = e^{-0.086643t}$

    $\displaystyle ln(0.007143) = -0.086643t$

    $\displaystyle -4.94164 = -0.086643t$

    $\displaystyle t = \frac{-4.94164}{-0.086643} = 57.0343\, days$

    -Dan
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  3. #3
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    Hello, Skor!

    Iodine-135 has a half-life of 8 days.
    How long would it take for 28 kg of Iodine-135 to decay to 200 g?

    The answer s 57.03 days.
    There is a half-life formula, but we can derive it ourselves.

    Assume an exponential function: $\displaystyle I\:=\:c\,e^{-kt}$
    . . where $\displaystyle I$ is the amount of I-135 at time $\displaystyle t$ (days).
    Let $\displaystyle I_o$ = original amount of I-135.

    At the very beginning $\displaystyle (t = 0)$, we have: .$\displaystyle I_o \:=\:c\,e^0\quad\Rightarrow\quad c\:=\:I_o$

    In 8 days, only half of the I-135 remains. .That is, when $\displaystyle t = 8,\;I = \frac{1}{2}I_o$
    So we have: .$\displaystyle \frac{1}{2}I_o\:=\:I_o\,e^{-8k}\quad\Rightarrow\quad e^{-8k} \:=\:\frac{1}{2}$

    Rewrite in logarithmic form: .$\displaystyle -8k \:=\:\ln\left(\frac{1}{2}\right) \:=\:\ln\left(2^{-1}\right) \:=\:-\ln 2$
    . . Hence:. . $\displaystyle k \:=\:\frac{\ln 2}{8}$

    Therefore, the function is: .$\displaystyle I \:=\:I_o\,e^{-\frac{\ln2}{8}t}$


    We are told that: .$\displaystyle I_o \:=\:28\text{ kg}\:=\:28,000\text{ g}$ and $\displaystyle I \:=\:200\text{ g}$

    So we have: .$\displaystyle 200 \:=\:28,000e^{-\frac{\ln 2}{8}t}$

    . . $\displaystyle e^{-\frac{\ln2}{8}t}\:=\:\frac{1}{140}\quad\Rightarrow \quad -\frac{\ln 2}{8}t \:=\:\ln\left(\frac{1}{140}\right) \:=\:-\ln140$

    Therefore: .$\displaystyle t\:=\:\frac{8\ln140}{\ln2} \:=\:57.03426414\:\approx\: 57.03\text{ days}$

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