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Math Help - Radioactive Decay Problem

  1. #1
    Aug 2011

    Radioactive Decay Problem

    An unknown radioactive element decays into non-radioactive substances. In days the radioactivity of a sample decreases by percent. What is the half life of the element in days?


    [IMG]file:///C:/Users/WILS58%7E1.003/AppData/Local/Temp/moz-screenshot.png[/IMG][IMG]file:///C:/Users/WILS58%7E1.003/AppData/Local/Temp/moz-screenshot-1.png[/IMG]How long will it take for a sample of 100 mg to decay down to 60 mg? How many days?
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  2. #2
    e^(i*pi)'s Avatar
    Feb 2009
    West Midlands, England

    Re: Radioactive Decay Problem

    First of all we can do some estimation. 51% is near half so we're expecting an answer of around 900 days. Other than that we'd need the decay constant (except if the assumption I make later about the initial condition is valid)

    Short Way
    Half life is related to the decay constant by the following equation: t_{1/2} = \dfrac{\ln(2)}{\lambda} where \lambda is the decay constant.

    Long Way
    Since you've posted this in differential equations we have the separable DE: \dfrac{dN}{dt} = -\lambda t

    As said, this is separable: \dfrac{dN}{N} = -\lambda \ dt

    \int \dfrac{dN}{N} = -\lambda \int \ dt \Leftrightarrow \ \ln |N| = -\lambda t + C_1 (where C_1 is the constant of integration)

    |N| = e^{-\lambda t + C_1} = e^{C_1}e^{-\lambda t} = Ce^{-\lambda t} where C = e^{C_1} (also a constant)

    I am also going to assume the initial condition of when t =0 then N=N_0 (I'm not sure how valid this assumption is but without being told the decay constant it's the only way to find it)

    Sub in the initial condition and you find that N = N_0e^{-\lambda t} and rearranged it is \lambda = -\dfrac{1}{t} \ln \left(\dfrac{N}{N_0}\right)

    In your case: \lambda = -\dfrac{1}{900} \ln \left(\dfrac{0.49N_0}{N_0}\right) \approx 7.926 \cdot 10^{-4} \text{  day}^{-1}

    Finally: t_{1/2} = \dfrac{\ln(2)}{7.926 \cdot 10^{-4}\text{  day}^{-1}} = 874.5 \text{  days}

    We can check this against what we'd expect and it's close enough to be correct. You'll need to sort out the attachment on your second one but try the same thing again.
    Last edited by e^(i*pi); August 26th 2011 at 11:02 AM.
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