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Math Help - Growth/Decay Question

  1. #1
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    Growth/Decay Question

    This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
    Isotope:14-C
    Half-life (years): 5715
    Initial Quantity: 15 g
    Amount After 4000 Years: ?? g

    Also unsure as to starting this one.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by BeSweeet View Post
    This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
    [i]Isotope:[/]14-C
    Half-life (years): 5715
    Initial Quantity: 15 g
    Amount After 4000 Years: ?? g

    Also unsure as to starting this one.
    Use the following two formulae:

    {\lambda} = \frac{ln2}{t_{1/2}}

    A(t) = A_0e^{-\lambda t}

    Where:

    • \lambda = Decay Constant
    • t_{1/2} = half life
    • A(t) = Amount left at time t
    • A_0 = Amount at t=0
    • t = time.
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Use the following two formulae:

    {\lambda} = \frac{ln2}{t_{1/2}}

    A(t) = A_0e^{-\lambda t}

    Where:

    • \lambda = Decay Constant
    • t_{1/2} = half life
    • A(t) = Amount left at time t
    • A_0 = Amount at t=0
    • t = time.
    Thanks for the quick reply.
    [tex]\lambda[/MATH - 15?
    t_{1/2} - 5715
    A(t) - ?
    A_0 - 15
    t - ?

    Help?
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  4. #4
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    Quote Originally Posted by BeSweeet View Post
    This problem uses radioactive isotopes. I need to find the amount of grams after 4000 years. I'm given:
    [i]Isotope:[/]14-C
    Half-life (years): 5715
    Initial Quantity: 15 g
    Amount After 4000 Years: ?? g

    Also unsure as to starting this one.
    Let t denote the time measured in years, A_0 the initial amount and A(t) the amount after t years. Then A(t) is given by:

    A(t)=A_0 \cdot e^{k\cdot t} where k is a constant which you have to calculate first.

    You already know: If the initial value is A_0 then A(5715) = \frac12 A_0. That means:

    \frac12 A_0 = A_0 \cdot e^{k \cdot 5715} . Solve for k. I've got k \approx -0.0001212856

    Now calculate A(4000) with A_0 = 15\ g
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  5. #5
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by BeSweeet View Post
    Thanks for the quick reply.
    [tex]\lambda[/MATH - 15?
    t_{1/2} - 5715
    A(t) - ?
    A_0 - 15
    t - ?

    Help?
    \lambda = \frac{ln2}{5715} (which gives the same value as k as indicated by earboth)

    A(t) is what you're trying to find.

    t = 4000
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  6. #6
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    I am so freaking confused right now.

    This may be asking a lot, but could someone find the answer, and show the steps? The best way for me, personally, to figure out how to do something, is to analyze steps that get an answer.
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