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Math Help - An exponential decay problem.

  1. #1
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    An exponential decay problem.

    A sample of uranium-239 (U-239) decays into neptunium-239 (Np-239) according to the standard decay function:

    After 10 minutes, the sample has decayed to 64% of its initial amount.

    a) The value of the disintegration constant, pi, is approximately 0.045/min

    b) The half-life of U-239 is 15.5 minutes approximately.
    (working on this at the moment)

    c) Write the equation that gives the amount of U-239 remaining as a function of time, in terms of its half-life.

    d) Suppose the initial sample had a mass of 25mg. How fast is the sample decaying after 15 minutes?
    Last edited by kmjt; June 5th 2010 at 01:05 PM.
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    Is this equation N(t)= N_0e^{-\pi t}

    N_0 is the initial amount of uranium at t= 0

    You are told that at t = 10 then N(10) = 0.65\times N_0

    So you have

    N(t)= N_0e^{-\pi t}

    at t = 10

    0.65\times N_0= N_0e^{-10\pi }

    Dividing N_0 from each side leaving

    0.65= e^{-10\pi }

    Can you now solve for \pi ?
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  3. #3
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    Solved for pi and got approximately 0.043. I believe that is correct.

    For part b how would I start?
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    Quote Originally Posted by kmjt View Post
    Solved for pi and got approximately 0.043. I believe that is correct.

    For part b how would I start?
    half-life ... time it takes for half of the original amount ( N_0) to decay.

    \frac{N_0}{2} = N_0 e^{-\pi t}

    solve for t
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  5. #5
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    Quote Originally Posted by kmjt View Post
    Solved for pi and got approximately 0.043. I believe that is correct.

    For part b how would I start?
    For N(t)= N_0e^{-0.043 t} solve for t when N(t) = \frac{N_0}{2}

    I.e \frac{N_0}{2}= N_0e^{-0.043 t}

    Go get em!
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    Hmm, this is how I tried solving for t:

    As you said: No/2 = Noe^-pi(t)

    I rewrote it as (0.5)( No) = Noe^-pi(t)

    Divided No out..

    0.5 = e^-pi(t)

    ln(0.5) = ln(e^-pi(t))

    ln(0.5) = -pi(t)

    ln(0.5)/-pi = t

    t = approximately 0.22

    Where did I go wrong? The answer is 15.5 minutes

    Just subbed in 0.043 for pi and the answer I got was -7..
    Last edited by kmjt; May 5th 2010 at 04:48 PM. Reason: trying again
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  7. #7
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    Quote Originally Posted by kmjt View Post
    Hmm, this is how I tried solving for t:

    As you said: No/2 = Noe^-pi(t)

    I rewrote it as (0.5)( No) = Noe^-pi(t)

    Divided No out..

    0.5 = e^-pi(t)

    ln(0.5) = ln(e^-pi(t))

    ln(0.5) = -pi(t)

    ln(0.5)/-pi = t

    t = approximately 0.22

    Where did I go wrong? The answer is 15.5 minutes

    Just subbed in 0.043 for pi and the answer I got was -7..
    you used the constant \pi = 3.14159...


    also, a correction ... \pi = \frac{\ln(.64)}{-10} \approx 0.0446...
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  8. #8
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    Ooooh, worked it out and got the right answer

    The next part asks for an equation that gives the amount of U-239 remaining as a function of time, in terms of its half-life.

    The answer is N( t) = N0(1/2)^(t/15.4)

    What are the steps one must take to get to this answer? And i'm looking at the exponent part.. t/15.4, and i'm wondering why is it not 15.5? Or is it something completely different?
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  9. #9
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    Write the equation that gives the amount of U-239 remaining as a function of time, in terms of its half life.
    The answer to c) is:



    But what steps would I have to go through to get to that answer?
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