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Math Help - Radioactive Decay Problem?

  1. #1
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    Radioactive Decay Problem?

    The problem I am having with this is that I learned how to solve this type of problem a long time ago and when I was taught it they used another type of formula with different letters so I am kind of lost: (Can you tell me what I got write and what I did wrong if the answer is incorrect-Thanks)

    Here is the question:
    I am given the equation dA/dt=-kA

    A(t) = the amount of material present(in grams)
    t = time (minutes)
    k= some positive constant

    a.) Derive an equation for the amount A(t) present at time t in terms of the constant k and the amount A(0) present at time t=0

    This is what I did: (However I used the letters I learned and so would this be correct)

    (- dN/N)= λ * dt

    So,
    dN(t)/dt =-λ N(t)
    dN(t)/N(t)= -λ dt
    In N(t)=-λ dt
    In N(t)= -λ t+C
    N(t)=e^C e^-λt= Noe^-λ t (the o is between the N and e but on the bottom)


    (did I do the problem correctly- if not what steps am I missing - Thanks)

    b.) If A(5)= 1/3 A(3), find k

    (What would I do to find the solution to this problem- could you tell me the steps/formula to take- Thanks)

    c.) At what time t will the amount A(t) be 1/4 A(0)

    (Could you help me with this problem as well I don't now where ton start- could you show me the steps to take- Thanks)

    Thank you for all the help I really appreciate it
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  2. #2
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    Your working looks good, I get.

    \displaystyle  \frac{dA}{dt}= -kA

    First you separate the variables

    \displaystyle  \frac{dA}{A}= -k ~dt

    now integrate

    \displaystyle \int \frac{dA}{A}= \int-k ~dt

    \displaystyle \ln A= -kt +C

    \displaystyle A= e^{-kt +C}

    \displaystyle A= e^{-kt} e^C

    \displaystyle  e^C is just a constant so \displaystyle  e^C = A_0

    \displaystyle A= A_0e^{-kt}

    For \displaystyle A(5)= \frac{1}{4}

    \displaystyle A= A_0e^{-kt}\implies  \frac{1}{4}= A_0e^{-5k}

    now solve for k
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  3. #3
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    I get that for a.) the equation will be A= Aoe^-kt (The o is meant to be between the A and e)

    However, I am still confused about part b.) If A(5)= 1/3 A(3), find k

    and

    c.)At what time t will the amount A(t) be 1/4 A(0)


    (What part was the A(5) = 1/4 for and the A= Aoe^-kt --> 1/4 = Aoe^-5k for?)

    Are you saying that for question c.) the answer is that when t is 5 it will be 1/4

    How do you solve for k and what part is this question to b.) or c.)

    Thanks for the help
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  4. #4
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    Quote Originally Posted by pickslides View Post

    \displaystyle A= A_0e^{-kt}

    For \displaystyle A(5)= \frac{1}{4}

    \displaystyle A= A_0e^{-kt}\implies  \frac{1}{4}= A_0e^{-5k}

    now solve for k
    Here's k

    \displaystyle   \frac{1}{4}= A_0e^{-5k}

    \displaystyle   \frac{1}{4A_0}= e^{-5k}

    \displaystyle   \ln \frac{1}{4A_0}= -5k

    \displaystyle    k= \frac{-1}{5}\ln \frac{1}{4A_0}
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  5. #5
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    So this is the answer to part b and c?

    Thank you so much for the help
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  6. #6
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    I get that for part c.) of my question the answer is that when t is 5 it will be 1/4 (that is correct right?)

    But for part b.) is the answer k=-1/5 ln 1/4Ao (the o is below the A) Is this correct?

    Thanks
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  7. #7
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    The A_0 is all one number. It means the initial amount of A

    Your understanding in part c) is correct.
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