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Math Help - Radioactive decay

  1. #1
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    Radioactive decay

    Quantity of radioactive material after 20 years decayed to 50 grams, after 40 years it decayed to 20 grams.What was the amount of radioactive material to start with.

    In problems like this usually we were finding half life or time but never initial amount.
    Radioactive decay is given by

    N=No e^kt

    I know that I have to find N at t=0. My problem is solving for k from the above equation.

    Thank you.
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  2. #2
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    For N=N_0 e^{kt}

    (t,N) =(20,50),(40,20) sub them in to find N_0,k
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  3. #3
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    I understand that.
    50=Noe^k(20)

    How to actually solve this.
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  4. #4
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    Simple answer, you don't solve that equation by itself. You need to use all of the data given, ie. 20 = N_0 e^{40k}
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  5. #5
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    50=N_0 e^{20k} ...(1)

    20=N_0 e^{40k} ...(2)

    Can you proceed?
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  6. #6
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    No not really.
    I don't understand what to do with equations.
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  7. #7
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    You have two equations and two unknowns, you need to eliminate one of them so that you can find the other. If you were going to do this which unknown factor do you think would be the easiest to eliminate?
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  8. #8
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    50=N_0 e^{20k} ...(1) \implies \frac{50}{ e^{20k}}= N_0

    20=N_0 e^{40k} ...(2) \implies \frac{20}{ e^{40k}}= N_0

    equating left hand sides \implies \frac{20}{ e^{40k}}=\frac{50}{ e^{20k}}

    Then

    20e^{20k} = 50e^{40k}

    \frac{20}{50} = \frac{e^{40k}}{e^{20k}}

    \frac{2}{5} = e^{40k-20k}

    \frac{2}{5} = e^{20k}

    \ln \frac{2}{5} = 20k

    \frac{1}{20}\ln \frac{2}{5} = k

    Now can you find N_0 ?
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  9. #9
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    So what is left to do is take k and plug it into one of the No equations?
    50/e^20(-.0458)=No
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  10. #10
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    Yep.
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  11. #11
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    Quote Originally Posted by pickslides View Post
    50=N_0 e^{20k} ...(1)

    20=N_0 e^{40k} ...(2)

    Can you proceed?
    Quote Originally Posted by ado061 View Post
    No not really.
    I don't understand what to do with equations.
    You take natural logs of both equations which reduces them to linear equations in L=\ln(N_0) and $$ k. These you solve in the usual manner and then N_0=\exp(L)


    CB
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