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    Futuremoose's Avatar
    Futuremoose Posts: 1, Reputation: 1
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    #1

    Apr 1, 2011, 07:46 PM
    half life equation solver?
    The half-life for the radioactive decay of C-14 is 5730 years.
    How long will it take for 30% of the C-14 atoms in a sample of C-14 to decay?
    That is all the information I am given, it doesn't give me an initial amount or anything, how do I go about solving this?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #2

    Apr 1, 2011, 11:40 PM

    What formula do you know which can help you?
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #3

    Apr 2, 2011, 05:09 AM

    An easy way to find your constant of proportionately is to use the formula for half-life:



    You are given the half-life decay time. Plug it in and solve for k.

    Then, you can use



    Where A is the amount remaining, and is the initial amount.

    If 30% decays, then 70% remains.

    You do not need to be given an initial amount.
    jcaron2's Avatar
    jcaron2 Posts: 986, Reputation: 204
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    #4

    Apr 2, 2011, 07:11 AM

    Here's one more way to think about it, which should give you exactly the same answer as the formulas Galactus gave you (or the formula(s) you can look up in your textbook, as Unknown suggested):

    The half-life is the average amount of time it takes for half of the remaining atoms to decay. So after one half-life, only half of the original atoms remain. After another half-life the number has been cut in half again (so one 1/4 of the original ones remain). After a third half-life, only 1/8 remain; after a fourth, only 1/16 remain.

    Do you see a pattern? Notice the denominator of the fraction goes 2, 4, 8, 16 after 1, 2, 3, 4 half-lives respectively? 2, 4, 8, 16, etc. is the same as 2^1, 2^2, 2^3, 2^4, etc. (the carat
    symbol ^ means "raised to the power of" in case you're not familiar). So the remaining portion of a sample is 1/(2^n), where n is the number of half-lives that have passed. Mathematically this is the same, of course, as saying 2^(-n).

    Keep in mind that even though we just figured out this formula by using whole numbers of half-lives (1, 2, 3, 4, etc.), there's no rule that says it has to be a whole number. We can use this same relationship to compute the remaining portion after 2.354 half-lives, for example, by simply finding 2^(-2.354).

    You're trying to find how long it will take before only 30% of the original C-14 atoms remain. So now just write out the equation that you need to solve:



    Solve for n, and that's your answer. BUT keep in mind that n is the number of half-lives, not the number of years. Since you should probably give your answer in years, just multiply the number of half-lives by 5730.

    And one last thing: You probably already know how to do this, but just in case you don't, here's the trick for computing a logarithm in a base other than 10 or e.

    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #5

    Apr 2, 2011, 12:09 PM

    And I still prefer the other way to write this formula :p



    h is half life and the rest, you should know.

    Then, you know that

    h = 5730

    Just solve for t

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