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Word Problem with Constant 'e'

  1. Jan 20, 2012 #1
    1. The problem statement, all variables and given/known data

    For living organic material, the ratio of the number of radioactive carbon isotopes (carbon-14) to the number of non-radioactive carbon isotopes (carbon-12) is about 1 to 10^12. When organic material dies, its carbon 12 content remains fixed whereas its radioactive carbon-14 bagins to decay with a half-life of about 5700 years. To estimate the age of dead organic material, scientists use the following formula which denotes the ratio of carbon-14 to carbon-12 present at any time 't' (in years).

    R=(1/10^12)e^(-t/8223)

    Estimate the age of newly discovered fossil in which the ratio of carbon-14 to carbon-12 is 1/10^13

    2. Relevant equations

    -None-

    3. The attempt at a solution

    I just substituted one thing, I got confused, do I plug in 5,700 for 'R' or 11,400 for 'R' then solve?

    From this: R=(1/10^12)e^(-t/8223)
    To this: R=(1/10^13)e^(-t/8223)

    That is as far as I got

    Please help! Thanks
     
  2. jcsd
  3. Jan 20, 2012 #2

    Simon Bridge

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    You have to understand what the different terms mean.
    The general equation would be:
    [tex]R=R_0e^{-t/\tau}[/tex] where
    [itex]R_0[/itex] is the ratio that it started out with (at t=0);
    [itex]\tau[/itex] is the "mean-life" of the process in question, related to the half-life[1] by [itex]\tau[/itex]=T1/2/ln(2); and
    [itex]R[/itex] is the ratio now (i.e. after some time t has passed).

    eg. for C14, the half-life is 5700yrs so the mean life is 8223yrs. So [itex]\tau=8223\text{yrs}[/itex].

    So - is 1/10^13 the ratio it started with or the ratio now?
    Where does it go?

    ----------------------------
    note: Also written R=R0eλt
    ... which should look familiar if you ever played "hλlf life" :)

    λ = ln(2)/T1/2
     
    Last edited: Jan 20, 2012
  4. Jan 20, 2012 #3
    Yeah I dont understand any of what you just said. I know you replace the 1/10^12 to 1/10^13 because they gave you a new ratio. But for what everthing is equal to, 'R,' do I put in 5700 or what?
     
  5. Jan 20, 2012 #4

    Simon Bridge

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    If you don't at least try you are going to fail your course!
    This is not correct.
    The answer is "what". Please be aware that I am not going to do your homework for you. You have to do it yourself.
    So you have to figure out for yourself what to put for R, but I can guide you.
    You figure it out by understanding what it means - what does R, in the equation stand for? What does it mean?
    What kind of thing is the 5700years? What is it called?

    Consider what the equation is telling you:
    If the creature had died today, then the ratio would be 1/10^12
    In 5700 years the ratio would have been 0.5/10^12 because e^(5700/8223)=0.5

    This also means that if the creature had dies 5700 years ago, the ratio back then would have been 1/10^12 and now it will read as 0.5/10^12.

    Get it yet?
     
    Last edited: Jan 20, 2012
  6. Jan 20, 2012 #5
    Oh, I think I get it.. So for the thing to decay fully we plug 11400 into 't' because it said it began to decay with a half-life of about 5700 years.

    Once we plug it in, we can put it in the calculator and found out what 'R' is right?

    I am saying I do this:

    R=(1/10^13)e^(-11400/8223)
    ?
     
  7. Jan 20, 2012 #6

    Simon Bridge

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    Please answer the questions.
    What does R stand for in the equation?
    What does t stand for in the equation?

    What is the quantity you are trying to find?

    Also please answer the questions in post #4.
    Those are not rhetorical - they are put there for you to tell me the answers.
     
  8. Jan 21, 2012 #7
    What I said in thread starter post, is all the problem gives me, that's why I am confused just like you.. Sorry.

    All I can answer for you is that 't' stands for time in years only because that is given.
     
  9. Jan 21, 2012 #8

    Simon Bridge

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    Time is correct for t.

    The answer to everything I asked you is in post #1 and post #2.

    Do you even know what the point of the equation is?
    My problem is that in order to guide you, I need to find some point in your course where you do understand something. But, if I take what you have said at face value, you don't even understand basic algebra. I'll see if I can get someone more experienced to see if they can figure out what to do.
     
  10. Jan 21, 2012 #9
    Can you please show me what the formula would look like with everything substituted in? Maybe then I might understand
     
  11. Jan 21, 2012 #10

    Simon Bridge

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    Nice try: that sort of thing has a special name, it is called "doing your homework for you". We don't do that :)

    I did give you some examples in post #4
    OK - so lets do the last one:
    A body is found that has a C14 ratio of 0.5/10^12
    How long ago did it die?

    We reason that if the current C14 ratio is R, then t will be the time since the C14 ratio was at maximum. (post #2) We write[tex]\frac{0.5}{10^{12}} = \frac{1}{10^{12}}e^{t/8332}[/tex]then solve for t.
     
  12. Jan 22, 2012 #11

    Redbelly98

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    Okay, they are saying that R:

    (1) is equal to the expression above,

    and

    (2) is the ratio of carbon-14 to carbon-12 present at any time 't' ...

    And here they are telling us that at some time 't', the ratio of carbon-14 to carbon-12 is equal to 1/1013.

    In other words:

    R is the ratio of carbon-14 to carbon-12,​
    and
    the ratio of carbon-14 to carbon-12 is 1/1013.​

    Therefore, R = _____?
     
  13. Jan 22, 2012 #12
    would i put 5700 in for 't' and change 1/10^12 to 1/10^13?

    Then solve for R?
     
  14. Jan 22, 2012 #13

    Redbelly98

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    No.

    There is a simple, logical connection you have to make here:
    Fill in the blank, use this value for R, then solve for t.
     
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