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Homework Help: Calculation of a half-life how?

  1. Apr 3, 2005 #1
    "A radioactive substance has a half-life of 5 days, initial mass of 12kg. How much the original isotope will remain after 10 days?"

    This is just an example, the question itself is not important, i just want to know how to solve this type of problems generally.. thank you!
  2. jcsd
  3. Apr 3, 2005 #2
    The general form for growth/decay is the model of:

    [tex]A(t) = A_{i}e^{kt}[/tex]

    where Ai is the initial amount, k the growth rate, and t is time.

    You need to solve for k... so. Set it up like this.

    [tex]6 = 12e^{5k}[/tex]

    Solve for k and use the new equation to find any other questions pertaining to this model.
  4. Apr 3, 2005 #3

    Doc Al

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    What's the definition of half-life? In this example, what happens after 5 days?
  5. Apr 3, 2005 #4
    If im not mistaken..

    Half life is the time taken for half the number of nuclides to decay. So, as far as I understand, if the half life is of 5 days, after this time the substance will have half its initial mass.. in this case.. 6kg?

    Still somewhat confused.. :confused:
  6. Apr 3, 2005 #5
    Look at my post... that's exactly what I have. Can you please indicate what is confusing?

    EDIT: The followings posts are probably an easier way to look at the problem. Since 10 days is conveniently twice the half-life, you don't need to use a growth/decay model.
    Last edited: Apr 3, 2005
  7. Apr 3, 2005 #6
    That is right. The half life is also independent of how many particles you have (within the scale we're discussing here - [itex]\mbox{kg}[/itex]). So how much will you have left after 10 days?

    Of course, that's just because this particular question is set up such that the answer is obvious (once you understand it). In general you will need to solve an equation to figure out how much remains.
  8. Apr 3, 2005 #7

    Doc Al

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    Staff: Mentor

    Right. And what about after another 5 days (for a total of 10 days)?
  9. Apr 3, 2005 #8
    Maybe because..

    ..you've confused a few numbers..

    Is it supposed to be:

    10 = 12e^(5k) ?
  10. Apr 3, 2005 #9


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    May want to check out the "radioactive decay" post in the Differential Equations section. Remember to use enough decimal digits to get an accurate answer (Naeem, are you reading this?).
  11. Apr 3, 2005 #10
    No.... A(t) is the amount of the substance. The initial amount was 12 kg and after 5 days (t), you have 6 left. I'm fairly certain my method is correct.

    This method isn't necessary... if you look at the above posts you'll see why.
  12. Apr 3, 2005 #11

    Would it be.. 1/4 of the initial mass?
  13. Apr 3, 2005 #12
    Yes, that's correct. :)

    Just to show you the method I used, I'll solve this problem for you.

    [tex]6 = 12e^{5k}[/tex]
    [tex]k = \frac{-\ln(2)}{5}[/tex]

    Now you have the function
    [tex]A(t) = 12e^{\frac{-t\ln{2}}{5}[/tex]

    Plugging in t = 10 you get

    [tex]A(t) = 12e^{\frac{-10\ln{2}}{5}[/tex]
    [tex] A(t) = 3[/tex]

    Which is 1/4 of the original amount.
    Last edited: Apr 3, 2005
  14. Apr 3, 2005 #13
    The equation for an exponentially decaying quantity [itex]I(t)[/itex] is

    [tex] I(t) = I_0e^{-kt}[/tex]

    where [itex]I_0 = I(0)[/itex]. If the half-life is [itex]T[/itex], what it means is that

    [tex]I(T) = \frac{I_0}{2}[/tex]

    In your case, you are given the quantity at time 0 as [itex]12 \mbox{kg}[/itex], so [itex]I_0 = 12 \mbox{kg}[/itex]. The half-life is 5 days, so (omitting units for brevity)

    [tex]I(5) = \frac{I_0}{2} = 6 = I_0 e^{-5k} = 12e^{-5k}[/tex]

    ie. you're solving

    [tex] 6 = 12e^{-5k}[/tex]

    for [itex]k[/itex]. Once you do that, you need to find [itex]I(10)[/itex]. All you need to do to do that is to evaluate the formula for [itex]I(t)[/itex] with [itex]t=10[/itex]:

    [tex]I(10) = 12e^{-10k}[/tex]

    which isn't a problem now, since you found [itex]k[/itex] in the last step!
    Last edited: Apr 3, 2005
  15. Apr 3, 2005 #14
    So in the case of this question it is pretty much an..

    ..arithmetic sequence.. :]

    Thank you all very much, now im sure i understood the idea and its definitely not as hard as i thought..
  16. Apr 3, 2005 #15
    No problem.
  17. Apr 3, 2005 #16
    Geometric sequence actually. An exponentially decaying quantity is one that is decaying continuously in analogy to the discrete ratio for successive terms of a geometric sequence.
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