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Radioactive decay questions

  1. Oct 15, 2008 #1


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    I'd like to know when the formula [tex]N(t)=N_0e^{-\lambda t}[/tex] is not valid anymore. By that I mean... since [tex]N_0[/tex] is the number of atoms at time [tex]t=0[/tex] and [tex]N(t)[/tex] is the number of atoms at time "[tex]t[/tex]", we see that [tex]N(t)[/tex] depends of [tex]N_0[/tex]. Now my question is : how do you know how many atoms should we take in count? Say we have 2 balls of plutonium, separated by 3 meters. How do you apply the formula given above? Is it still valid? Do you have to take [tex]N_0[/tex] as the number of atoms in the 2 balls, or you can apply the formula for each ball?
    To be more precise, what is the minimum density of radioactive elements we can consider to have a decent approximation using the formula?
    What is the "error" of the formula?
    I'm sorry if this makes a lot of questions and if they're not precise enough.
  2. jcsd
  3. Oct 15, 2008 #2


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    It doesn't matter.
    The number of decays in a period of time doesn't depend physically on the number of atoms present - it's really just another way of writing the probability of a given atom decaying within a certain time.

    Obviously it breaks down when there are only a few atomspresent since it is only an average rate.
  4. Oct 16, 2008 #3
    I'd generally agree with mgb's post...radioactive decay is a quantum tunneling statistical phenom so if you've got a glob of matter big enough to see likely there are enough atoms in it for statistical reasoning to apply....
  5. Oct 16, 2008 #4
    I think you guys are making this harder than the original question. The answer is, you apply the equation to either or both balls of decaying material. If you apply it to both, you get the "N" vs time that is the total number in both balls. If you apply it to one ball, N is the number in that ball. The two balls don't have to know anything about each other... and ultimately, there is a probability of each individual atom decaying (thats what the lambda is in the equation)

    Sorry I don't know how to write equations here, but think about this:

    N = No exp(-lambda t)

    so the time derivative of N (the rate of decay) is

    N-dot = (-lambda)No exp(-lambda t)

    then N-dot divided by N(t) (or the probability of decay) is

    ((-lambda)No exp(-lambda t))/(No exp(-lambda t))

    see, the "No" drops out

    Have I done that right? Does it help??

    Now, can someone point me to the instructions for entering equations here?
  6. Oct 16, 2008 #5


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    Radioactive decay is often described in terms of the 'number of atoms present' this makes it sound like there is some hidden/quantum communication between the atoms as they decide how many are going to decay in the next second. From this explanation it's resonable to ask how you define a group of atoms - does the rate change if you split the group in two etc.

    The important thing is that the half life and decay laws have nothing to do with radioactivity - they work for any process where a large number of things have the same probability of doing some event within a certain time frame.

    ps. Click the 'go advanced' button in the reply - then the Sigma sign for help with latex.
  7. Oct 16, 2008 #6


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    I think I got it. It means that it doesn't matter if I take into consideration one ball or 2 balls, the formula would work in either cases. Nice to know.
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