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Exponential decay

  1. Jan 26, 2004 #1

    chroot

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    From a member called "Banana":

    Now I'm trying to do one with radioactive decay. Do you think I would use those same formulas? The only confusing thing is that it's not presented the same. It says that a material decays so that it is 99 percent gone in 6.65 half-lives (so would you double that?). After how many half-lives is the material 99.9 percent gone? I tried to piece together some ideas I found in other text books, but I don't think it makes sense. Here's what I tried:

    - ln (N/No)/ln2
    - ln (.001/.01)/.693 = 3.3 (I don't think this is right because I didn't even find anywhere to insert the 6.65)

    Is that wrong? Thanks for your help, I REALLY appreciate it. Banana
     
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  3. Jan 26, 2004 #2

    chroot

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    The key equation is the same. In this case, after one half-life has elapsed, the population of remaining radioactive atoms is one-half what it was at time t=0.

    Thus

    [tex]0.5 N_0 = N_0 e^{k t_{hl}}[/tex]

    Where [itex]t_{hl}[/itex] is the half-life.

    If one-half the population remains after one half-life, you get this equation:

    [tex]0.5 = e^{k \cdot 1}[/tex]

    Thus, k always equals [itex]\ln{0.5}[/itex], or about -0.69, in problems dealing with half-life.

    Now, using this k, we can solve for the situation where 99.9% of the original population is gone (therefore 0.1% remains):

    [tex]0.001 N_0 = N_0 e^{\ln{0.5} t}[/tex]

    Solve for t. Let me know what you get.

    - Warren
     
  4. Jan 27, 2004 #3
    [tex] \frac{a}{a-x}=2^{\frac{t}{\tau}} [/tex]

    tau : half life

    And a is initial amt and a-x is the remaining
    from the data

    [tex] log(1000) =log2*\frac{t}{\tau}[/tex]

    So u get t={log1000/log2}*tau =9.96 [tex]\tau[/tex]
     
    Last edited: Jan 27, 2004
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