Radioactive decay formula

  1. What lead them to use e and the natural log of 2 in the decay formula? A much simpler (to me at least) method would is:

    N=No*.5^(time/half life)
  2. jcsd
  3. rock.freak667

    rock.freak667 6,231
    Homework Helper

    Well since

    [tex]N=N_0 e^{- \lambda t}[/tex]

    when t=half-life(T); N=[itex]\frac{N_0}{2}[/itex]

    [tex]\frac{N_0}{2}=N_0 e^{- \lambda T}[/tex]

    simplify that by canceling the N_0 and then take logs and you'll eventually get

  4. I'm curious on why they chose to use [tex]N=N_0 e^{- \lambda t}[/tex] instead of [tex]N = N_O .5^{\frac{t}{half-life}}[/tex] The 2nd one is one that I figured out, and it makes more sense to me; it is based off the idea of half-lives. (I'm not saying it's original or hasn't been done before, just was never shown to me)
  5. Vanadium 50

    Vanadium 50 18,489
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    It's the same question as "why log base e and not log base 2"? It happens to make some calculations easier. (Note that your calculator has a ln(x) button but probably not a log2(x) button)
  6. I've never had any issues using it...I can see where you're coming from though. Your point is for when you're solving for the time or half-life. But then all you have to do is take the log10 and divide.

    It could also be I use it just to make my teacher grade things two ways xD
  7. malawi_glenn

    malawi_glenn 4,725
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    the differential equation is:

    [tex] \frac{dN}{dt} = \lambda N [/tex]

    Solve it.
  8. Wouldn't there have to be a negative sign in there somewhere >_>

    I believe you have to use integrals to solve that, which I haven't done yet.
  9. malawi_glenn

    malawi_glenn 4,725
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    yeah it should have a minus sign, good! :-)

    Solving this:

    [tex] \int N ^{-1}dN = - \int \lambda dt [/tex]

    [tex] \ln(N(t)) - \ln(N(0)) = -\lambda t [/tex]

    [tex] \ln(N(t)/N(0)) = -\lambda t [/tex]

    [tex] N(t)/N(0) = e^{-\lambda t } [/tex]

    [tex] N(t) = N(0) e^{-\lambda t } [/tex]

    Lambda is the number of decays per unit time, is related to half life by:
    [tex] \lambda = \frac{\ln 2}{T_{1/2}} [/tex]
  10. So if you used based base .5 instead of base e, you'd get what I worked out on my own. The main thing that would change then would be the [tex]\lambda[/tex]
  11. malawi_glenn

    malawi_glenn 4,725
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    it is easier working with base e when solving the differential eq.

    Then if you think it is easier to work in basis 0.5 when you calculate, then it is up to you.
  12. malawi_glenn

    malawi_glenn 4,725
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    and practically, it is easier to measure the decay constant lambda then the half life.
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