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## Main Question or Discussion Point

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)

N=No*.5^(time/half life)

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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)

N=No*.5^(time/half life)

- #2

rock.freak667

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[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

[tex]T=\frac{ln2}{\lambda}[/tex]

- #3

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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)

[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

[tex]T=\frac{ln2}{\lambda}[/tex]

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Vanadium 50

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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

- #6

malawi_glenn

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the differential equation is:

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

Solve it.

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

Solve it.

- #7

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Wouldn't there have to be a negative sign in there somewhere >_>the differential equation is:

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

Solve it.

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

- #8

malawi_glenn

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yeah it should have a minus sign, good! :-)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.

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]

- #9

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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]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

malawi_glenn

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

- #11

malawi_glenn

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

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