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Radioactive decay formula

by SReinhardt
Tags: decay, formula, radioactive
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SReinhardt
#1
Jun5-08, 03:49 PM
P: 5
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)
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rock.freak667
#2
Jun5-08, 07:26 PM
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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

[tex]T=\frac{ln2}{\lambda}[/tex]
SReinhardt
#3
Jun5-08, 07:48 PM
P: 5
Quote Quote by rock.freak667 View Post
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

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

Vanadium 50
#4
Jun5-08, 09:40 PM
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Radioactive decay formula

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)
SReinhardt
#5
Jun5-08, 09:51 PM
P: 5
Quote Quote by Vanadium 50 View Post
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)
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
malawi_glenn
#6
Jun6-08, 02:51 AM
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the differential equation is:

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

Solve it.
SReinhardt
#7
Jun6-08, 07:36 AM
P: 5
Quote Quote by malawi_glenn View Post
the differential equation is:

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

Solve it.
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.
malawi_glenn
#8
Jun6-08, 07:58 AM
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Quote Quote by SReinhardt View Post
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.
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]
SReinhardt
#9
Jun6-08, 08:07 AM
P: 5
Quote Quote by malawi_glenn View Post
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]
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]
malawi_glenn
#10
Jun6-08, 08:11 AM
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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.
malawi_glenn
#11
Jun6-08, 03:12 PM
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and practically, it is easier to measure the decay constant lambda then the half life.


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