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Radioactive decay formula 
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#1
Jun508, 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) 


#2
Jun508, 07:26 PM

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P: 6,202

Well since
[tex]N=N_0 e^{ \lambda t}[/tex] when t=halflife(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
Jun508, 07:48 PM

P: 5




#4
Jun508, 09:40 PM

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P: 16,458

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)



#5
Jun508, 09:51 PM

P: 5

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


#6
Jun608, 02:51 AM

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the differential equation is:
[tex] \frac{dN}{dt} = \lambda N [/tex] Solve it. 


#7
Jun608, 07:36 AM

P: 5

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


#8
Jun608, 07:58 AM

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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
Jun608, 08:07 AM

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#10
Jun608, 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. 


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