Why is e and ln(2) used in the radioactive decay formula?

[SReinhardt]

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)

 

[rock.freak667]

Well since

N=N_0 e^{- \lambda t}

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

\frac{N_0}{2}=N_0 e^{- \lambda T}

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

T=\frac{ln2}{\lambda}

 

[SReinhardt]

Well since

N=N_0 e^{- \lambda t}

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

\frac{N_0}{2}=N_0 e^{- \lambda T}

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

T=\frac{ln2}{\lambda}

I'm curious on why they chose to use N=N_0 e^{- \lambda t} instead of N = N_O .5^{\frac{t}{half-life}} 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]

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]

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]

the differential equation is:

\frac{dN}{dt} = \lambda N

Solve it.

 

[SReinhardt]

the differential equation is:

\frac{dN}{dt} = \lambda N

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]

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:

\int N ^{-1}dN = - \int \lambda dt

\ln(N(t)) - \ln(N(0)) = -\lambda t

\ln(N(t)/N(0)) = -\lambda t

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

N(t) = N(0) e^{-\lambda t }

Lambda is the number of decays per unit time, is related to half life by:
\lambda = \frac{\ln 2}{T_{1/2}}

 

[SReinhardt]

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

Solving this:

\int N ^{-1}dN = - \int \lambda dt

\ln(N(t)) - \ln(N(0)) = -\lambda t

\ln(N(t)/N(0)) = -\lambda t

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

N(t) = N(0) e^{-\lambda t }

Lambda is the number of decays per unit time, is related to half life by:
\lambda = \frac{\ln 2}{T_{1/2}}

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 \lambda

 

[malawi_glenn]

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]

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