• SReinhardt

#### 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)

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

$$T=\frac{ln2}{\lambda}$$

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)

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)

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

the differential equation is:

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

Solve it.

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.

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:
$$\lambda = \frac{\ln 2}{T_{1/2}}$$

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 $$\lambda$$

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.

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