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Given the following function, how does one rewrite the exponential part of the equation into $e^{-L/L_{0}}$, where $L_{0}$ is the decay constant

$f(L)=16\frac{a}{b}\left(1-\frac{a}{b}\right)\exp\left(\frac{-2L}{c}\bigl(2d(b-a)\bigr)^{1/2}\right)$

I know this problem (demi-vie) from school and it is solved usually by setting $f(x)$ to $\frac{1}{2}$.

So for $f(L)=\frac{1}{2}$ this gives: $ \begin{align*} \frac{1}{32}\frac{b^{2}}{a(b-a)} &=\frac{1}{32}\frac{b}{a}\left(1-\frac{a}{b}\right)^{-1}\\ &= \exp\left(\frac{-2L_{0}}{c}\bigl(2d(b-a)\bigr)^{1/2}\right)\\ \end{align*} $

$ \Rightarrow \frac{c}{-2(2d(b-a))^{1/2}}\log\left(\frac{b^{2}}{32a(b-a)}\right) = L_{0} $

Wikipedia gives another definition of decay constant: $t_{1/2}=\frac{\log(2)}{\lambda},$ where $\lambda$ is the decay constant. So the "correct" $L_{0}$ would be:$\frac{L_{0}}{\log(2)} = \frac{c}{-2(2d(b-a))^{1/2}\log(2)}\log\left(\frac{b^{2}}{32a(b-a)}\right)= L_{0corrected} $

Now the exp part in $f(L)$ should be rewritten as $e^{-L/L_{0}}$. I don't see how to achieve that. Does somebody see how this is possible? Merci.

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    Sorry for the co$n$fusio$n$.2011-12-05

1 Answers 1

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I think the problem is that you're including the constants in front of the exponential in your calculations. Decay constants, half-lives etc. only relate to the exponential decay, not to the amplitude. So in this case you simply have

$f(L)=16\frac{a}{b}\left(1-\frac{a}{b}\right)\exp\left(-L/L_0\right)$

with the characteristic decay length

$L_0=\frac c{2\left(2d(b-a)\right)^{1/2}}\;,$

or

$f(L)=16\frac{a}{b}\left(1-\frac{a}{b}\right)\exp\left(-\lambda L\right)$

with the decay rate

$\lambda=\frac{2\left(2d(b-a)\right)^{1/2}}c\;.$