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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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You write $f(x)$, but I can't quite tell where $x$ is in your exponential expression... –  J. M. Dec 5 '11 at 1:00
    
Sorry for the confusion. –  VVV Dec 5 '11 at 1:03

1 Answer 1

up vote 1 down vote accepted

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

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

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Thank you, joriki. –  VVV Dec 5 '11 at 10:30

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