Why is diffusion coefficient = 1/2?

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Homework Statement


I'm working with a 0ne-dimensional random walk and looking at Fick's second law, below. All I've read seems to take D = 1/2 as a given for ordinary diffusion, but where does this come from? Is there a way to derive it?


Homework Equations


<br /> \frac{\partial }<br /> {{\partial t}}p\left( {x,t} \right) = D\frac{{\partial ^2 }}<br /> {{\partial x^2 }}p\left( {x,t} \right)<br />


The Attempt at a Solution

 
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Essentially D defines your units, so you are allowed to choose any numerical value. I guess people choose 1/2 because then the "usual" solution is a Gaussian with variance t (instead of √(2Dt)).
 
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Thank you, but can you expand on your answer a bit? Not sure I understand what you mean in your second sentence.
 
You know how to solve diffusion equation with Fourier transforms, right? The result is
p = \frac{1}{\sqrt{4\pi Dt}} \exp(-x^2/4Dt)
so if you choose D = 1/2, this becomes a normal distribution with mean 0 and variance t.
 
Of course, now I get it! Thank you.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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