Distance a density will propagate in time

[Phyisab****]

Homework Statement

I just got a correct answer to a problem using the equation for the distance a density will propagate in time t below. Does anyone know where this comes from? I've already solved the homework problem, I just am not comfortable without understanding where this came from.

Homework Equations

x = 2 \sqrt{D t}

Where x is the distance the density has propagated.

The Attempt at a Solution

So the diffusion equation can be solved for a 1-d situation giving (according to wikipedia I admit)

n=n_{0}erfc(\frac{x}{2\sqrt{D t}})

Where erfc is the complementary error function, not sure how to make that look better in tex. Essentially to come up with the equation for diffusion length, the argument of erfc must be equal to 1. But I can't find a way to convince myself that is the case. Can anyone shed some light on this? Or maybe I'm way off, any help would be appreciated.

 

[Matterwave]

For all kinds of things like "diffusion length" it just means a defined length or some other parameter such that something happens.

A good analogy would be like Half-life. It is the time spent such that half the original radioactive material is left. In this way, the diffusion length is something like the length such that X% of the material has diffused through.

It's just a definition thing, it's not anything absolute.

 

[Phyisab****]

Hmmm. Well the problem is: A small amount of N15 gas is introduced into a container of N14 gas. The container is spherical, 1m in diameter, at room temp and pressure. Make a rough estimate of how long it will be until you can be pretty sure the molecules are uniformly mixed?

How would you go about this problem? I used the equation above, but I can't justify it I just found it on wikipedia. This is a problem from Reif and he makes no mention of that equation. He doesn't really address how to do any calculations, all he does is derive the diffusion coefficient by two methods, and never says anything about how to use it. I guess that means I should be able to figure that out myself.