Derive an expression for mean free path from survival eqn exp(-x/λ)

[P4Penguin]

Homework Statement

The probability of gas molecules surviving a distance x without collision is given by exp(-x/λ), use this expression to prove λ = 1/nπσ² where σ = mean diameter and n = number density.

Relevant Equations

f(x) = exp(-x/λ)

If the distance between the centres of two molecules is σ, then imagining a a cylinder with radius σ the number of molecules can be given by πσ²cn where c = average velocity.
So mean free path can be given by λ = c/πσ²cn = 1/nπσ². But do I derive it from exp(-x/λ)?

 

[Charles Link]

I'll give you a hint to get started: The probability of no collision in distance ## \Delta x ## is ##1-n (\pi \sigma^2) \Delta x ##. What is the probability it survives a finite distance ## x ## with no collision? The rest is just a little calculus with the exponential ## e ##.

Edit: Note: ## n (\pi \sigma^2) \Delta x ## is the probability of a collision in a very small distance ##\Delta x ##. Observe that ## (\pi \sigma^2) \Delta x ## is the volume ## \Delta V ## that gets spanned in a distance ## \Delta x ##, and for small volumes, the probability of another particle in that volume will be ## n \Delta V ##.

 

[Charles Link]

and a follow-on: Once you show that ## P(X>x)=e^{-x/\lambda} ##, then (what is called a probability distribution function) ##F(x)=P(X \leq x) =1-e^{-x/\lambda} ##. You then take a derivative to get the probability density function ## f(x) =F'(x) ##, and it is routine from there to calculate the mean free path.

Note: It is a little unclear from the statement of the problem exactly what they are looking for, but with what I gave you, you should be able to tie it all together. The problem is somewhat incomplete if all they are wanting is what I gave you here in post 3...

I suggest working it starting with post 2 above=that seems to be the more logical way of proceeding, rather than beginning with ## P(X>x)=e^{-x/\lambda} ##. With the hint of post 2, you can show that ## P(X>x)=e^{-x/\lambda} ##, etc. (Note: ## X ## is the random variable for where the first collision occurs).

 

[Charles Link]

We=the Homework Helpers aren't supposed to furnish answers, but we could use some feedback from the OP @P4Penguin here: Were you able to show from post 2 that ## P(X>x)=e^{-x/\lambda} ##?

If you followed the hint, you then needed to use a little mathematics of the exponential function ## e^x=\lim_{N \, \rightarrow \, \infty} (1+\frac{x}{N})^N ##. This problem is a good mathematical exercise, but it might be somewhat advanced for a Biology or Chemistry student.

 

[P4Penguin]

I was able to get the expression of probability P = ##e^{-x/\lambda}##. I followed the approach that the probability of no collision over ##f(x+dx) = f(x)(1-\frac{dx}{\lambda})## then expanded LHS using Taylor series. Then by taking the first two terms and neglecting the higher order terms, I get a differential equation, upon solving which ##e^{-x/\lambda}## is obtained. But I didn't understand this part: "You then take a derivative to get the probability density function f(x)=F′(x), and it is routine from there to calculate the mean free path."

 

[Charles Link]

"You then take a derivative to get the probability density function f(x)=F′(x), and it is routine from there to calculate the mean free path."

That is something they teach in a first year class in probability theory: ##F(x)= P(X \leq x)=1-e^{-x/\lambda} ##. Then ## F(x+dx)=P(X \leq x+dx) ##, so that ## P(x<X \leq x+dx)=F(x+dx)-F(x)=F'(x) \, dx =f(x) \, dx ##.
Then the mean free path ## \bar{X}=\int x \, f(x) \, dx ##.
The probability density function ## f(x) ## and the mean free path ## \bar{X} ## are readily computed.

also, from post 2 (the hint), the probability ## P(X>x)=(1-n (\pi \sigma^2) \Delta x)^{x/\Delta x} ##.
Here you could use the mathematics of ## e^x ## from the second paragraph of post 4 to get
## P(X>x)=e^{-x/\lambda } ##, with ## \lambda=\frac{1}{n (\pi \sigma^2)} ##.

Your way of solving it was rather clever, but you might find it good reading to see how the problem is more often approached, as I have done above. They normally don't give you ## P(X>x)=e^{-x/\lambda} ## as the starting point=that is something that we are able to derive.