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.