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

In summary, the conversation discusses the calculation of the mean free path, which is a measure of the average distance a molecule can travel without colliding with another molecule. The probability of no collision in a given distance is calculated using the formula 1-n(πσ²)Δx, and taking the limit as the distance approaches infinity yields the exponential function e^-x/λ. The probability density function and mean free path can then be calculated using this function.
  • #1
P4Penguin
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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)?
 
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  • #2
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 ##.
 
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  • #3
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).
 
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  • #4
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.
 
  • #5
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."
 
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  • #6
P4Penguin said:
"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.
 
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1. What is the mean free path?

The mean free path is the average distance traveled by a particle before it collides with another particle or object.

2. How is the mean free path related to the survival equation exp(-x/λ)?

The mean free path can be calculated from the survival equation exp(-x/λ) by taking the inverse of the coefficient λ, which represents the probability of a particle surviving without a collision.

3. What is the significance of the survival equation in determining the mean free path?

The survival equation is used to calculate the probability of a particle surviving without a collision, which is essential in determining the mean free path. It also takes into account the size and number of particles present in a given space.

4. How does the mean free path affect the behavior of particles in a system?

The mean free path determines the average distance between collisions for particles, which can affect their overall behavior and movement within a system. It is also used in various calculations and models to understand the dynamics of a system.

5. Can the mean free path be altered or controlled?

The mean free path can be altered or controlled by changing the properties of the system, such as the size or number of particles present, or by altering the temperature or pressure. It can also be influenced by external factors such as electric or magnetic fields.

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