Mean Free Path — Question 3

Section 19-5 · Physics pages 3116–3117

Question 3 An exponential free-path model predicts survival and quantiles In a supplied model, the probability of traveling at least distance L without collision is S(L) = e−L/λ, with λ = 80.0 nm. Flights are statistically independent and the medium is uniform. L S(L) Axes and labels specify the model; the sketch is schematic. (a) Find the probability of no collision over 200 nm and the distance exceeded by half of all free flights. (b) Find the distance over which 95.0% of trajectories have suffered at least one collision. Is the mean free path a maximum possible flight?
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Question 3 - Solution An exponential free-path model predicts survival and quantiles L S(L) (a) Method and calculation S(200 nm) = e−200/80 = e−2.5 ≈.0820850. S(L50) = .500 ⇒L50 = λ ln 2 ≈55.4518 nm. (b) Interpretation and further calculation 1 −S(L95) = .950 ⇒L95 = −λ ln(.0500) = 80 ln 20 ≈239.659 nm. S(λ) = e−1 ≈.367879. Many flights exceed the mean, and the ideal exponential distribution has no finite upper cutoff. Physical check The median is smaller than the mean because the distribution has a long positive tail.

Question 3

An exponential free-path model predicts survival and quantiles

In a supplied model, the probability of traveling at least distance LL without collision is S(L)=e−L/λS(L)=e^{-L/\lambda}, with λ=80.0\lambda=80.0 nm. Flights are statistically independent and the medium is uniform.

Diagram shown in the original SVG worksheet.

  1. Find the probability of no collision over 200 nm and the distance exceeded by half of all free flights.

  2. Find the distance over which 95.0% of trajectories have suffered at least one collision. Is the mean free path a maximum possible flight?

Question 3 - Solution

An exponential free-path model predicts survival and quantiles

Diagram shown in the original SVG worksheet.

(a) Method and calculation

S(200nm)=e−200/80=e−2.5≈.0820850.S(200\,\mathrm{nm})=e^{-200/80}=e^{-2.5}\approx.0820850.S(L50)=.500⇒L50=λln⁡2≈55.4518nm.S(L_{50})=.500\ \Rightarrow\ L_{50}=\lambda\ln2\approx55.4518\,\mathrm{nm}.

(b) Interpretation and further calculation

1−S(L95)=.950⇒L95=−λln⁡(.0500)=80ln⁡20≈239.659nm.1-S(L_{95})=.950\ \Rightarrow\ L_{95}=-\lambda\ln(.0500)=80\ln20\approx239.659\,\mathrm{nm}.S(λ)=e−1≈.367879.S(\lambda)=e^{-1}\approx.367879.Many flights exceed the mean, and the ideal exponential distribution has no finite upper cutoff.

Physical check

The median is smaller than the mean because the distribution has a long positive tail.