Consider the collisions between the molecules of a gas. The free-path length of

Question

Consider the collisions between the molecules of a gas. The free-path length of a molecule is the distance that it travels between successive collisions. It is shown in the kinetic theory of gases that the probability for the free-path length to have a value between x and x + dx is directly proportional to e^-x/lambda dx, where lambda is a positive constant. Show that the average free-path length (called the mean free path) is lambda. Find the probability of a free-path length being greater than or equal to 2 lambda. Quantum Mechanics

Explanation / Answer

given that probability of mean free path to have a value between x and x + dx is proportional to e^(-x)dx, where lambda is the constant
so, let probability of the mean free path to be between x and dx be p
then p = lambda*e^(-x)/dx
so average mean free path = [integration]{(p*dx)dx}/[integrate]{p} = [integrate]{lambda*e^(-x)dx}
average mean free path = [integrate]{lambda*e^(-x)dx} = lambda[integrate]{e^(-x)dx}
integrating from x = 0 to x = infinity
average mean free path = lambda*[e^(-0) - e^(-infinity)] = lambda

probability of mean free path to be greater than or equal to 2*lambda is
p = [integrate]((p*dx)dx)/[integrate](dx) = [INTEGRATE] [lambda*e^(-X)dx]/integrate (dx)
integrating from x = 0 to x = 2*lambda, and then finding 1 - p
1 - p = 1 - lambda*(e^(-0) - e^(-2*lambda))/2*lambda = 1 - (1 + e^(-2lambda))/2 = 1 - e^(-2*lambda)

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