Probability that a molecule will travel a distance at least equal to the mean free pa

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The discussion focuses on calculating the probability that a molecule will travel a distance equal to or greater than the mean free path before colliding. The exponential probability distribution function f(r) represents the likelihood of a molecule traveling a distance r before a collision occurs. Participants confirm that the probability can be derived from the ratio of molecules traveling greater than the mean free path to the total number of molecules. Normalization of the function f(r) is emphasized to ensure that the total probability integrates to one. The calculations for both parts of the homework are acknowledged as correct, with clarification on the meaning of f(r).
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Homework Statement



a) Find the probability that a molecule will travel a distance at least equal to the mean free path before its next collision.
b)After what distance of travel since the last collision is the probability of having suffered the next collision equal to 1/2?

Homework Equations


Exponential probability distribution
f(r) = Ae-r/\lambda

where A = a constant, \lambda = mean free path


The Attempt at a Solution



P = Integral (limits \lambda to \infty)f(r) dr / Integral (limits 0 to \infty) f(r) dr
I will do the calculations of both parts but not sure about the probability. I have found probability by number of molecules with distance between collision greater than lambda / total number of molecules. Is this probability correct?
 
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What exactly is f(r) supposed to represent? You say it's a probability distribution but of what exactly?
 


vela said:
What exactly is f(r) supposed to represent? You say it's a probability distribution but of what exactly?

f(r) is a continuous function that gives the probability for a molecule to travel a distance r before having a collision.
 


OK, your calculation is correct. Typically, though, you normalize f(r). In other words, you set the constant A, which is a normalization constant, so that

\int_0^\infty f(r)\,dr = 1

You're effectively doing this in your calculation by dividing by the integral from 0 to infinity.
 


@ vela
thank you.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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