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Marivi Fernandez-Serra

]]>Question 2.25 c is difficult and I'm giving here the answer:

In page 42, the book shows that the mean free path l in air is ~ 150 nm, and the average collision time is s. In 1 second we have that N~ steps, and the net distance traveled is =80000 steps or 12 nm.

From here we can see that N increases with time linearly by the distance traveled goes as , so goes as . At higher T the mean free path, which is proportional to V/N (we saw this in class) would increase in proportion to T. But the velocity is also larger, in proportion to . so the collision time increases only in proportion to . The expected distance traveled is proportional to times the mean free path, which makes the expected distance traveled to increase as . This is the correct result.

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