# Problem 2.25

This is part of the HW set. The help you need has been given in the previous problem. You basically need to use the gaussian distribution. For part B the answer is very similar to part d of the the previous problem. Now x in the gaussian represents the excess of forward(or backward) steps over N/2. So in terms of x and l (l is the length of a step), the net distance traveled should be 2xl, (the excess of forward steps of N). In this case, the gaussian distribution is $~e^{\frac{-x^2}{2l^2N}}$ . The width is $\sqrt{2N}l$ , that is $\sqrt{2N}$ steps of length l. Now you can answer question 2.25 b.

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 $3\times10^{-10}$ s. In 1 second we have that N~ $3\times10^{9}$ steps, and the net distance traveled is $\sqrt{2N}$ =80000 steps or 12 nm.

From here we can see that N increases with time linearly by the distance traveled goes as $\sqrt{2N}$ , so goes as $\sqrt{t}$ . 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 $\sqrt{T}$ . so the collision time increases only in proportion to $1/\sqrt{T}$ . The expected distance traveled is proportional to $\sqrt{N}$ times the mean free path, which makes the expected distance traveled to increase as $T^{3/4}$ . This is the correct result.