The mathematics of the previous problem can also be applied to a one-dimensional random walk: a journey consisting of $N$steps, all the same sic, cache chosen randomly to be cither forward or backward. (The usual mental image is that of a drunk stumbling along an alley.)

(a) Where are you most likely to find yourself, after the end of a long random walk?

(b) Suppose you take a random walk of $10,000$steps (say each a yard long). About how far from your starting point would you expect to be at the end?

(c) A good example of a random walk in nature is the diffusion of a molecule through a gas; the average step length is then the mean free path, as computed in Section $1.7.$Using this model, and neglecting any small numerical factors that might arise from the varying step size and the multidimensional nature of the path, estimate the expected net displacement of an air molecule (or perhaps a carbon monoxide molecule traveling through air) in one second, at room temperature and atmospheric pressure. Discuss how your estimate would differ if the clasped time or the temperature were different. Check that your estimate is consistent with the treatment of diffusion in Section$1.7.$

(a) This method is technically analogous to the coin turning operation in difficulty $2.24$, with a step here to right yielding in a very head or a move to the left yielding in an exceedingly tail. As a corollary, the range of possible step patterns in an exceedingly chaotic system of $N$steps is $2N$, and also the risk of coming away n steps first from start is

$\begin{array}{c}P=\frac{\mathrm{\Omega }(N,n)}{{\mathrm{\Omega }}_{max}}\\ \end{array}$

$\mathrm{\Omega }(N,n)=\left(\begin{array}{c}n+N-1\\ n\end{array}\right){\mathrm{\Omega }}_{max}=2^N$

$\mathrm{\Omega }(N,n)\approx \left(\begin{array}{c}n+N\\ n\end{array}\right)=\frac{N!}{n!(N-n)!}$

$P=\frac{1}{2^N}\frac{N!}{n!(N-n)!}$

$\mathrm{\Omega }\approx 2^N{e}^{-\frac{2x^2}{N}}$

$x_{\text{width}}=\sqrt{\frac{N}{2}}$

$\mathrm{\Omega }(N,n)=\left(\begin{array}{c}n+N-1\\ n\end{array}\right){\mathrm{\Omega }}_{max}=2^N$

(b) for less than a variate of $N=10000$steps, we are able to find yourself later territory:

$\begin{array}{r}-\sqrt{\frac{N}{2}}\le x\le +\sqrt{\frac{N}{2}}\end{array}$

$\begin{array}{r}-\sqrt{\frac{10000}{2}}\le x\le +\sqrt{\frac{10000}{2}}\\ \end{array}$

$-70\le x\le +70$

(c) The diffusion rate in a perfect gas is one instance of a stochastic process . We calculate the mean free path of a gas increases of radius $r$ and used the identical model as in section $1.7$:

$\ell \approx \frac{1}{4\pi r^2}\frac{V}{V_{mo}^{\gamma }}$

where could we be? The frequency of air molecules during a volume $V$is denoted by $N_{\text{mol}}$. The underlying speed determined through kinetic law is employed to live the common movement of either a particle, which is:

$\overline{v}=\sqrt{\frac{3KT}{m}}$

$\begin{array}{c}\ell =1.5\times {10}^{-7}\mathrm{m}\\ \end{array}$

$\overline{v}=500\mathrm{m}\cdot 8^{-1}$

We must always count the number of steps at a time $t$because the number of steps in in an exceedingly chaotic system to depict diffusion as a random process.

$N=\frac{\overline{v}t}{\ell }$

$d\approx x_{\text{width}}\ell =\sqrt{\frac{N}{2}}\ell$

$d\approx \sqrt{\frac{\overline{v}t}{2\ell }}\ell =\sqrt{\frac{1}{2}\overline{v}t\ell }$

$\begin{array}{c}d\approx \sqrt{\frac{1}{2}\times 500\times \left(1.5\times {10}^{-7}\right)}=0.0061\mathrm{m}\\ \end{array}$

$d\approx 6.1\mathrm{mm}$

$\begin{array}{c}D\approx \frac{1}{2}\ell \overline{v}\\ \end{array}$

$(\mathrm{\Delta }x{)}^2\approx D\mathrm{\Delta }t\to \mathrm{\Delta }x\approx \sqrt{\frac{1}{2}\overline{v}\mathrm{\Delta }t\ell }$

it is also merit note that when a gas molecule diffuses $6mm$in $1$second doesn't indicate it's doing so at a continuing speed of $6mm$ per second. Since this distance diffused is adequate the inverse of your time, the diffusion speed slows over time. A substance takes roughly $4$months to permeate the length of a $10mm$room, which is noticeably slower than $6mm$per second, as seen in problem $1.68$. If we warm the earth while keeping everything else constant, the speed rises from $\overline{v}=\sqrt{\frac{3KT}{m}}$, but still the phase velocity remains constant, therefore the the amount of moves during a given moment rises, implying that the molecules moves.