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I'm reading semiconductor physics by Neamen and when the disscusion about the diffusion current density he wrote this (I'll cite the image from the book because it has an important graph that will help in clarifying the question)

enter image description here

My question is: How did he say that half of electrons at $x=-l$ will pass the the area at $x=0$? if the diffusion goes from high concentration to low concentration ,the how electrons from $x=-l$ will pass at $x=0$ and it's obviously that $n(x=0)>n(x=-l)$?

correct me if I said anything wrong

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  • $\begingroup$ Diffusion going from high concentration to low concentration is the result of a large number of random walks throughout the material. If each step is random, half will be in one direction and half in the other (for a 1-D problem). $\endgroup$
    – Jon Custer
    Commented Mar 28, 2023 at 20:49
  • $\begingroup$ @Joncuster I understand this but how the lower in concentration will go to the higher ? $\endgroup$
    – Abdelrahman_200
    Commented Mar 28, 2023 at 20:52
  • $\begingroup$ In a random walk, atoms in the lower concentration region go both left and right. The net movement is from high to low concentration, but that does not imply that no atoms move from low to high. $\endgroup$
    – Jon Custer
    Commented Mar 28, 2023 at 20:55
  • $\begingroup$ Diffusion current is a result of the fact that there are more carriers randomly making their way from $x = \ell$ toward $x = 0$ than from $x = -\ell$ toward $x = 0$, just by virtue of there being more carriers at $x = \ell$ than at $x = -\ell$. It's not because the random motion of the carriers at $x = \ell$ is preferentially toward the left, or vice versa. $\endgroup$
    – Puk
    Commented Mar 28, 2023 at 22:11

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