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In my textbook, it is given that drift velocity $$v=(eE/m)t$$ where $t$ is the relaxation time. But this is the maximum velocity an electron achieves before collision and hence the average velocity should be $$(1/2)*(eE/m)t.$$ Can someone tell me how my logic is incorrect?

Qmechanic
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D. Drake
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    the relaxation time is the average time between two collisions, not the exact time between them. So the v given by your textbook is the average electron velocity not the maximum electron velocity as you have stated. – Zarathustra May 21 '18 at 20:28
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    Duplicate? https://physics.stackexchange.com/q/323959/ – Farcher May 21 '18 at 21:33

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There's a distribution of drift times. Longer drift times have a larger effect on drift speed, and some of them are really long.

More specifically, the drift time distribution is exponential:

$P(t) = \frac{1}{\tau} e^{-t/\tau}$

Then the average speed is:

$ <v> = \int_0^\infty a t P(t) dt = \int_0^\infty a t \frac{1}{\tau} e^{-t/\tau} dt$

$ = a \tau [-(t/\tau + 1) e^{-t/\tau} \rvert_0^{\infty} $

$ = a \tau $

There's no factor of two because, averaging over the long tail of the exponential, the longer times outweigh the shorter ones.

Bob Jacobsen
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