Does the number of electrons colliding in wire get double as the length of wire gets doubled? Does the number of electrons colliding in a wire get doubled when length of wire is made twice with its area of cross-section remain constant.
My calculations for this are:-
For wire of length $L$ and area of cross-section:
Let the average distance after travelling which electron will collide $=d$
The number of free electrons in wire $=n $
Time taken between two successive collisions of electron $=τ$
Electric field across it $=E_1$
And Voltage apllied across wire $=V$
Average velocity of electron $= E_1eτ/2m$ The number of collision of electron in time $t =  E_1eτtn/2md$
When using a wire of length 2L but with same area of cross-section and same voltage apllied across it:
Number of free electrons $ =2n$
Electric field inside wire $= E_2=V/2L=E_1/2$
Average velocity of electrons $= E_2eτ/2m=E_1eτ/2×2m$
And at last, the number of collisions $= E_2eτt×2n/2md=E_1eτtn/2md$
We can see here that number of collision remains same when length of wire is changed.
*Edit: There are lot of errors above. So I am writing correct version down( correct in my sense)
For wire of length $L$:
The number of free electrons in wire $=n$
Time taken between two successive collisions of electron (it is constant for a material for reason I don't know)$=τ$
Electric field across it $=E_1$
And Voltage apllied across wire $=V$
The number of collision of electron in time $t =  tn/ \tau $
When using a wire of length 2L and area of cross section A but with same area of cross-section and same voltage apllied across it:
Number of free electrons $ =2n $
Average velocity of electrons $= t2n/τ$
Finally, why is the $\tau$ uneffected by Electric field, $E = V/L$? ($L$is the length of wire).
 A: I think that the main observation is that in a typical metal the velocity of the conduction electrons (the Fermi velocity) is much, much bigger than the drift velocity. The Fermi velocity is on the order of $10^6$ m/s. The drift velocity depends on many details, like the mobility and the actual field, but a reasonable number is $\sim 1$ m/s. This means that the mean free time and the mean free path are completely unaffected by the field, the size of the wire, etc. The same goes for the number of collisions per unit time and conduction electron. This means that if you double the sample, you get twice as many collisions.
Postcript: Consider the case without a field. There is a distribution function $f(\vec{v},\vec{x},t)$ for the electrons in the metal. The distribution function is isotropic, and the mean velocity and the current are zero. In a typical metal the distribution is close a sharp Fermi distribution, so that states below the Fermi velocity are occupied, and states above the Fermi velocity are empty. The Fermi edge is smeared out because of thermal effects. The mean (absolute) velocity of electrons that scatter is close to the Fermi velocity $v_F$. In a typical metal the Fermi velocity is large, on the order of $10^6$ m/s.
Now we turn on a field $E$. This shifts the equilibrium distribution $f\to f+\delta f$. The distribution $\delta f$ is anisotropic. In particular, the mean velocity is $\vec{v}_d$, called the drift velocity. The drift velocity is proportional to the electric field, $\vec{v}_d=\gamma/e \vec{E}$ ($\gamma$ is called the mobility). The mean current is also not zero, $\vec{\jmath}=\sigma \vec{E}$ ($\sigma$ is the conductivity).
The point is that $\delta f\ll f$, and $|\vec{v}_d|\ll v_F$, despite the fact that the total current is governed by $\vec{v}_d$. As a consequence the collision rate is determined by $f$, and is approximately idependent of $E$.
