The Exponential decay.

Question

I am studying semiconductor physics. there is a paragraph about Drude model in E.spenke's book "Electronic semiconductors" page 259 in art §9:
"if on the average, a time $τ$ elapses between two collisions of an electron, $dt/τ$ terminal points of such times fall on average into the time interval $dt$. thus an electron suffers in the interval $dt$ on the average $dt/τ$ collisions, and $N$ electrons suffer $Ndt/τ$ collisions. thus collisions eliminate in an interval $dt$ on the average
$$dN=N.dt/τ$$ electrons from a group with uniform velocity and direction hence decrease exponentialy with time: $$N=N(0).e^{−t/τ}$$ ..."

I think $dt/τ$ is the probability of an electron to collide. normally in exponential decays mean free time is found in a different manner usualy it is assumed that decay rate is propotional to the concentration i.e $dN/dt∝N$ hence $$dN/dt=−kN⟹τ=\frac{∫_0^∞t (kN dt )}{N(0)}⟹τ=1/k$$ but in above paragraph it is assumed that if $τ$ is mean time then it would directly imply an exponential profile.

(Mathematicaly $$τ=\frac{τ_1+τ_2+τ_3+...τ_n}{n}$$ here $n$ is a dummy variable and has no connection with the problem mentioned. $τ_n$ represents time taken by an electron in making $n$th collision with lattice atom)

Question: Can we show mathematically from the definition of $τ$ that in time interval $dt$ probability of electron to collide is $dt/τ$? And hence the Exponential decay law.

Answer 1

I think it might help to think of the problem in an 'inverse' manner. Let $1/\tau$ be the probability that an electron suffers a collision in unit time (as a density). For now, assume $\tau$ is just some positive constant. So if you assume a uniform distribution of collision probabilities in a small time interval $\Delta t$, the probability that there is a collision in this time is $\Delta t/\tau$, for any single electron.

Now take a finite time interval $t$ and split it into $n$ parts and take $\Delta t = t/n$. The probability that a single electron had no collision in this time $\Delta t$ is $(1 - \Delta t/\tau)$. So if you go $n$ such intervals, the probability that there have been no collisions is $\left(1 - \frac{\Delta t}{\tau}\right)^n$. Using $t$ for $\Delta t$ and taking $n$ to be very large, in the limit

$$ \lim_{n\to \infty} \left(1 - \frac{t}{\tau n}\right)^n = \exp(-t/\tau) $$

So the probability that in a finite time $t$, a single electron has not suffered any collisions is $\exp(-t/\tau)$. Similarly, this is also the probability that there is no collision in the next $t$ seconds. In this, the uniformity of the probability is strictly true only if $\Delta t$ is small, so taking the limit only reaffirms this idea.

As for what $\tau$ means, consider the expectation value of the time between collisions. Firstly, the probability that no collisions have occurred at time $t$, starting from $t = 0$ is $\exp(-t/\tau)$. The expected time between collisions is hence given by: $$ \langle t_{coll}\rangle = \frac{\int_0^\infty t \exp(-t/\tau) \, dt}{\int_0^\infty \exp(-t/\tau) \, dt} = \tau $$

Now we can relate $\tau$ (which we initially just took to be a constant) to the expected or mean time between collisions. From this fact, I guess the exponential decay for $N$ in your question should follow.