Consider a group of $N$ electrons, all of which experience a collision with lattice ions at $t=0 .$ One can show that the number of electrons that suffer their next collision between $t$ and $t+d t$ follows the exponentially decreasing distribution $$n(t) d t=\frac{N e^{-t / \tau}}{\tau} d t$$ where $\tau$ is the mean free time. (a) Show that $$\int_{0}^{\infty} n(t) d t=N$$ as expected. (b) Show that this distribution leads to $\bar{t}=\tau$. Hint: $$\bar{t}=\frac{n_{1} t_{1}+n_{2} t_{2}+\cdot \cdot+n_{f} t_{f}}{N}=\frac{1}{N} \int_{0}^{\infty} t \cdot n(t) d t$$ (c) Show, similarly, that $t^{2}=2 \tau^{2},$ as stated in the derivation of Equation 12.9.