CBSE Class 12 · 2007 · Current ElectricityShort Answer

$\rho =\frac{m}{n{e}^2\tau }$. In a \textbf{metal} the carrier density $n$ is essentially constant; a rise in $T$ increases lattice vibrations and collision frequency, lowering $\tau$, so $\rho \propto 1/\tau$ increases. In a \textbf{semiconductor} a rise in $T$ frees many more carriers (breaking covalent bonds), so $n$ increases strongly and $\rho \propto 1/n$ decreases — the fall due to $n$ dominates the small change in $\tau$.