Question

Temperature of a pure semiconductor is $0\ \mathrm{K$. Comment on its conductivity.}

Hint:

Use $n_i \propto e^{-E_g/(2k_BT)}$. As $T\downarrow 0$, $n_i\!\downarrow 0 $\Rightarrow$ \sigma=q(n\mu_n+p\mu_p)\!\downarrow 0$. Only at $T>0$ do intrinsic semiconductors conduct via thermally generated $e^-$–hole pairs.

Answer 1

Step 1: Intrinsic (pure) semiconductor at $T=0\ \mathrm{K$.}
A pure/intrinsic semiconductor has a completely filled valence band (VB) and an empty conduction band (CB) separated by an energy gap $E_g$ (e.g., $\sim 1.12\ \mathrm{eV}$ for Si). At absolute zero, the Fermi–Dirac distribution becomes a step function: \[ f(E)=\begin{cases} 1, & E<E_F
0, & E>E_F \end{cases} \] For an intrinsic semiconductor, $E_F$ lies near the middle of the band gap; since $E_C>E_F$ and $E_V<E_F$, all VB states are occupied and all CB states are empty.

Step 2: Carrier concentrations vanish at $0\ \mathrm{K$.}
The intrinsic carrier concentration \[ n_i(T)=\sqrt{N_C N_V}\, \exp\!\left(-\frac{E_g}{2k_BT}\right) \] satisfies $n_i\to 0$ as $T\to 0$ (because the exponential $\to 0$).
$\Rightarrow$ Electron concentration in CB, $n\to 0$; hole concentration in VB, $p\to 0$.

Step 3: Conductivity expression.
For a semiconductor, \[ \sigma = q\,(n\mu_n + p\mu_p) \] where $q$ is the electronic charge and $\mu_n,\ \mu_p$ are mobilities. At $T=0$, $n=p=0$ (mobilities are finite but irrelevant).
$\Rightarrow$ $\sigma = q\,(0\cdot \mu_n + 0\cdot \mu_p)=0$.

Step 4: Physical interpretation.
With no thermally excited carriers, current cannot flow under an applied electric field. Hence the material behaves as an insulator at $0\ \mathrm{K}$ (formally $\rho=1/\sigma\to\infty$).

Final Answer: At $0\ \mathrm{K}$, a pure (intrinsic) semiconductor has zero conductivity and acts like a perfect insulator.