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 $T\downarrow 0$, $n_i\!\downarrow 0 ⇒ \sigma=q(n\mu_n+p\mu_p)\!\downarrow 0$. 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.