Past exam of the chemistry A course of the University of Cambridge / 2023 / Paper 2 / A10 / a

The density of states $D(E)$ counts available one-electron states: $D(E)dE$ is the number with energies in $[E,E+dE]$. It describes the spectrum, irrespective of whether those states contain electrons.

The density of occupied states additionally weights each state by the Fermi-Dirac distribution:Thus $Z(E)dE$ is the mean number of electrons in that energy interval, and $N_e=\int Z(E)dE$. State and electron densities must use the same normalization, including whether they are extensive or per volume and whether $D$ already includes spin degeneracy. The symbol $Z(E)$ here denotes occupied density, not a canonical partition function.

The finite-temperature Fermi level lies at the middle of the gap in the stipulated symmetric intrinsic model:Here $E_v$ is the valence-band maximum and $E_c$ the conduction-band minimum. Thermal promotion creates one conduction electron and one electron hole, so charge neutrality requires $n=p$. Equal constant densities of states in broad, symmetric bands give equal populations when the distances $E_c-\mu$ and $\mu -E_v$ are equal.

This is a statement about the chemical potential, not an available electron state in the gap. Unequal band densities of states would shift the intrinsic Fermi level away from exact midgap; the equality follows from the symmetry assumed here.

Heating promotes electrons across the band gap, creating mobile conduction electrons and electron holes. Their contributions add:where $u_e,u_h$ are the electron mobility and hole mobility, and $e$ is the positive elementary-charge magnitude. The increase in carrier density gives the rising electrical conductivity in the intrinsic activation regime. Mobility need not itself increase with temperature; the model attributes the dominant increase to carrier generation.

Let $D$ be the equal constant density of states per unit volume and energy in each band. In the nondegenerate semiconductor regime, the conduction-band occupation isFor bands broad compared with $k_{B}T$, integrating above the conduction edge givesThe valence-hole integral similarly gives $p=D{k}_{B}T{e}^{-(\mu -E_v)/(k_{B}T)}$. Midgap $\mu$ therefore impliesInserting the carrier densities into the transport expression yieldsOn treating the algebraic and mobility prefactor as approximately constant, the requested leading activation law isThe factor $1/2$ is essential: an intrinsic excitation creates an electron–hole pair, and charge neutrality places the chemical potential halfway across the symmetric gap. The conduction-electron population is governed by $E_c-\mu =E_g/2$, not by replacing this separation with the full gap.

There is a qualification to the proportionality as printed. A literal constant-density-of-states intrinsic semiconductor with temperature-independent mobilities retains the additional factor $T$. Indeed, using the full Fermi-Dirac distribution giveswhich reduces to the expression above when $E_g\gg k_{B}T$. Thus the pure exponential is an activation approximation, not an exact consequence of constant DOS alone. A mobility sum proportional to $T^{-1}$ would cancel the leading algebraic factor.

Using the requested exponential law with a temperature-independent prefactor,The conductivity ratio is two, soThe denominator is positive, as required for a positive gap and rising conductivity. If one instead retains the constant-DOS factor $T$ while keeping mobility constant, the inferred gap becomesThat is a different prefactor model; $1.00\mathrm{eV}$ is the estimate from the relation specified for this part.