 Solution

The plateau in both source plots is approximately $10^{18.5}{\mathrm{m}}^{-3}$, soThese are graphical estimates, not high-precision fitted material parameters. To extract activation energies consistently with part (a), retain the temperature powers in the effective density of states factors.

On the semiconductor carrier freeze-out tail, put $x=1/T$. ThenFor example, the right-hand graph gives roughly ${\log }_{10}[n/({\mathrm{m}}^{-3})]\simeq 17.7$ at $x=0.10{\mathrm{K}}^{-1}$ and $13.0$ at $x=0.30{\mathrm{K}}^{-1}$. Subtract the known prefactor contribution before interpreting the slope:Thus $\boxed{E_i\simeq 9\mathrm{meV}}$, to the accuracy of the drawing. Ignoring the weak prefactor gives about $9.3\mathrm{meV}$, a similar rough estimate.

For the intrinsic branch, useThe left-hand graph gives $n(100\mathrm{K})\simeq 2.5\times 10^{21}{\mathrm{m}}^{-3}$ and $n(500\mathrm{K})\simeq 1.5\times 10^{24}{\mathrm{m}}^{-3}$, both far above $N_D$. ThereforeHere $k_B=8.6173\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$. Omitting the $T^{3/2}$ prefactor over this wide temperature interval would instead give approximately $0.14\mathrm{eV}$, so it is not a negligible correction for the band-gap extraction from intrinsic carrier activation. The source graph warrants an approximate narrow band gap, rather than several significant figures.