= Solution
\b[Band edges and optical transitions.] A <direct band gap> has the conduction minimum and valence maximum at the same <crystal momentum>. A near-edge <photon> then connects the two almost vertically on an energy-versus-wavevector plot, because its <momentum> is small on the <Brillouin zone> scale. An <indirect band gap> has the extrema at different momenta, so an additional <momentum> transfer, commonly supplied by <phonon> absorption or emission, is required for a near-edge radiative transition. This reduces its probability relative to an allowed direct transition.
Near the zone centre, an isotropic direct-gap model has $E_c(k)=E_c+\hbar^2k^2/(2m_e^*)$ and downward valence dispersions $E_{v,j}(k)=E_v-\hbar^2k^2/(2m_{h,j}^*)$. The <heavy-hole and light-hole bands> are the upper branches with large and small positive hole masses: the heavy branch is flatter. In common cubic <semiconductors> these arise from the $j=3/2$ manifold, while <spin-orbit coupling> places the $j=1/2$ <split-off valence band> lower by $\Delta_{\mathrm{so}}$. Its thermally generated hole population is suppressed by approximately $e^{-\Delta_{\mathrm{so}}/(k_BT)}$ relative to upper-valence holes, apart from density-of-states factors. Neglect is justified when $\Delta_{\mathrm{so}}\gg k_BT$; it is not a universal exclusion of this band at arbitrary <temperature>.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2017-semiconductor-bands.png]
{title=Original schematic direct and indirect band edges, including heavy, light and split-off hole branches}
For an indirect material such as Si, the conduction minimum need not be at $\Gamma$. In applying the carrier-density formulae, $E_c$ must mean the actual minimum of the included conduction valley. The supplied scalar effective masses and densities of states constitute a simplified band-edge model; they do not specify Si's full valley anisotropy or multiplicity.
\b[Carrier populations.] Let $f(E)$ be the <Fermi-Dirac distribution>. In the nondegenerate conduction tail, $f(E)\simeq e^{-(E-\mu)/(k_BT)}$. Set $x=E-E_c$ and integrate the given <density of states>:
$$
n=\int_{E_c}^\infty g_e(E)f(E)\,dE=\frac1{2\pi^2}\left(\frac{2m_e^*}{\hbar^2}\right)^{3/2}e^{-(E_c-\mu)/(k_BT)}\int_0^\infty x^{1/2}e^{-x/(k_BT)}\,dx.
$$
The integral is $\Gamma(3/2)(k_BT)^{3/2}=\sqrt\pi(k_BT)^{3/2}/2$. Defining the <effective density of states> gives
$$
\boxed{n=N_c e^{-(E_c-\mu)/(k_BT)},\qquad N_c=2\left(\frac{m_e^*k_BT}{2\pi\hbar^2}\right)^{3/2}.}
$$
The factor two is already included in the supplied <density of states>; it must not be added a second time.
A hole is an unoccupied valence state. For $E$ sufficiently below $\mu$, $1-f(E)\simeq e^{(E-\mu)/(k_BT)}$. Substitution of $x=E_v-E$ gives in exactly the same way
$$
\boxed{p=N_v e^{-(\mu-E_v)/(k_BT)},\qquad N_v=2\left(\frac{m_h^*k_BT}{2\pi\hbar^2}\right)^{3/2}.}
$$
Their product eliminates the <chemical potential>:
$$
\boxed{np=N_cN_v e^{-E_g/(k_BT)}=n_i^2.}
$$
This is the <semiconductor mass-action law> at <thermal equilibrium>. In an <intrinsic semiconductor>, <charge neutrality> additionally gives
$$
\boxed{n=p=n_i=\sqrt{N_cN_v}\,e^{-E_g/(2k_BT)}=2\left(\frac{k_BT}{2\pi\hbar^2}\right)^{3/2}(m_e^*m_h^*)^{3/4}e^{-E_g/(2k_BT)}.}
$$
In a doped <semiconductor> the product still applies in this nondegenerate equilibrium model, but the two populations need not be equal.
\b[Intrinsic <chemical potential>.] Taking logarithms of $n=p$ yields the <intrinsic-semiconductor chemical potential>
$$
\boxed{\mu_i=\frac{E_c+E_v}{2}+\frac{k_BT}{2}\ln\frac{N_v}{N_c}=E_v+\frac{E_g}{2}+\frac{3k_BT}{4}\ln\frac{m_h^*}{m_e^*}.}
$$
For unequal band degeneracies, use the first form with the full $N_c,N_v$. For the supplied mass ratio $m_h^*/m_e^*=1/2$, the shift from midgap at $300\,\mathrm K$ is approximately $-13.4\,\mathrm{meV}$, toward the valence edge.
\b[Three <temperature> regimes.] The original plot runs from high <temperature> on the left to low <temperature> on the right. At high <temperature>, thermal excitation across the gap dominates and $n\simeq p\simeq n_i$ rises rapidly with <temperature>. In the middle, donors and acceptors are effectively ionized, the material is extrinsic, and $n\simeq N_D-N_A$ makes the plateau. At low <temperature>, <dopant freeze-out> binds increasing numbers of <Electrons> to donor levels, reducing the mobile-electron population as the sample cools. Its activation scale is a donor <binding energy>, not the full <band gap>. Compensation and incomplete ionization affect that regime.
\b[Gap estimate from the original figure.] On the dashed intrinsic branch, $n_i=C T^{3/2}e^{-E_g/(2k_BT)}$. Hence <band-gap extraction from intrinsic carrier activation> uses the slope of $\ln(n_i/T^{3/2})$ versus $1/T$. With $x=1000\,\mathrm K/T$ and $k_B$ expressed in eV/K, two points give
$$
E_g=\frac{2000\,k_B\mathrm K}{x_2-x_1}\left[\ln\frac{n_{i,1}}{n_{i,2}}-\frac32\ln\frac{x_2}{x_1}\right].
$$
The very compressed intrinsic branch is only approximately readable: $n_i\sim10^{15}\,\mathrm{cm^{-3}}$ near $x\sim1.8$ and $n_i\sim10^{13}\,\mathrm{cm^{-3}}$ near $x\sim2.5$. These give \b[$E_g\sim1.0\,\mathrm{eV}$], rather than a precision determination. Neglecting the thermal prefactor gives about $1.13\,\mathrm{eV}$ from those same rough points. Thus \b[a gap of order $1$–$1.1\,\mathrm{eV}$] is justified by this schematic plot; its resolution does not support distinguishing a few hundredths of an electronvolt. The declining freeze-out branch must not be used to infer the full gap.
\b[Room-temperature densities.] At $300\,\mathrm K$, the supplied masses give
$$
N_c=2.43\times10^{19}\,\mathrm{cm^{-3}},\qquad N_v=8.61\times10^{18}\,\mathrm{cm^{-3}}.
$$
In the fully ionized extrinsic plateau, put $N=N_D-N_A=10^{15}\,\mathrm{cm^{-3}}$. <Charge neutrality> and the mass-action law imply
$$
n-p=N,\qquad np=n_i^2,\qquad n=\frac{N+\sqrt{N^2+4n_i^2}}2,\qquad p=\frac{n_i^2}{n}.
$$
Since $n_i\ll N$, \b[$n\simeq10^{15}\,\mathrm{cm^{-3}}$] is insensitive to the imprecise gap. The minority density is much more sensitive:
$$
\boxed{p\simeq\frac{N_cN_v}{N}e^{-E_g/(k_BT)}=2.10\times10^{23}e^{-E_g/(0.02585\,\mathrm{eV})}\,\mathrm{cm^{-3}}.}
$$
Using $E_g=1.0\,\mathrm{eV}$ gives $n_i\simeq5.8\times10^{10}\,\mathrm{cm^{-3}}$ and $p\simeq3.3\times10^6\,\mathrm{cm^{-3}}$. Using $1.1\,\mathrm{eV}$ instead gives $n_i\simeq8.3\times10^9\,\mathrm{cm^{-3}}$ and $p\simeq7.0\times10^4\,\mathrm{cm^{-3}}$. These are explicitly conditional estimates within the stated mass model. The plot alone cannot fix a precise minority density because the uncertain gap occurs inside an exponential.
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