= Solution
Let $\alpha=\hbar^2/(2m^*)$, so the initial conduction-electron kinetic <energy> is $E_1=\alpha k_1^2$. In the normal momentum-conserving process with the final hole fixed at zero <wavevector>, the two final conduction electrons obey $\boldsymbol k_1=\boldsymbol k_2+\boldsymbol k_3$. Their total kinetic <energy> satisfies
$$
\alpha(k_2^2+k_3^2)=\frac{E_1}2+\frac\alpha2|\boldsymbol k_2-\boldsymbol k_3|^2\geq\frac{E_1}2.
$$
Energy conservation for <impact ionization> is $E_1=E_g+\alpha(k_2^2+k_3^2)$. Consequently $E_1\geq E_g+E_1/2$, giving
$$
\boxed{E_{1,\min}=2E_g.}
$$
The bound is attained when $\boldsymbol k_2=\boldsymbol k_3=\boldsymbol k_1/2$; opposite spins permit the two electrons to share that <wavevector>. Fixing the hole momentum is essential: a mobile final hole gives a different threshold. This is the threshold for <impact ionization with a hole at zero momentum>.
The two hot electrons subsequently redistribute <energy> by electron-electron collisions and transfer it to the lattice by <electron-phonon scattering>. Optical-phonon emission can provide large discrete losses while the electrons are hot; acoustic phonons dominate later small <energy> exchanges. Electron-electron collisions alone conserve the <electron> system's total <energy> and do not cool it into the lattice. The electron-hole pair eventually recombines radiatively or through nonradiative channels, such as defect-assisted recombination or <Auger recombination>. Recombination is distinct from the preceding intraband cooling.
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