 Solution

Let $\alpha ={\hslash }^2/(2m^{\ast })$, 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 ${\mathbf{k}}_1={\mathbf{k}}_2+{\mathbf{k}}_3$. Their total kinetic energy satisfiesEnergy conservation for impact ionization is $E_1=E_g+\alpha (k_2^2+k_3^2)$. Consequently $E_1\ge E_g+E_1/2$, givingThe bound is attained when ${\mathbf{k}}_2={\mathbf{k}}_3={\mathbf{k}}_1/2$; 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.