# Recent Advances in Laser-Induced Surface Damage of KH\({}_{2}\)Po\({}_{4}\) [PERSON]
## 2 Mechanism of Pulsed Laser-Induced Surface Damage
### Ionization Damage
#### 2.2.2 Avalanche Ionization  
Actually, the laser will interact with the lattice, electron, and phonon in the crystal material when the incident laser irradiates the crystal material. As a result, the crystal material will absorb the incident laser. There are two types of electrons in the materials: bound electrons and free electrons. Generally, the ionization potential of the bound electron is much larger than the photon energy of the incident laser, so the bound electron cannot generally absorb photons. However, the normal impurity-free KDP crystals are dielectric materials that do not have free electrons in the body. Therefore, the source of free electrons is very important during laser irradiation. It is clear that the KDP crystal materials will not absorb the laser energy greatly. Based on the understanding of the mechanism of photoionization, it can be found that the free electrons generated by photoionization in the early stage will be used as seed electrons to absorb photon energy by inverse bremsstrahlung under the subsequent laser irradiation. When the absorbed energy reaches a certain standard, the seed electrons will collide with the bound electrons in the valence band (VB) inside the material, causing it transition to the conduction band (CB) [54]. These new free electrons with lower energy will be generated, which will continue to absorb energy and collide with the atoms. With the increase of the density of free electrons in KDP crystal materials, the crystal materials with dielectric properties will gradually become the metal-like materials. Furthermore, with the subsequent irradiation of laser, the energy is continuously absorbed, and the density of free electrons inside the materials will increase rapidly. Researchers have successively constructed and optimized the corresponding models for avalanche ionization to determine the density of free electrons generated by avalanche ionization. Figure 4: Correlations between incident laser intensity \(I\) and PI rate \(W_{PI}\) with the consideration of full Keldysh’s ionization, individual multiphoton ionization (MPI) and tunneling ionization (TI) ionizations under 800 nm lasers irradiation [29]. [PERSON] [26; 55; 56] criticized the classical avalanche model with the imperfection of calculating the avalanche ionization rate and built the Sparks model for the avalanche ionization. \[W_{AI-Sparks}=\frac{0.693e^{2}E^{2}\tau_{K}}{m_{e}^{*}E_{g}(1+\omega^{2}\tau_{K} ^{2})}-\frac{0.693\hbar\omega_{p}}{E_{g}\tau_{L}} \tag{8}\]  
where \(\tau_{K}\) represents the electro-acoustic relaxation time in the case of large angle scattering, \(\tau_{K}=1.36\times 10^{-15}\) s. \(\tau_{L}\) is the electro-acoustic relaxation time when scattering at large and small angles, and \(\tau_{L}\)= \(8.77\times 10^{-16}\) s. \(E\) is the radiation electric field amplitude. \(\omega_{p}\) is the average phonon frequency, there is \(\hbar\omega_{p}=0.025\) eV. The model assumes that the collision frequency and collision energy loss are constant during collisions between electrons. Sparks indicated that only when the average electron energy is greater than the energy required for electron impact ionization, can electrons reach the ionization limit, triggering avalanche ionization. [PERSON] et al. constructed the avalanche ionization model based on the heating phenomenon of free electrons in the conduction band [57]. \[W_{AI-Drude}=\frac{\sigma}{E_{g}}I(t)=\frac{3.2e^{2}\pi\varepsilon_{0}\sqrt{m_ {e}^{*}}.n(t)}{2\sqrt{2}cn_{0}m_{e}^{*}\Big{[}n^{2}(t)+0.08m_{e}^{*}(\pi \varepsilon_{0}{}^{2}\omega E_{g})^{2}\Big{]}}I(t) \tag{9}\]  
where \(\sigma\) represents the absorption cross section of the photon, and \(\tau_{c}\) is the time of the electron collision. [PERSON] et al. [58] proposed an avalanche ionization model based on the semi-empirical formula, which can be applied to all electrical intensities. \[W_{AI-Thorber}=\frac{v_{s}eE}{E_{g}}\exp(-\frac{E_{I}}{E(1+E/E_{P})+E_{KT}}) \tag{10}\]  
where \(e\) is the electronic charge.