Si polarizability, local field and resonant frequency

Silicon is a covalent, non-polar crystal, so its permittivity comes only from electronic polarization of the valence electrons. Use the Clausius-Mossotti equation and the Lorentz local field.

Given: $\varepsilon_r = 11.9$, $N = 5 \times 10^{28}\ \text{m}^{-3}$, $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}$. Each Si atom has $Z = 4$ valence electrons.

Electronic polarizability per Si atom

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0} \;\Rightarrow\; \alpha_e = \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2}$$ $$\alpha_e = \frac{3 \times 8.854 \times 10^{-12}}{5 \times 10^{28}} \times \frac{10.9}{13.9} = 4.17 \times 10^{-40}\ \text{F m}^2$$

Local field versus applied field

Lorentz field: $E_{loc} = E + \dfrac{P}{3\varepsilon_0}$ with $P = \varepsilon_0(\varepsilon_r - 1)E$:

$$E_{loc} = E + \frac{(\varepsilon_r - 1)E}{3} = \frac{\varepsilon_r + 2}{3}E = \frac{13.9}{3}E = 4.63E$$

So the local field is about 4.63 times the applied field (greater by $3.63E$).

Resonant frequency

In the simple model the electron cloud ($Ze$, mass $Zm_e$) is bound to the nucleus by a spring, and

$$\alpha_e = \frac{Ze^2}{m_e\omega_0^2} \;\Rightarrow\; \omega_0 = \sqrt{\frac{Ze^2}{m_e\alpha_e}}$$ $$\omega_0 = \sqrt{\frac{4 \times (1.602 \times 10^{-19})^2}{9.109 \times 10^{-31} \times 4.17 \times 10^{-40}}} = 1.64 \times 10^{16}\ \text{rad/s}$$ $$f_0 = \frac{\omega_0}{2\pi} = 2.62 \times 10^{15}\ \text{Hz}$$

This lies in the ultraviolet region, so electronic polarization in Si follows the field up to optical frequencies.

Answer: $\alpha_e = 4.17 \times 10^{-40}\ \text{F m}^2$; $E_{loc} = 4.63E$; $\omega_0 = 1.64 \times 10^{16}\ \text{rad/s}$ ($f_0 \approx 2.6 \times 10^{15}$ Hz, taking $Z = 4$).