Si electronic polarizability and local field

Pure Si has only electronic polarization (covalent, no permanent dipoles, cubic structure), so the Clausius-Mossotti equation with the Lorentz local field applies.

Given: $\varepsilon_r = 11.9$, $N = 5 \times 10^{28}\ \text{m}^{-3}$, $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$.

(i) Electronic polarizability per Si atom

Clausius-Mossotti equation:

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}$$ $$\begin{aligned} \alpha_e &= \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2} \\ &= \frac{3 \times 8.85 \times 10^{-12}}{5 \times 10^{28}} \times \frac{11.9 - 1}{11.9 + 2} \\ &= 5.31 \times 10^{-40} \times 0.7842 = 4.16 \times 10^{-40}\ \text{F m}^2 \end{aligned}$$

(If the local field were ignored, $\alpha_e = \varepsilon_0(\varepsilon_r - 1)/N = 1.93 \times 10^{-39}$ F m², about 4.6 times too large; this shows why the local field matters.)

(ii) Local field compared with applied field

The Lorentz local field is

$$E_{loc} = E + \frac{P}{3\varepsilon_0}, \qquad P = \varepsilon_0(\varepsilon_r - 1)E$$ $$E_{loc} = E\left(1 + \frac{\varepsilon_r - 1}{3}\right) = \frac{\varepsilon_r + 2}{3}E = \frac{13.9}{3}E = 4.63E$$

So the field acting on each Si atom is about 4.63 times the applied field, i.e. greater than it by $3.63E$ (363%).

Answer: (i) $\alpha_e \approx 4.16 \times 10^{-40}\ \text{F m}^2$; (ii) $E_{loc} = 4.63E$.