Si polarizability with 6×10²² electrons/cm³

Use the Clausius–Mossotti relation for a covalent (non-polar) solid, where only electronic polarization exists.

Data and assumptions

• $N = 6\times10^{22}\ \text{cm}^{-3} = 6\times10^{28}\ \text{m}^{-3}$, taken as the number of polarizable Si units per m³ (as given).
• Relative permittivity of Si, $\varepsilon_r = 11.9$ (standard value, not given in the question).
• $\varepsilon_0 = 8.854\times10^{-12}$ F/m.

i) Electronic polarizability

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0} \;\Rightarrow\; \alpha_e = \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2}$$ $$\begin{aligned} \frac{3\varepsilon_0}{N} &= \frac{3\times8.854\times10^{-12}}{6\times10^{28}} = 4.427\times10^{-40} \\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{10.9}{13.9} = 0.7842 \\ \alpha_e &= 4.427\times10^{-40}\times0.7842 = 3.47\times10^{-40}\ \text{F m}^2 \end{aligned}$$

Answer (i): $\alpha_e \approx 3.47\times10^{-40}\ \text{F m}^2$ per Si atom.

(If the given figure is read strictly as valence electrons, the atom density is $6\times10^{28}/4 = 1.5\times10^{28}$ m⁻³ and the polarizability per atom becomes $1.39\times10^{-39}$ F m².)

ii) Local field compared with applied field

For a sample with electrodes on opposite faces, the Lorentz local field is

$$E_{loc} = E + \frac{P}{3\varepsilon_0} = E + \frac{(\varepsilon_r - 1)E}{3} \;\Rightarrow\; \frac{E_{loc}}{E} = \frac{\varepsilon_r + 2}{3}$$ $$\frac{E_{loc}}{E} = \frac{11.9 + 2}{3} = \frac{13.9}{3} = 4.63$$

Answer (ii): the local field is about 4.63 times the applied field.