---
title: "Si polarizability, local field and resonant frequency"
subject: "Electrical Engineering Material"
chapter: 3
key: "semester_3/electrical_engineering_material:si-polarizability-resonant-frequency"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-polarizability-resonant-frequency
site: BCT Engineering
---

# Si polarizability, local field and resonant frequency

Electrical Engineering Material · Chapter 3: Dielectric Materials

- Asked in: 2081 Bhadra, 4 marks; 2080 Bhadra, 6 marks (asked 2 times)
- Video: [Derivation of Electronic Polarizability|Dielectrics|Applied Physics|B.Sc Physics](https://www.youtube.com/watch?v=eRJU_IaUKY8) — Vignana Saraswathi
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/si-polarizability-resonant-frequency
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-3

## Question

Determine electronic polarizability due to valence electrons per Si-atoms. If the sample is supplied by a voltage on its electrode, by how much is the local field greater than the applied field? Also determine the resonant frequency. Take εr = 11.9 and number of Si-atoms per unit volume = 5×10²⁸ m⁻³.

## Short answer

Use Clausius-Mossotti: α_e = (3ε0/N)(εr-1)/(εr+2) = 4.17×10⁻⁴⁰ F m². Local field E_loc = (εr+2)/3 E = 4.63E. Resonant frequency ω0 = √(Ze²/(m_eα_e)) = 1.64×10¹⁶ rad/s (f0 ≈ 2.6×10¹⁵ Hz).

## Answer

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 $Z = 4$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which relation gives the electronic polarizability of Si from eps_r and N?
   A. alpha_e = (3 eps0/N)(eps_r - 1)/(eps_r + 2)
   B. alpha_e = (eps0/N)(eps_r + 2)/(eps_r - 1)
   C. alpha_e = N eps0 (eps_r - 1)
   D. alpha_e = (3 eps0/N)(eps_r + 2)/(eps_r - 1)
   Answer: A. The Clausius-Mossotti equation rearranged for alpha_e. (Key point: Formula for alpha_e)

2. For eps_r = 11.9, how many times is the Lorentz local field greater than the applied field?
   A. 3.63
   B. 11.9
   C. 4.63
   D. 13.9
   Answer: C. E_loc/E = (eps_r + 2)/3 = 13.9/3 = 4.63. (Key point: Local field ratio)

3. In the spring model used, which expression gives the resonant angular frequency from alpha_e?
   A. omega0 = alpha_e m_e/(Z e^2)
   B. omega0 = sqrt(alpha_e/(Z e^2 m_e))
   C. omega0 = Z e^2/(m_e alpha_e)
   D. omega0 = sqrt(Z e^2/(m_e alpha_e))
   Answer: D. From alpha_e = Z e^2/(m_e omega0^2), omega0 = sqrt(Z e^2/(m_e alpha_e)). (Key point: Resonant frequency model)

4. The resonant frequency of about 2.6 x 10^15 Hz lies in the ultraviolet. What does this imply?
   A. Electronic polarization in Si fails at radio frequency
   B. Si has no polarization at all
   C. Electronic polarization in Si follows the field up to optical frequencies
   D. Si shows ionic polarization at this frequency
   Answer: C. The resonance is above visible light, so the electron cloud follows the field through the optical range. (Key point: Result and meaning)

## Related in Electrical Engineering Material

- [Si electronic polarizability and local field](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-local-field) — Electrical Engineering Material, chapter 3, asked 2×
- [Si polarizability with 6×10²² electrons/cm³](https://www.bctengineer.com/question.md?key=semester_3%2Felectrical_engineering_material%3Asi-polarizability-6e22) — Electrical Engineering Material, chapter 3