Question

\(\dfrac{n\alpha}{3\epsilon_0} = \dfrac{(\epsilon_r - 1)}{(\epsilon_r + 2)}\) is known as ____________ relation.

• A. Debye
• B. Clausius-Mossotti
• C. Einstein-Debye
• D. Bose-Einstein

Hint:

The formula \(\tfrac{\epsilon_r-1}{\epsilon_r+2} = \tfrac{n\alpha}{3\epsilon_0}\) links dielectric constant to polarizability via the local field.

Answer 1

The Correct Option is B

Step 1: Identify the physical quantities. Here \(n\) is the number density of atoms/molecules, \(\alpha\) is the atomic (electronic) polarizability, \(\epsilon_0\) is the permittivity of free space, and \(\epsilon_r\) is the relative permittivity (dielectric constant).

Step 2: The relation links the macroscopic dielectric constant \(\epsilon_r\) to the microscopic polarizability \(\alpha\) by accounting for the local (Lorentz) field inside the dielectric: \[\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{n\alpha}{3\epsilon_0}.\]

Step 3: This bridge between micro and macro properties is the Clausius-Mossotti relation. (Its optical form, using refractive index \(n_r^2 = \epsilon_r\), is the Lorentz-Lorenz equation.)

Step 4: The other names refer to unrelated results (Debye relaxation, Einstein-Debye specific heat, Bose-Einstein statistics), so they are ruled out.\[\boxed{\text{Clausius-Mossotti relation}}\]