Q. 3. Derive Clausius-Mossotti expression for the molecule
polarisation
Answers
Answer:
the Clausius-Mossotti relation connects the relative permittivity εr of a dielectric to the polarizability α of the atoms or molecules constituting the dielectric.
Answer:
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Explanation:
The Clausius–Mossotti relation expresses the dielectric constant (relative permittivity, εr) of a material in terms of the atomic polarizability, α, of the material's constituent atoms and/or molecules, or a homogeneous mixture thereof. It is named after Ottaviano-Fabrizio Mossotti and Rudolf Clausius. It is equivalent to the Lorentz–Lorenz equation. It may be expressed as:[1][2]
{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}={\frac {N\alpha }{3\varepsilon _{0}}}}{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}={\frac {N\alpha }{3\varepsilon _{0}}}}
where
{\displaystyle \varepsilon _{r}=\epsilon /\epsilon _{0}}{\displaystyle \varepsilon _{r}=\epsilon /\epsilon _{0}} is the dielectric constant of the material, which for non-magnetic materials is equal to {\displaystyle n^{2}}n^{2} where {\displaystyle n}n is the refractive index
{\displaystyle \varepsilon _{0}}\varepsilon _{0} is the permittivity of free space
{\displaystyle N}N is the number density of the molecules (number per cubic meter), and
{\displaystyle \alpha }\alpha is the molecular polarizability in SI-units (C·m2/V).
In the case that the material consists of a mixture of two or more species, the right hand side of the above equation would consist of the sum of the molecular polarizability contribution from each species, indexed by i in the following form:[3]
{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}=\sum _{i}{\frac {N_{i}\alpha _{i}}{3\varepsilon _{0}}}}{\displaystyle {\frac {\varepsilon _{\mathrm {r} }-1}{\varepsilon _{\mathrm {r} }+2}}=\sum _{i}{\frac {N_{i}\alpha _{i}}{3\varepsilon _{0}}}}
In the CGS system of units the Clausius–Mossotti relation is typically rewritten to show the molecular polarizability volume {\displaystyle \alpha '=\alpha /(4\pi \varepsilon _{0})}\alpha '=\alpha /(4\pi \varepsilon _{0}) which has units of volume (m3).[2] Confusion may arise from the practice of using the shorter name "molecular polarizability" for both {\displaystyle \alpha }\alpha and {\displaystyle \alpha '}\alpha ' within literature intended for the respective unit system.