Determine the percentage of ionic polarizability in the sodium chloride crystal which has the optical index of refraction and the static dielectric constant as 1.5 and 5.6 respectively.
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Answer:
ionic polarizability in the sodium chloride crystal which has the optical index of refraction and the static dielectric constant as 1.5 and 5.6=3.7⋅10−40 Fm²
Explanation:
For the ionic polarizability in the sodium chloride crystal Clausius-Mossoti condition can be composed as:
εr−1εr+2=N3ε0(αe++αe−+αi)'................(1)
where αe+ and αe− are electron polarizabilities of positive and negative particles in gem grid. Their aggregate can be resolved utilizing a Lorentz-Lorenz condition, which is gotten from the way that the impact of an outside electromagnetic field with moderately high frequencies comparing to the apparent and bright scope of the range brings about the uprooting of just the electron shells comparative with nuclear cores, while the more enormous particles (molecules and particles) lack the opportunity to move from the involved positions. Correspondingly, just electronic polarization adds to the polarization of the medium (for example αi≈0), and the refractive record is connected with the electron polarizability of the particles by the Lorentz recipe:
n2−1n2+2=N3ε0(αe++αe−)................(2)
By subtracting (2) from (1),
εr−1εr+2−n2−1n2+2=N3ε0αi................(3)
ionic polarizability in the sodium chloride crystal is :
αi=3ε0N((εr−1/εr+2)−(n2−1/n2+2))=((3⋅8.85⋅10−12 Fm−1/2.24⋅1028 m−3)(5.62−15.62+2−1.52−11.52+2))
=3.7⋅10−40 Fm²
Therefore ionic polarizability in the sodium chloride crystal (αi)=3.7⋅10−40 Fm²
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