Question

The static dielectric constant of water is 8.1 and its index of refraction is 1.33. What is the percentage contribution of ionic polarizability ?​

Answer 1

Answer:

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Answer 2

Answer:

The polarizability contribution by ions is $29.02$ %.

Explanation:

• Ionic polarization occurs when electricity is applied to an ionic substance and then the cation and anion move from different sources resulting in a moment of the dipole.
• Ionic polarization is due to the movement of cation and anion in their original position toward different species.

The formula used is given as:

$\frac{K-1}{K+2} =\frac{N(\alpha e+\alpha i)}{3eo}$

where, $K$ is water dielectric constant.

            $\alpha e$ is electronic polarization factor.

            $\alpha i$ is ionic polarization factor.

Step 1:

It is given that the dielectric constant of water, $Er=8.1$

Putting values in the above formula:

$\frac{8.1-1}{8.1+2} =\frac{N(\alpha e+\alpha i)}{3eo}$

$\frac{7.1}{10.1} =\frac{N(\alpha e+\alpha i)}{3eo}$

$\frac{71}{101} =\frac{N(\alpha e+\alpha i)}{3eo}$

Also,  $\frac{n^{2} -1}{n^{2} +2} =\frac{N(\alpha e)}{3eo}$

Here, $n=$ the index of refraction $=1.33$

$\frac{1.33^{2} -1}{1.33^{2} +2} =\frac{N(\alpha e)}{3eo}$

$\frac{0.7689}{3.7689} =\frac{N(\alpha e)}{3eo}$

$0.204 =\frac{N(\alpha e)}{3eo}$

Step 2:

Dividing the above two relations:

$\frac{71}{101*0.204} =\frac{\alpha e+\alpha i}{\alpha e}$

$\frac{71}{20.604} =\frac{\alpha e+\alpha i}{\alpha e}$

The percentage contribution = $=\frac{\alpha e}{\alpha e+\alpha i}{}*100$ %

                                                $= \frac{20.604}{71}*100%$ %

                                                $=29.02$ %

So, the ionic polarizability percentage is $29.02$ %.