91Ó°ÊÓ

The given data is listed below as:

• The value of the atomic polarizability is, $\mathrm{Î\pm }=1×{10}^6\mathrm{N}/\mathrm{C}$
• The value of the large electric field is,$E=3×{10}^6\mathrm{N}/\mathrm{C}$
• The charge of one proton is,$q=e=1.6×{10}^{-19}\mathrm{C}$

Polarization of material is a product of polarizability and electric field of atom.

The polarization is described as the product of the atomic polarizability and electric field of atom.

The equation of the polarization is expressed as:

$p=\mathrm{Î\pm }E$

Here,$\mathrm{Î\pm }$ is the atomic polarizability, and $E$is the electric field of atom.

Substitute the values in the above equation.

$$\begin{gathered} p=\left(1×{10}^{-40}\mathrm{C}·\mathrm{m}/\left(\mathrm{N}/\mathrm{C}\right)\right)×\left(3×{10}^6\mathrm{N}/\mathrm{C}\right) \\ =3×{10}^{-34}\frac{\left(\mathrm{C}·\mathrm{m}\right)·\left(\mathrm{N}/\mathrm{C}\right)}{\left(\mathrm{N}/\mathrm{C}\right)} \\ =3×{10}^{-34}\mathrm{C}·\mathrm{m} \end{gathered}$$

The relation between the charge separation and polarization is expressed as:

$$\begin{gathered} p=-qs \\ s=\frac{p}{q} \end{gathered}$$

Here, $p$is the polarization, $q$is the charge of the proton and$s$ is the charge separation.

Substitute the values in the above equation.

$$\begin{gathered} s=\frac{3×{10}^{-34}\mathrm{C}·\mathrm{m}}{1.6×{10}^{-19}\mathrm{C}} \\ =2×{10}^{-15}\mathrm{m} \end{gathered}$$

Thus, the charge separation produced in a typical atom is $2×{10}^{-15}\mathrm{m}$.