91Ó°ÊÓ

The electron cloud for a hydrogen atom in the ground state has a charge density

$\mathrm{ÒÏ}\mathit{\left(}r\mathit{\right)}\mathit{=}\frac{q}{\mathrm{Ï€}{a}^{\mathit{3}}}{e}^{\frac{\mathit{-}\mathit{2}r}{a}}$

The electric field on a spherical Gaussian surface of radius r is

$\mathbf{E}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{enc}}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, ${\mathrm{Q}}_{\mathrm{enc}}$is the charge enclosed by the Gaussian surface.

The expression for the charge enclosed inside the Gaussian surface is

Substitute the expression for $\mathrm{ÒÏ}$and get

$$\begin{gathered} Q_{enc}=\frac{4\mathrm{π}q}{\mathrm{π}{a}^3}{∫}_0^r{e}^{\frac{-2r\text{'}}{a}}r{\text{'}}^{2}dr\text{'} \\ =\frac{4q}{a^3}{\left[-\frac{a}{2}{e}^{\frac{-2r}{a}}\left(r{\text{'}}^2+ar\text{'}+\frac{a^2}{2}\right)\right]}_0^r \\ =q\left[1-e^{\frac{-2r}{a}}\left(1+2\frac{r}{a}+2\frac{r^2}{a^2}\right)\right] \end{gathered}$$

Substitute this expression in equation (1) and get

$$\begin{gathered} E_e=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}q\left[1-e^{\frac{-2r}{a}}\left(1+2\frac{r}{a}+2\frac{r^2}{a^2}\right)\right] \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}q\left[1-\left(1-2\frac{r}{a}+2\frac{r^2}{a^2}-.....\right)\left(1+2\frac{r}{a}+2\frac{r^2}{a^2}\right)\right] \\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0{r}^2}q\left[\frac{4}{3}{\left(\frac{r}{a}\right)}^3+.....\right] \\ ≈\frac{qr}{3\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3} \end{gathered}$$

But qr = p , the dipole moment of the atom.

The expression for atomic polarizability is

$\mathrm{Î\pm }=\frac{p}{E}$

Substitute the expressions in the above equation and get

$$\begin{gathered} \mathrm{Î\pm }=\frac{qr}{{\displaystyle \frac{qr}{3\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3}}} \\ =3\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3 \end{gathered}$$

Thus, the atomic polarizability is$3\mathrm{Ï€}{\mathrm{Î\mu }}_0{a}^3$