According to quantum mechanics, the electron cloud for a hydrogen

atom in the ground state has a charge density

$\mathbf{\rho }\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{=}\frac{\mathbf{q}}{{\mathbf{\tau \tau a}}^{\mathbf{3}}}{\mathbf{e}}^{\frac{\mathbf{-}\mathbf{2}\mathbf{r}}{\mathbf{a}}}$

where qis the charge of the electron and ais the Bohr radius. Find the atomic

polarizability of such an atom. [Hint:First calculate the electric field of the electron cloud, ${\mathbf{E}}_{\mathbf{e}}\left(r\right)$ then expand the exponential, assuming $\mathit{r}\mathit{\ll }\mathit{a}$.

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

$\rho \mathit{\left(}r\mathit{\right)}\mathit{=}\frac{q}{\pi a^{\mathit{3}}}{e}^{\frac{\mathit{-}\mathit{2}r}{a}}$

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

$\mathbf{E}\mathbf{=}\frac{{\mathbf{Q}}_{\mathbf{enc}}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

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 $\rho$and get

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

Substitute this expression in equation (1) and get

$$\begin{gathered} E_e=\frac{1}{4\pi {\epsilon }_0{r}^2}q\left[1-e^{\frac{-2r}{a}}\left(1+2\frac{r}{a}+2\frac{r^2}{a^2}\right)\right] \\ =\frac{1}{4\pi {\epsilon }_0{r}^2}q\left[1-\left(1-2\frac{r}{a}+2\frac{r^2}{a^2}-.....\right)\left(1+2\frac{r}{a}+2\frac{r^2}{a^2}\right)\right] \\ =\frac{1}{4\pi {\epsilon }_0{r}^2}q\left[\frac{4}{3}{\left(\frac{r}{a}\right)}^3+.....\right] \\ \approx \frac{qr}{3\pi {\epsilon }_0{a}^3} \end{gathered}$$

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

The expression for atomic polarizability is

$\alpha =\frac{p}{E}$

Substitute the expressions in the above equation and get

$$\begin{gathered} \alpha =\frac{qr}{{\displaystyle \frac{qr}{3\pi {\epsilon }_0{a}^3}}} \\ =3\pi {\epsilon }_0{a}^3 \end{gathered}$$

Thus, the atomic polarizability is$3\pi {\epsilon }_0{a}^3$