Hydrogen Atom Model Context

The question models a hydrogen atom where the electronic charge ($-e$) is spread uniformly within a sphere of radius $a$, centered on the proton ($+e$). When this atom is exposed to an external electric field $E$, the center of the electron distribution shifts relative to the proton.

Polarizability Formula for Uniform Electron Distribution

For an atom model with uniformly distributed electronic charge in a sphere of radius $a$, the polarizability ($\alpha$) is determined by its structure, not the external field strength. The formula is:

$ \alpha = 4\pi\epsilon_0 a^3 $

This formula relates the induced dipole moment ($\mathbf{p}$) to the applied electric field ($\mathbf{E}$) through $\mathbf{p} = \alpha \mathbf{E}$. It is derived from the equilibrium condition of forces acting on the displaced electron cloud. The polarizability $\alpha$ is a measure of how easily the atom's electron cloud can be distorted.

Calculation of Hydrogen Atom Polarizability

The given parameters are:

• Radius of the uniformly distributed electron charge, $a = 0.5 \times 10^{-10} \text{ m}$
• Permittivity of free space, $\epsilon_0 \approx 8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2$

Substitute these values into the polarizability formula:

$ \alpha = 4\pi\epsilon_0 a^3 $

First, calculate $a^3$:

$ a^3 = (0.5 \times 10^{-10} \text{ m})^3 = (0.5)^3 \times (10^{-10})^3 \text{ m}^3 $

$ a^3 = 0.125 \times 10^{-30} \text{ m}^3 $

Now, compute $\alpha$:

$ \alpha = 4\pi \times (8.854 \times 10^{-12} \text{ C}^2/\text{Nm}^2) \times (0.125 \times 10^{-30} \text{ m}^3) $

$ \alpha = (4 \times \pi \times 8.854 \times 0.125) \times (10^{-12} \times 10^{-30}) \text{ C}^2\text{m}/\text{N} $

Using $\pi \approx 3.14159$:

$ \alpha \approx (12.566 \times 8.854 \times 0.125) \times 10^{-42} \text{ C}^2\text{m}/\text{N} $

$ \alpha \approx 13.896 \times 10^{-42} \text{ C}^2\text{m}/\text{N} $

Rounding to two significant figures gives:

$ \alpha \approx 1.4 \times 10^{-41} \text{ C}^2\text{m}/\text{N} $