6.6.28. The dielectric permittivity of argon at temperature $0 \, ^\circ C$ and pressure $1 \, \text{atm}$ is $1.00056$. Estimate the radius of the argon atom, assuming that the charge of the electrons is uniformly distributed throughout the volume of the atom, and that the nucleus is at the centre of the atom.
Solution
Concentration of atoms: $$P = n k T \quad\Rightarrow\quad n = \frac{P}{kT}.$$ When an atom is placed in an external electric field $E$, the nucleus is displaced relative to the centre of the electron cloud by a distance $x$ under the action of a force $F = eZ \cdot E$.
This force is balanced by the force from the field of the electron cloud. This field can be found from Gauss's theorem: $$\oint \vec E_e \, d\vec S = \varepsilon_0 \sum Q \ \Rightarrow\ E_e \cdot 4\pi x^2 = \varepsilon_0 Z e \frac{x^3}{R^3}.$$ The equilibrium condition for the nucleus is: $$\vec E = -\vec E_e.$$ Then, due to the displacement, the atom acquires a dipole moment $$p = Z e x = 4\pi \varepsilon_0 R^3 E.$$ The total polarization of the medium is: $$n p = (\varepsilon - 1) \varepsilon_0 E \ \Rightarrow\ 4\pi R^3 \frac{P}{kT} = \varepsilon - 1.$$ $$R \approx \sqrt[3]{\frac{(\varepsilon - 1) k T}{4\pi P}} \approx 1.2 \cdot 10^{-10} \ \text{m} = 0.12 \ \text{nm}.$$ This is of the order of one angstrom – quite plausible. Interestingly, the specific chemical element was not needed in the solution.