Edit to “Solution”

jzmicer edited Jun 28, 2026, 5:09 PM

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revision #19342 parent #19318 ← older

en/6.6.28.md +1 −1

			### Statement

			$6.6.28.$ 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 $F = eZ \cdot E$.
		@@ -12,7 +12,7 @@Solution

			This force is balanced by the force from the field of the electron cloud. This field can be found from Gauss's theorem:
			$$
		−	\oint \vec E_e \, d\vec S ~~= \varepsilon~~_0 ~~\sum Q \ \Rightarrow~~\ E_e \cdot 4\pi x^2 ~~= \varepsilon~~_0 ~~Z e~~ \frac{x^3}{R^3}.
		+	\oint\vec E_e d\vec S=\frac{ \sum Q}{\varepsilon_0} \ \to \ E_e\cdot 4\pi x^2=\frac{Ze}{\varepsilon_0 } \frac{x^3}{R^3}
			$$
			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.

			#### Answer
			$$
			\boxed{R \approx \sqrt[3]{\frac{(\varepsilon - 1) k T}{4\pi P}} \approx 0.12 \ \text{nm}}
			$$
		unchanged lines 19

	### Statement		### Statement

	$6.6.28.$ 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.		$6.6.28.$ 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		### Solution

	Concentration of atoms:		Concentration of atoms:
	$$		$$
	P = n k T \quad\Rightarrow\quad n = \frac{P}{kT}.		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 $F = eZ \cdot E$.		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 $F = eZ \cdot E$.
	@@ -12,7 +12,7 @@Solution

	This force is balanced by the force from the field of the electron cloud. This field can be found from Gauss's theorem:		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}.		\oint\vec E_e d\vec S=\frac{ \sum Q}{\varepsilon_0} \ \to \ E_e\cdot 4\pi x^2=\frac{Ze}{\varepsilon_0 } \frac{x^3}{R^3}
	$$		$$
	The equilibrium condition for the nucleus is:		The equilibrium condition for the nucleus is:
	$$		$$
	\vec E = -\vec E_e.		\vec E = -\vec E_e.
	$$		$$
	Then, due to the displacement, the atom acquires a dipole moment		Then, due to the displacement, the atom acquires a dipole moment
	$$		$$
	p = Z e x = 4\pi \varepsilon_0 R^3 E.		p = Z e x = 4\pi \varepsilon_0 R^3 E.
	$$		$$
	The total polarization of the medium is:		The total polarization of the medium is:
	$$		$$
	n p = (\varepsilon - 1) \varepsilon_0 E \ \Rightarrow\ 4\pi R^3 \frac{P}{kT} = \varepsilon - 1.		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}.		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.		This is of the order of one angstrom – quite plausible. Interestingly, the specific chemical element was not needed in the solution.

	#### Answer		#### Answer
	$$		$$
	\boxed{R \approx \sqrt[3]{\frac{(\varepsilon - 1) k T}{4\pi P}} \approx 0.12 \ \text{nm}}		\boxed{R \approx \sqrt[3]{\frac{(\varepsilon - 1) k T}{4\pi P}} \approx 0.12 \ \text{nm}}
	$$		$$
	unchanged lines 19