Правка раздела «Solution»
en/6.6.28.md
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| ### 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: | |||
| $$ | |||
| − | \oint | ||
| + | \oint\vec E_e d\vec S=\frac{ \sum Q}{\varepsilon_0} \ \to \ E_e\cdot 4\pi x^2=\frac{Ze}{\varepsilon_0 } \frac{x^3}{R^3} | ||
| $$ | |||
| 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 | |||
| $$ | |||
| \boxed{R \approx \sqrt[3]{\frac{(\varepsilon - 1) k T}{4\pi P}} \approx 0.12 \ \text{nm}} | |||
| $$ | |||
| ещё строк без изменений 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 |
\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}} | ||
| $$ | $$ | ||
| ещё строк без изменений 19 | |||