| @@ -1,8 +1,69 @@ |
| ### Problem |
| ### Problem |
| 6.6.2. The dielectric constant of helium at temperature $0^\circ C$ and pressure 1 atm is $1.000074$. Find the dipole moment of a helium atom in a uniform electric field of strength 300 V/cm. |
| 6.6.2. The dielectric constant of helium at temperature $0^\circ C$ and pressure 1 atm is $1.000074$. Find the dipole moment of a helium atom in a uniform electric field of strength 300 V/cm. |
| |
| |
| ### Solution |
| ### Solution |
| |
| |
| |
| |
| |
| ### 1st Solution Option |
| |
| |
| |
| #### 0. Write down what we have. |
| |
| \[ |
| |
| \varepsilon = 1.000074 |
| |
| \] |
| |
| |
| |
| \[ |
| |
| t = 0^\circ C \Leftrightarrow T = 273\ \text{K} |
| |
| \] |
| |
| |
| |
| \[ |
| |
| E = 300\ \text{V/cm} \Rightarrow 3 \times 10^{4}\ \text{V/m} |
| |
| \] |
| |
| |
| |
| \[ |
| |
| P = 1\ \text{atm} \approx 1.01 \times 10^{5}\ \text{Pa} |
| |
| \] |
| |
| |
| |
| #### 1. Find the concentration of atoms. |
| |
| |
| |
| Write down the ideal gas equation of state and express $n$: |
| |
| |
| |
| \[ |
| |
| P = nKT \Rightarrow n = \frac{P}{KT} |
| |
| \] |
| |
| |
| |
| #### 2. Relationship between $\varepsilon$ and polarization $P_{\text{pol}}$ |
| |
| |
| |
| For an isotropic dielectric: |
| |
| \[ |
| |
| \boxed{P_{\text{pol}} = \varepsilon_{0}(\varepsilon - 1)E} |
| |
| \] |
| |
| |
| |
| #### 3. Dipole moment of one atom. |
| |
| |
| |
| By definition: |
| |
| \[ |
| |
| P_{\text{pol}} := n \cdot p |
| |
| \] |
| |
| |
| |
| \[ |
| |
| \Downarrow |
| |
| \] |
| |
| |
| |
| \[ |
| |
| p = \frac{P_{\text{pol}}}{n} |
| |
| = \frac{\varepsilon_{0}(\varepsilon - 1)E \cdot KT}{P} |
| |
| = \frac{8.85 \times 10^{-12} \cdot (1.000074 - 1) \cdot 3 \times 10^{4} \cdot 1.38 \times 10^{-23} \cdot 273}{1.01 \times 10^{5}} |
| |
| \approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m} |
| |
| \] |
| |
| ### Answer |
| |
| \[ |
| |
| \boxed{p \approx 7.3 \times 10^{-37}\ \text{C} \cdot \text{m}} |
| |
| \] |
| |
| |
| |
| |
| |
| ### 2nd Solution Option |
| |
| |
| |
| |
| #### 0. Write down what we have |
| #### 0. Write down what we have |
| \[ |
| \[ |
| \varepsilon = 1.000074 |
| \varepsilon = 1.000074 |
| \] | | \] |
| \[ | | \[ |
| t = 0^\circ C \Leftrightarrow T = 273\ \text{K} | | t = 0^\circ C \Leftrightarrow T = 273\ \text{K} |
| \] | | \] |
| \[ | | \[ |
| E = 300\ \text{V/cm} \Rightarrow 3 \times 10^{4}\ \text{V/m} | | E = 300\ \text{V/cm} \Rightarrow 3 \times 10^{4}\ \text{V/m} |
| \] | | \] |
| \[ | | \[ |
| P = 1\ \text{atm} \approx 1.01 \times 10^{5}\ \text{Pa} | | P = 1\ \text{atm} \approx 1.01 \times 10^{5}\ \text{Pa} |
| \] | | \] |
| | | |
| #### 1. First, write down what we need to find. | | #### 1. First, write down what we need to find. |
| \[ | | \[ |
| p = \alpha E | | p = \alpha E |
| \] | | \] |
| where $\alpha$ is the polarizability of the atom, which can be found using the Clausius-Mossotti formula (an explanation of this formula will be given before the final answer, in case you are encountering it for the first time). | | where $\alpha$ is the polarizability of the atom, which can be found using the Clausius-Mossotti formula (an explanation of this formula will be given before the final answer, in case you are encountering it for the first time). |
| | | |
| #### 2. Finding the polarizability of the atom. | | #### 2. Finding the polarizability of the atom. |
| | | |
| Write down the Clausius--Mossotti formula, then express $\alpha$: | | Write down the Clausius--Mossotti formula, then express $\alpha$: |
| \[ | | \[ |
| \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n\alpha}{3\varepsilon_{0}}} | | \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{n\alpha}{3\varepsilon_{0}}} |
| \] | | \] |
| \[ | | \[ |
| \alpha = \frac{3\varepsilon_{0}}{n}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) | | \alpha = \frac{3\varepsilon_{0}}{n}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) |
| \] | | \] |
| where $n$ is the concentration of helium atoms, which we can find from the ideal gas equation of state: | | where $n$ is the concentration of helium atoms, which we can find from the ideal gas equation of state: |
| \[ | | \[ |
| P = nKT \Rightarrow n = \frac{P}{KT} | | P = nKT \Rightarrow n = \frac{P}{KT} |
| \] | | \] |
| where $K$ is the Boltzmann constant. | | where $K$ is the Boltzmann constant. |
| | | |
| Now we have what we were looking for: | | Now we have what we were looking for: |
| \[ | | \[ |
| \alpha = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) | | \alpha = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) |
| \] | | \] |
| | | |
| #### 3. Substitute $\alpha$ into the dipole moment formula and find the dipole moment of the helium atom. | | #### 3. Substitute $\alpha$ into the dipole moment formula and find the dipole moment of the helium atom. |
| \[ | | \[ |
| \boxed{p = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) \cdot E} | | \boxed{p = \frac{3KT\varepsilon_{0}}{P}\left(\frac{\varepsilon - 1}{\varepsilon + 2}\right) \cdot E} |
| \] | | \] |
| \[ | | \[ |
| p = \frac{3 \cdot 1.38 \times 10^{-23} \cdot 273 \cdot 8.85 \times 10^{-12}}{1.01 \times 10^{5}} | | p = \frac{3 \cdot 1.38 \times 10^{-23} \cdot 273 \cdot 8.85 \times 10^{-12}}{1.01 \times 10^{5}} |
| \times \left(\frac{1.000074 - 1}{1.000074 + 2}\right) | | \times \left(\frac{1.000074 - 1}{1.000074 + 2}\right) |
| \times 3 \times 10^{4} | | \times 3 \times 10^{4} |
| \approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m} | | \approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m} |
| \] | | \] |
| | | |
| #### Theory on the Clausius- Mossotti Formula | | #### Theory on the Clausius- Mossotti Formula |
| | | |
| The Clausius- Mossotti formula describes the relationship between the static dielectric constant of a dielectric and the polarizability of its constituent particles. It was derived independently by Ottaviano F. Mossotti in 1850 and by Rudolf J. E. Clausius in 1879. In cases where the substance consists of particles of one kind, in the Gaussian system of units the formula is: | | The Clausius- Mossotti formula describes the relationship between the static dielectric constant of a dielectric and the polarizability of its constituent particles. It was derived independently by Ottaviano F. Mossotti in 1850 and by Rudolf J. E. Clausius in 1879. In cases where the substance consists of particles of one kind, in the Gaussian system of units the formula is: |
| \[ | | \[ |
| \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} N \alpha} | | \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} N \alpha} |
| \] | | \] |
| where $\varepsilon$ is the dielectric constant, $N$ is the number of particles per unit volume, and $\alpha$ is their polarizability. | | where $\varepsilon$ is the dielectric constant, $N$ is the number of particles per unit volume, and $\alpha$ is their polarizability. |
| | | |
| If the substance consists of particles of different kinds, the formula takes the following form: | | If the substance consists of particles of different kinds, the formula takes the following form: |
| \[ | | \[ |
| \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} \sum_{i=1}^{j} N_{i} \alpha_{i}} | | \boxed{\frac{\varepsilon - 1}{\varepsilon + 2} = \frac{4\pi}{3} \sum_{i=1}^{j} N_{i} \alpha_{i}} |
| \] | | \] |
| | | |
| For more details, you can read here: (https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation) | | For more details, you can read here: (https://en.wikipedia.org/wiki/Clausius%E2%80%93Mossotti_relation) |
| | | |
| #### Answer | | #### Answer |
| \[ | | \[ |
| \boxed{p \approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m}} | | \boxed{p \approx 7.3 \times 10^{-38}\ \text{C} \cdot \text{m}} |
| \] | | \] |
| | | |
