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| | | ### Statement |
| | | |
| | − | $12.1.18.$ |
| | + | $12.1.18^\ast$. Estimate the depth of penetration of an electromagnetic wave perpendicular to its surface into the conductor. The wave frequency $\nu = 10^{15}\text{ Hz}$, the number of conduction electrons per unit volume $n_e = 10^{22}\text{ cm}^{-3}$. |
| | | |
| | − | Estimate the depth of penetration of an electromagnetic wave perpendicular |
| | − | to its surface into the conductor. The wave frequency v = 1015Hz, the number |
| | − | of conduction electrons per unit volume ne= 1022cm−3. |
| | − | |
| | | ### Solution |
| | | |
| | − | To estimate the penetration depth of an electromagnetic wave, the free electron plasma model is used. |
| | + | To estimate the penetration depth of a high-frequency electromagnetic wave, the free electron plasma model is used. |
| | + | At optical frequencies, the penetration depth of the field (skin depth) is characterized by the plasma frequency of the conductor $\omega_p$, which in SI is: |
| | + | $$\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$$ |
| | | |
| | − | At optical frequencies, the permittivity of the conductor is |
| | + | Substitute the known values ($e \approx 1.6 \cdot 10^{-19}\text{ C}$, $\varepsilon_0 \approx 8.85 \cdot 10^{-12}\text{ F/m}$, $m_e \approx 9.1 \cdot 10^{-31}\text{ kg}$) and the electron concentration $n_e = 10^{22}\text{ cm}^{-3} = 10^{28}\text{ m}^{-3}$: |
| | + | $$\omega_p^2 = \frac{10^{28} \cdot (1.6 \cdot 10^{-19})^2}{8.85 \cdot 10^{-12} \cdot 9.1 \cdot 10^{-31}} \approx 3.18 \cdot 10^{31}\text{ rad}^2/\text{s}^2$$ |
| | + | $$\omega_p \approx 5.6 \cdot 10^{15}\text{ rad/s}$$ |
| | | |
| | − | $\varepsilon(\omega) = 1 - \frac{\omega_p^2}{\omega^2}, |
| | − | \qquad \omega = 2\pi\nu$, |
| | + | The angular frequency of the incident wave is $\omega = 2\pi\nu = 2\pi \cdot 10^{15} \approx 6.3 \cdot 10^{15}\text{ rad/s}$. Since the wave frequency is of the same order of magnitude as the plasma frequency, a rough estimate for the characteristic field attenuation depth $\delta$ in the metal is given by the plasma length: |
| | + | $$\delta \approx \frac{c}{\omega_p}$$ |
| | | |
| | − | where the plasma frequency in SI is |
| | + | Numerically: |
| | + | $$\delta \approx \frac{3 \cdot 10^8\text{ m/s}}{5.6 \cdot 10^{15}\text{ s}^{-1}} \approx 5.4 \cdot 10^{-8}\text{ m}$$ |
| | + | Converting to convenient units, we get $\delta \approx 5 \cdot 10^{-6}\text{ cm}$, which equals $50\text{ nm}$. |
| | | |
| | − | $\omega_p = \sqrt{\frac{n_e e^2}{\varepsilon_0 m_e}}$. |
| | − | |
| | − | With e = $1.6\times10^{-19}\,\text{C}, \varepsilon_0 = 8.85\times10^{-12}\,\text{F/m}, m_e = 9.1\times10^{-31}\,\text{kg}$: |
| | − | |
| | − | $\omega_p^2 = \frac{10^{28}(1.6\times10^{-19})^2}{8.85\times10^{-12}\times9.1\times10^{-31}} \approx 3.2\times10^{31}\,\text{rad}^2/\text{s}^2, |
| | − | \quad \omega_p \approx 5.6\times10^{15}\,\text{rad/s}$ |
| | − | |
| | − | $Since \omega = 2\pi\times10^{15} \approx 6.3\times10^{15}\,\text{rad/s} > \omega_p$ |
| | − | the wave propagates with attenuation. The penetration depth \delta in a conductor is given by the damping length of the field intensity: |
| | − | |
| | − | $\delta = \frac{c}{\sqrt{\omega_p^2 - \omega^2}} \approx \frac{c}{\omega_p} |
| | − | \quad (\text{since } \omega \sim \omega_p)$ |
| | − | |
| | − | Numerically, |
| | − | |
| | − | $\delta \approx \frac{3\times10^{10}\,\text{cm/s}}{5.6\times10^{15}\,\text{s}^{-1}} \approx 5.4\times10^{-6}\,\text{cm}$ |
| | − | |
| | − | $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. |
| | − | |
| | | #### Answer |
| | − | |
| | − | $\boxed{\delta \approx 5\times10^{-6}\,\text{cm} \; (= 50\,\text{nm})}$. |
| | + | $\delta \approx 5 \cdot 10^{-6}\text{ cm}$ ($50\text{ nm}$). |
