Question 7.3: ELECTRONIC POLARIZABILITY OF COVALENT SOLIDS Consider a pure...
ELECTRONIC POLARIZABILITY OF COVALENT SOLIDS Consider a pure Si crystal that has \varepsilon _r = 11.9.
a. What is the electronic polarizability due to valence electrons per Si atom (if one could portion the observed crystal polarization to individual atoms)?
b. Suppose that a Si crystal sample is electroded on opposite faces and has a voltage applied across it. By how much is the local field greater than the applied field?
c. What is the resonant frequency f_o corresponding to \omega _{o}?
From the density of the Si crystal, the number of Si atoms per unit volume, N, is given as 5 \times 10^{28} m^{−3} .
a. Given the number of Si atoms, we can apply the Clausius–Mossotti equation to find \alpha _{e}
\alpha_{e}=\frac{3 \varepsilon_{o}}{N} \frac{\varepsilon_{r}-1}{\varepsilon_{r}+2}=\frac{3\left(8.85 \times 10^{-12}\right)}{\left(5 \times 10^{28}\right)} \frac{11.9-1}{11.9+2}=4.17 \times 10^{-40} F m ^{2}
This is larger, for example, than the electronic polarizability of an isolated Ar atom, which has more electrons. If we were to take the inner electrons in each Si atom as very roughly representing Ne, we would expect their contribution to the overall electronic polarizability to be roughly the same as the Ne atom, which is 0.45 \times 10^{−40} F m^{2}.
b. The local field is
E_{ loc }=E+\frac{1}{3 \varepsilon_{o}} P
But, by definition,
P=\chi_{e} \varepsilon_{o} E=\left(\varepsilon_{r}-1\right) \varepsilon_{o} E
Substituting for P,
E_{ loc }=E+\frac{1}{3}\left(\varepsilon_{r}-1\right) E
so the local field with respect to the applied field is
\frac{E_{ loc }}{E}=\frac{1}{3}\left(\varepsilon_{r}+2\right)=4.63
The local field is a factor of 4.63 greater than the applied field.
c. Since polarization is due to valence electrons and there are four per Si atom, we can use Equation 7.7,
\alpha_{e}=\frac{Z e^{2}}{m_{e} \omega_{o}^{2}} [7.7]
\omega_{o}=\left(\frac{Z e^{2}}{m_{e} \alpha_{e}}\right)^{1 / 2}=\left[\frac{4\left(1.6 \times 10^{-19}\right)^{2}}{\left(9.1 \times 10^{-31}\right)\left(4.17 \times 10^{-40}\right)}\right]^{1 / 2}=1.65 \times 10^{16} rad s ^{-1}
The corresponding resonant frequency is \omega_{o} / 2 \pi \text { or } 2.6 \times 10^{15} Hz , which is typically associated with electromagnetic waves of wavelength in the ultraviolet region.