Question 7.18: IONIC POLARIZATION RESONANCE IN KCl Consider a KCl crystal w...

At optical frequencies the dielectric constant \varepsilon _{rop} is determined by electronic polarization. At low frequencies and under dc conditions, the dielectric constant \varepsilon_{rdc} is determined by both electronic and ionic polarization. If N_{i} is the concentration of negative and positive ion pairs, then Equation 7.94 becomes

\frac{\varepsilon_{r}(\omega)-1}{\varepsilon_{r}(\omega)+2}-\frac{\varepsilon_{\text {rop }}-1}{\varepsilon_{\text {rop }}+2}=\frac{N_{i} \alpha_{i}}{3 \varepsilon_{o}}=\frac{N_{i} Q^{2}}{3 \varepsilon_{o} M_{r}\left(\omega_{I}^{2}-\omega^{2}+j \gamma_{I} \omega\right)}                [7.94]

\frac{\varepsilon_{\text {rdc }}-1}{\varepsilon_{\text {rdc }}+2}=\frac{\varepsilon_{\text {rop }}-1}{\varepsilon_{\text {rop }}+2}+\frac{1}{3 \varepsilon_{o}} N_{i} \alpha_{i}(0)

There are four negative and positive ion pairs per unit cell, and the cell dimension is a. The concentration of negative and positive ion pairs N_{i} is

N_{i}=\frac{4}{a^{3}}=\frac{4}{\left(0.629 \times 10^{-9}  m \right)^{3}}=1.61 \times 10^{28}  m ^{-3}

Substituting \varepsilon _{rdc} = 4.84 and \varepsilon _{rop} = 2.19 and N_{i} in Equation 7.94

\alpha_{i}(0)=\frac{3 \varepsilon_{o}}{N_{i}}\left[\frac{\varepsilon_{\text {rdc }}-1}{\varepsilon_{\text {rdc }}+2}-\frac{\varepsilon_{\text {rop }}-1}{\varepsilon_{\text {rop }}+2}\right]=\frac{3\left(8.85 \times 10^{-12}\right)}{1.61 \times 10^{28}}\left[\frac{4.84-1}{4.84+2}-\frac{2.19-1}{2.19+2}\right]

we find

\alpha_{i}(0)=4.58 \times 10^{-40}  F  m ^{2}

The relationship between \alpha _{i}(0) and the resonance absorption frequency involves the reduced mass M_{r} of the K^{+} −Cl^{−} ion pair,

M_{r}=\frac{M_{+} M_{-}}{M_{+}+M_{-}}=\frac{(39.09)(35.45)\left(10^{-3}\right)}{(39.09+35.45)\left(6.022 \times 10^{23}\right)}=3.09 \times 10^{-26}  kg

At \omega = 0, the polarizability is given by Equation 7.91, so the resonance absorption frequency \omega _{I} is

\alpha_{i}(0)=\frac{Q^{2}}{M_{r} \omega_{I}^{2}}                 [7.91]

\omega_{I}=\left[\frac{Q^{2}}{M_{r} \alpha_{i}(0)}\right]^{1 / 2}=\left[\frac{\left(1.6 \times 10^{-19}\right)^{2}}{\left(3.09 \times 10^{-26}\right)\left(4.58 \times 10^{-40}\right)}\right]^{1 / 2}=4.26 \times 10^{13}  rad  s ^{-1}

or                                                 f_{I}=\frac{\omega_{I}}{2 \pi}=6.8 \times 10^{12}  Hz

This is about a factor of 1.5 greater than the observed resonance absorption frequency of 4.5 \times  10^{12}  Hz. Typically one accounts for the difference by noting that the actual ionic charges may not be exactly +e on K^{+} and −e on Cl^{−}, but Q is effectively 0.76e. Taking Q = 0.76e makes f_{I} = 5.15 \times  10^{12} Hz, only 14 percent greater than the observed value. A closer agreement can be obtained by refining the simple theory and considering how many effective dipoles there are in the unit cell along the direction of the applied field.