Question 7.2: ELECTRONIC POLARIZABILITY OF A VAN DER WAALS SOLID The elect...
ELECTRONIC POLARIZABILITY OF A VAN DER WAALS SOLID The electronic polarizability of the Ar atom is 1.7 \times 10^{−40} F m^{2}. What is the static dielectric constant of solid Ar (below 84 K) if its density is 1.8 g cm^{−3}?
To calculate \varepsilon_{r} we need the number of Ar atoms per unit volume N from the density d. If M_{at} = 39.95 is the relative atomic mass of Ar and N_{A} is Avogadro’s number, then
N=\frac{N_{A} d}{M_{ at }}=\frac{\left(6.02 \times 10^{23} mol ^{-1}\right)\left(1.8 g cm ^{-3}\right)}{\left(39.95 g mol ^{-1}\right)}=2.71 \times 10^{22} cm ^{-3}
with N=2.71 \times 10^{28} m ^{-3} \text { and } \alpha_{e}=1.7 \times 10^{-40} F m ^{2} , we have
\varepsilon_{r}=1+\frac{N \alpha_{e}}{\varepsilon_{o}}=1+\frac{\left(2.71 \times 10^{28}\right)\left(1.7 \times 10^{-40}\right)}{\left(8.85 \times 10^{-12}\right)}=1.52
If we use the Clausius–Mossotti equation, we get
\varepsilon_{r}=\frac{1+\frac{2 N \alpha_{e}}{3 \varepsilon_{o}}}{1-\frac{N \alpha_{e}}{3 \varepsilon_{o}}}=1.63
The two values are different by about 7 percent. The simple relationship in Equation 7.14 underestimates the relative permittivity.
\varepsilon_{r}=1+\frac{N \alpha_{e}}{\varepsilon_{o}} [7.14]