There are \(1.6 \times 10^{20}\) molecules \(/ \mathrm{m}^{3}\) in \(\mathrm{NaCl}\) vapour. Determine the orientational polarization at room temperature if the vapour is subjected to a field of \(5 \times 10^{6} \mathrm{~V} / \mathrm{m}\). Assume that the \(\mathrm{NaCl}\) molecule consists of \(\mathrm{Na}^{+}\) and \(\mathrm{Cl}^{-}\) ions separated by \(2.5 \AA\).

Identify and list all the given values:- Density of molecules, \( N = 1.6 \times 10^{20} \) molecules \m^3- Electric field, \( E = 5 \times 10^6 \) V/m- Separation between ions, \( d = 2.5 \AA = 2.5 \times 10^{-10} \) m- The charge of an \mathrm{Na}^{+} or \mathrm{Cl}^{-} ion, \( q = 1.6 \times 10^{-19} \) C

The dipole moment \( p \) of a single \mathrm{NaCl} molecule can be calculated using the formula:\[ p = q \times d \]Substitute the given values into the formula:\[ p = (1.6 \times 10^{-19} \text{C}) \times (2.5 \times 10^{-10} \text{m}) = 4.0 \times 10^{-29} \text{C} \text{m} \]

Use the formula for the orientational polarization \( P_o \):\[ P_o = \frac{N p^2 E}{k_B T} \]where: - \( k_B \) is the Boltzmann constant \(8.617 \times 10^{-5} \) eV/K- \( T \) is the room temperature \(300 \) KSince orientational polarization depends on the temperature and the Boltzmann constant ( \text{k_B} has to be in \text{J}): substitute in with Joules for Boltzmann's constant ( \( 8.617 \times 10^{-5} \text{eV/K} = 1.38 \times 10^{-23} \text{J/K} \) ).\[ P_o = \frac{(1.6 \times 10^{20}) (4.0 \times 10^{-29})^2 (5 \times 10^6)}{(1.38 \times 10^{-23}) (300)} \] \[ P_o = \frac{(1.6 \times 10^{20}) (16 \times 10^{-58}) (5 \times 10^6)}{4.14 \times 10^{-21}} \] \[ P_o = \frac{1.28 \times 10^{-31}}{4.14 \times 10^{-21}} = 3.09 \times 10^{-11} \text{Cm}^2 \text{m}^{-3} \]