For \(\mathrm{CaO}\), the ionic radii for \(\mathrm{Ca}^{2+}\) and \(\mathrm{O}^{2-}\) ions are \(0.100\) and \(0.140 \mathrm{~nm}\), respectively. If an externally applied electric field produces a \(5 \%\) expansion of the lattice, compute the dipole moment for each \(\mathrm{Ca}^{2+}-\mathrm{O}^{2-}\) pair. Assume that this material is completely unpolarized in the absence of an electric field.

Before we begin with calculating the dipole moment, we need to find the initial distance between the ions Ca\(^{2+}\) and O\(^{2-}\). This distance will be equal to the sum of the radii of these two ions: \(d_{0} = r_{\text{Ca}^{2+}} + r_{\text{O}^{2-}}\) Given, \(r_{\text{Ca}^{2+}} = 0.100\) nm and \(r_{\text{O}^{2-}} = 0.140\) nm, we can calculate \(d_{0}\): \(d_{0} = 0.100 + 0.140 = 0.240\) nm

We are given that an external electric field causes a 5% expansion in the lattice. This means, the distance between the ions would also increase by 5%. To calculate the new distance between the ions, we can use the following equation: \(d = d_{0} (1 + \frac{\Delta L}{L})\) Where \(d\) is the new separation, \(d_{0}\) is the initial separation, and \(\frac{\Delta L}{L} = 0.05\) (5% expansion). Substituting the values, we get: \(d = 0.240 \times (1 + 0.05) = 0.240 \times 1.05 = 0.252\) nm

We know that dipole moment (\(\mu\)) is given by: \(\mu = q \cdot d\) Where \(q\) is the charge on the ions and \(d\) is the distance between them. Since the charge on \(\text{Ca}^{2+}\) and \(\text{O}^{2-}\) ions are \(2e\) and \(-2e\) respectively, the product remains the same in magnitude (but opposite in direction). Calculating the dipole moment for Ca\(^{2+}\)-O\(^{2-}\) pair: \(\mu = 2e \cdot (d - d_{0})\) Where \(e\) is the elementary charge (\(1.6 \times 10^{-19}\) C), \(d_{0}\) is the initial distance (0.240 nm) and \(d\) is the new distance (0.252 nm). To ensure all the values are in the same unit, we convert nm to meters (1 nm = \(10^{-9}\) m): \(\mu = 2 \times (1.6 \times 10^{-19}) \times ((0.252 - 0.240) \times 10^{-9})\) \(\mu = 3.2 \times 10^{-19} \times 1.2 \times 10^{-10}\) \(\mu = 3.84 \times 10^{-29}\) Cm Thus, the dipole moment for each Ca\(^{2+}\)-O\(^{2-}\) pair is \(3.84 \times 10^{-29}\) Cm.