Question

18.53 For CaO, the ionic radii for Ca2 and O2 ions are 0.100 and 0.140 nm, respectively. If an externally applied electric field produces a 5% expansion of the lattice, compute the dipole moment for each Ca2O2pair. Assume that this material is completely unpolarized in the absence of an electric field.

Answer 1

Answer:

$= 1.92*10^-^3^0 C.m$

Explanation:

Given:

$r_ca_^2_^+ = 0.100 nm$

$r_o_^2_^- = 0.14 nm$

Let's find the distance of separation between cation and anion when there is no applied electric field with the formula:

$d = r_ca_^2_^+ + r_o_^2_^-$

d = 0.100 nm + 0.140 nm

= 0.240 nm

Let's also calculate the distance of separation between anion and cation when there is an applied electric field.

We use the formula:

∆d = 5%d => 0.05d

= 0.05 * 0.024 nm

∆d = 0.0120 nm

$0.120* 10^-^9m$

$= 1.20*10^-^1^1m$

Given magnitude of each dipole= $1.602*10^-^1^9 C$

Let's find the dipole moment, with the formula:

p = q∆d

Substituting figures in the formula, we have:

$p = 1.602*10^-^1^9 * 12*10^-^2^1$

$= 1.92*10^-^3^0 C.m$

Answer 2

Answer:

Explanation:

Answer:

= 1.92*10^-^3^0 C.m

Explanation:

Given:

r_ca_^2_^+ = 0.100 nm

r_o_^2_^- = 0.14 nm

Let's find the distance of separation between cation and anion when there is no applied electric field with the formula:

d = 0.100 nm + 0.140 nm = 0.240 nm

Let's also calculate the distance of separation between anion and cation when there is an applied electric field.

We use the formula:

∆d = 5%d => 0.05d

= 0.05 * 0.024 nm

∆d = 0.0120 nm

0.120* 10^-^9m

= 1.20*10^-^1^1m

Given magnitude of each dipole= 1.602*10^-^1^9 C

Let's find the dipole moment, with the formula:

p = q∆d

Substituting figures in the formula, we have:

p = 1.602*10^-^1^9 * 12*10^-^2^1

= 1.92*10^-^3^0 C.m