Question

From the ionic radii given in the figure, calculate the potential energy of a single Ca−O ion pair that is just touching.

Answer 1

The potential energy of a single Ca−O ion pair that is just touching is approximately
`\(1.50 * 10^(-18) \, \text{J}\).`

To calculate the potential energy (U) of a single Ca−O ion pair just touching, you can use Coulomb's law, which is given by the formula:

`\[ U = \frac{q_1 \cdot q_2}{{r}} \]`

Where:

- k is Coulomb's constant (
`\(8.99 * 10^9 \, \text{Jm/C}^2\)`)

-
`\( q_1 \) and \( q_2 \)` are the charges of the ions (in Coulombs)

- r is the distance between the centers of the ions (in meters)

Given ionic radii:

-
`\( r_{\text{Ca}} = 1.76 \)` (ionic radius of Ca in Ångströms)

-
`\( r_{\text{O}} = 0.66 \)` (ionic radius of O in Ångströms)

The distance (r) is the sum of the ionic radii when the ions are just touching:

`\[ r = r_{\text{Ca}} + r_{\text{O}} \]`

Calculate the potential energy (U):

`\[ U = \frac{{k \cdot |q_{\text{Ca}} \cdot q_{\text{O}}|}}{{r}} \]`

Now, considering that
`\( q_{\text{Ca}} \) and \( q_{\text{O}} \)` are the charges of Ca and O ions (which are typically +2 and -2 respectively), calculate U.

`\[ U = \frac{{(8.99 * 10^9 \, \text{Jm/C}^2) \cdot |(+2) \cdot (-2)|}}{{r_{\text{Ca}} + r_{\text{O}}}} \]`

`\[ U \approx \frac{{8.99 * 10^9 * 4}}{{1.76 + 0.66}} \, \text{J} \]`

`\[ U \approx 1.50 * 10^(-18) \, \text{J} \]`

The probable question may be:

From the ionic radii given in the figure (O= 0.66, Ca= 1.76), calculate the potential energy of a single Ca−O ion pair that is just touching.

e=1.60 x 10^-19C

k=8.99 x 10^9 Jm/C^2)