Question

Calculate the force of attraction between a Ca^2+ and an 0^2- irons whose centers are sep by a distance of 1.25 nm.

Answer 1

Given:

Calcium and Oxygen ions each having valence = 2 or 2 valence electrons

seperation distance, r = 1.25nm =
`1.25* 10^(-9)` m

Coulombian force for two point charges seperated by a distance 'r' is given by:

F =
`\frac{1}{4\pi\epsilon _{_(o)} }* (q_(1)q_(2))/(r^(2))` (1)

Now, we know that

`(1)/(4\pi \epsilon _(o)) = 9* 10^{^(9)}`

q = ne

where, n = no. of electrons

e = charge of an electron =
`1.6* 10^(-19)` C

`q_(1)` =
`n_(1)`e = 2e

`q_(2)` =
`n_(2)`e = 2e

Using Eqn (1)

F =
`(9* 10^{^(9)})`
`* \frac{4* (1.6* 10^(-19))^{^(2)}}{(1.25* 10^(-9))^(2)}`

F =
`5.89*10^(-10)` N

Since, the force is between two opposite charged ions, it is attractive in nature.