Question

Consider two different ions. The anion has a valence of -2. The cation has a valence of +2. The two ions are separated by a distance of 1 nm. Please calculate the force of attraction between the anion and cation. The force of attraction is given by: (9 x 109 V/C) (Z)(2)(e2) FA valence of the ions, e = charge of an electron 1.602 x 10-19 C Where Z1 and Z2 r distance between ions 1N 1 (V C/m)

Answer 1

Force of attraction = 35.96
`* 10^(27)`N

Step-by-step explanation:

Given: charge on anion = -2

Charge on cation = +2

Distance = 1 nm =
`10^(-9)` m

To calculate: Force of attraction.

Solution: The force of attraction is calculated by using equation,

`F = (k * q_1 q_2)/( \r^2)` ---(1)

where, q represents the charge and the subscripts 1 and 2 represents cation and anion.

k =
`8.99 * 10^9 \ Nm^(2)C^(-2)`

F = force of attraction

r = distance between ions.

Substituting all the values in the equation (1) the equation becomes

`F = (8.99 * 10^9 * 2 * 2)/( \left ( 10^-9 \right )^2 )`

Force of attraction = 35.96
`* 10^(27)`N