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

Answer:

Force of attraction = 35.96 $\times 10^{27}$N

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,

$F = \dfrac{k \times q_1 q_2}{ \r^2}$ ---(1)

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

k = $8.99 \times 10^9 \ Nm^{2}C^{-2}$

F = force of attraction

r = distance between ions.

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

$F = \dfrac{8.99 \times 10^9 \times 2 \times 2}{ \left ( 10^-9 \right )^2 }$

Force of attraction = 35.96 $\times 10^{27}$N