Question

Calculate the force of attraction between a cation with a valence of +2 and an anion with a valence of -1, the centers of which are separated by a distance of 2.6 nm.

Answer 1

Force of attraction between cation and anion

charge on cation (q1) = +2 = 2 x 1.9 x 10^-19 C = 3.8 x 10^-19 C

charge on anion (q2) = -1 = 1 x 1.9 x 10^-19 C = 1.9 x 10^-19 C

distance (d) between the cation and anion = 2.6 nm = 2.6 x 10^-9 m

Using relation,

force of attraction (F)

F = k.q1.q2/d^2

with,

k being a constant = 8.99 x 10^9 N.m^2/C^2

Feeding the above values,

force of attraction (F) = (8.99 x 10^9) x (1.9 x 10^-19) x (3.8 x 10^-19)/(2.6 x 10^-9)^2

                                  = 9.602 x 10^-11 N

[N is for Newton units]