Answer:
A) F_a = 45.19 × 10^(-12) N
B) F_repulsive = -45.19 × 10^(-12)
Explanation:
We are given;
atomic radii of a divalent cation = 0.97 nm
atomic radii of monovalent anion = 0.135 nm
Distance between the 2 atomic centres; R = 0.97 + 0.135 = 1.105 nm = 1.105 × 10^(-9) m
A) Now, Force of attraction will be given by the formula;
F_a = [(z1 × e)•(z2 × e)]/(4πε_o•R²)
Where;
z1 is the valency of the divalent cation = 2
z2 is the valency of the monovalent anion = 1
e is electron charge = 1.602 × 10^(-19) C
ε_o is vacuum of permeability with a constant value of 8.85 × 10^(-12) C²/N.m²
Plugging in the relevant values;
F_a = [(2 × 1.602 × 10^(-19))•(1 × 1.602 × 10^(-19))]/(4π × 8.85 × 10^(-12) × (1.105 × 10^(-9))²)
F_a = (513.2808 × 10^(-40))/(11.35793096213 × 10^(-28))
F_a = 45.19 × 10^(-12) N
B) At same separation distance which is equilibrium;
F_repulsive = -F_a
F_repulsive = -45.19 × 10^(-12)
