Explanation:
Given that,
Radius of divalent cation, [tex]a_1=0.19\ nm[/tex]
Radius of monovalent anion, [tex]a_2=0.126\ nm[/tex]
Charge for divalent cation, [tex]Z_1=+2[/tex]
Charge for monovalent cation, [tex]Z_2=-1[/tex]
The sum of radii of both cations and anion is, [tex]a_o=0.19+0.126=0.316\ nm=0.316\times 10^{-9}\ m[/tex]
(a) The force of attraction between two ions is given by :
[tex]F=k\dfrac{-Z_1Z_2e^2}{a_o^2}[/tex]
[tex]F=-9\times 10^9\times \dfrac{2\times (-1)\times (1.6\times 10^{-19})^2}{(0.316\times 10^{-9})^2}[/tex]
[tex]F=4.61\times 10^{-9}\ N[/tex]
(b) Let F' is the force of repulsion at this same distance such that,
F + F' = 0
So, [tex]F'=-4.61\times 10^{-9}\ N[/tex]
Hence, this is the required solution.
