Respuesta :
The attractive force between two isolated ions is given by the formula:
[tex]F = \frac{1}{4\pi \xi _o}\frac{(z_1e)(z_2e)}{r^{2}}[/tex] -(1)
where,
[tex]F[/tex] is force of attraction, [tex]\xi _o[/tex] is permittivity of vacuum = [tex]8.85 \times 10^{12} F/m[/tex], [tex]z_1[/tex] and [tex]z_2[/tex] are valence of two ions, [tex]e[/tex] is the electron charge = [tex]1.62 \times 10^{19} C[/tex] and [tex]r[/tex] is the inter-atomic distance.
The inter-atomic distance, [tex]r = r_c+r_a[/tex]
where, [tex]r_c[/tex] is the radius of cation and [tex]r_a[/tex] is the radius of anion.
Since value of radius of cation is given so,
[tex]r = 0.096 + r_a[/tex] -(2)
The cation and anion are divalent so the value of [tex]z_1[/tex] and [tex]z_2[/tex] is 2 and -2 of cation and anion respectively.
Substituting the values in formula (1):
[tex]1.49\times 10^{-8} N = \frac{1}{4\pi\times 8.85\times10^{-12}F/m}\frac{(2)\times 1.69\times 10^{-19}C\times (-2)\times 1.69\times 10^{-19}C}{r^{2}}[/tex]
[tex]1.49\times 10^{-8} = \frac{9.231\times 10^{-28}}{r^{2}}[/tex]
[tex]r^{2} = \frac{9.231\times 10^{-28}}{1.49\times 10^{-8}}[/tex]
[tex]r = 2.489\times 10^{-10} m[/tex] = [tex]0.2489 nm[/tex]
Substituting this value of [tex]r[/tex] in equation 2:
[tex]0.2489 nm = 0.096 + r_a[/tex]
[tex]r_a = 0.1529 nm[/tex]
Hence, the radius of anion is [tex]0.1529 nm[/tex].
