Answer:
The answer is "1.29".
Explanation:
Formula for electrostatic potential energy:
[tex]\to V=\frac{K Q_1 \times Q_2}{r}[/tex]
where
[tex]\to K = \text{electrostatic constant}\\\\\to Q= \text{Charge on Cation}\\\\[/tex]
The Charge on Cation is [tex]C_{s}^{+}[/tex] is cation in both [tex]C_sI[/tex] and [tex]C_sF[/tex] were same:
[tex]\to V\ \alpha \ \frac{1}{r}[/tex]
[tex]\to \frac{V(C_sF)}{V(C_sI)} = \frac{r(C_sI)}{r(C_sF)}[/tex]
[tex]= \frac{r(C_s^{+}) +r(I^{-})}{r(C_s^{+}) +r(F^{-})} \\\\=\frac{170+220}{170+133} \\\\=\frac{390}{303} \\\\ =1.2871 \\\\ =1.29[/tex]
