91Ó°ÊÓ

1. The edge length of the cube,$\mathrm{a}=0.40\text{nm}$.
2. The ${\mathrm{Cs}}^{+}$ions are each deficient by one electron (and thus each has a charge of $+e$), and the $C{l}^{−}$ion has one excess electron (and thus has a charge of $−e$).

Using the same concept of Coulomb's law and the formula of net charge, we can get the net force applied on the chlorine ion respective to the given number of ions present.

Formula:

The magnitude of the electrostatic force between any two particles, $\mathrm{F}=\mathrm{k}\frac{\left|{\mathrm{q}}_1\right|\left|{\mathrm{q}}_2\right|}{{\mathrm{r}}^2}$ (i)

Since, there are two${\mathrm{Cs}}^{+}$ ions in such a way that the pair exerts forces that have the same magnitude but are oppositely directed; thus, the net force due to them on the chlorine ion is sum to zero.

Similarly for every cesium ion, it can be paired in this way; the total force on the chlorine ion is zero.

The length of a body diagonal of a cube is $\sqrt{3a}$, where a is the length of a cube edge.

Thus, the distance from the centre of the cube to a corner (that is the distance between the cesium and chlorine ions) is $\mathrm{r}=(\sqrt{3}/2)\mathrm{a}$

Thus, the magnitude of net force using equation (i) is

$\begin{array}{c}\mathrm{F}=\frac{{\mathrm{ke}}^2}{(3/4){\mathrm{a}}^2}\\ \text{(asallpairscanceleachother'sforceandremaining}\\ \text{onecesiumiongivesthenetforce)}\\ =\frac{(9.0×{10}^9{\text{Nm}}^{\text{2}}{\text{/C}}^{\text{2}}){(1.6×{10}^{−19}\text{C})}^2}{(3/4){(0.4×{10}^{−9}\text{m})}^2}\\ =1.9×{10}^{−9}\text{N}\end{array}$

Since both the added charge and the chlorine ion are negative, the force is repulsive. The chlorine ion is pushed away from the site of the missing cesium ion.

Hence, the value of the force is $1.9×{10}^{-9}\text{ N}$.