Question

I cannot figure out the final answer

Caesium chloride (CsCI), used in DNA separation and nuclear medicine, forms salt crystals built from the unit cells shown in the figure below. CI ions form the corners of a cube and a Cs ion is at the centre of the cube. The edge length of the cube, which is called the lattice constant, is 0.4 nm 0.4 nm Cst (a) What is the magnitude of the net force (in N) exerted on the cesium ion by its eight nearest Cl neighbours? (b) If the CI in the lower left corner is removed, what is the magnitude of the net force (in N) exerted on the cesium ion at the centre by the seven remaining nearest chlorine ions? 1.92 In what direction does this force act on the cesium ion? O away from the cormer with the removed CI O toward the corner with the removed Cl

Answer 1

Electrostatic force is given by:

F = k*q1*q2/R^2

Force will be attractive if both charge have opposite sign and Force will be repulsive if both charge has same sign.

Now Net force on the Cesium ion will be,

due to the 6 Cl^- ions (on the opposite corners) net force on Cs+ will be zero, now from the one corner we have removed Cl- ion, So due the opposite corner of this removed Cl- ion there will be net electrostatic force, which will be

q1 = charge on Cl^- ion = -e = -1.6*10^-19 C

q2 = charge on Cs⁺ ion = +e = +1.6*10^-19 C

R = distance between Center and one corner of cube = d*[sqrt (3)/2]

R = 0.4*(sqrt 3)/2 = 0.346 nm

k = electrostatic constant = 9*10^9

So,

F = 9*10^9*(1.6*10^-19)^2/(0.346*10^-9)^2

F = 1.92*10^-9 N

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