In crystals of the salt cesium chloride, cesium ions Cs^+ form the eight corners

Question

In crystals of the salt cesium chloride, cesium ions Cs^+ form the eight corners of a cube and a chlorine ion Cl^- is at the cube's center (see the figure). The edge length of the cube is L 0.35 nm. The Cs^+ ions are each deficient by one electron (and thus each has a charge of +e), and the Cl^- ion has one excess electron (and thus has a charge of -e). (a) What is the magnitude of the net electrostatic force exerted on the Cl^- ion by the eight Cs^+ ions at the corners of the cube? (b) If one of the Cs^+ ions is missing, the crystal is said to have a defect; what is the magnitude of the net electrostatic force exerted on the Cl^- ion by the seven remaining Cs^+ ions?

Explanation / Answer

(a) Net electrostatic force due to eight Cs ions on Cl ion will be zero, As opposite Cs ions will attract Cl ion with equal and opposite forces along the four body diagonals
(b) The seven remaining ions will exert a force of only one Cs ion opposite to the missing ion. the effect of the other six paired ions will be again zero.
Specifically the force, F in this case = ke^2/[([(sq rt(3))/2]^2)*(0.35*10^-9)^2], where k = 9*10^9 and e = 1.6*10^-19
or F = 2.5*10^-9 N

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