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 = 0.46 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) every Cs+ ion will exert same magnitude force on Cl- .

and for every Cs+ ion forces there be a force in opposite direction due to another diagonally opposite ion.

hence F_net = (F - F) + (F - F) + (F - F) + (F - F) = 0

(B) Now F_net = (F) + (F - F) + (F - F) + (F - F)

= F

and F = k e e / r^2

r = half of diagonal length = sqrt(0.46^2 + 0.46^2 + 0.46^2)/2

r = 0.398 m

F = (9 x 10^9)(1.6 x 10^-19)^2 / (0.398^2)

F = 1.45 x 10^-27 N

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