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 = 0.47 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

Given,

L = 0.47 nm = 0.47 x 10-9 m

a) Net net force will ne ZERO in this case, since every Cs+ ion at the corner of the cube will exert the force with the same magnitude on Cl- ion but these force will have an opposing force of the same magnitude from the charge present in apposite side.

b)We know that, electrostsic force is given by:

F = k q1q2/r2

where, k is the constant, q1 and q2 are the magnitudes of charges and r is the distance between them.

Lets assume that a -e charge has been intoduced to one of the corner, this ultimately nutralizes the Cs+ ion at that location.Now, as discussed in (a), the net force of three pairs( 6 Cs+) charges on Cl- will be zero. So the only force on Cl- is that due to the -e.

Distance between the two will be, r = sqrt(3)/2 L

F = k e2/ [sqrt(3)/2 L)]2 = 4 k e2/ 3 L2

F = 4 x 9 x 109 x 1.6 x 10-19 x 1.6 x 10-19/ 3 x 0.47 x 10-9 x 0.47 x 10-9 = 1.39 x 10-9 N =  1.4 x 10-9 N

Hence, F = 1.4 x 10-9 N

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