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 (Fig. 21-35). The edge length of the cube is L = 0.42 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) force of the same magnitude on the chlorine ion at the cube centeris exerted by Every cesium ion at a corner of the cube Each force is a due to the attraction phenomena and is directed towardthe cesium ion that causes its exertion along the cube's diagonal.One can pair each cesium ion with other, diametrically lined up atthe opposite corner. Because of the fact that the two ions in thuspair formed exert forces that have the same magnitude but are inthe opposite direction,the net force is zero. and for all pairswith zero force. net force is 0. b) F= ke2/d2= ke2/(3/4)a2 =( 9 x109Nm2/C2)(1.6x10-19C)2 / (3/4)(0.4x10-9m)2 = 1.9*10-9 N

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