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 = 0.40 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) from the figure

every cesium ion at a corner of the cube exerts a force of the same magnitude on the chlorine ion at the cube center. each force is a force of attraction and is directed toward the cesium ion that exerts it , along the body diagonal of the cube . we can pair every cesium ion with another, diametrically positioned at the opposite corner of the cube. since the ions in such a pair exert forces that have the same magnitude but are oppositely directed, teh two forces sum ti zero and , since every cesium ion can be paired so total force on the chlorine ion is zero.

b)   we supposse electron at the posotion of one cesium ion . this neutralizes ion , and as the electric force on the chlorine ion is concerned.

the length of a body diagonal of a cube is 3a

teh distance from the center of the cube to a corner is d = 3/2 * a

the force magnitde is

F = k ( -e) ( -e)/d2

    = ke2/d2

   = 9*109 * ( 1.6 * 10-19C)2/ 3/4 *( 0.4 * 10-9)2

    =1.92 * 10-9N

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