In crystals of the salt cesium chloride, cesium ions Cs+ form the eight corners
ID: 249705 • Letter: I
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 = 0.30 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) Each cesium ion at a corner of the cube exerts a force of the same magnitude on the cl 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 each cesium ion with another, diametrically positioned at the opposite corner of the cube. Since the 02 ions in such a pair exert forces which have the same magnitude but are oppositely directed, the two forces sum to zero and, since every cesium ion can be paired in this way, the total force on the chlorine ion is zero.
(b). Rather than remove a cesium ion, we superpose charge –e at the position of one cesium ion. This neutralizes the ion, and as far as the electrical force on the chlorine ion is concerned, it is equivalent to removing the ion. The forces of the eight cesium ions at the cube corners sum to zero, so the only force on the chlorine ion is the force of the added charge.
The length of a body diagonal of a cube is sqrt(3)a , where a is the length of a cube edge.
F in this case = ke^2/[([(sq rt(3))/2]^2)*(30*10^-9)^], where k = 9*10^9 and e = 1.6*10^-19
or F = [[(9*10^9)*(1.6*10^-19)^2]*(4/3)]*[(10^18/900)]
= 3.4133*10^-13 N ..............Ans.
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