CBSE Class 12-science Answered

Figure shows the CsCl lattice that is made in a cube of side 40 nm . Positively charged Cs ions ( deficient in one electron )

are at corners of cube and negatively charged chlorine ion ( excess of one electron ) is at centre O of cube.

Attractive electrostatic force between Cl ion and Cs ion at A is nullified by attractive electrostatic force between Cl ion and Cs ion at H .

Similarly , Attractive electrostatic force between Cl ion and Cs ion at D is nullified by attractive electrostatic force between Cl ion and Cs ion at E .

Attractive electrostatic force between Cl ion and Cs ion at G is nullified by attractive electrostatic force between Cl ion and Cs ion at B

Attractive electrostatic force between Cl ion and Cs ion at F is nullified by attractive electrostatic force between Cl ion and Cs ion at C .

Hence net electrostatic force on Cl ion is zero .

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If Cs ion at A is removed , then attractive electrostatic force between Cl ion and Cs ion at H as shown in above fifure is the only force that is unbalanced .

Other forces between Cl ion and Cs ion at oter corners are nullified as explained in previous part.

Hence net force on Cl ion is given by

where K = 1/(4πε_o ) = 9 × 10⁹ N m² C^-2 is Coulomb's constant , q is magnitude of charge on each ion

and d is distance between Cl ion and Cs ion .

d² = (3/4) [ 40 × 10^-9 ]² = 1.2 × 10^-15 (nm)²

Hence net force on Cl ion is given by