If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?(a) 1.32 × 10^6 pm^2(b) 1.32 × 10^-18pm^2(c) 2.20 × 10^5 pm^2(d) 2.20 × 10^-19 pm^2The question was asked in an international level competition.This is a very interesting question from Solid State in division Solid State of Chemistry – Class 12

If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC

1.	If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?(a) 1.32 × 10^6 pm^2(b) 1.32 × 10^-18pm^2(c) 2.20 × 10^5 pm^2(d) 2.20 × 10^-19 pm^2The question was asked in an international level competition.This is a very interesting question from Solid State in division Solid State of Chemistry – Class 12
Answer» The correct answer is (a) 1.32 × 10^6 pm^2 To EXPLAIN: ABCABC arrangement is found in CCP. In closed CUBIC packing, relation between edge length of unit CELL, a, and RADIUS of particle, r, is given as a=2\(\sqrt{2}\)r. Surface area (S.A.) = 6a^2 From the RELATIONSHIP, a^2 = 8r^2 S.A. = 6a^2 = 48r^2 When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10^6 pm^2.

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