If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?(a) S = 32d^4/3(b) S = 2d^2(c) S = 4d^2(d) S = 8d^2I have been asked this question in semester exam.My question is from Solid State in portion Solid State of Chemistry – Class 12

If the body-centered unit cell is assumed to be a cube of edge length

1.	If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?(a) S = 32d^4/3(b) S = 2d^2(c) S = 4d^2(d) S = 8d^2I have been asked this question in semester exam.My question is from Solid State in portion Solid State of Chemistry – Class 12
Answer» Correct answer is (d) S = 8d^2 Explanation: For BCC unit CELL the relation between radius of a particle ‘r’ and edge length of unit cell, a, is r = \(\FRAC{\sqrt{3}}{4}\)a. We know that diameter, d = 2r = \(\frac{\sqrt{3}}{2}\)a Implying d^2 = \(\frac{3}{4}\)a^2 Therefore, 4d^2/3=a^2 Multiplying by 6 on both sides gives S = 6a^2 = 8d^2, where S is the surface area of the CUBE = 6a^2.

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