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In a crystal, the tetrahedral voids are occupied by atoms of radius 0.1 nm. The edge length of the cubic unit cell is 0.5 nm. What is the packing efficiency  percentage of tetrahedral voids occupied by atoms?

 

Option: 1

25%


Option: 2

50%


Option: 3

75%


Option: 4

100%


Answers (1)

best_answer

The radius of the tetrahedral void, r = 0.1 nm

The distance between two opposite faces of the cubic unit cell, the diagonal of the cube, d = a \times \sqrt 3, where a is the edge length of the cube.

Here, a = 0.5 nm

So, d = 0.5\ nm \times \sqrt 3 = 0.866\ nm 

The diameter of the tetrahedral void,

D = \sqrt 2 \times r = 0.141\ nm

The fraction of tetrahedral voids occupied by atoms

\\\\= (4/3) \times \pi \times (r/D)^3\\ \\ = (4/3) \times \pi \times (0.1/0.141)^3 \\\\\approx 0.5

Therefore, the percentage of tetrahedral voids occupied by atoms = 50%.

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vinayak

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