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A metal crystallises in a face-centred cubic (fcc) structure. If the radius of the metal atom is r , what is the shortest distance between the centres of two nearest tetrahedral voids in the lattice?

Options

  1. A2 r
  2. B2r
  3. C2 2 r
  4. Dr 2

Correct answer

A. 2 r

Step-by-step solution

In an fcc unit cell, there are 8 tetrahedral voids located on the body diagonals. These voids form a smaller simple cube of edge length a 2 within the unit cell. Thus, the shortest distance between the centres of two nearest tetrahedral voids is a 2 . For an fcc lattice, the relationship between the edge length a and the atomic radius r is given by: 2 a = 4r a = 2 2 r Substituting the value of a into the distance expression: Shortest distance = a 2 = 2 2 r 2 = 2 r Answer: 2 r

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