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#### In a face-centred cubic (FCC) structure, what is the fraction of octahedral voids occupied by particles?Option: 1 $1/2$Option: 2 $1/2$Option: 3 $3/4$Option: 4 $4/5$

In a face-centred cubic (FCC) structure, each unit cell contains 4 particles located at the corners and one particle located at the centre of each face of the cube. The coordination number of each particle in an FCC structure is 12.An octahedral void is a space between six particles arranged in the shape of an octahedron. In an FCC structure, each particle at the centre of a face of the cube is surrounded by six nearest neighbour particles located at the corners of adjacent unit cells.

Thus, each face-centred particle contributes $1/2$ an octahedral void to the structure.Since there are 6 face-centred particles per unit cell in an FCC structure, the total number of octahedral voids per unit cell is $6 x 1/2 = 3.$

Now, we need to determine how many of these octahedral voids are occupied by particles. Each octahedral void can be occupied by one atom or ion. In an ideal FCC structure, all octahedral voids are occupied. Therefore, the fraction of octahedral voids occupied by particles is given by:

$\mathrm{Fraction\ of \ octahedral \ voids\ occupied = \frac{Number\ of\ particles }{ Total\ number\ of \ octahedral\ voids}}$

The number of particles in an FCC unit cell is 4, and the total number of octahedral voids is 4.

Therefore, the Fraction  of octahedral voids occupied = 4 / 4

Simplifying this fraction, we get: Fraction  of octahedral voids occupied = 1 + 1/3

Thus, the fraction of octahedral voids occupied by particles in an FCC structure is 4/ 3, which is equivalent to option (D)