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In FCC lattice of a metal solid a metal atom on a face centre is in contact with how many other face centred atoms?.....plz explain in detail




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In a face centered cubic lattice crystal,  FCC :  there is 1 atom of the metal at each of the 8 corners.  Only 1/8 of the atom is inside the unit cube.

There are 6 faces of the unit cube.  On each face, there is an atom at the center of the face.  So 1/2 of each atom is inside the cube.    Thus there are 4 atoms belonging to one unit cell or cube.

An atom at the corner of the cube : has a coordination number of 12.  It has 12 neighbours.   There are 12 face centered atoms, which are its neighbours at distance a/√2.  The 6 atoms at the 6 neighbouring corners are at a distance "a".  a = edge of the unit cube.

For an atom at the center of the face,  it is in contact with 4 atoms at the 4 corners of the face it is part of.   The distance between the face center and the corner is a/√2,  where a = edge or side of the unit cube/cell.

The distance between adjacent face centered atoms = √ [(a/2)²+(a/2)² ] = a/√2.
There are 4 edges to the face that the face centered atom is part of.  On each of the edges there two faces. Atoms Centered on those faces are close to the face centered atom.  The distance between them is  a/√2.    So there are 8 such atoms.

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