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In a face-centred cubic lattice, a unit cell is shared equally by how many unit cells :-
A
4
B
2
C
6
D
8
Detailed Solution
In a crystal lattice the unit cells are packed face to face, so a cubic unit cell has six faces in common with six neighbouring unit cells (one on each face).
In a face-centred cubic unit cell there is one atom at the centre of each of these six faces, and each face-centre atom is shared between the given cell and the neighbouring cell on that face.
So the unit cell shares its face-centred atoms with six surrounding unit cells, which is the answer given by the key (6).
Contribution of each face-centre atom to one cell $= \dfrac{1}{2}$, so the six faces give $6 \times \dfrac{1}{2} = 3$ atoms; the eight corners give $8 \times \dfrac{1}{8} = 1$ atom.
Total number of atoms per fcc unit cell $= 3 + 1 = 4$.
Note: each face-centre atom itself is shared by 2 unit cells and each corner atom by 8; the wording of the question is ambiguous, and the key takes the six face-sharing neighbours.
In a face-centred cubic unit cell there is one atom at the centre of each of these six faces, and each face-centre atom is shared between the given cell and the neighbouring cell on that face.
So the unit cell shares its face-centred atoms with six surrounding unit cells, which is the answer given by the key (6).
Contribution of each face-centre atom to one cell $= \dfrac{1}{2}$, so the six faces give $6 \times \dfrac{1}{2} = 3$ atoms; the eight corners give $8 \times \dfrac{1}{8} = 1$ atom.
Total number of atoms per fcc unit cell $= 3 + 1 = 4$.
Note: each face-centre atom itself is shared by 2 unit cells and each corner atom by 8; the wording of the question is ambiguous, and the key takes the six face-sharing neighbours.