
How many lattice points are there in one unit cell of each of the following lattice?
(i)Face-centered cubic
(ii)Face-centred tetragonal
(iii)Body centred.
Answer
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Hint: An array of unit cells makes up the crystal lattice in a solid. In a crystal lattice, the lattice points in a unit cell are shared by adjacent unit cells.
Complete answer:
-A set of infinite, arranged points which are related to each other by translational symmetry is called a crystal lattice. In simple terms, it is the arrangement of atoms in a solid. Crystal lattice is made up of a unit cell. When this unit cell is in cube shape, then the whole crystal is said to be a cubic crystal system. There are three types or varieties of cubic system. These are primitive cubic, body centered cubic and face-centered cubic systems.
-Face-centered cubic systems have their atoms present in each face as well as in the corners of a unit cell. For a unit cell, there are 8 corners and each corner has an atom. Each atom in the corner is shared by 8 adjacent unit cells. So, for a unit cell only $\frac{1}{8}th$ of an atom contributed. Atoms at the corners per unit cell=$8\times \frac{1}{8}=1$ atom.
Similarly, each atom present at the face of a unit cell is shared by an adjacent unit cell. So, atoms present at the face contribute only half in a unit cell. There are 6 faces in a unit cell, so,
Atoms at the face-center per unit cell=$6\times \frac{1}{2}=3$
Thus, total number of atoms present in a unit cell of a face-centered lattice is = $8\times \frac{1}{8}+6\times \frac{1}{3}=1+3=4$
Total number of atoms is the same as the number of lattice points.
-In face-centered tetragonal system, the atoms are present in each corner and are shared by 8 adjacent unit cells. So, lattice points contributed by the corners are =$8\times \frac{1}{8}=1$ atom.
Similarly, there are 6 faces and each lattice point is shared by 2 unit cells. So, the lattice points contributed by the faces in a unit cell $6\times \frac{1}{2}=3$
Total number of lattice points per unit cell=$8\times \frac{1}{8}+6\times \frac{1}{3}=1+3=4$
-In body-centered, one lattice point is present in the body centre of the unit cell other than the points present in the corners. SO contribution of the lattice point in a unit cell by the corners=$8\times \frac{1}{8}=1$ and at the point center it is 1 lattice point.
So, the total number of lattice points contributed per unit cell=1+1=2.
Thus, the number of lattice points contributed per unit cell by
(i)Face-centered cubic is 4.
(ii)Face-centered tetragonal is 4..
(iii)Body-centered is 2.
Note:
Face centered cubic system and face-centered tetragonal system unit cells are different. In a face-centered cubic system, the unit cell is cubic and in face-centered tetragonal the unit cell is tetragonal shape.
Complete answer:
-A set of infinite, arranged points which are related to each other by translational symmetry is called a crystal lattice. In simple terms, it is the arrangement of atoms in a solid. Crystal lattice is made up of a unit cell. When this unit cell is in cube shape, then the whole crystal is said to be a cubic crystal system. There are three types or varieties of cubic system. These are primitive cubic, body centered cubic and face-centered cubic systems.
-Face-centered cubic systems have their atoms present in each face as well as in the corners of a unit cell. For a unit cell, there are 8 corners and each corner has an atom. Each atom in the corner is shared by 8 adjacent unit cells. So, for a unit cell only $\frac{1}{8}th$ of an atom contributed. Atoms at the corners per unit cell=$8\times \frac{1}{8}=1$ atom.
Similarly, each atom present at the face of a unit cell is shared by an adjacent unit cell. So, atoms present at the face contribute only half in a unit cell. There are 6 faces in a unit cell, so,
Atoms at the face-center per unit cell=$6\times \frac{1}{2}=3$
Thus, total number of atoms present in a unit cell of a face-centered lattice is = $8\times \frac{1}{8}+6\times \frac{1}{3}=1+3=4$
Total number of atoms is the same as the number of lattice points.
-In face-centered tetragonal system, the atoms are present in each corner and are shared by 8 adjacent unit cells. So, lattice points contributed by the corners are =$8\times \frac{1}{8}=1$ atom.
Similarly, there are 6 faces and each lattice point is shared by 2 unit cells. So, the lattice points contributed by the faces in a unit cell $6\times \frac{1}{2}=3$
Total number of lattice points per unit cell=$8\times \frac{1}{8}+6\times \frac{1}{3}=1+3=4$
-In body-centered, one lattice point is present in the body centre of the unit cell other than the points present in the corners. SO contribution of the lattice point in a unit cell by the corners=$8\times \frac{1}{8}=1$ and at the point center it is 1 lattice point.
So, the total number of lattice points contributed per unit cell=1+1=2.
Thus, the number of lattice points contributed per unit cell by
(i)Face-centered cubic is 4.
(ii)Face-centered tetragonal is 4..
(iii)Body-centered is 2.
Note:
Face centered cubic system and face-centered tetragonal system unit cells are different. In a face-centered cubic system, the unit cell is cubic and in face-centered tetragonal the unit cell is tetragonal shape.
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