Table of contents 
Bravais Lattices 
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Bravais (1848) showed from geometrical considerations that there are only seven shapes in which unit cells can exist.
These are
Moreover, he also showed that there are basically four types of unit cells depending on the manner in which they are arranged in a given shape. These are:
He also went on to postulate that out of the possible twentyeight unit cells (i.e. seven shapes, four types in each shape = 28 possible unit cells), only fourteen actually would exist. These he postulated based only on symmetry considerations. These fourteen unit cells that actually exist are called Bravais Lattices.
The seven crystal systems are given below.
➤ Cubic
➤ Orthorhombic
➤ Tetragonal
➤ Monoclinic
➤ Triclinic
➤ Hexagonal
➤ Rhombohedral
The table given below can be used to summarize types of lattice formation.
Q.1: Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8 Å, b = 4.4 Å and C = 7.2 Å. If the molar mass is 21.76 g. Calculate the density of the crystal.
Solution:
Since, Density, Here z = 4, Av. No = 6.023 x 10^{23} &
Volume = V = a x b x c
= 6.8 x 10^{8} x 4.4 x 10^{8} x 7.2 x 10^{8} cm^{3 }
= 2.154 x 10^{22} cm^{3}
^{}
= 0.6708 gm/cm^{3}
Q.2. An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom in each corner of the cube and two atoms on one of its face diagonals. If the volume of this unit cell is 24 x 10^{24 }cm^{3} and density of the element is 7.20 gm/cm^{3}, calculate no. of atoms present in 200 gm of the element.
Ans: Number of atoms contributed in one unit cell = one atom from the eight corners + one atom from the two face diagonals = 1+1 = 2 atoms
Mass of one unit cell = volume × its density
= 24 × 10^{–24} cm^{3} × 7.2 gm cm^{3}
= 172.8 × 10^{–24} gm
∴ 172.8 10^{–24} gm is the mass of one – unit cell i.e., 2 atoms
∴ 200 gm is the mass = 2 × 200 / 172.8 × 10^{–24} atoms = 2.3148 × 10^{24} atoms
Q.3. Calculate the void fraction for the structure formed by A and B atoms such that A form hexagonal closed packed structure and B occupies 2/3 of octahedral voids. Assuming that B atoms exactly fitting into octahedral voids in the HCP formed by A.
Ans: Total volume of A atom = 6 × 4 / 3 πrA^{3}
Total volume of B atoms = 4 × 4/3 πrA^{3}
=4 × 4/3 π(0.414rA)^{3}
Since rB/rA as B is in octahedral void of A
Volume of HCP = 24√2rA^{3}
Packing fraction = 6 × 4/3 πrA^{3} + 4 × 4/3 π (0.414rA)^{3} / 24√2rA^{3} = 0.7756
Void fraction = 10.7756 = 0.2244
Q.4: An element with molar mass 2.7×10^{}^{2}kg mol^{1} forms a cubic unit cell with edge length 405 pm. If its density is 2.7×10^{3}kg^{3}, what is the nature of the cubic unit cell?
Ans: d = 2.7×10^{3}kg^{3}
M = 2.7×10 ^{}^{2 }kg mol^{1}
a= 405 pm = 405 X 10^{12}
N_{A}= 6.023 X 10^{23}
Using the formula
Z=4
Unit cell is fcc unit cell.
Volume of 54 g of the element = 0.054/(2.7 x 10^{3}) = 2 X 10^{6}
Number of unit cell in this volume = volume of 554 g of element/volume of each unit cell = 2 x 10^{6}/(405 x 10^{12})^{3} = 3.012 x 10^{22}
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