1. PH0101 UNIT 4 LECTURE 4 RECIPROCAL LATTICE
FEATURES OF RECIPROCAL LATTICE
GRAPHICAL REPRESENTATION
GENERAL PROCEDURE
SIMPLE CUBIC STRUCTURE
BODY CENTERED CUBIC STRUCTURE
PH 0101 UNIT 4 LECTURE 4

2. The concept of ‘reciprocal lattice’ provides a device for
tabulating both the slopes and the interplanar
spacings of the planes of a crystal lattice.
In a crystal, there exist many sets of planes with
different orientations and spacing. These planes can
cause diffraction.
If we draw normals to all sets of planes, from a
common origin, then the end points of normals form a
lattice which is called as ‘reciprocal lattice’.
PH 0101 UNIT 4 LECTURE 4

3. FEATURES OF RECIPROCAL LATTICE
Each point in the reciprocal lattice preserves the
characteristics of the set of planes which it
represents.
Its direction with respect to the origin represents
the orientation of the planes.
Its distance from the origin represents the
interplanar spacing of the planes.
PH 0101 UNIT 4 LECTURE 4

4. GRAPHICAL REPRESENTATION
Consider all the planes belonging to a single zone.
Since all the planes to be considered are parallel to a
common line, known as zone line, the normals to these
planes lie in the same plane (normal to the zone axis).
In this way, these planes can be represented by a two
dimensional reciprocal lattice.
PH 0101 UNIT 4 LECTURE 4

5. FORMATION OF RECIPROCAL LATTICE
PH 0101 UNIT 4 LECTURE 4

6. The figure shows the unit cell of a monoclinic
RECIPROCAL LATTICE
The figure shows the unit cell of a monoclinic
crystal looking along its unique axis designated by C.
The cell edges are ‘a’ and ‘b’.
The example also shows the edge view of four (h k l)
planes, namely (100), (110), (120) and (010).
PH 0101 UNIT 4 LECTURE 4

7. GENERAL PROCEDURE FOR LOCATING THE RECIPROCAL LATTICE
A lattice point is taken as common origin.
From the common origin, draw a normal to each
plane.
Place a point on the normal to each plane (h k l) at a
distance from the origin equal to
Such points form a periodic array called reciprocal
lattice.
PH 0101 UNIT 4 LECTURE 4

8. CRYSTAL STRUCTURE OF MATERIALS
NUMBER OF ATOMS PER UNIT CELL (n)
The total number of atoms present in an unit cell is
known as number of atoms per unit cell.
COORDINATION NUMBER (CN)
It is the number of nearest neighboring atoms to a
particular atom.
ATOMIC RADIUS (r)
It is the radius of an atom. It is also defined as half the
distance between two nearest neighboring atoms in a
crystal.
PH 0101 UNIT 4 LECTURE 4

9. CRYSTAL STRUCTURE OF MATERIALS
ATOMIC PACKING FACTOR (APF)
It is the ratio of volume occupied by the atoms or molecules
in an unit cell (v) to the total volume of the unit cell (V).
APF =
APF = No. of atoms present in the unit cell x Volume of the atom
Volume of the unit cell
PH 0101 UNIT 4 LECTURE 4

10. SIMPLE CUBIC STRUCTURE
PH 0101 UNIT 4 LECTURE 4

11. SIMPLE CUBIC STRUCTURE CORDINATION NUMBER (CN)
Each and every corner atom is shared by eight adjacent
unit cells. The contribution of each and every corner atom
to one unit cell is 1/8.
The total number of atoms present in a unit cell =1/8 x 8 =1.
CORDINATION NUMBER (CN)
For SCC atom, there are four nearest neighbours in its own
plane.
There is another nearest neighbour in a plane which lies
just above this atom and yet another nearest neighbour in
another plane which lies just below this atom.
Therefore the co-ordination number is 6.
PH 0101 UNIT 4 LECTURE 4

12. SIMPLE CUBIC STRUCTURE ATOMIC PACKING FACTOR (APF)
ATOMIC RADIUS (R)
Since the atoms touch along cube edges, the atomic radius for a
simple cubic unit cell is, r =
(where a = 2r, is the lattice constant)
ATOMIC PACKING FACTOR (APF)
APF =
v = 1  4/3 r3
PH 0101 UNIT 4 LECTURE 4

13. SIMPLE CUBIC STRUCTURE
V = a3
 APF =
APF =
Substituting r = , we get,
PH 0101 UNIT 4 LECTURE 4

14. SIMPLE CUBIC STRUCTURE
Therefore packing density = /6 =
APF = 0.52
Thus 52 percent of the volume of the simple
cubic unit cell is occupied by atoms and the
remaining 48 percent volume of the unit cell is
vacant or void space.
PH 0101 UNIT 4 LECTURE 4

15. BODY CENTERED CUBIC STRUCTURE
A body centred cubic structure has eight comer
atoms and one body centred atom.
The atom at the centre touches all the eight
corner atoms.
The BCC structure is shown in the figure.
PH 0101 UNIT 4 LECTURE 4

16. BODY CENTERED CUBIC STRUCTURE
PH 0101 UNIT 4 LECTURE 4

17. BODY CENTERED CUBIC STRUCTURE
PH 0101 UNIT 4 LECTURE 4

18. BODY CENTERED CUBIC STRUCTURE
In BCC unit cell, each and every corner atom is shared by eight
adjacent unit cells.
So, the total number of atoms contributed by the corner atoms is
× 8 = 1.
A BCC unit cell has one full atom at the centre of the unit cell.
The total number of atoms present in a BCC unit cell = 1+1 = 2.
PH 0101 UNIT 4 LECTURE 4

19. BODY CENTERED CUBIC STRUCTURE
CO-ORDINATION NUMBER (CN)
Let us consider a body centred atom. The nearest neighbour for a
body centred atom is a corner atom. A body centred atom is
surrounded by eight corner atoms.
Therefore, the co-ordination number of a BCC unit cell = 8.
ATOMIC RADIUS (R)
For a BCC unit cell, the atomic radius can be calculated
From figure AB = BC = AD = ‘a’ and CD = 4r.
PH 0101 UNIT 4 LECTURE 4

20. BODY CENTERED CUBIC STRUCTURE
From the triangle, ACD,
CD2 = AC2 + AD2
CD2 = 2a2 + a2
(4r)2 = 3a2
16r2 = 3a2
i.e. r2 =
PH 0101 UNIT 4 LECTURE 4

21. BODY CENTERED CUBIC STRUCTURE ATOMIC PACKING FACTOR (APF)
atomic radius r = a
ATOMIC PACKING FACTOR (APF)
APF =
The number of atoms present in an unit cell = 2
 v = 2 x 4/3 r3
V = a3
PH 0101 UNIT 4 LECTURE 4

22. BODY CENTERED CUBIC STRUCTURE
APF =
Substituting r = we get
APF = = 0.68
PH 0101 UNIT 4 LECTURE 4

23. BODY CENTERED CUBIC STRUCTURE
Thus 68 percent of the volume of the BCC unit cell is occupied by atoms and the remaining 32 percent volume of the unit cell is vacant or void space.
PH 0101 UNIT 4 LECTURE 4

24. Physics is hopefully simple but Physicists are not
PH 0101 UNIT 4 LECTURE 4