Engineering Physics-I
Crystal Physics- Atomic rad., Coord. No.,APF for SC and BCC
Introduction
Most of the materials in solid state are crystalline. Among these many are in the
polycrystalline state. To obtain single crystal one has to employ a suitable crystal growth method.
This may vary from one material to another. To understand the different material properties, the
structure of the material is very essential. The unit cell which is the building block of a crystalline
material can be made of the basic units namely the basis or the pattern unit or the motif. In this
section we study the number of atoms per unit cell, how they are arranged in simple structures to
render the atomic radius, the number of first nearest neighbours and how the basic units are packed
to form different crystal structures.
Learning Objectives
On completion of this chapter you will be able to:
1. calculate the number of atoms per unit cells of simple structures
2. define atomic radius and coordination number
3. derive the packing factors for SC and BCC
Calculation of number of atoms per unit cell
The number of atoms per unit cell depends on the crystal structure and the type of basic
elements (basis or pattern unit or motif) that form the crystal. For example: The number of atoms
in a simple cubic (SC) crystal is one in the case of polonium. The number of atoms in a body
centred cubic (BCC) crystal is two in the case of sodium.
In SC unit cell, the atoms occupy only the corners. In BCC unit cell, the atoms occupy the
corners as well as the body centre of the unit cell. In FCC unit cell, the atoms occupy the corners
as well as the face centre of the unit cell. In the case of hexagonal closed packed (HCP) structure,
atoms occupy the 12 corners, the centre of the bottom and top faces and the body centre of the
alternate one sixth of the unit cell. The atoms available at the corners and in the face centres are
shared by more than one unit cells. Only the atoms present well within the unit cell solely belong
to the unit cell.
Thus the number of atoms per unit cell is equal to the product of 'the number of atoms per
basis' and 'the number of basis per unit cell'. In general, if the basis has n number of atoms and the
number of basis per unit cell is m, then the number of atoms in the unit cell will be “mn”.
The number of atoms in the FCC, HCP and diamond unit cells are respectively 4, 6, and 8.
Atomic radius
In solid state physics, the atomic radius is half the distance between two nearest neighbour
atoms. In the case of SC all the adjacent corner atoms in a cubic unit cell are touching each other
along the edges (Fig.1). Let 'a' and 'r' be the side of the cubic unit cell and radius of the atoms in
the unit cell respectively.
The atomic radius in the case of FCC, and HCP are
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 2 a / 4 and a/2 respectively.
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Engineering Physics-I
Crystal Physics- Atomic rad., Coord. No.,APF for SC and BCC
a
Figure 1
In SC, 'a' and 'r' are related by,
r = a/2
Figure 1
In BCC the atoms along the body diagonal are touching each other as shown below (Fig.2).
Figure 2
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Engineering Physics-I
Crystal Physics- Atomic rad., Coord. No.,APF for SC and BCC
In BCC, 'a' and 'r' are related by,
r = 3 a /4
Co-ordination number:
It is the number of nearest neighbour atoms with respect to any atom in the lattice. It varies with
one structure to another. As this number increases the density of packing of the
atoms/ions/molecules in the available space to form a crystal structure increases.
The coordination numbers for SC, BCC, FCC, HCP and the diamond structure are 6, 8, 12, 12, and
4 in the same order.
Packing factor:
Definition: packing factor =
volume occupied by the atoms of the unit cell  v
volume of the unit cell V 
Packing factor for SC:
v = Number of atoms in the unit cell x volume of one atom
=1x
4r
3
3
For SC,
r = a/2
This is obtained by considering the atoms along a cube edge (Fig.3).
2r = a
r
r
Figure 3
V =a
3
Substituting the expressions for v and V in the packing factor expression and simplifying, one gets

packing factor = =0.52
6
This indicates that 52 % of the available space is occupied by the atoms forming the SC lattice.
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Session No: < 3 >
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Engineering Physics-I
Crystal Physics- Atomic rad., Coord. No.,APF for SC and BCC
Packing factor for BCC:
v = Number of atoms in the unit cell x volume of one atom
=2x
4r
3
3
r
2r
r
3 a
Figure 4
For BCC,
r = 3 a /4
This is obtained by considering the atoms along a body diagonal (Fig.4).
V =a 3
Substituting the expressions for v and V in the packing factor expression and simplifying, one gets
3
packing factor = 
=0.68
8
This indicates that 68 % of the available space is occupied by the atoms forming the BCC lattice.
Thus BCC is more dense than SC.
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Engineering Physics-I
Crystal Physics- Atomic rad., Coord. No.,APF for SC and BCC
Check your understanding
1. State true or false: The coordination number and packing factor are connected
someway.
2. State true or false: Does the coordination number decide the mechanical
strength of the material?
3. State true or false: Do the atomic radius of a material dependent on structure?
4. What decides the density of a material?
5. Can one have different crystal structures with the same starting material? If not,
why?
Summary
On completion of this chapter you have learned :
1. how to calculate the number of atoms in SC, BCC, FCC and HCP
2. to define atomic radius, coordination number and packing factor
3. how to calculate the packing factors for SC and BCC crystals.
Suggested Reading
1. Engineering Physics- P.K.Palanisamy (SCITECH PUBLICATIONS (INDIA) PVT. LTD. , Chennai)
2. Foundations of materials science and engineering – William F. Smith (McGraw Hill)
3. Elements of materials science and engineering- Lawrence Van Vlack (Pearson Education India).
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Session No: < 3 >
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