Physics - topic 2.pdf, physics sem1 ece

physics

1. Unit -1
Properties of Different Crystal Structures

2. Ø Learn about effective no. of atoms in different unit cells, relation
connecting radius of atom and the lattice parameter
ØAtomic Packing Fraction of different cubic systems, HCP and
Diamond structures
Objectives

3. Coordination number
Coordination number is the number of nearest neighbour to a particular atom in the crystal
Simple Cubic :
The coordination number of SC crystal :
6
6 atoms touch each atom in Simple Cube

4. Body Centered Cubic :
Coordination number
Coordination number is the number of nearest neighbour to a particular atom in the crystal
The coordination number of BCC crystal :
8
The body centered atom is in contact with all the eight corner atoms. Each
corner atom is shared by eight unit cells and hence, each of these atoms is in
touch with eight body centered atoms

5. Face Centered Cubic :
FCC coordination number : 12
For example, the face centered atom
in the side face is in contact with four
corner atoms and
four other face-centered atoms behind it
(two sides, top and bottom) and
is also touching four face-centered atoms
of the unit cell in front and bcak faces
Coordination number
Let See that .....
Coordination number
Coordination number is the number of nearest neighbour to a particular atom in the crystal

6. Centered atom at the back
face
Coordination number
FCC
Let us consider two unitcell to
visuvalize

7. Coordination number
FCC
Øcontact with four
corner atoms

8. Coordination number
FCC
Øcontact with four
corner atoms
Øcontact with four
face centered atoms-
(two sides, top and bottom)

9. Coordination number
Co-ordination number
= 4 + 4 + 4
=12
FCC
Øcontact with four
corner atoms
Øcontact with four
face centered atoms-
two sides, top and bottom faces
Øtouching four face-centered
atoms of the unit cell - two cells
front and back faces

10. Coordination number
Hexagonal Close Packed Structure (HCP ):
Hexagonal (HCP ) lattice Coordination
number is 12.
The center atom of the top face is in touch
with six corner atoms,
three atoms of the mid layer and
other three atoms of the mid layer of the
unit cell above it.

11. Let us consider two unitcell to
visuvalize
Coordination number HCP

12. Coordination number
center atom (green) of the top face is in touch
with six corner atoms (red), three atoms (red) of the
mid layer and other three atoms (red)
of the mid layer of the unit cell above it
= 6+3+3 =12
HCP

13. Atomic packing fraction
Atomic packing factor (APF) or packing efficiency indicates how closely
atoms are packed in a unit cell and is given by the ratio of volume of
atoms in the unit cell and volume of the unit cell
Atomic Packing Fraction = (No. of atoms in unit cell)  ( Volume of atom)
Volume of unit cell

14. Atomic packing fraction
Why Atomic Packing Fraction is important ?
oDensity of the material
oexplains many properties of materials
For example, metals with a high atomic packing factor will have a higher
malleability or ductility
3
3
4
r
Considering the atoms are rigid spheres,
Volume of the atom =
Where r is the radius of the atom

15. Number of Lattice Points in Cubic Crystal Systems

If there is only one atom located at each lattice point,
Sharing of corner atoms

16. In the Simple Cubic unit cell:
lattice point / unit cell = (8 corners)1/8 = 1
In BCC unit cells:
lattice point / unit cell
= (8 corners)1/8 + (1 center)(1) = 2
In FCC unit cells:
lattice point / unit cell
= (8 corners)1/8 + (6 faces)(1/2) = 4
The number of atoms per unit cell would be 1, 2, and 4, for the simple cubic,
body-centered cubic, and face-centered cubic, unit cells, respectively.
Number of Lattice Points in Cubic Crystal Systems

17. Relationship between Atomic Radius and Lattice Parameters
Simple Cubic :
we find that atoms touch along the edge
of the cube in SC structures
ra2

18. Relationship between Atomic Radius and Lattice Parameters
In BCC structures, atoms touch along the body
diagonal. There are two atomic radii from the
center atom and one atomic radius from each
of the corner atoms on the body diagonal, so
3
4r
a
Body Centered Cubic :
a
a2

19. Relationship between Atomic Radius and Lattice Parameters
Face Centered Cubic :
a
In FCC structures, atoms touch along the face
diagonal of the cube.
There are four atomic radii along this
length—two radii from the face-centered atom
and one radius from each corner, so:
2
4r
a

20. Atomic packing factor
Simple Cubic :
52.0
6)2(
)
3
4
(1)(
Factor Packing Atomic
2r cell,unit cubic simplefor Since,
)
3
4
)(atoms/cell of. (No.
cellunit theof olume
)atom theof olume)(atoms/cell of . (No
Factor Packing Atomic
3
3
3
3







r
r
a
a
r
v
v
Considering the atoms as hard spheres of radius, r
Volume of the atom =
3
3
4
r

21. Atomic packing factor
Body Centered Cubic :
68.0
643
338
)
3
4
)
3
4
(2)(
Factor Packing Atomic
4r 3a cell,unit BCCfor Since,
)
3
4
1)( 8
8
1
(
cellunit theof olume
) atom theof olume( ) atoms/cell of . (No
Factor Packing Atomic
3
3
3
3
3
3















r
r
r
r
a
r
v
v




22. Atomic packing factor
Face Centered Cubic :
74.0
18)2/4(
)
3
4
(4)(
Factor Packing
24r/ cells,unit FCCfor Since,
)
3
4
)(atoms/cell (4
Factor Packing Atomic
3
3
3
0
3






r
r
a
a
r
In the FCC unit cell effective number of atoms =
8 corner atoms x (1/8) (each atom is shared by 8 unit cells)
+ 6 face centeredatoms x1/2 (each shared by two unit cells) =
4

23. Atomic packing factor
Hexagonal Close Packed (
HCP
) structure
In the Hexagonal unit cell,
number of atoms = 12 corner atoms × 1/6 (shared by six unit cells)
+ Two face atoms × 1/2 + 3 interior = 6.
2r = a

24. Atomic packing factor
Hexagonal Close Packed (HCP) structure
Unit cell volume = Area of the hexagon × height
= Area of the six triangles × height (c)
= ( 6 × Area of  AGB) ×
c
=
(6 × ½ × a × h) × c

=
(3 × a × a sin60) × c
= 3a
2
c sin60

25. Atomic packing factor
Hexagonal Close Packed (HCP) structure
For an ideal hexagonal crystal c/a ratio = 1.633
eunitcellvolumeofth
r)
3
4
)(atoms/cell (6
Factor Packing Atomic
3



26. 1. Each basal plane has nearest neighbor atoms making equilateral triangles.
So, a=2r (where r is the sphere radius).
2. Each atom at height c/2 above the basal plane is positioned directly above the centroid
of the triangles in the base plane. For an equilatreral triangle, the distance from a vertex to
the centroid is two-thirds the length of the median, and is hence
( 2 / 3 ) × ( a √ 3 / 2 ) = a / √ 3
3. Each atom in the base plane has a nearest neighbor in this middle plane. So, the
distance from the corner atom in the base plane to the nearby atom in the mid-plane is 2r.

4. This distance can also be calculated from Pythagoras, giving:
4r
2
= a
2
= ( a / √ 3 )
2
+ ( c / 2 )
2

Give a Try : c/a = 1.633
Hexagonal Close Packed (HCP) structure

27. Atomic packing factor
Diamond

structure
:
Projection View
The diamond unit cell is cubic. The cell has eight atoms.
Elemental semiconductors Ge and Si have this crystal structure.

28. Diamond

structure
:
Atomic packing factor
There are different ways, one can calculate
the packing fraction of diamond structure.
We can use the unfilled space of the unit cell ..
in our calculation
unfilled space are called voids

29. Atomic packing factor
Diamond structure :
To get the relation between the radius of the atom
and the lattice parameter,
add two voids (equal to the atomic dimensions)
diagnally in the unit cell at
(1/2,1/2,1/2) and (3/4,3/4,3/4) positions
as shown in the figure

30. Atomic packing factor
Diamond structure :
Diamond Structure : FCC unit cell + 4 extra atoms

No. of atoms in the unit cell = 1/8 × 8 (corner atoms) + 1/2 × 6 (face centered atoms)
+ 4 (atoms inside the unit cell)
= 1 + 3 + 4
=
8

31. Atomic packing factor
Diamond structure :
34.0
)3/8(
)
3
4
)(8(
)
3
4
)(atoms/cell (8
factor Packing Atomic
83
3
3
3
0
3
0




r
r
a
r
ra



32. The Zinc blende (ZnS) cubic crystal structure. Many important compound crystal
Structures have the zinc blende structure. Examples: AlAs, GaAs, Gap, GaSb, InAs, InP,
InSb, ZnS, ZnTe.
From Principles of Electronic Materials and
Devices, Fourth Edition, S.O. Kasap (©
McGraw-Hill Education, 2018)
Zinc Blende Cubic Crystal Structure
Same Diamond sturcture with two different atoms

34. FCCHCP
Note the difference between stacking sequence and unit cell definition
HCP – basal plane (closed packed plane)
FCC – (111) plane (closed packed plane)

36. Calculate the atomic concentration (number of atoms per unit volume) in Cu and the
density of the crystal given that the atomic mass of Cu is 63.55 g mol
−1
and the radius
of the Cu atom is 0.128 nm.
Solution:
Problems

37. Problems

38. Problems
The lattice constant of Si is 5.43 Å . What will be the radius of a silicon atom?
Calculate the theoretical density of silicon. The atomic mass of Si is 28.1 gm/mol.
SOLUTION:
For the diamond cubic structure,
Therefore, substituting a = 5.43 Å,
the radius of silicon atom = 1.176 Å .
There are eight Si atoms per unit cell.
ra830
3
38
23
3
/33.2
cm) 1043.5(
10023.6
1.28
8
volume
mass
density cmg
a
N
M
n
A
at









39. Problems
Gallium Arsenide (GaAs) has the zinc blende crystal structure with lattice parameters of
a = 0.565 nm and the atomic masses Ga and As as 69.73 g/mol, and 74.92 g/mol
respectively. In GaAs, the number of Ga will be in eqaul number to As in the unit cell.
Calculate the density of GaAs. What is the atomic concentration in GaAs crystal?

40. In Zinc blende structure, no. of atoms in the unit cell =
8
In the case of GaAs, there are
4 Ga and 4 As
atoms in the unit cell. Gallium
concentration in the unit cell,
Total atomic concentration = 2  n
Ga
= 4.44
×
10
28
m
-3
(or) There are 2.22
×
10
28
Ga-As pairs per m
3
.
328
393
1022.2
)10565.0(
44




 m
a
n
Ga Problems
Gallium Arsenide (GaAs) has the zinc blende crystal structure with lattice parameters of
a = 0.565 nm and the atomic masses Ga and As as 69.73 g/mol, and 74.92 g/mol
respectively. In GaAs, the number of Ga will be in eqaul number to As in the unit cell.
Calculate the density of GaAs. What is the atomic concentration in GaAs crystal?
Soluton :

41. Density of GaAs,
 
3
33
23
3
28
3
33
3
/33.5
/1033.5
10022.6
)10(92.7473.69
1022.2
)10()(
)10(
cmg
mkg
N
MM
a
n
a
N
kgM
n
A
AsGaGaAsA
at






















Problems