# Physics Numerical Cha2

```Solved Problems in Solid State Physics -I
Chapter No. 1 and 2
Dr. Ram Chand
Physics Department, University of Hail
6 Feb, 2019
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
1/8
Chapter No. 1
Crystal Structure
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
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Problem 1: (Tetrahedral Angles)
The angle between the tetrahedral bonds of diamond are the same as the
angles between the body diagonals of a cube, as shown in Fig. 10. Use
elemnetary vector analysis to …nd the value of the angle.
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
3/8
Solution to Problem 1: (Tetrahedral Angles)
Referring to Fig. 10, the angles between the tetrahedral bonds of diamond
are equal to those between !
a1 and !
a2 . !
a2 and !
a3 , or !
a3 and !
a1 .So
1 2
a ( 1 1 +1 )
a1 a2
= 14a2 (12 +12 +12 ) =
ka1 kka2 k
4
= cos 1 ( 1/3) = 109.49o
cos θ =
θ
Ram C (UoH)
Department of Physics, UoH
1
3
6 Feb, 2019
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Problem 2: (Indice of Planes)
Consider the planes with indices (100) and (001); the lattice is fcc, and
the indices refer to the conventional cubic cell. What are the indices of
these planes when referred to the primitive axes of Fig. 11?
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
5/8
Solution to Problem 2: (Indice of Planes)
Referring to Fig 11, the plane with index (100) is the plane which parallel
to y z plane abd cuts x axis at x = a. and this plane intercepts !
a1 ,
!
!
!
!
!
!
a2 , a3 axes at 2 a1 , ∞ a2 (does not intercept a2 axis) and 2 a3
respectively. 21 : 0 : 21 = 1 : 0 : 1. The index referred to the primitive axes
!
a1 , !
a2 , !
a3 is then (1, 0, 1).
Similarly, the plan with index (001) referred to cubic cell. The plane is
parallel to x y plane and cuts z axis at z = a. And this plane
intercepts !
a1 , !
a2 , !
a3 axes at ∞ !
a1 , 2 !
a2 ,2 !
a3 . Hence the index referred to
the primitive axes is (011).
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
6/8
Problem 3: (hcp Structure)
Show that the c/a ratio for an ideal hexagonal close-packed structure is
8 1/2
= 1.633. If c/a is signi…cantly larger than this value, the crystal
3
structure may be thought of as composed of planes of closely packed
atoms, the planes being lossely packed.
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
7/8
Solution to Problem 3: (hcp Structure)
Suppose the radius of an atom is r . Since it’s an ideal hexagonal
close-packed structure, see Fig 21, c = 2r ( the two atoms touch) and a1 or
a2 = 2r ( the two atoms touch). Also, from ther
geometry the distance
between the center layer atom and top atom is
pa
3
2
+
c 2
2
= 2r
(the two atoms touch)= a, so we obtain
c 2
a
=
a2
c2
a2
3 + 4 =
q
8
c
8
3 or a =
3
= 1.633
q
1.633, the atoms on the top do not touch the atoms on the
If 83
center layer. And this means, the crystal structure is composed of planes
of closely packed atoms ( atoms on the each layer still touch each other),
the plane being loosely stacked.
Ram C (UoH)
Department of Physics, UoH
6 Feb, 2019
8/8
```
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