LUMS School of Science and Engineering
Solution of assignment 3
PH-332
March, 30, 2011
Bravais Lattice
Answer: 1
We have given that
~c = 1.5(î + ĵ + k̂)
15
=
(î + ĵ + k̂)
10
3
=
(î + ĵ + k̂)
2
2~c = 3(î + ĵ + k̂)
=
3î + 3ĵ + 3k̂
= ~a + ~b + c~0
1
=
~a + ~b + c~0
2
0 ~
∵ ~a = 3î, ~b = 3ĵ and let c = 3k̂
Now we have primitive translational vectors
1
~a = 3î, ~b = 3ĵ, ~c = ~a + ~b + c~0
2
Where ~c = 12 ~a +~b + c~0 is the body centered position of the cubic unit cell defined
by ~a, ~b and ~c. The Bravais lattice is therefore bcc.
Volume of conventional unit cell will be,
vol of conventional cell
=
3 × 3 × 3 Å
3
3
= 27 Å
Since the volume of a primitive unit cell is half of the volume of the bcc (since
the conventional unit cell contains two lattice points while primitive unit cell
contains only one), thus volume of primitive unit cell will be,
vol.of primitivecell
vol. of conventional cell
2
3
27 Å
=
2
3
= 13.5 Å
=
Lattice Types in 3-D
Answer: 2
(i) Since a 6= b 6= c and γ 6= β 6= γ = 90o , i.e. three primitive vectors are unequal
and there is a lack of symmetry. This is a characteristic of triclinic structure.
Therefore the unit cell belongs to a triclinic system.
(ii) Since a = b 6= c, γ = β = 90o and γ = 120o , This is a characteristic of
1
Hexagonal structure. Therefore the unit cell belongs to a Hexagonal system.
Tetrahedral Angles
Answer: 3
a2
a1
a3
The body diagonals of the cubes shown in the figure are vectors that connect
the lattice points at the origin to the lattice points at the body centers. The
primitive cell obtained on completing the rhombohedron. In terms of the cube
edge a, the primitive translational vectors are,
a
(x̂ + ŷ − ẑ)
2
a
a~2 = (ŷ + ẑ − x̂)
2
a
a~3 = (ẑ + x̂ − ŷ)
2
a~1 =
Let the angle between ~a1 and ~a2 be θ, then
µ
~a1 · ~a2 =
¶µ
¶
a
a
(x̂ + ŷ − ẑ) .
(ŷ + ẑ − x̂)
=
2
2
|a1 ||a2 | cos θ
¯
¯¯
¯
¯a
¯¯
¯
¯ (x̂ + ŷ − ẑ)¯¯ a (ŷ + ẑ − x̂)¯ cos θ
¯2
¯¯ 2
¯
sµ ¶
µ ¶2 µ
¶2 sµ
¶2 µ ¶2 µ ¶2
2
a
a
a
a
a
a
a2
−
(−1 + 1 − 1) =
+
+ −
+
+
cos θ
4
2
2
2
2
2
2
√ a √ a
a2
−
=
3 × 3 cos θ
4
2
2
1
−
= cos θ
3
µ
¶
1
θ = cos−1 −
3
= 109.5o
Wigner Seitz Cell
Answer: 4
(a). By construction every vector in the Bravais lattice has an inverse vector in
the same Bravais lattice. Similarly following the Wigner Seitz cell’s construction
in which we use perpendicular bisectors,we find that for every side of the Wigner
Seitz cell there is an inverse of that side i.e.Wigner Seitz cell consists of pairs
2
of opposite sides. Thus the number of sides of the Wigner Seitz cell are even.
That is why a triangular Wigner Seitz cell is meaningless i.e a Wigner Seitz
cell cannot have three sides. Since Wigner Seitz cell is an area enclosed by the
perpendicular bisectors it cannot have two sides because two sides cannot form
a closedarea.Thus Wigner Seitz cell has minimum 4 sides. Number of sides of a
Wigner Seitz cell are always even. Therefore Wigner Seitz cell be a closed area
with 4 sides(rectangle), 6 sides (Hexagon) or more.
Now let us prove it by constructing the Wigner Seitz cell. Let us consider Bravais
lattice shown in figure (a).
B
-A
B
A
-A
-B
O
C
A
O -C
-B
fig(a)
( b)
Take a Bravais lattice point as origin O. The nearest Bravais lattice point to the
right side of the origin point is A. There is a negative or inverse point −A on the
left side of origin, such that measure of OA is equal to the measure of O(−A).
similarly we have points B and −B that are opposite to each other. Draw
perpendicular lines joining A to −A and B to −B through O. The perpendicular
bisectors to these lines will give us the Wigner Seitz cell that is rectangular.
Now consider figure (b). Take a Bravais lattice point as origin O. The nearest
Bravais points are A and −A. Take the closest point to O not on the line OA,
call this point B. Draw perpendicular to OA through O. Let point B is on left
side of this perpendicular line. Now there will be a Bravais lattice point C on
¯ = OA.
¯ Any other Bravais
the line through B and parallel to OA, such that BC
¯ will not contribute to this side of Wigner Seitz cell, since it
lattice point on BC
¯
has to be further out than A, B, C or −A. Similarly on the other side of OA
we will find the points −B and −C. The Wigner Seitz cell is a Hexagon, unless
B is on the normal line , in which case the cell is a rectangle.
(b). The diagonal of a parallelogram is the line joining the opposite corners of
the parallelogram. The opposite corners of a parallelogram face of a Wigner
Seitz cell for the fcc lattice have points (1/4, 1/4, 0) and (1/2, 0, 1/2). While the
other diagonal connects the points (1/4, 1/4, 1/4) and (1/4, 1/4, 3/4). If a is the
3
length of one side of the given fcc, then,
¯
|AC|
=
=
|a(1/2, 0, 1/2) − (1/4, 1/4, 0)|
a√
2
2
Similarly,
¯
|BC|
=
=
|a(1/4, 1/4, 3/4) − (1/4, 1/4, 1/4)|
a
2
Thus ratio will be,
a√ a
2:
2 √ 2
2:1
HCP Structure
Answer: 5
C
a2
A
a2
a1
B
a1 +a2
a2
a1
a1+a2
a1
Let us consider an ideal hcp structure in which a1 = a2 = a and a3 = c with
α = β = 90o while γ = 120o . Let use spheres of radius R on each lattice
point of hcp structure such that |a1 | = |a2 | = a = 2R. Since hcp consists of
two interpenetrating hexagonal Bravais lattices, displaced from one another by
a¯1 /3 + a¯2 /3 + a¯3 /2, thus
Thus
¯
¯
¯ a¯1
¯
¯ + a¯2 + a¯3 ¯
¯3
3
2¯
Also
c
¯
¯
¯ ¯
¯ 1¯ ¯
¯1
¯ (a¯1 + a¯2 )¯ + ¯a¯3 ¯
¯ 2¯ ¯
¯3
¯
¯
¯
¯1
¯ (a¯1 + a¯2 )¯ + 1 c
¯ 2
¯3
Also from the figure it is clear that,
¯2 ¯ ¯2
¯
¯
¯ ¯
¯1
¯ (a¯1 + a¯2 )¯ + ¯ 1 c¯
¯
¯2 ¯
¯3
¯
¯2
¯1
¯
¯ (a¯1 + a¯2 )¯ + 1 c2
¯3
¯
4
4
= 2R
= |a¯3 |
= 2R
= 2R
=
(2R)2
=
4R2
Since angle between a¯1 and a¯2 is 60o , from the figure it is clear that,
¯
AC
¯ + BC
¯
= AB
¯ = BC
¯ and AB
¯ can be calculated as follows.
Where AB
Consider triangle ABD,
tan 30o
⇒
=
x =
=
⇒
¯ = BC
¯
AB
=
⇒
¯
AC
=
=
⇒
a¯1 + a¯2
=
a/2
x
a
o
2
tan
√ 30
3
a
2
√
3
a
√2
√
3
3
a+
a
2
2
√
3a
√
3a
Using the value of (a¯1 + a¯2 ) we have,
µ
¶2
1 √
1
× 3a + c2
3
4
2
a
c2
+
3
4
c2
4
c2
a2
c
a
= 4R2
= a2
=
=
2a2
3
8
3
= 1.632
Answer: 6
In the simple cubic each atom is surrounded by 8 other atoms. There are 8
corners in a simple cube. Each atom contributes its 1/8 part in a simple cube.
Therefore,
total number of atoms
1
8
= 1 atom
=
8×
Let lattice constant is a, then radius of each atom will be a/2.
volume of one atom in simple cube
total volume of cube
5
4
π(a/2)3
3
4 a3
=
π
3 8
πa3
=
6
= a3
=
Thus the coverage of simple cubic lattice will be,
coverage of SC
=
volume occupied by an atom
total volume
=
πa3
6
a3
= 0.52
%coverage =
0.52 × 100
= 52 %
6