# Solution manual - Chapter 4 4.2 Jens Zamanian March 10, 2014

```Solution manual - Chapter 4
Jens Zamanian
March 10, 2014
4.2
From Fig. 1 below we clearly see that for (n1 , n2 , n3 )
a) if the ni are either all odd or all even we get a BCC lattice and
b) if the sum of ni is even we get a FCC lattice.
Figure 1: BCC and FCC lattices. The BCC is obtained by only including points where all ni ’s
are all even or all odd an the FCC by considering the points where the sum of the ni ’s is even.
4.5
a) For the hexagonal close-packing structure we may calculate the c/a ratio by considering Fig.
2. We see that the value of c/a can be obtained by calculating the height of the pyramid formed
between two layers and multiplying this by two. The side of the pyramid is a. By first considering
the green triangle in the figure we quickly get that
x=
a/2
a
=p .
cos(30 )
3
(1)
We have here used that the sides of the triangle are all the same so that angles are all 60 , and
that the hypothenuse of the green triangle bisects this angle. Then, by considering the red triangle
and using that the hypothenuse is a we get
r
r
p
1
2
2
2
h= a
x =a 1
=a
.
(2)
3
3
1
Finally we use that c = 2h to get
c
=
a
r
8
.
3
(3)
Figure 2: BCC and FCC lattices. The BCC is obtained by only including points where all ni ’s
are all even or all odd an the FCC by considering the points where the sum of the ni ’s is even.
Alternative solution: The primitive vectors of a simple hexagonal Bravais lattice can be taken
to be
p
a
3a
ˆ+
ˆ , a3 = cˆ
a1 = aˆ
x , a2 = x
y
z
(4)
2
2
and a close-packed hexagonal structure is obtained by adding the basis
v1 = 0,
v2 =
a1 + a2
a3
+ .
3
2
(5)
As can be seen in Fig. 3 the three vectors a1 , a2 and v2 forms a tetrahedron. For an ideal close-
Figure 3: Tetrahedron formed by the three vectors a1 , a2 and v2 .
packed hexagonal structure these vectors must point on the centers of touching perfects spheres.
Hence we must have that the length of the vectors are the same. We know that |a1 | = |a2 | = a.
We further have
2
|v2 | = v2 · v2 =
a21 + 2a1 · a2 + a22
a2
a2 + a2 + a2 /4 + 3a2 /4 c2
a2
c2
+ 3 =
+
=
+ .
9
4
9
4
3
4
2
(6)
Equating the this length with a now gives
c=
r
p
2 2a
8
=
a,
3
3
(7)
as before.
b) The volume of the primitive cell of a simple hexagonal lattice is given by
p
3 2
vH = |a1 · (a2 ⇥ a3 )| =
a c.
2
For the close-packed hexagonal structure the primitive cell is half this volume
p
3 2
vHCP =
a c.
4
(8)
(9)
If we have an ideal hcp we get
1
vHCP = p a3
2
The volume of the unit cell of the BCC lattice is obtained by using the primitive vectors
b1 = aˆ
x,
b2 = aˆ
y,
a3 =
a
ˆ+z
ˆ) .
(ˆ
x+y
2
We then get the volume as
vBCC = |(b1 ⇥ b2 ) · b3 | =
(10)
(11)
a3
.
2
(12)
If we start with a BCC structure with aBCC = 4.23 ˚
A and transition to an ideal HCP without
chaining the density, then the volume of the primitive cell must be conserved. We hence get
vHCC = vBCC
)
a3HCC = 2
1/2 3
aBCC
)
aHCC = 2
1/6
aBCC .
(13)
In the specific case we will have aHCC = 3.77 ˚
A.
4.6
To calculate the packing fractions of the di↵erent structures, we take a unit cell and put spheres at
the lattice sites. These spheres are then scaled up so that they touch each other and we calculate
which fraction of the unit cell that is filled by these.
FCC: From Fig. 4 we see that the spheres will have a radius which
p is 1/4 of the diagonal of
the side of the cube. The sides of the cube is a and the diagonal hence 2a and we get r = a/23/2 .
The cube contains in total 4 full spheres. The packing fraction is then the volume of these four
spheres divided by the volume of the cube, i.e.
p
4 ⇥ 4⇡r3 /3
16⇡ a3
2⇡
= 3 9/2 =
.
(14)
3
a
3a 2
6
Diamond: The diamond structure is obtained by taking an FCC lattice and adding another
ˆ+z
ˆ), i.e. we have the basis
FCC lattice which is displaced by w2 = (a/4)(ˆ
x+y
w1 = 0,
w2 =
a
ˆ+z
ˆ) .
(ˆ
x+y
4
(15)
When we scale up the spheres in a diamond structure we have, according to Fig. 4, the radius of
the spheres will this time be
p
1
3a
r = |w2 | =
.
(16)
2
8
3
Figure 4: Left: Close-packed FCC structure. Right: Close-packed diamond structure. The diamond structure consists of two intersecting FCC lattices (blue and green) displaced by the vector
ˆ+z
ˆ) (red line).
w2 = (a/4)(ˆ
x+y
In the diamond structure each cube of volume a3 contains 8 full spheres and we hence get the
packing fraction
p
4⇡r3 1
3⇡
=
8⇥
.
(17)
3 a3
16
BCC: Use that the radius of the spheres will be 1/4 of the space diagonal of the cube.
Simple cubic: This time the radius will be a/2.
4
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