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# Download 3.4 For the HCP crystal structure, show that the ideal c/a ratio is

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```3.4
For the HCP crystal structure, show that the ideal c/a ratio is 1.633.
Solution
A sketch of one-third of an HCP unit cell is shown below.
Consider the tetrahedron labeled as JKLM, which is reconstructed as
The atom at point M is midway between the top and bottom faces of the unit cell--that is MH = c/2.
since atoms at points J, K, and M, all touch one another,

JM = JK = 2R = a
where R is the atomic radius.
Furthermore, from triangle JHM,

(JM ) 2 = ( JH) 2  ( MH) 2
or
2


c 
a 2 = ( JH ) 2 +  
2 
And,
Now, we can determine the JH length by consideration of triangle JKL, which is an equilateral triangle,

cos 30 =
a /2
=
JH
3
2
and
JH =

a
3
Substituting this value for JH in the above expression yields

 a 2 c 2
a2
c2
a 2 =   +   =
+
 3 
2 
3
4

and, solving for c/a

c
=
a

8
= 1.633
3
3.5
Show that the atomic packing factor for BCC is 0.68.
Solution
The atomic packing factor is defined as the ratio of sphere volume to the total unit cell volume, or
APF =
VS
VC
Since there are two spheres associated with each unit cell for BCC

4R3  8R3
VS = 2 (sphere volume) = 2 
 3 
=
3


Also, the unit cell has cubic symmetry, that is VC = a3.

But a depends on R according to Equation 3.4, and
4R 3 64 R 3
VC =   =
 3 
3 3
Thus,

APF =

VS
VC
=
8 R 3 /3
64 R 3 /3 3
= 0.68
3.7
Tungsten has a BCC crystal structure, an atomic radius of 0.137 nm, and an atomic weight of 183.84
g/mol.
Compute and compare its theoretical density with the experimental value found inside the front cover
of the book.
Solution
This problem calls for a computation of the density of
 =
For BCC, n = 2 atoms/unit cell, and
tungsten.
According to Equation 3.7
nAFe
VC N A

3
4 R 
VC =  
 3 
Thus,

 =
=
nAFe
4 R 3
  N A
 3 
(2 atoms/unit cell)(183.84 g/mol)

3
(4) (0.137  10-7 cm) / 3  / (unit cell ) (6.022  10 23 atoms/mol)


= 19.28 g/cm3
The value given inside the front cover is 19.3 g/cm3.
3.9
Calculate the radius of a vanadium atom, given that V has a BCC crystal structure, a density of 5.96 g/cm 3,
and an atomic weight of 50.9 g/mol.
Solution
For BCC, n = 2 atoms/unit cell,
and
4 R 3
64 R 3
VC =   =
 3 
3 3
Since, from Equation 3.7

 =

=
nAV
VC N A
nAV
64 R 3 

N A
 3 3 
and solving for R the previous equation

3 3nA 1/3
V
R = 

64  N A 
and incorporating values of parameters given in the problem statement

 (3 3) (2 atoms/unit cell)(50.9 g/mol)
1/3
R = 

(64) (5.96 g/cm3)(6.022  10 23 atoms/mol) 

= 1.32  10-8 cm = 0.132 nm
3.27
The unit cell for Cr2O3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm.
If the density of this material is 5.22 g/cm3, calculate its atomic packing factor.
ionic radii of 0.062 nm and 0.140 nm, respectively for Cr
3+
For this computation assume
2-
and O .
Solution
This problem asks for us to calculate the atomic packing factor for chromium oxide given values for
the a and c lattice parameters, and the density. It first becomes necessary to determine the value of n' in
Equation 3.8. This necessitates that we calculate the value of VC, the unit cell volume. In Problem 3.6 it was
shown that the area of the hexagonal base (AREA) is related to a as
AREA = 6R 2
a 2
3  6  3  1.5a 2
2 
inasmuch as, for HCP, a = 2R (where R is the atomic radius).

the problem statement into the above expression leads to
3
Now, incorporating the value of a provided in
AREA = (1.5) (4.961  10 -8 cm) 2 ( 3)  6.39  1015 cm2
The unit cell volume now is just

VC = (AREA)(c) = (6.39  10 -15 cm2 )(1.360  10 -7 cm)
= 8.70  10-22 cm3

Now, solving for n' (Equation 3.8) yields
n' =


(5.22 g/cm3)(6.022

N AVC
 AC
+
 AA
 10 23 formula units/mol)(8.70  10 -22 cm3/unit cell)
(2)(52.00 g/mol) + (3)(16.00g/mol)
= 18.0 formula units/unit cell
Or, there are 18 Cr2O3 units per unit cell, or 36 Cr3+ ions and 54 O2- ions.
As given in the problem statement,
the radii of these two ion types are 0.062 and 0.140 nm, respectively.
Thus, the total sphere volume in
Equation 3.3 (which we denote as VS), is just
4 
4 
VS = (36)   (rCr 3 ) 3 + (54)   (rO 2 ) 3
3 
3 

4 
4 
= (36)   (6.2  109 cm) 3 + (54)   (1.4  108 cm) 3
3 
3 
= 6.57  10-22 cm3

Finally, the APF is just
APF 
VS
6.57  10-22 cm 3

 0.755
VC
8.70  10-22 cm 3
3.34
A hypothetical AX type of ceramic material is known to have a density of 2.65 g/cm 3 and a unit cell of
cubic symmetry with a cell edge length of 0.43 nm.
40.3 g/mol, respectively.
possible for this material:
The atomic weights of the A and X elements are 86.6 and
On the basis of this information, which of the following crystal structures is (are)
rock salt, cesium chloride, or zinc blende?
Solution
We are asked to specify possible crystal structures for an AX type of ceramic material given its
density (2.65 g/cm3), that the unit cell has cubic symmetry with edge length of 0.43 nm (4.3  10-8 cm), and the
atomic weights of the A and X elements (86.6 and 40.3 g/mol, respectively).
Using Equation 3.8 and solving
for n' yields
n' 

(2.65 g/cm3) (4.30


VC N A
 AC
+
 AA

 10 -8 cm) 3 /unit cell (6.022  10 23 formula units/mol)
(86.6 + 40.3) g/mol
= 1.00 formula units/unit cell
Of the three possible crystal structures, only cesium chloride has one formula unit per unit cell, and therefore, is
the only possibility.
3.53
Determine the indices for the directions shown in the following cubic unit cell:
Solution
Direction A is a [01 1 ]direction, which determination is summarized as follows.
position the origin of the coordinate system at the tail of the direction vector;
coordinate system
We first of all
then in terms of this new

x
y
z
Projections
0a
–b
–c
Projections in terms of a, b, and c
0
–1
–1
Reduction to integers
not necessary
[01 1 ]
Enclosure
Direction B is a [210] direction, which determination is summarized as follows. We first of all
position the origin of the coordinate system at the tail 
of the direction vector; then in terms of this new
coordinate system

x
y
Projections
–a
Projections in terms of a, b, and c
–1
b
2
1
Reduction to integers
–2
Enclosure
2


1
z
0c
0
0
[210]
Direction C is a [112] direction, which determination is summarized as follows. We first of all

position the origin of the coordinate system at the tail of the direction vector; then in terms of this new
coordinate system
Projections
Projections in terms of a, b, and c
Reduction to integers

Enclosure

x
y
a
2
1
b
2
1
2
2
1

1

[112]
z
c
1
2
Direction D is a [112] direction, which determination is summarized as follows.
position the origin of the coordinate system at the tail of the direction vector;
coordinate system
then in terms of this new

Projections
Projections in terms of a, b, and c
Reduction to integers

Enclosure

x
y
z
a
2
1
b
2
1
–c
2
2

1


1
[112]
We first of all
–1
–2
3.63
Determine the Miller indices for the planes shown in the following unit cell:
Solution
For plane A we will leave the origin at the unit cell as shown; this is a (403) plane, as summarized
below.
Intercepts
x
y
z
a
b
2c
2
1
Intercepts in terms of a, b, and c
3
2

2
Reciprocals of intercepts

2
0

Reduction

4
0

(403)

Enclosure
3
3
2
3
For plane B we will move the origin of the unit cell one unit cell distance to the right along the y axis,
and one unit cell distance parallel to the x axis;
thus, this is a (1 1 2) plane, as summarized below.
y
z
Intercepts
x

–a
–b
c
Intercepts in terms of a, b, and c
–1
–1
1
Reciprocals of intercepts
–1
–1
Reduction
2

(not necessary)

(1 1 2)
Enclosure

2
2
3.75
(a) Derive linear density expression for FCC [100] direction in terms of the atomic radius R.
(b) Compute linear density value for the same direction for lead. Round your answer to 3 significant
digits.
Solution
(a)
In the figure below is shown a [100] direction within an FCC unit cell.
For this [100] direction there is one atom at each of the two unit cell corners, and, thus, there is the equivalent of
1 atom that is centered on the direction vector.
The length of this direction vector is just the unit cell edge
length, 2R 2 (Equation 3.1). Therefore, the expression for the linear density of this plane is
LD100 =

number of atoms centered on [100] direction vector
length of [100] direction vector


1 atom
1

2R 2 2R 2
(b)
From the table inside the front cover, the atomic radius for lead is 0.175 nm.

linear density for the [100] direction is
LD100 ( Pb) 
1

1
2 R 2 (2)(0.175 nm) 2
 2.02 nm 1  2.02 10 9 m 1
Therefore, the
3.78
(a) Derive planar density expressions for BCC (100) and (110) planes in terms of the atomic radius R.
(b) Compute and compare planar density values for these same two planes for vanadium.
Solution
(a)
A BCC unit cell within which is drawn a (100) plane is shown below.
For this (100) plane there is one atom at each of the four cube corners, each of which is shared with four
Thus, there is the equivalence of 1 atom associated with this BCC (100) plane.
The
planar section represented in the above figure is a square, wherein the side lengths are equal to the unit cell edge
4R 2
16 R 2
4R
length,
(Equation 3.4); and, thus, the area of this square is just 
=
. Hence, the planar



3
3
 3 
density for this (100) plane is just

PD100 =


number of atoms centered
 on (100) plane
area of (100) plane

1 atom
3

2
16R
16R2
3
A BCC unit cell within which is drawn a (110) plane is shown below.
For this (110) plane there is one atom at each of the four cube corners through which it passes, each of which is
shared with four adjacent unit cells, while the center atom lies entirely within the unit cell.
Thus, there is the
equivalence of 2 atoms associated with this BCC (110) plane.
The planar section represented in the above
figure is a rectangle, as noted in the figure below.
From this figure, the area of the rectangle is the product of x and y. The length x is just the unit cell edge
4R
length, which for BCC (Equation 3.4) is
. Now, the diagonal length z is equal to 4R. For the triangle
3
bounded by the lengths x, y, and z

y
z 2  x2
Or

4R 2
4R 2
(4 R) 2  
 3 
 
3
 
y
Thus, in terms of R, the area of this (110) plane is just

4 R 4 R 2  16 R2 2
Area(110)  xy  

 3 



3
  3 
And, finally, the planar density for this (110) plane is just

PD110 =

number of atoms centered on (110) plane
area of (110) plane

2 atoms
16 R 2
2

3
8 R2
2
3
(b)
From the table inside the front cover, the atomic radius for vanadium is 0.132 nm.

the planar density for the (100) plane is
Therefore,
PD100 (V) 
3
3

 10.76 nm2  1.076  1019 m2
2
16 R
16 (0.132 nm) 2
While for the (110) plane

PD110 (V) 

3
8 R2
2

3
 15.22 nm2  1.522  1019 m2
8 (0.132 nm) 2 2
```
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