7/31/2019 Tutorial 2- Questions Answer Scheme

1/10

-TUTORIAL QUESTIONS ANSWER SCHEME (BMFB 2213)-

TUTORIAL 2: THE STRUCTURE OF CRYSTALLINE SOLIDS

1.

a. Atomic structure: Each atom consists of a very small nucleus composed of

protons and neutrons, which is encircled by moving electrons.

Crystal structure: for crystalline materials, the manners in which atoms or

ions are arrayed in space. It is defined in terms of the unit cell geometry and

the atom positions within the unit cell

b. Crystal structure: for crystalline materials, the manners in which atoms or

ions are arrayed in space. It is defined in terms of the unit cell geometry and

the atom positions within the unit cell

Crystal system: a scheme which crystal classifies structures according to unitcell geometry. This geometry is specified in terms of the relationships between

edge lengths and interaxial angles. There are seven different crystal systems

c. Single crystal: a crystalline solid for which the periodic and repeated

atomic pattern extends throughout its entirely without interruption

Polycrystalline materials: refers to crystalline materials that are composed of

more than one crystal or grain

d. Isotropy: having identical values of a property in all crystallographic directions

Anisotropy: exhibiting different values of a property in differentcrystallographic directions.

2.

a. Crystalline solid: the state of a solid material characterized by a

periodic and repeating three- dimensional array of atoms, ions or

molecules

b. Space lattice: a three- dimensional array of points each of which

has identical surroundings

c. Lattice point: one point in an array in which all the points have

identical surroundings

d. Unit cell: a convenient repeating unit of a space lattice. The axial

lengths and axial angles are the lattice constants of the unit cell

3. APF = volume of atoms in BCC unit cell

Volume of BCC unit cell

Since there are two atoms per BCC unit cell, the volume of atoms in the unit cell of radius R

is

BMFB 2213- SEM. 1 0809 1

7/31/2019 Tutorial 2- Questions Answer Scheme

2/10

V atoms = (2) (4/3 R3) = 8.373 R3

The volume of the BCC unit cell is

V unit cell = a3

Where a is the lattice constant. The relationship between a and R is obtained from figure below,which shows that the atoms in the BCC unit cell touch each other across the cubic diagonal.

Thus

(3a) = 4R or a = 4R

3

Thus

V unit cell = a3 = 12.32 R3

The atomic packing factor for the BCC unit cell is, therefore,

APF = V atoms/ unit cell = 8.373 R3 = 0.68

V unit cell 12.32 R3

4.

(a) Total sphere and unit cell volumes: calculated in terms of atomic radius(R)

(b) Volume for sphere: 4/ 3 R

(c) Since there are 4 atoms per FCC unit cell, the total FCC sphere volume is:Vs = (4) 4/ 3 R = 16/ 3 R

(d) Total unit cell volume:

Vc = a = (2R(2) ) = 16 R (2) , a = 2R(2)Therefore,

APF = Vs = 16/ 3 R = 0.74

Vc 16 R (2)

BMFB 2213- SEM. 1 0809 2

APF = fraction of solid sphere volume in a unit cell = total sphere volume = Vs

total unit cell volume Vc

7/31/2019 Tutorial 2- Questions Answer Scheme

3/10

5. For this problem, we are asked to calculate the volume of a unit cell of aluminum. Aluminum

has an FCC crystal structure. The FCC unit cell volume may be computed as below:

6. The volume of the zinc HCP unit cell can be obtained by determining the area of the base of

the unit cell and then multiplying this by its height (see figure below).

The area of the base of the unit cell is area ABDEFG of figure (a) and (b). This total area

consists of the areas of six equilateral triangles of area ABC of figure (b). From figure (c),

Area of triangle ABC= (base) (height)

= (a) (a sin 60) = a2 sin 60

From figure (b),

Total area of HCP base = (6) ( a2

sin 60)= 3 a2 sin 60

From figure (c),

Volume of zinc HCP unit cell = (3 a2 sin 60) ( c)

= (3) (0.2665 nm) 2 (0.8660) (0.4947 nm)

= 0.0913 nm3

7. From the BCC figure, we can see that the atoms in the BCC unit cell touch across the cube

diagonals. Thus, if a is the length of the cube edge, then

(3a) = 4R

where R is the radius of the iron atom. Therefore

a = 4R = 4 (0.124 nm) = 0.2864 nm

(3) (3)

8. Refer Callister

9. Notes: n = number of atoms associated with each unit cell

A = atomic weight

Vc = volume of the unit cellNA = Avogrados number i.e 6.023 x 1023 atoms/ mol

BMFB 2213- SEM. 1 0809 3

7/31/2019 Tutorial 2- Questions Answer Scheme

4/10

Equation:

For BCC: - n = 2 atoms/ unit cell and

Thus,

The value given inside the front cover of the Callister book is 7.87 g/ cm3.

10. (a) The volume of the Ti unit cell may be computed using Equation (3.5) as

Now, for HCP, n = 6 atoms/unit cell, and for Ti, A Ti= 47.9 g/mol. Thus,

= 1.058 x 10-22 cm3/unit cell = 1.058 x 10-28 m3/unit cell

(b) From the solution to Problem 3.7, since a = 2R, then, for HCP

but, since c = 1.58a

Now, solving fora

BMFB 2213- SEM. 1 0809 4

= nA Fe

Vc NA

7/31/2019 Tutorial 2- Questions Answer Scheme

5/10

= 2.96 x 10-8 cm = 0.296 nm

And finally,c = 1.58a = (1.58)(0.296 nm) = 0.468 nm

11. For each of these three alloys we need to, by trial and error, calculate the density and

compare it to the value cited in the problem.

For SC, BCC, and FCC crystal structures, the respective values of n are 1, 2, and 4,

whereas the expressions fora (since VC = a3) are 2R, 2R (2), and 4R/ (3).

For alloy A, let us calculate assuming a BCC crystal structure.

= 6.40 g/cm3

Therefore, its crystal structure is BCC.

For alloy B, let us calculate assuming a simple cubic crystal structure.

= 12.30g/cm3

Therefore, its crystal structure is simple cubic.

For alloy C, let us calculate assuming a BCC crystal structure.

= 9.60 g/ cm3.

Therefore, its crystals structure is BCC.

12.

BMFB 2213- SEM. 1 0809 5

=

7/31/2019 Tutorial 2- Questions Answer Scheme

6/10

13.

BMFB 2213- SEM. 1 0809 6

7/31/2019 Tutorial 2- Questions Answer Scheme

7/10

14.

BMFB 2213- SEM. 1 0809 7

7/31/2019 Tutorial 2- Questions Answer Scheme

8/10

15.

16.

BMFB 2213- SEM. 1 0809 8

7/31/2019 Tutorial 2- Questions Answer Scheme

9/10

BMFB 2213- SEM. 1 0809 9

7/31/2019 Tutorial 2- Questions Answer Scheme

10/10

17.

BMFB 2213- SEM. 1 0809 10