Ejercicios Capítulo 3 La Ciencia De Los Materiasles Askeland

Exercises chapter 3:
3.23.- A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the num-ber of unit cells and (b) the number of iron atoms in the paper clip. (See AppendixAfor required data).
Solution: the lattice parameter for BCC iron is 2.866 x 10 exp -8 cm. therefore:
V unit cell= -(2.88 x 10 exp-8 cm)ᶟ = 2.354 x 10 exp. -23 cmᶟ.
The density is 7.87 g/mᶟ. thenumber of unit cell is:
Number= 0.59 g / (7.87 g/cmᶟ) (2.354 x 10 exp. -23 cmᶟ/cell) = 3.185 x 10 exp. 21 cells.
There are two atoms / cell in BCC iron. The number of atoms is:
Number = (3.185 x 10exp. 21 cells) (2 atoms/ cell) = 6.37 x 10exp. 21 atoms.
3.24 Aluminum foil used to package food is approximately 0.001 inch thick. Assume thatall of the unit cells of the aluminum are arranged so thata sub cero is perpendicular to the foil surface. For a 4 in x 4 in. square of the foil, determine (a) the total number of unit cells in the foil and (b) the thickness of the foil in number of unitcells. (SeeAppendix A.).
Solution:
The lattice parameter for aluminium is 4.04958 x 10 exp. -8 cm. therefore:
V unit cell= (4.04958 x 10exp -8) ³ = 6.6409 x 10 exp. -23 cm³
The volume of the foilis:
V foil = (4in.)(4 in.)(0.001 in.) = 0.016 in. ³ = 0.262 cm³.
The number of unit cells in the foil is:

Number = 0.262 cm³/ 6.6409 x 10 exp. -23 cm³/cell = 3.945 x 10 exp. 21 cells.
Thethickness of the foil, in number of unit cells, is:
Number = (0.001 in.)(2.54 cm/in.)/4.04958 x 10 exp. -8 = 6.27 x 10exp 4cells.


3.37. determine the miller indices for the directions in the cubicunit cell shown in figure.
Solution:
A: 0, 1, 0 - 0, 1, 1= [00-1].
B: ½, 0, 0 – 0, 1, 0= ½, -1, 0= [1,-2, 0]
C: 0, 1, 1 – 1, 0, 0= [111]
D: 1, 0, 1, ½ -0, ½, 1= 1, -1/2, -1/2= [2-1-1]
3.38.determine the indices for the directions in the cubic unit cell shown in figure.
Solution:
A: 0, 0, 1 – 0, 1, 1= -1, 0, 1 = [-101]
B: 1, 0, 1 – 1/2, 1, 0= ½ , -1, 1 = [1-22]
C: 1, 0, 0 – 0, 3/4, 1 =...