Lesson 2 Homework Solutions Extra PDF

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MATSE 259 Lesson 2: Crystalline Structures and Defects Practice Problem Solutions 9th Edition 3.11 3.22 3.30 3.40 4.1 4.8 4.34

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3.11 Zirconium has an HCP crystal structure and a density of 6.51 g/cm3. (a) What is the volume of its unit cell in cubic meters? (b) If the c/a ratio is 1.593, compute the values of c and a. Solution (a) The volume of the Zr unit cell may be computed using Equation 3.5 as VC 

nAZr N A

Now, for HCP, n = 6 atoms/unit cell, and for Zr, AZr = 91.22 g/mol. Thus, VC 

(6 atoms/unit cell)(91.22 g/mol) (6.51 g/cm3)(6.022  10 23 atoms/mol)

= 1.396  10-22 cm3/unit cell = 1.396  10-28 m3/unit cell

(b) From Equation 3.S1 of the solution to Problem 3.6, for HCP VC = 6 R2 c 3

But, since a = 2R, (i.e., R = a/2) then

a 2 3 3 a2c VC = 6   c 3  2  2

but, since c = 1.593a

VC =

3 3 (1.593) a 3 = 1.396  10-22 cm3/unit cell 2

Now, solving for a (2)(1.396  10-22 cm3 ) 1/3 a =   (3) ( 3) (1.593)  

2

= 3.23  10-8 cm = 0.323 nm

And finally c = 1.593a = (1.593)(0.323 nm) = 0.515 nm

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3.22 List the point coordinates for all atoms that are associated with the FCC unit cell (Figure 3.1). Solution From Figure 3.1b, the atom located of the origin of the unit cell has the coordinates 000. Coordinates for other atoms in the bottom face are 100, 110, 010, and

11 22

0 . (The z coordinate for all these points is

zero.) For the top unit cell face, the coordinates are 001, 101, 111, 011, and

11 22

1.

Coordinates for those atoms that are positioned at the centers of both side faces, and centers of both front and back faces need to be specified. For the front and back-center face atoms, the coordinates are 1 and 0 1 2

0

1 2

11 22

11 22

, respectively. While for the left and right side center-face atoms, the respective coordinates are

and

1 1

1 .

2 2

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3.30 Within a cubic unit cell, sketch the following directions: (a)

[1 10] ,

(e) [1 1 1] ,

(b)

[1 2 1],

(f) [1 22],

(c)

[01 2],

(g) [12 3 ],

(d)

[13 3] ,

(h) [1 03].

Solution The directions asked for are indicated in the cubic unit cells shown below.

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3.40 Sketch within a cubic unit cell the following planes: (a) (01 1 ) ,

(e) (1 11 ) ,

(b) (112 ) ,

(f) (12 2 ) ,

(c) (102 ) ,

(g) (1 23 ) ,

(d) (13 1) ,

(h) (01 3 )

Solution

The planes called for are plotted in the cubic unit cells shown below.

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4.1 Calculate the fraction of atom sites that are vacant for lead at its melting temperature of 327°C (600 K). Assume an energy for vacancy formation of 0.55 eV/atom. Solution In order to compute the fraction of atom sites that are vacant in lead at 600 K, we must employ Equation 4.1. As stated in the problem, Qv = 0.55 eV/atom. Thus,    Q  0.55 eV / atom Nv = exp  v = exp   5 N  kT   (8.62  10 eV / atom - K) (600 K) 

= 2.41  10-5

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4.8 What is the composition, in weight percent, of an alloy that consists of 6 at% Pb and 94 at% Sn? Solution In order to compute composition, in weight percent, of a 6 at% Pb-94 at% Sn alloy, we employ Equation 4.7 as

CPb =

=

' A CPb Pb ' A C ' A CPb Pb Sn Sn

 100

(6)(207.2 g / mol)  100 (6)(207.2 g / mol)  (94)(118.71 g / mol)

= 10.0 wt%

CSn =

=

' A CSn Sn ' A C ' A CPb Pb Sn Sn

 100

(94)(118.71 g / mol)  100 (6)(207.2 g / mol)  (94)(118.71 g / mol)

= 90.0 wt%

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4.34 For an ASTM grain size of 8, approximately how many grains would there be per square inch at (a) a magnification of 100, and (b) without any magnification? Solution (a) This part of problem asks that we compute the number of grains per square inch for an ASTM grain size of 8 at a magnification of 100. All we need do is solve for the parameter N in Equation 4.16, inasmuch as n = 8. Thus N  2 n 1

= 28 1 = 128 grains/in.2

(b) Now it is necessary to compute the value of N for no magnification. In order to solve this problem it is necessary to use Equation 4.17:  M 2 N M    2 n1 100 

where NM = the number of grains per square inch at magnification M, and n is the ASTM grain size number. Without any magnification, M in the above equation is 1, and therefore,  1 2 N 1    28 1  128 100 

And, solving for N1, N1 = 1,280,000 grains/in.2.

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