CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS

Transcription

1 4-1 CHAPTER 4 IMPERFECTIONS IN SOLIDS PROBLEM SOLUTIONS Vacancies and Self-Interstitials 4.1 In order to compute the fraction of atom sites that are vacant in copper at 1357 K, we must employ Equation 4.1. As stated in the problem, Q v = 0.90 ev/atom. Thus, N v N = exp Q v = exp kt 0.90 ev /atom ( ev/atom- K) (1357 K) = 4.56 x 10-4

2 4-36 Dislocations Linear Defects 4.6 The Burgers vector and dislocation line are perpendicular for edge dislocations, parallel for screw dislocations, and neither perpendicular nor parallel for mixed dislocations.

3 In the drawing below is shown the atoms on the (100) face of an FCC unit cell; the interstitial site is at the center of the edge. The diameter of an atom that will just fit into this site (r) is just the difference between that unit cell edge length (a) and the radii of the two host atoms that are located on either side of the site (R); that is r = a R However, for FCC a is related to R according to Equation 3.1 as a = R ; therefore, solving for r from the above equation gives r = a R A (100) face of a BCC unit cell is shown below. = R R = 0.41R

4 4-6 The interstitial atom that just fits into this interstitial site is shown by the small circle. It is situated in the plane of this (100) face, midway between the two vertical unit cell edges, and one quarter of the distance between the bottom and top cell edges. From the right triangle that is defined by the three arrows we may write a + a 4 = (R + r) However, from Equation 3.3, a = 4R, and, therefore, making this substitution, the above equation takes the form 3 4R 3 + 4R 4 3 = R +Rr + r After rearrangement the following quadratic equation results: r + Rr 0.667R = 0 And upon solving for r: r = (R) ± (R) (4)(1)( 0.667R )

5 4-7 = R ±.58R And, finally r(+) = r( ) = R +.58R = 0.91R R.58R =.91R Of course, only the r(+) root is possible, and, therefore, r = 0.91R. BCC. Thus, for a host atom of radius R, the size of an interstitial site for FCC is approximately 1.4 times that for

6 4-47 DESIGN PROBLEMS Specification of Composition 4.D1 This problem calls for us to compute the concentration of lithium (in wt%) that, when added to aluminum, will yield a density of.47 g/cm 3. Solution of this problem requires the use of Equation 4.10a, which takes the form ρ ave = 100 C Li ρ Li C Li ρ Al inasmuch as C Li + C Al = 100. According to the table inside the front cover, the respective densities of Li and Al are and.71 g/cm 3. Upon solving for C Li from the above equation, we get C Li = 100 ρ Li (ρ Al ρ ave ) ρ ave (ρ Al ρ Li ) = (100)(0.534 g /cm3 )(.71 g /cm 3.47 g /cm 3 ) (.47 g /cm 3 )(.71 g /cm g /cm 3 ) =.38 wt%

7 5-3 Diffusion Mechanisms 5.3 (a) With vacancy diffusion, atomic motion is from one lattice site to an adjacent vacancy. Selfdiffusion and the diffusion of substitutional impurities proceed via this mechanism. On the other hand, atomic motion is from interstitial site to adjacent interstitial site for the interstitial diffusion mechanism. (b) Interstitial diffusion is normally more rapid than vacancy diffusion because: (1) interstitial atoms, being smaller, are more mobile; and () the probability of an empty adjacent interstitial site is greater than for a vacancy adjacent to a host (or substitutional impurity) atom.

8 (a) The driving force is that which compels a reaction to occur. (b) The driving force for steady-state diffusion is the concentration gradient.

9 This problem calls for computation of the diffusion coefficient for a steady-state diffusion situation. Let us first convert the carbon concentrations from weight percent to kilograms carbon per meter cubed using Equation 4.9a. For wt% C C C " = C C ρ C C C + C Fe ρ Fe 10 3 = g /cm g /cm kg C/m 3 Similarly, for wt% C C C " = g /cm g /cm = kg C/m 3 Now, using a rearranged form of Equation 5.3 x D = J A x B C A C B = ( kg/m 10 -s) 3 m 1.18 kg /m kg /m 3 =.3 x m /s

10 This problem calls for an estimate of the time necessary to achieve a carbon concentration of 0.35 wt% at a point 6.0 mm from the surface. From Equation 5.6b, x Dt = constant But since the temperature is constant, so also is D constant, and or x t = constant x 1 = x t 1 t Thus, (.0 mm) 15 h = (6.0 mm) t from which t = 135 h